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Luna2uno
2005-Jun-30, 01:03 AM
Hypothetically, per Equivalence Principle, what would kilograms be for any given mass in a variable G?

Asking it this way may be illustrated as follows:

We are in a region of space where G is much higher than here, say beyond the solar system somewhere. Let's say it's 10G. Then, per equivalence, what would the kilograms measuring mass, or inertia, be in that region? Would kilograms be 10 times greater than here, 10kg? Or perhaps 100 times greater?

Think of this, and why I am bringing up this hypothetical question:

If G is 10 times what we know as the universal Newton's G, and the equivalence requires that inertial mass measured in kg is also 10 times, what happens to the kg in terms of what we know as measurement of mass here? So per equivalence, 10G gives us 10kg, but this may be only a local pehonomenon, meaning that 10kg in our kilograms may be 10 times that, viz. 100kg.

Is there an issue here, or kill the thread now? :roll:

papageno
2005-Jun-30, 09:26 AM
This should be in the "Against The Mainstream" forum.

Hypothetically, per Equivalence Principle, what would kilograms be for any given mass in a variable G?
The Equivalence Principle says that the gravitational mass equals dynamical (a.k.a. inertial) mass.
It has nothing to do with the value of G.

Newtons' formula for gravitation:
F = G * (m*M) / r^2 (1)

Newton's second law:
F = M * a (2)

Equivalence principle: M in (1) is the same as M in (2).

Asking it this way may be illustrated as follows:

We are in a region of space where G is much higher than here, say beyond the solar system somewhere. Let's say it's 10G. Then, per equivalence, what would the kilograms measuring mass, or inertia, be in that region? Would kilograms be 10 times greater than here, 10kg? Or perhaps 100 times greater?
As I said, 1 kg gravitational mass is 1 kg inertial mass, whatever the value of G.
What would change is the gravitational force on the mass.

Think of this, and why I am bringing up this hypothetical question:

If G is 10 times what we know as the universal Newton's G, and the equivalence requires that inertial mass measured in kg is also 10 times, what happens to the kg in terms of what we know as measurement of mass here? So per equivalence, 10G gives us 10kg, but this may be only a local pehonomenon, meaning that 10kg in our kilograms may be 10 times that, viz. 100kg.

Is there an issue here, or kill the thread now? :roll:
Well, I suggest you look up the threads in the "Against The Mainstream" forum where Lunatik and Jerry try to argue in favour of a variable G.

Luna2uno
2005-Jun-30, 04:42 PM
Thanks papageno for your response. My purpose for entering this hypothetical question on the Equivalence Principle here (rather than Against the Mainstream, per Lunatik &amp; Jerry) is not to argue for a variable G, which would be speculative, but to consider how such a (hypothetical future) discovery would affect our measure of mass in kilograms. Which kilograms would we use, and how would they be affected? Your response addresses how kilograms work at 1 G, which is known, but how would this change if we found a variable G, at 10G for example? Or would it not change at all, and still preserve equivalence?

To my thinking (and I must admit I really do not know the answer to this hypothetical question on measuring mass under a variable G scenario), the kilograms we developed in our 1G universe are in part a function of gravity, mainly Earth's gravity, so we can weigh things in kilograms. The Equivalence is that this same kilograms applies to F = Ma, as you pointed out, so we can measure inertial mass with the same unit. I believe it was Einstein who thus resolved that gravity and inertial mass are linked, which we know as the Equivalence Principle. So the question remains, in a hypothetical variable G, would the kilogram units remain the same, or forced to change?

I would think this is a valid astronomy-physics question, in anticipation of some point in the future that our distant space probes, or other observations, yield a variable G. To date, this has not been observed, to my knowledge. Perhaps this question of measurement in kilograms (at this point a merely philosophical question since we have not confirmed any change in Newton's G from its universal constant) should be explored in the event we find the universal G is something else. We must allow for nature to be a tricky place, so she might throw us a surprise. Would we know what to do with our units of measure of mass at that point if she did? :-?

papageno
2005-Jun-30, 05:03 PM
Thanks papageno for your response. My purpose for entering this hypothetical question on the Equivalence Principle here (rather than Against the Mainstream, per Lunatik &amp; Jerry) is not to argue for a variable G, which would be speculative, but to consider how such a (hypothetical future) discovery would affect our measure of mass in kilograms. Which kilograms would we use, and how would they be affected? Your response addresses how kilograms work at 1 G, which is known, but how would this change if we found a variable G, at 10G for example? Or would it not change at all, and still preserve equivalence?
Equivalence principle and the value of G are independent from each other.
Do not confound mass and weight.

To my thinking (and I must admit I really do not know the answer to this hypothetical question on measuring mass under a variable G scenario), the kilograms we developed in our 1G universe are in part a function of gravity, mainly Earth's gravity, so we can weigh things in kilograms. The Equivalence is that this same kilograms applies to F = Ma, as you pointed out, so we can measure inertial mass with the same unit. I believe it was Einstein who thus resolved that gravity and inertial mass are linked, which we know as the Equivalence Principle. So the question remains, in a hypothetical variable G, would the kilogram units remain the same, or forced to change?
It would be the same.
The Equivalence principle is based on experimental results. Einstein decided to elevate to the status of postulate.

I would think this is a valid astronomy-physics question, in anticipation of some point in the future that our distant space probes, or other observations, yield a variable G. To date, this has not been observed, to my knowledge.
I explained it to Lunatik.
If G depended on positions, the mass of an object would not be affected, but the gravitational force would be.
A different force would give a different acceleration, because the inertial mass has not changed.
But this problem is no more exotic than a variable dielectric constant in electromagnetism (which gives us refraction, and lenses).

Perhaps this question of measurement in kilograms (at this point a merely philosophical question since we have not confirmed any change in Newton's G from its universal constant) should be explored in the event we find the universal G is something else. We must allow for nature to be a tricky place, so she might throw us a surprise. Would we know what to do with our units of measure of mass at that point if she did? :-?
Yes, because the experiments used for the units would not change their outcome if we found something new.

Luna2uno
2005-Jul-01, 01:48 AM
I explained it to Lunatik.
If G depended on positions, the mass of an object would not be affected, but the gravitational force would be.
A different force would give a different acceleration, because the inertial mass has not changed.
But this problem is no more exotic than a variable dielectric constant in electromagnetism (which gives us refraction, and lenses).
I see this really as a question referring to our units of measure, what we call kilograms. Can the same kilograms be used if G is different from what we know it to be as a universal constant?

In yours you said: "The Equivalence Principle says that the gravitational mass equals dynamical (a.k.a. inertial) mass. It has nothing to do with the value of G."

Granted, given that G is universally the same, it has nothing to do with it, though G is part of the function describing Newton's formula for gravitation, as per yours above:

F = G * (m*M) / r^2 , which is related to Newton's second law:

F = M * a

Now, this equivalence can be also shown as:

F = M * a = M * (G*m) / r^2, where by default a = (G*m) / r^2

which also means: G = (r^2 * a) / m

Now assume that both a and r^2 are fixed, same values, but G is greater, viz. G1 = 10G. So we have:

G1 = (r^2 * a) / m1, except now of necessity, m1 = 1/10th of m, if G1 = 10G.

However the mass had not changed, same mass (same atomic composition and volume), so the mass did not suddenly shrink to a tenth of its original form. What changed instead was that the measures in kilograms had changed, to where now the kilograms are 10 times greater than the kilograms used earlier, to match up with G ten times Newton's G.

Can you see how this could be a problem? Though for now, given that G is universal, we don't have a problem. But if it were discovered that G is different, something might have to be adjusted in the measure of our (Earth derived) kilograms.

(That said, I still think that the answer above, kg1 = 10kg is wrong, but I'm not sure of what the right answer is. I suspect a is in fact not fixed as assumed, for a variable G. Hypothetically, the real answer may be more like kg1 = 100 kg, if G1 = 10G, or its squared. It may take 10 times as much acceleration to move the same mass in 10G, so a is not fixed, but rather a1 = 10X a. But I don't know this.)

So you can see why I am frustrated, and I don't like my own answers! :( There must be a better way to see this.

Interesting if this might not apply as well to a " variable dielectric constant in electromagnetism", since it might impact how light bends around stars, which would impact gravitational lensing. :-?

Actually, now that I re-read this, I can almost begin to appreciate the frustration Galileo must have had trying to prove why the Earth is not standing still with the heavens going around, but instead it is spinning. :)

papageno
2005-Jul-01, 09:52 AM
I explained it to Lunatik.
If G depended on positions, the mass of an object would not be affected, but the gravitational force would be.
A different force would give a different acceleration, because the inertial mass has not changed.
But this problem is no more exotic than a variable dielectric constant in electromagnetism (which gives us refraction, and lenses).
I see this really as a question referring to our units of measure, what we call kilograms. Can the same kilograms be used if G is different from what we know it to be as a universal constant?
The definition of the unit kilogram has nothing to do with G.

In yours you said: "The Equivalence Principle says that the gravitational mass equals dynamical (a.k.a. inertial) mass. It has nothing to do with the value of G."

Granted, given that G is universally the same, it has nothing to do with it,...
No, it has nothing to do with G, whether it is constant or not.

... though G is part of the function describing Newton's formula for gravitation, as per yours above:

F = G * (m*M) / r^2 , which is related to Newton's second law:

F = M * a

Now, this equivalence can be also shown as:

F = M * a = M * (G*m) / r^2, where by default a = (G*m) / r^2

which also means: G = (r^2 * a) / m
As you can see, G does not affect M, which is the mass the Equivalence has been applied to, but affects the acceleration a the mass M is subjected to.

Now assume that both a and r^2 are fixed, same values, but G is greater, viz. G1 = 10G. So we have:

G1 = (r^2 * a) / m1, except now of necessity, m1 = 1/10th of m, if G1 = 10G.
But we applied the Equivalence principle to M.
By changing the value of G, you changed the force M and m are subjected to.

If the only mean we had to measure the mass m, was from the acceleration of M due to its gravitational interaction with m, then changing G would affect our measured m because the acceleration is different (assuming that we did not know that G has a different value).

However the mass had not changed, same mass (same atomic composition and volume), so the mass did not suddenly shrink to a tenth of its original form. What changed instead was that the measures in kilograms had changed,...
No. What changed is the gravitational force between the two masses, and hence the acceleration changed.

...to where now the kilograms are 10 times greater than the kilograms used earlier, to match up with G ten times Newton's G.
This has nothing to do with the units.

Can you see how this could be a problem? Though for now, given that G is universal, we don't have a problem. But if it were discovered that G is different, something might have to be adjusted in the measure of our (Earth derived) kilograms.
No. The units would not need to be adjusted.

We don't need to change the unit of electric charge because the dielectric constant is not universal in materials.

(That said, I still think that the answer above, kg1 = 10kg is wrong, but I'm not sure of what the right answer is. I suspect a is in fact not fixed as assumed, for a variable G.
This is the point.

Hypothetically, the real answer may be more like kg1 = 100 kg, if G1 = 10G, or its squared. It may take 10 times as much acceleration to move the same mass in 10G, so a is not fixed, but rather a1 = 10X a. But I don't know this.)

So you can see why I am frustrated, and I don't like my own answers! :( There must be a better way to see this.
You just need to realize that the value of the constant G does not affect the mass of an object.

Interesting if this might not apply as well to a " variable dielectric constant in electromagnetism", since it might impact how light bends around stars, which would impact gravitational lensing. :-?
Refraction is very common phenomenon, which is due to a change in dielectric constant: this is why lenses work.
A G dependent on position would not be more exotic than refraction.

Luna2uno
2005-Jul-01, 08:53 PM
Let's see if this anecdotal illustration better explains how I see it:

I live on planet X (not a real planet we know) where gravity's proportional G is ten times what we know here as G, so Xian's gravity is 10G (in Earth terms). I very carefully measure this Gx, set up my weights of measure in kilograms per this Gx, then work out the Equivalence Principle per F = Ma = G*(Mm)/ r^2 (everyone knows gravity and inertial mass are related), so my mass Mx (and mx) is measured in the kilograms I developed. Now I am content, since I worked it all out, where my units of measure for weight on X are measured in kilograms, for which I then established an equivalence with F = Ma, where Mx is measured in kgx. Confident, I now teach at a prominent Xian university and (since I never traveled off world) merrily accept that my Gx and kilograms kgx are universal. :)

Four hundred years go by and in a very fancy space ship arrive people (to every Xian's surprise) who say they're from some far off place called Earth. Now these Earthians (all descendents of a prominent university where physics had been taught with confidence for the past 400 years) are very eager to impress their newfound Xians, so they too go and measure G and kilograms. To their surprise, they discover that the Xians are using a different unit of measure for kilograms than the Earthian measure. So they carefully explain to the Xians that G is not what they thought it is (since it is universal), but it is 10 times less, and that only the "acceleration" derived from the greater gravity of their planet is 10 times greater. Kilograms cannot change. They further explain that what they had done wrong was make a tenfold mistake (or hundredfold?) in estimating their kilograms. They made the error of thinking that their Gx (which is 10G in Earthian terms) is the correct G, so the kilograms they developed was based on this error. Since, as your Earthian student descendents take great pains to explain, only "aceleration" is ten times what it should be, so mass measured in kilograms has to be the same, so their Xian kilograms are obviously wrong. The Xians challenge this, saying no, that the Earthian kilograms are wrong, because they are only a tenth of what they should be for G, as everyone who studied at the Xian universsity can tell you, and that their kilograms are correct, since the acceleration works out exactly for their equivalence principle F = Ma. In fact, they (barely) tolerantly explain, the Earthians had got it wrong. In thinking that G is only a tenth of Gx (everyone knows this is a universal constant), it is the Earthians who should adjust their kilograms to reflect the correct G. And that correct measure of mass is tenfold Earthian kilograms, so obviously Xian kilograms are the correct measure for mass. :(

Well, this heated discussion goes on for some time, and as their appears no solution to this problem, with both Earthians and Xians convinced their measures for mass are correct, it appears the two worlds are in danger of declaring war. Knowing that we really don't know, my (long descendent) student politely (diplomatically) reminds his friends at the Xian academy that it was smart Earthians who came to Xian, and not the otherway around, so perhaps Earthian measured kilograms of mass should be adopted (though they are different from Xian kilograms), just to keep the peace. But he gets shouted down because they say the Earthians are on X now, so must use Xian kilograms instead. :-?

Who is right?

nutant gene 71
2005-Jul-04, 04:51 AM
DISCLAIMER

Let it be know to all who post here and read, that "Lunatik" (http://www.badastronomy.com/phpBB/search.php?search_author=Lunatik) had been retired, put into permanent "safing" with post # 555 (http://www.badastronomy.com/phpBB/viewtopic.php?p=483557&amp;sid=436cd39f7f853765c0f9baf e8d166d8f#483557), and that my lame attempt to revive him with "Luna2uno" had been in violation of BA rules (prior unbeknownst to me, but Phil made the point), so neither name shall henceforth be shown. All future posts will now default to my other (unwitting alias), from here on in my legitimate handle: "nutant gene 71" (http://www.badastronomy.com/phpBB/search.php?search_author=nutant%20gene%2071).

I fully accept any and all criticisms, scorn, ridicule, shunning, or wisecracks, for I am truly repentant.

And I would not be here were it not the high level with which I hold the participants of this board, myself excepted.

Mea culpa. :oops:

Tobin Dax
2005-Jul-04, 06:18 AM
Luna2uno, I still don't understand your question. I'm pretty confused about what you're asking right now. I've only skimmed the last half of the thread, but let me attempt a response.

As papageno said, don't confuse mass and weight. Mass is an intrinsic property to matter, where as weight is force caused by gravity and determined by mass, distance, and the value of G. (The equivalence principle does not apply here, as a = G*M/R^2 [your mass, m, remains the same].) The amount of mass here where G=G would be the same amount where G'=2G.

Now, different systems of units are a completely different beast. Your story seems to discuss different systems of units. The amount of mass is the amount of mass is the amount of mass, no matter how it is defined. A mass of 10 kg is the same amount of mass whether measured in kg, g, slugs, or whatever. 10 miles is the same distance if measured in miles, feet, meters, or parsecs, in just the same way as above. It's also true that G has different numerical values in units of kg,m,s, or g,cm,s, or slugs,feet,fortnights, but these are all the same value.

Mass doesn't change if G does.

papageno
2005-Jul-04, 01:13 PM
So, Luna2uno, you were Lunatik.
And you still don't get the distinction between mass as physical quantity and the unit of measurement to express that quantity in numbers.

Now, different systems of units are a completely different beast. Your story seems to discuss different systems of units. The amount of mass is the amount of mass is the amount of mass, no matter how it is defined. A mass of 10 kg is the same amount of mass whether measured in kg, g, slugs, or whatever. 10 miles is the same distance if measured in miles, feet, meters, or parsecs, in just the same way as above. It's also true that G has different numerical values in units of kg,m,s, or g,cm,s, or slugs,feet,fortnights, but these are all the same value.

Mass doesn't change if G does.
Exactly my point.

Whether G is a universal constant or not, it has absolutely no bearing on the unit kilogram.

nutant gene 71
2005-Jul-04, 06:17 PM
So, Luna2uno, you were Lunatik.
And you still don't get the distinction between mass as physical quantity and the unit of measurement to express that quantity in numbers.

Now, different systems of units are a completely different beast. Your story seems to discuss different systems of units. The amount of mass is the amount of mass is the amount of mass, no matter how it is defined. A mass of 10 kg is the same amount of mass whether measured in kg, g, slugs, or whatever. 10 miles is the same distance if measured in miles, feet, meters, or parsecs, in just the same way as above. It's also true that G has different numerical values in units of kg,m,s, or g,cm,s, or slugs,feet,fortnights, but these are all the same value.

Mass doesn't change if G does.
Exactly my point.

Whether G is a universal constant or not, it has absolutely no bearing on the unit kilogram.
Yes, I agree with you and Tobin Dax, that it doesn't matter what units of measuring mass are used (it could be stones), since mass is mass, always the same. Nothing changed as far as the physical properties of the object of mass is concerned, whether weighed or accelerated. What I had been trying to show, why this is a problem for me, is that a unit of measure for mass derived in one G is going to be different for the same unit of measure in another G. In other words, 1 kg doesn not equal 10 kg for the same mass. So this unit of measure chosen is dependent on G where it is derived.

Let me show it another way. As shown above, G can be defined as G = r^2 * a/ m. But then it must also be: 10G = r^2 * 10a/ m, so mass is still the same, only G' is ten times what we know. Now, if the Xians (per illustration above) think their 10G is merely Gx (one unit of G'), then of necessity their equation would be: Gx = r^2 * 10a/ (?m). This is the problem I'm trying to show. Should (?m) now not be, in Xian kilograms, 10m? So per "their" equivalence, kgx = 10kg in ours.

Why is this important, since it appears a rather mundane problem? I can see it as a problem when it comes to estimating the size and density of a foreign body should G there be different from ours. Back to planet X, if Gx = 10G, and we're using our kilograms, then Gx = r^2 * 10a/ m, but if our kilograms are used, then "mx", planet X's mass, is 10 kg in our terms, but one kgx in theirs. I interpret this as us thinking their planet X should be either 10 times the size of Earth (which it is not) or 10 times the density. Another way is to say that their planet, given its known size parameters, is actually ten times gravity denser than it should be. In fact, if Gx is ten times G, the density of the planet need not be affected, only the results of what things would weigh there, and by equivalence, how things would respond to acceleration (and perhaps also affect their centripetal force, so affect their planetary spin).

Can you see where this is taking me? If, for example, a neutron star (so called) has a great mass equivalent (in our G terms) to several solar masses, we of necessity must think it is very dense for the amount of gravity it displays (hence the resultant spin is very great), that its composition must of necessity be something we do not have here in our vicinity of space, though we know it is very small by comparrison to our Sun. But if it is the G of the neutron star that is so great that it "appears" as if it were several solar masses (I'll skip for now the reasoning why this might be), then its size may be even larger than we estimate (as if made of only neutrons), and density not necessarily so compact. In fact, it may be not too much denser than our own Sun, and perhaps only comparatively smaller if its internal radiation pressure is less, meaning it was a small star to start with.

Take another example. I read somewhere that Jupiter may have a rocky core about two or three Earth masses. Whether or not this is true, I can't confirm since I never saw how this was arrived at, whether through radar probing of Jupiter's interior, or derived from atmospheric occultation, or from ephemeris spin data(?). But if true, given a constant G, how could a small rocky core hold such a vast atmosphere? Unless the G is much greater than supposed, it is virtually impossible as a gas. (This may be another reason why speculations on Jupiter's atmosphere is that it has a liquid core?) I know all the arguments against why this cannot be true, how the springs on Huygens worked properly, etc. (in fact I have no way of knowing whether or not my hypothetical planetary G' calculations are right, as shown earlier), but if mass is measured in Earth's kg, then Jupiter's atmopshere cannot be possible for such a small rocky core. A small rocky core can hold a very large atmopshere only if the acceleration towards the center of mass, the gravity, is much greater for the size and density of the planet would otherwise allow. This is why I think the kilograms used is important, because if they are not adjusted for local G conditions, like in the neutron star example above, we may be overestimating density versus what it really is. A neutron star may not be so dense, only its mass (due to much higher G) acts as if it were.

There is also a practical side to this question (on hypothetical mass in a hypothetical variable G), and that has to do with how space probes will behave near any under-over estimated planetary body. If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments. In effect, it cleans up our engineering task for space flight with a better physics. It's not that we fail to get there, since using gravitational assist trajectories are of necessity self correcting (G * M as a product value is still the same, even if G and M are wrong), but that we may be handicapped with a constant G. A better way may be to use local kilograms (as opposed to Earth kilograms) to work out the dynamics of how a spaceprobe will behave in the vicinity of another planet, hypothetically.

So, can a (variable) measure of mass size and density improve on Newton's physics for a variable G? That ultimately is the question. Can we better understand very distant bodies, such as neutron stars with variable kg, adjusted for local G? Would our overall understanding of cosmology be improved, if G is found to be a variable (something we still do not know)? These were my reasons for bringing up this question. (At this point, however, I don't even want to get close to what this means for Einstein's General Relativity theory.) For this reason, I titled this question as "hypothetical" only, until such time that we find G to be otherwise than now postulated.

And I still do not know what the Xian's kilograms should be, ten fold or a hundred fold. :)

Sorry about the identity mixups, it may be due to a "multiple personality" syndrome. :oops: I never like "Lunatik" anyway.

Tassel
2005-Jul-04, 06:55 PM
If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.
Can you provide references for these "numerous" presumably unexplained and unplanned "inflight adjustments" you claim are happening? Also, what are "adjustment tables" and can you provide a reference?

nutant gene 71
2005-Jul-04, 07:57 PM
If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.
Can you provide references for these "numerous" presumably unexplained and unplanned "inflight adjustments" you claim are happening? Also, what are "adjustment tables" and can you provide a reference?
Alas, I cannot at this time, so take my statement under advisement. I can only refer you to this "gravity anomaly" (http://www.cage.curtin.edu.au/~will/grav_anoms.htm) page for now, only very indirectly related. This ESA page on trajectory corrections (http://sci.esa.int/science-e/www/object/index.cfm?fobjectid=33877&amp;fbodylongid=1437) is better, but still not it. If I find different, I'll report. I read something somewhere, but can't put my fingers on it now... :(

Tassel
2005-Jul-04, 08:16 PM
Alas, I cannot at this time, so take my statement under advisement.
Since you cannot provide any evidence for "numerous inflight adjustments" or "adjustment tables", I will take your statements as if you just made them up for your own benefit.

I can only refer you to this "gravity anomaly" (http://www.cage.curtin.edu.au/~will/grav_anoms.htm) page for now, only very indirectly related.
The subject matter on that page is in no way related to your claims.

This ESA page on trajectory corrections (http://sci.esa.int/science-e/www/object/index.cfm?fobjectid=33877&amp;fbodylongid=1437) is better, but still not it.

If I find different, I'll report. I read something somewhere, but can't put my fingers on it now... :(
Yes, cerainly, if you can find any evidence that your claims have any validity whatsoever, I'd be happy to see it. Until then, I'm not sure why you would make statements about "adjustment tables" and "numerous inflight adjustments".

papageno
2005-Jul-05, 09:48 AM
Yes, I agree with you and Tobin Dax, that it doesn't matter what units of measuring mass are used (it could be stones), since mass is mass, always the same.
So what is the problem?

Nothing changed as far as the physical properties of the object of mass is concerned, whether weighed or accelerated. What I had been trying to show, why this is a problem for me, is that a unit of measure for mass derived in one G is going to be different for the same unit of measure in another G. In other words, 1 kg doesn not equal 10 kg for the same mass. So this unit of measure chosen is dependent on G where it is derived.
It is not.
As long as the Equivalence Principle holds, one can use purely dynamical measurements to determine a unit of mass, which is valid for gravitational measurements as well.
It has nothing to do with the value of G.

Let me show it another way. As shown above, G can be defined as G = r^2 * a/ m.
Not defined, but measured.
The formula you just wrote is a way to measure G.

But then it must also be: 10G = r^2 * 10a/ m, so mass is still the same, only G' is ten times what we know.
You measure 10G because the acceleration is 10a.
And the acceleration is 10a because the gravitational force is ten times stronger.
And the gravitational force is ten times stronger because G' = 10 G.
Do you now understand that what you wrote enables to measure G from the acceleration of the mass? And that this relies on the Equivalence Principle, and does not affect the unit for mass?

Now, if the Xians (per illustration above) think their 10G is merely Gx (one unit of G'), then of necessity their equation would be: Gx = r^2 * 10a/ (?m). This is the problem I'm trying to show. Should (?m) now not be, in Xian kilograms, 10m? So per "their" equivalence, kgx = 10kg in ours.
And how is that different from using another system of units?

Why is this important, since it appears a rather mundane pro
blem? I can see it as a problem when it comes to estimating the size and density of a foreign body should G there be different from ours.
If G has a different value from what we expect, and we relied only on G to determine the mass, then the measurement of the mass would give a different value within the same system of units.
It does not change the definition of units, becuase the definition does not rely on one particular value for G.

Back to planet X, if Gx = 10G, and we're using our kilograms, then Gx = r^2 * 10a/ m, but if our kilograms are used, then "mx", planet X's mass, is 10 kg in our terms, but one kgx in theirs.
In Europe I am 180 cm tall.
In the US I am 71 inches tall.
Did my height change? No.

I interpret this as us thinking their planet X should be either 10 times the size of Earth (which it is not) or 10 times the density.
Europeans do not think that the US are 2.5 times taller than Europe!

Another way is to say that their planet, given its known size parameters, is actually ten times gravity denser than it should be.
The force of gravity would be stronger because G is larger, but dynamical measurements would still yield the same masses.

If I take with me a ruler that uses cms, I am still 180 cm even if I am in the US.
If I take a ruler that uses inches, I am 71 inches in Europe.

In fact, if Gx is ten times G, the density of the planet need not be affected, only the results of what things would weigh there, and by equivalence, how things would respond to acceleration (and perhaps also affect their centripetal force, so affect their planetary spin).
Wrong.
The dynamical measurements would not be affected by a different value of G.
Only the gravitational force would be different, not the mass.

Can you see where this is taking me?
You start from a very serious and deep misunderstanding of measurement unit, force, mass and acceleration.
This won't take you anywhere.

If, for example, a neutron star (so called) has a great mass equivalent (in our G terms) to several solar masses, we of necessity must think it is very dense for the amount of gravity it displays (hence the resultant spin is very great), that its composition must of necessity be something we do not have here in our vicinity of space, though we know it is very small by comparrison to our Sun. But if it is the G of the neutron star that is so great that it "appears" as if it were several solar masses (I'll skip for now the reasoning why this might be), then its size may be even larger than we estimate (as if made of only neutrons), and density not necessarily so compact. In fact, it may be not too much denser than our own Sun, and perhaps only comparatively smaller if its internal radiation pressure is less, meaning it was a small star to start with.

Take another example. I read somewhere that Jupiter may have a rocky core about two or three Earth masses. Whether or not this is true, I can't confirm since I never saw how this was arrived at, whether through radar probing of Jupiter's interior, or derived from atmospheric occultation, or from ephemeris spin data(?). But if true, given a constant G, how could a small rocky core hold such a vast atmosphere?
The atmosphere has amss of its own, and it interacts gravitationally.
How do you think the Sun formed: it does not even have a rocky core!

Unless the G is much greater than supposed, it is virtually impossible as a gas.
Why?
On what are your expectations based?

(This may be another reason why speculations on Jupiter's atmosphere is that it has a liquid core?) I know all the arguments against why this cannot be true, how the springs on Huygens worked properly, etc. (in fact I have no way of knowing whether or not my hypothetical planetary G' calculations are right, as shown earlier), but if mass is measured in Earth's kg, then Jupiter's atmopshere cannot be possible for such a small rocky core. A small rocky core can hold a very large atmopshere only if the acceleration towards the center of mass, the gravity, is much greater for the size and density of the planet would otherwise allow.
You still assume that a gas does not interact gravitationally.
On what is this "idea" based?

This is why I think the kilograms used is important, because if they are not adjusted for local G conditions, like in the neutron star example above, we may be overestimating density versus what it really is. A neutron star may not be so dense, only its mass (due to much higher G) acts as if it were.

There is also a practical side to this question (on hypothetical mass in a hypothetical variable G), and that has to do with how space probes will behave near any under-over estimated planetary body.
Weird... Huygens landed as planned, didn't it?

If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.
What makes you think that we did not get that right?

In effect, it cleans up our engineering task for space flight with a better physics. It's not that we fail to get there, since using gravitational assist trajectories are of necessity self correcting (G * M as a product value is still the same, even if G and M are wrong), but that we may be handicapped with a constant G. A better way may be to use local kilograms (as opposed to Earth kilograms) to work out the dynamics of how a spaceprobe will behave in the vicinity of another planet, hypothetically.
Dynamics does not need "local" kilograms, because the kilogram is the same.

So, can a (variable) measure of mass size and density improve on Newton's physics for a variable G?
You assume that Newtonian physics relies on a universal G.
I already explained to you several times that taking into account a variable G is mathematically trivial, and it would be obvious from the orbits of objects like comets.
You have to invoke a magical re-adjustment of mass to keep the orbits consistent with a constant G, in which the change of mass would be obvious from measurements of the inertial mass and moment of inertia (in objects like the probes we sent out).
All the observations disprove your idea of a "variable G + mass compensation"

That ultimately is the question. Can we better understand very distant bodies, such as neutron stars with variable kg, adjusted for local G? Would our overall understanding of cosmology be improved, if G is found to be a variable (something we still do not know)?
None seem to have given us a groundbreaking new understanding of the Universe.

These were my reasons for bringing up this question. (At this point, however, I don't even want to get close to what this means for Einstein's General Relativity theory.) For this reason, I titled this question as "hypothetical" only, until such time that we find G to be otherwise than now postulated.
The postulate is based on a wealth of experimental evidence, which you simply ignored.

Maksutov
2005-Jul-05, 11:11 AM
If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.
Can you provide references for these "numerous" presumably unexplained and unplanned "inflight adjustments" you claim are happening? Also, what are "adjustment tables" and can you provide a reference?
Alas, I cannot at this time, so take my statement under advisement. I can only refer you to this "gravity anomaly" (http://www.cage.curtin.edu.au/~will/grav_anoms.htm) page for now, only very indirectly related. This ESA page on trajectory corrections (http://sci.esa.int/science-e/www/object/index.cfm?fobjectid=33877&amp;fbodylongid=1437) is better, but still not it. If I find different, I'll report. I read something somewhere, but can't put my fingers on it now... :(
There will always be course corrections. Among other real factors, this is a result of metrology. All instruments used to measure something have intrinsic errors. These errors start with accuracy and precision.

In the case of measurements of a body's location in space, typically the farther away from the measuring instrument the body is, the greater the error factor that subtracts from the results reported by the measuring instrument. In the case of dynamic systems, the persons responsible for keeping the object on the intended path will use the data based on nominal values or values corrected for known or estimated errors. There is no absolute way to perform these adjustments precisely when based on data which have error factors.

The only feasible method is to spread the measurements and adjustments out over time such that the results of the corrections can be seen as a cumulative change either toward the planned path or away from it. This is fundamental to navigation.

Just as when the body was near the measuring instruments, when the body is approaching the target, the target affords a calibration standard that allows for finer and finer adjustments, since the adjustments can be readily compared to a known standard, in this case the target.

Of course the "burns" performed to achieve course corrections are variable too, and become another of the many factors which, if considered here, would overextend this post.

That is a simple overview of how course corrections work, and why they are necessary.

Course corrections have nothing to do with such non-factors as variable G, and other imaginary concepts.

worzel
2005-Jul-05, 02:32 PM
I don't think I've seen the actual value of G used anywhere in this thread, only the symbol. Doesn't that suggest to you, mutant gene, that the actual value is irrelevant to your argument, depsite your argument being about its value changing?

nutant gene 71
2005-Jul-06, 12:30 AM
I don't think I've seen the actual value of G used anywhere in this thread, only the symbol. Doesn't that suggest to you, mutant gene, that the actual value is irrelevant to your argument, depsite your argument being about its value changing?
Worzel, I treat this question as "hypothetical" because to date no hard evidence had been found to doubt Newton's G, which is G = 6.67E-11 Nm^2 kg^-2 (or m^3 kg^-1 s^-2). As far as we know, that is what G is, so the question of measurement in any kind of variable kilograms had not come up before, to my knowledge.

What is being explored is that IF Earth's G were instead tenfold, i.e., G' = 66.7E-11 Nm^2 kg^2, what would be the equivalence (inertial) mass of a cubic decimeter of water in kilograms? Would that same cube, which would not change in size, still be the same kilogram as before and still maintain the equivalence principle?

That essentially is the real question of this "hypothetical variable mass in a hypothetical variable G". I realize the new artefact for one kilograms is now a platinum-iridium rod kept under highly controlled conditions in France at the International Bureau of Weights and Measures, and that there had been proposals to replace it with something more scientific than a matter prototype, such as a count of atoms in a perfect crystal. At present, nothing else replaces the definition of one Kilogram of mass, a scalar, dimensionless, and represented by this artefact.

nutant gene 71
2005-Jul-06, 12:48 AM
If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.
Can you provide references for these "numerous" presumably unexplained and unplanned "inflight adjustments" you claim are happening? Also, what are "adjustment tables" and can you provide a reference?
Alas, I cannot at this time, so take my statement under advisement. I can only refer you to this "gravity anomaly" (http://www.cage.curtin.edu.au/~will/grav_anoms.htm) page for now, only very indirectly related. This ESA page on trajectory corrections (http://sci.esa.int/science-e/www/object/index.cfm?fobjectid=33877&amp;fbodylongid=1437) is better, but still not it. If I find different, I'll report. I read something somewhere, but can't put my fingers on it now... :(
There will always be course corrections. Among other real factors, this is a result of metrology. All instruments used to measure something have intrinsic errors. These errors start with accuracy and precision.

In the case of measurements of a body's location in space, typically the farther away from the measuring instrument the body is, the greater the error factor that subtracts from the results reported by the measuring instrument. In the case of dynamic systems, the persons responsible for keeping the object on the intended path will use the data based on nominal values or values corrected for known or estimated errors. There is no absolute way to perform these adjustments precisely when based on data which have error factors.

The only feasible method is to spread the measurements and adjustments out over time such that the results of the corrections can be seen as a cumulative change either toward the planned path or away from it. This is fundamental to navigation.

Just as when the body was near the measuring instruments, when the body is approaching the target, the target affords a calibration standard that allows for finer and finer adjustments, since the adjustments can be readily compared to a known standard, in this case the target.

Of course the "burns" performed to achieve course corrections are variable too, and become another of the many factors which, if considered here, would overextend this post.

That is a simple overview of how course corrections work, and why they are necessary.

Course corrections have nothing to do with such non-factors as variable G, and other imaginary concepts.
I believe what you are describing is a "chaotic effect" of computations over time and distance. Totally valid, and I take it for granted that it is so. What I had mentioned in my above regarding trajectoral corrections would have more to do with variability of the inertial mass itself.

Again, as I said to worzel above, there is no hard evidence this is so, and any anecdotal evidence of possible variable G, such as gravity anomalies, dark matter, or the Pioneers Anomaly, does not count as hard evidence, only possible reasons to research this further. The hypothetical question raised in this thread is whether or not our measurement of kilograms would be affected if G were something other than what we now know as a universal constant: G = 6.67E-11 Nm^2 kg^-2. Also bear in mind that should inertial mass be different (such as our space probes going into a higher G region) the overwhelming mass of solar system bodies, planets and moons, would still totally dominate per the G*M effect (meaning if G were different, M would be different too), so we would not necessarily see the variable inertial mass effect (of the probes), since this would be like comparing a speck of dust on the back of an elephant. The elephant controls the spec of dust, not the other way around.

The more pressing issue would then follow that if G were different, hypothetically, would it explain gravity anomalies such as neutron stars, giant planet atmospheres, dark matter, and possibly the Pioneers anamolous constant acceleration towards the Sun as well?

nutant gene 71
2005-Jul-06, 02:40 AM
Yes, I agree with you and Tobin Dax, that it doesn't matter what units of measuring mass are used (it could be stones), since mass is mass, always the same.
So what is the problem?
See my response to worzel above.

Nothing changed as far as the physical properties of the object of mass is concerned, whether weighed or accelerated. What I had been trying to show, why this is a problem for me, is that a unit of measure for mass derived in one G is going to be different for the same unit of measure in another G. In other words, 1 kg doesn not equal 10 kg for the same mass. So this unit of measure chosen is dependent on G where it is derived.
It is not.
As long as the Equivalence Principle holds, one can use purely dynamical measurements to determine a unit of mass, which is valid for gravitational measurements as well.
It has nothing to do with the value of G.
What is the Equivalence Principle for G' = 66.7E-11 Nm^2 kg^-2?

Let me show it another way. As shown above, G can be defined as G = r^2 * a/ m.
Not defined, but measured.
The formula you just wrote is a way to measure G.
Right. Once measured, that had been its "definition" as Newton's G, as a universal constant.

But then it must also be: 10G = r^2 * 10a/ m, so mass is still the same, only G' is ten times what we know.
You measure 10G because the acceleration is 10a.
And the acceleration is 10a because the gravitational force is ten times stronger.
And the gravitational force is ten times stronger because G' = 10 G.
Do you now understand that what you wrote enables to measure G from the acceleration of the mass? And that this relies on the Equivalence Principle, and does not affect the unit for mass?
Yes, I understand that, but there is a difference. What you had not understood, it seems, is that if G' were different, the inertial mass is affected, per Equivalence.

Since this is such a challenging question, let's spend some time here, and compare apples to apples. Let's say I have a basket of apples which weigh about 2.2 pounds (US) and I put them on a (European) scale in kilograms. What I find is this same basked of apples now weighs 1 kilogram. (I will address the difference between "weight" and "mass" in a minute.) Now put this 1 Kg. on a balance scale where on the other side is a cubic decimeter of water (or one kilogram platinum-irdidum rod) and what do we have? They balance, at least in Earth's gravity they balance. Now suppose that G' were different, let's say tenfold, for Earth. Would the scale still balance? Yes, the cubic decimeter of water would balance against the 1 Kg. apples (and the same for the scale using pounds), but with one difference: if you "weighed" the cubic decimeter of water in G' = 66.7E-11 Nm^2 kg^2, (meaning the Earth's gravitational acceleration was pulling on this cube with ten times the force), would it still be one kilogram compared to G where it was only a tenth, viz. G = 6.67E-11 Nm^2 kg^-2? If you put the basket of apples in 10 G (were it possible) and the cubic decimeter of water in 1 G, which would "weigh" more? Same mass.

To "weigh" mass is merely to subject it to Earth's gravitational acceleration, a = 9.8 m/s^2, so its "weight" could be said to be 9.8 kg m s^-2, but it's still the same kilogram. The apples still "weigh" one kilogram on the balance scale, their mass had not changed. If per equivalence you accelerated the basket of apples by the same rate, they would show the same "weight", but they are still one kilogram of mass, mass had not changed.

The question then remains, that if this one "kilogram" of apples were accelerated at 98 m/s^2 (where Earth's G' is tenfold), would it still be the same "kilogram"? No, the "weight" would change to 98 kg m/s^2, but the mass is still the same (1 kg) basket of apples, but now they weight 10 kgs. Same 1 kg. basket of apples, same mass, but Earth changed its G. What happened? Does the balance scale tip towards the apples rather than towards the one cubic decimeter of water? No, it does not. Now the cubic water mass is 10 kg. Does 1 kg = 10 kg? No sir, it does not.

Now can you see why I find this hypothetical question so challenging? I realize this is merely a "what if" question, but what if we find that other worlds or regions of space have a different G? If we find this, then it is truly exciting!

Now, if the Xians (per illustration above) think their 10G is merely Gx (one unit of G'), then of necessity their equation would be: Gx = r^2 * 10a/ (?m). This is the problem I'm trying to show. Should (?m) now not be, in Xian kilograms, 10m? So per "their" equivalence, kgx = 10kg in ours.
And how is that different from using another system of units?
Who is right, the Xians or the Earthians?

Why is this important, since it appears a rather mundane problem? I can see it as a problem when it comes to estimating the size and density of a foreign body should G there be different from ours.
If G has a different value from what we expect, and we relied only on G to determine the mass, then the measurement of the mass would give a different value within the same system of units.
It does not change the definition of units, becuase the definition does not rely on one particular value for G.
See my answer above regarding how 1 kg does not equal 10 kg.

Back to planet X, if Gx = 10G, and we're using our kilograms, then Gx = r^2 * 10a/ m, but if our kilograms are used, then "mx", planet X's mass, is 10 kg in our terms, but one kgx in theirs.
In Europe I am 180 cm tall.
In the US I am 71 inches tall.
Did my height change? No.
Your "height" would be equivalent to "r" in G = r^2 * a/ m, different issue.

I interpret this as us thinking their planet X should be either 10 times the size of Earth (which it is not) or 10 times the density.
Europeans do not think that the US are 2.5 times taller than Europe!
Rightly so. :)

Another way is to say that their planet, given its known size parameters, is actually ten times gravity denser than it should be.
The force of gravity would be stronger because G is larger, but dynamical measurements would still yield the same masses.

If I take with me a ruler that uses cms, I am still 180 cm even if I am in the US.
If I take a ruler that uses inches, I am 71 inches in Europe.
Of course.

In fact, if Gx is ten times G, the density of the planet need not be affected, only the results of what things would weigh there, and by equivalence, how things would respond to acceleration (and perhaps also affect their centripetal force, so affect their planetary spin).
Wrong.
The dynamical measurements would not be affected by a different value of G.
Only the gravitational force would be different, not the mass.
Wrong. Yes, only "the gravitational force would be different", but so also would the effective "kilograms" for mass. Same mass, but in 10 G, equivalence is now 10 kg (for the 1 kg basket of apples). Remember that it is G we're talking about for a planet that had not changed in size or volume, only the G changed. The cubic decimeter of water had not changed, only its effective "weight" had changed. Where m = 1 kg before (at 1 G), the mass (same mass, still cubic decimeter) is now m' = 10 kg (at 10 G). I see that as a change in the mass's "dynamic measurement", where a new "kilogram" defines (measures) the cubic decimeter of water. (Remember, 1 kg does not equal 10 kg for the same mass.)

Can you see where this is taking me?
You start from a very serious and deep misunderstanding of measurement unit, force, mass and acceleration.
This won't take you anywhere.
I am beginning to see you have a serious conceptual disconnect with what is being discussed here. It looks to me like you're still thinking 1 G, but ten times more powerful, in a 1 G universe. The conceptual adjustment necessary is to think in terms of a 10 G (or any variable G) universe, and then measure the new "kilograms" from there.

If, for example, a neutron star (so called) has a great mass equivalent (in our G terms) to several solar masses, we of necessity must think it is very dense for the amount of gravity it displays (hence the resultant spin is very great), that its composition must of necessity be something we do not have here in our vicinity of space, though we know it is very small by comparrison to our Sun. But if it is the G of the neutron star that is so great that it "appears" as if it were several solar masses (I'll skip for now the reasoning why this might be), then its size may be even larger than we estimate (as if made of only neutrons), and density not necessarily so compact. In fact, it may be not too much denser than our own Sun, and perhaps only comparatively smaller if its internal radiation pressure is less, meaning it was a small star to start with.

Take another example. I read somewhere that Jupiter may have a rocky core about two or three Earth masses. Whether or not this is true, I can't confirm since I never saw how this was arrived at, whether through radar probing of Jupiter's interior, or derived from atmospheric occultation, or from ephemeris spin data(?). But if true, given a constant G, how could a small rocky core hold such a vast atmosphere?
The atmosphere has amss of its own, and it interacts gravitationally.
How do you think the Sun formed: it does not even have a rocky core!
I am refering to the current need to invent "exotic matter" to explain the very high gravity in neutron stars, namely a neutron only composition. This new way of seeing mass no longer requires anything so exotic. The Sun's mass is very great, though not totally understood, in my opinion. Neither is Jupiter's composition well understood. Much of what we think we know on these matters are only one step removed from speculation. Steven Hawking at one time said the Sun's interior has a mini-black hole, for example. I realize the mass of the atmosphere plays into the equation of the planet's total mass, but I question how a "bubble" of atmosphere can hold itself together in its orbit around the Sun. I suspect, my spec, that Jupiter has a "small" rocky core.

Seismically, we're not even sure what the Earth's core is made of, though for now we have it as "solidified" iron due to the planet's gravitational pressure, but seismic waves don't go there (which gives the "hollow Earth" people cause to cheer, wrongly in my opinion). Why are seismic waves deflected from it? How much gravity is there at the center of the Earth, if mass is totally ambient and distributed evenly in all directions away from the center? Anyway, it looks like we really don't know what's at the center, so to explain the magnetic field, we gave it an iron core. Okay, I'm cool with that, though in the future we may discover otherwise. Ditto for the Sun and Jupiter, etc. For now, I don't think we really know these things, so only speculative theories.

Unless the G is much greater than supposed, it is virtually impossible as a gas.
Why?
On what are your expectations based?
Small rocky core with greater G would be one possibility, but I don't know either.

(This may be another reason why speculations on Jupiter's atmosphere is that it has a liquid core?) I know all the arguments against why this cannot be true, how the springs on Huygens worked properly, etc. (in fact I have no way of knowing whether or not my hypothetical planetary G' calculations are right, as shown earlier), but if mass is measured in Earth's kg, then Jupiter's atmopshere cannot be possible for such a small rocky core. A small rocky core can hold a very large atmopshere only if the acceleration towards the center of mass, the gravity, is much greater for the size and density of the planet would otherwise allow.
You still assume that a gas does not interact gravitationally.
On what is this "idea" based?
A "bubble" of gas would consolidate gravitationally into something at its center, much like our Earth's molten core consolidates into a solid core, if this is true. How would such a buble of gas exist otherwise? We're not talking about soap bubbles here (where the outer surface is held together by water tension), but vast collections of molecules put into immense spin, around what? More gas? Since I obviously don't know, I'd be curious as to your idea of what's inside Jupiter, really! :lol:

This is why I think the kilograms used is important, because if they are not adjusted for local G conditions, like in the neutron star example above, we may be overestimating density versus what it really is. A neutron star may not be so dense, only its mass (due to much higher G) acts as if it were.

There is also a practical side to this question (on hypothetical mass in a hypothetical variable G), and that has to do with how space probes will behave near any under-over estimated planetary body.
Weird... Huygens landed as planned, didn't it?
Yes. Though, if I may add, Jerry has reasons to think it landed with a clang, but that thread's been closed, so no further comment.

If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.
What makes you think that we did not get that right?
It was just an idea, not too attached to it. We may have it right, with gravitation assist, but we may not know why we have it right. I'm actually rather intrigued by how we maneuver these things to get to where we want them. I think Huygens is a great success, and so is Deep Impact. =D>

In effect, it cleans up our engineering task for space flight with a better physics. It's not that we fail to get there, since using gravitational assist trajectories are of necessity self correcting (G * M as a product value is still the same, even if G and M are wrong), but that we may be handicapped with a constant G. A better way may be to use local kilograms (as opposed to Earth kilograms) to work out the dynamics of how a spaceprobe will behave in the vicinity of another planet, hypothetically.
Dynamics does not need "local" kilograms, because the kilogram is the same.
Ah, "kilograms are the same", so we're back to square one! "Who was right?" The Xians or the Earthians?

So, can a (variable) measure of mass size and density improve on Newton's physics for a variable G?
You assume that Newtonian physics relies on a universal G.
I already explained to you several times that taking into account a variable G is mathematically trivial, and it would be obvious from the orbits of objects like comets.
You have to invoke a magical re-adjustment of mass to keep the orbits consistent with a constant G, in which the change of mass would be obvious from measurements of the inertial mass and moment of inertia (in objects like the probes we sent out).
All the observations disprove your idea of a "variable G + mass compensation"
This revives the "why are comets not showing orbital anomalies" question. I don't know if they are or not, since we don't have two way communications with them. By all "appearances" they seem to be where they're supposed to be, around the Sun in a very large elliptical orbit. Remember, comets are only specs of dust on the Sun's back.

That ultimately is the question. Can we better understand very distant bodies, such as neutron stars with variable kg, adjusted for local G? Would our overall understanding of cosmology be improved, if G is found to be a variable (something we still do not know)?
None seem to have given us a groundbreaking new understanding of the Universe.
What are you refering to? MOND? Dark Matter? Testing for Einstein's Relativity from Jupiter using the Sun's gravitational lensing? Like I said before, there is no "hard" evidence for a variable G, only anecdotal soft evidence, and even that is contested. Ibid. We don't know.

These were my reasons for bringing up this question. (At this point, however, I don't even want to get close to what this means for Einstein's General Relativity theory.) For this reason, I titled this question as "hypothetical" only, until such time that we find G to be otherwise than now postulated.
The postulate is based on a wealth of experimental evidence, which you simply ignored.
No, not ignored, but only considering as possible theoretical explanations without final verdict.

So, who was right? The Xians or the Earthians? (http://www.badastronomy.com/phpBB/viewtopic.php?p=495886&amp;sid=97f7daa21d2ad8dd1e6f64f b88c4697f#495886) :)

2005-Jul-06, 02:55 AM
Heres my two cents, mass is the amount of stuff in an object, weight is the effect that two objects have on one another say the earth and a person which is F in both equations.
f=m*a or sometimes written f=m*g is the force of acceraleration in gravitional feild of the the earth and can change if the mass of the larger object changed to like a star or blackhole or another planet and the f= m*m*G/(r^2) is just two particles in isolated system like deep
space but this G is a constant of the universe. At least that my version of whats going on i could be could wrong I am often. In metric or SI the unit of weight is the newton in imperial system it is the lb so if you did change the G the weight of the objects may change but the amount of material inside the objects has not changed. and by the way since I am in canada my 2 cents is worth about a penny us currency. sorry for any grammer or spelling mistakes and if I had scientific I could spit the numbers for up above I am at work so I dont have the calculator and hopefully i do not have re edit again.

Celestial Mechanic
2005-Jul-06, 03:40 AM
A 10 kg mass weighs 98 Newtons on Earth. It weighs about 16 Newtons on the Moon, but still has a mass of 10 kg.

If somehow G were to be made 10 times bigger than it is now and if the Earth does not compress and get smaller then that 10 kg mass will still be 10 kg but it will weigh 980 Newtons.

This is elementary physics. Please review a good elementary physics text on the distinction between mass and weight.

worzel
2005-Jul-06, 09:07 AM
I don't think I've seen the actual value of G used anywhere in this thread, only the symbol. Doesn't that suggest to you, mutant gene, that the actual value is irrelevant to your argument, depsite your argument being about its value changing?
Worzel, I treat this question as "hypothetical" because to date no hard evidence had been found to doubt Newton's G, which is G = 6.67E-11 Nm^2 kg^-2 (or m^3 kg^-1 s^-2). As far as we know, that is what G is, so the question of measurement in any kind of variable kilograms had not come up before, to my knowledge.

What is being explored is that IF Earth's G were instead tenfold, i.e., G' = 66.7E-11 Nm^2 kg^2,
Well simply quoting the number over and over now doesn't change my point. You haven't used the value in any of your arguments.

what would be the equivalence (inertial) mass of a cubic decimeter of water in kilograms? Would that same cube, which would not change in size, still be the same kilogram as before
Depends on what you mean by kilogram. If you mean the amount of mass then yes, if you mean how much does it weigh then no. The former is the correct usage of the term kilogram, the latter is a convenience because we live in an almost constant gravitational field where a 1kg "weight" is the amount of mass that happens to weigh 9.8N.

and still maintain the equivalence principle?
In Newtonian mechanics it was a bit of mystery why the inertial mass equals the gravitional mass - but given that they're equal, I don't see any reason to suppose they wouldn't be if G changed because that would require a big coincidence: that they are equal now!

Einstein's resolution was that spacetime is warped so that free falling masses are following Newton's first law (they keep going in a straight line) as best they can in warped spacetime (they follow geodesics). For the equivalence to break there would have to be a different warping of spacetime for different masses starting out on the same geodesic.

As I see it, a more interesting question would be: if intertia is resistance to the gravitional field (http://chaos.fullerton.edu/~jimw/general/inertia/) of the universe as a whole (Mach, Lense Thirring), then is there even any meaning to the quesion "what if G changed?" ?

papageno
2005-Jul-06, 09:24 AM
Again, as I said to worzel above, there is no hard evidence this is so, and any anecdotal evidence of possible variable G, such as gravity anomalies, dark matter, or the Pioneers Anomaly, does not count as hard evidence, only possible reasons to research this further. The hypothetical question raised in this thread is whether or not our measurement of kilograms would be affected if G were something other than what we now know as a universal constant: G = 6.67E-11 Nm^2 kg^-2.
And this question has been thoroughly answered with no.

Also bear in mind that should inertial mass be different...
It is not.

...(such as our space probes going into a higher G region) the overwhelming mass of solar system bodies, planets and moons, would still totally dominate per the G*M effect (meaning if G were different, M would be different too), so we would not necessarily see the variable inertial mass effect (of the probes), since this would be like comparing a speck of dust on the back of an elephant. The elephant controls the spec of dust, not the other way around.
So, why do you think anomalies like the Pioneer anomlies, have anything to do with G?

papageno
2005-Jul-06, 10:19 AM
Yes, I agree with you and Tobin Dax, that it doesn't matter what units of measuring mass are used (it could be stones), since mass is mass, always the same.
So what is the problem?
See my response to worzel above.

Nothing changed as far as the physical properties of the object of mass is concerned, whether weighed or accelerated. What I had been trying to show, why this is a problem for me, is that a unit of measure for mass derived in one G is going to be different for the same unit of measure in another G. In other words, 1 kg doesn not equal 10 kg for the same mass. So this unit of measure chosen is dependent on G where it is derived.
It is not.
As long as the Equivalence Principle holds, one can use purely dynamical measurements to determine a unit of mass, which is valid for gravitational measurements as well.
It has nothing to do with the value of G.
What is the Equivalence Principle for G' = 66.7E-11 Nm^2 kg^-2?
What part of the Equivalence Principle does not depend on the value of G is not clear?
To the best of our knowledge, the actual value of G, in any units, has no effect whatsoever on the equivalence of inertail and gravitational mass:
m(inertial) = m(gravit.), whatever the units and whatever the value of G.

Let me show it another way. As shown above, G can be defined as G = r^2 * a/ m.
Not defined, but measured.
The formula you just wrote is a way to measure G.
Right. Once measured, that had been its "definition" as Newton's G, as a universal constant.
Wrong.
The definition of G as physical quantity is in Newton's law for gravitation.
The formula above is a way to measure the value of G.

But then it must also be: 10G = r^2 * 10a/ m, so mass is still the same, only G' is ten times what we know.
You measure 10G because the acceleration is 10a.
And the acceleration is 10a because the gravitational force is ten times stronger.
And the gravitational force is ten times stronger because G' = 10 G.
Do you now understand that what you wrote enables to measure G from the acceleration of the mass? And that this relies on the Equivalence Principle, and does not affect the unit for mass?
Yes, I understand that, but there is a difference. What you had not understood, it seems, is that if G' were different, the inertial mass is affected, per Equivalence.
No.
You still do not understand that the Equivalence Principle is independent of the value of G.
How many times do I have to refer you to the equations?

Newtons' formula for gravitation:
F = G * (m*M) / r^2 (1)

Newton's second law:
F = M * a (2)

Equivalence principle: M in (1) is the same as M in (2).

M(1) = M(2): where does G enter?

Since this is such a challenging question, let's spend some time here, and compare apples to apples. Let's say I have a basket of apples which weigh about 2.2 pounds (US) and I put them on a (European) scale in kilograms. What I find is this same basked of apples now weighs 1 kilogram. (I will address the difference between "weight" and "mass" in a minute.) Now put this 1 Kg. on a balance scale where on the other side is a cubic decimeter of water (or one kilogram platinum-irdidum rod) and what do we have? They balance, at least in Earth's gravity they balance. Now suppose that G' were different, let's say tenfold, for Earth. Would the scale still balance? Yes, the cubic decimeter of water would balance against the 1 Kg. apples (and the same for the scale using pounds), but with one difference: if you "weighed" the cubic decimeter of water in G' = 66.7E-11 Nm^2 kg^2, (meaning the Earth's gravitational acceleration was pulling on this cube with ten times the force), would it still be one kilogram compared to G where it was only a tenth, viz. G = 6.67E-11 Nm^2 kg^-2? If you put the basket of apples in 10 G (were it possible) and the cubic decimeter of water in 1 G, which would "weigh" more? Same mass.
Weight is a force!
Changing G changes the force, hence the weight. It does not affect the mass.

To "weigh" mass is merely to subject it to Earth's gravitational acceleration, a = 9.8 m/s^2, so its "weight" could be said to be 9.8 kg m s^-2, but it's still the same kilogram. The apples still "weigh" one kilogram on the balance scale, their mass had not changed. If per equivalence you accelerated the basket of apples by the same rate, they would show the same "weight", but they are still one kilogram of mass, mass had not changed.

The question then remains, that if this one "kilogram" of apples were accelerated at 98 m/s^2 (where Earth's G' is tenfold), would it still be the same "kilogram"? No, the "weight" would change to 98 kg m/s^2, but the mass is still the same (1 kg) basket of apples, but now they weight 10 kgs.
kilogram is a unit of mass, not weight.
The weight you get is (1 kg)*(local g) N, where "local g" is the local gravitational acceleration on the surface of the planet; on Earth local g = 9.8 m/s^2, so 1 kg -> 9.8 N of weight.
On your planet, local g = 98 m/s^2, so 1 kg -> 98 N of weight.

Same 1 kg. basket of apples, same mass, but Earth changed its G. What happened? Does the balance scale tip towards the apples rather than towards the one cubic decimeter of water? No, it does not. Now the cubic water mass is 10 kg. Does 1 kg = 10 kg? No sir, it does not.
I see that you are confusing weight and mass.
1 kg(weight on Earth) = 1 kg(mass)*g = 9.8 N.
On your planet, g is different: g' = 10 g, hence
1 kg(weight on planet) = 1 kg(mass)*g' = 98 N.

Now can you see why I find this hypothetical question so challenging? I realize this is merely a "what if" question, but what if we find that other worlds or regions of space have a different G? If we find this, then it is truly exciting!
I see only a confusion about the unit kilogram.
It is not for weight, but for mass.

Now, if the Xians (per illustration above) think their 10G is merely Gx (one unit of G'), then of necessity their equation would be: Gx = r^2 * 10a/ (?m). This is the problem I'm trying to show. Should (?m) now not be, in Xian kilograms, 10m? So per "their" equivalence, kgx = 10kg in ours.
And how is that different from using another system of units?
Who is right, the Xians or the Earthians?
They are both right: they are just using different units.
Europeans are right when they say that I am 180 cm tall;
US citizens are right when they say that I am 71 inches tall.

Why is this important, since it appears a rather mundane problem? I can see it as a problem when it comes to estimating the size and density of a foreign body should G there be different from ours.
If G has a different value from what we expect, and we relied only on G to determine the mass, then the measurement of the mass would give a different value within the same system of units.
It does not change the definition of units, becuase the definition does not rely on one particular value for G.
See my answer above regarding how 1 kg does not equal 10 kg.

Back to planet X, if Gx = 10G, and we're using our kilograms, then Gx = r^2 * 10a/ m, but if our kilograms are used, then "mx", planet X's mass, is 10 kg in our terms, but one kgx in theirs.
In Europe I am 180 cm tall.
In the US I am 71 inches tall.
Did my height change? No.
Your "height" would be equivalent to "r" in G = r^2 * a/ m, different issue.
Your issue is a misunderstanding of the conventional units for wieght and mass.

I interpret this as us thinking their planet X should be either 10 times the size of Earth (which it is not) or 10 times the density.
Europeans do not think that the US are 2.5 times taller than Europe!
Rightly so. :)
So, you are worng if you think that mass changes if G is different.

Another way is to say that their planet, given its known size parameters, is actually ten times gravity denser than it should be.
The force of gravity would be stronger because G is larger, but dynamical measurements would still yield the same masses.

If I take with me a ruler that uses cms, I am still 180 cm even if I am in the US.
If I take a ruler that uses inches, I am 71 inches in Europe.
Of course.
So, why do you say that the unit kg for mass is different on your planet?

In fact, if Gx is ten times G, the density of the planet need not be affected, only the results of what things would weigh there, and by equivalence, how things would respond to acceleration (and perhaps also affect their centripetal force, so affect their planetary spin).
Wrong.
The dynamical measurements would not be affected by a different value of G.
Only the gravitational force would be different, not the mass.
Wrong. Yes, only "the gravitational force would be different", but so also would the effective "kilograms" for mass.
In this case, there is no such thing as "effective kilograms".
The kg unit for mass odes not change.
What changed is the gravitational acceleration at the surface, hence the gravitational force is difeerent = different weight.

Same mass, but in 10 G, equivalence is now 10 kg (for the 1 kg basket of apples). Remember that it is G we're talking about for a planet that had not changed in size or volume, only the G changed. The cubic decimeter of water had not changed, only its effective "weight" had changed. Where m = 1 kg before (at 1 G), the mass (same mass, still cubic decimeter) is now m' = 10 kg (at 10 G).
Only if you use kg for weight, which is wrong.
You should use N, the unit for force.

I see that as a change in the mass's "dynamic measurement", where a new "kilogram" defines (measures) the cubic decimeter of water. (Remember, 1 kg does not equal 10 kg for the same mass.)
And you see wrong.
Becuase the inertial mass has not changed.

Can you see where this is taking me?
You start from a very serious and deep misunderstanding of measurement unit, force, mass and acceleration.
This won't take you anywhere.
I am beginning to see you have a serious conceptual disconnect with what is being discussed here.
You do not grasp the difference between weight and mass, confusing the unit kg(mass) with the unit kg(weight on Earth) = 1 kg(mass)*9.8 m/s^2.

It looks to me like you're still thinking 1 G, but ten times more powerful, in a 1 G universe. The conceptual adjustment necessary is to think in terms of a 10 G (or any variable G) universe, and then measure the new "kilograms" from there.
Wrong.
I already explaiend to you uncountable times that a variable G is not an exotic concept, but unsupported by observations.

If, for example, a neutron star (so called) has a great mass equivalent (in our G terms) to several solar masses, we of necessity must think it is very dense for the amount of gravity it displays (hence the resultant spin is very great), that its composition must of necessity be something we do not have here in our vicinity of space, though we know it is very small by comparrison to our Sun. But if it is the G of the neutron star that is so great that it "appears" as if it were several solar masses (I'll skip for now the reasoning why this might be), then its size may be even larger than we estimate (as if made of only neutrons), and density not necessarily so compact. In fact, it may be not too much denser than our own Sun, and perhaps only comparatively smaller if its internal radiation pressure is less, meaning it was a small star to start with.

Take another example. I read somewhere that Jupiter may have a rocky core about two or three Earth masses. Whether or not this is true, I can't confirm since I never saw how this was arrived at, whether through radar probing of Jupiter's interior, or derived from atmospheric occultation, or from ephemeris spin data(?). But if true, given a constant G, how could a small rocky core hold such a vast atmosphere?
The atmosphere has amss of its own, and it interacts gravitationally.
How do you think the Sun formed: it does not even have a rocky core!
I am refering to the current need to invent "exotic matter" to explain the very high gravity in neutron stars, namely a neutron only composition.

This new way of seeing mass no longer requires anything so exotic. The Sun's mass is very great, though not totally understood, in my opinion. Neither is Jupiter's composition well understood. Much of what we think we know on these matters are only one step removed from speculation. Steven Hawking at one time said the Sun's interior has a mini-black hole, for example. I realize the mass of the atmosphere plays into the equation of the planet's total mass, but I question how a "bubble" of atmosphere can hold itself together in its orbit around the Sun. I suspect, my spec, that Jupiter has a "small" rocky core.

Seismically, we're not even sure what the Earth's core is made of, though for now we have it as "solidified" iron due to the planet's gravitational pressure, but seismic waves don't go there (which gives the "hollow Earth" people cause to cheer, wrongly in my opinion). Why are seismic waves deflected from it? How much gravity is there at the center of the Earth, if mass is totally ambient and distributed evenly in all directions away from the center? Anyway, it looks like we really don't know what's at the center, so to explain the magnetic field, we gave it an iron core. Okay, I'm cool with that, though in the future we may discover otherwise. Ditto for the Sun and Jupiter, etc. For now, I don't think we really know these things, so only speculative theories.
Your opinion does not seem to be based on a real understanding.

Unless the G is much greater than supposed, it is virtually impossible as a gas.
Why?
On what are your expectations based?
Small rocky core with greater G would be one possibility, but I don't know either.

(This may be another reason why speculations on Jupiter's atmosphere is that it has a liquid core?) I know all the arguments against why this cannot be true, how the springs on Huygens worked properly, etc. (in fact I have no way of knowing whether or not my hypothetical planetary G' calculations are right, as shown earlier), but if mass is measured in Earth's kg, then Jupiter's atmopshere cannot be possible for such a small rocky core. A small rocky core can hold a very large atmopshere only if the acceleration towards the center of mass, the gravity, is much greater for the size and density of the planet would otherwise allow.
You still assume that a gas does not interact gravitationally.
On what is this "idea" based?
A "bubble" of gas would consolidate gravitationally into something at its center, much like our Earth's molten core consolidates into a solid core, if this is true.[/quote]
What makes you think that the material composing Jupiter should behave exactly as the material composing Earth?

How would such a buble of gas exist otherwise? We're not talking about soap bubbles here (where the outer surface is held together by water tension), but vast collections of molecules put into immense spin, around what? More gas? Since I obviously don't know, I'd be curious as to your idea of what's inside Jupiter, really! :lol:
I see no numbers to support your idea.
Can you show that the mass of Jupiter cannot be held together by gravity?

This is why I think the kilograms used is important, because if they are not adjusted for local G conditions, like in the neutron star example above, we may be overestimating density versus what it really is. A neutron star may not be so dense, only its mass (due to much higher G) acts as if it were.

There is also a practical side to this question (on hypothetical mass in a hypothetical variable G), and that has to do with how space probes will behave near any under-over estimated planetary body.
Weird... Huygens landed as planned, didn't it?
Yes. Though, if I may add, Jerry has reasons to think it landed with a clang, but that thread's been closed, so no further comment.

If we can get that right, then we can have a more direct physics to plot flight paths without having to use adjustment tables, and then numerous inflight adjustments.
What makes you think that we did not get that right?
It was just an idea, not too attached to it. We may have it right, with gravitation assist, but we may not know why we have it right. I'm actually rather intrigued by how we maneuver these things to get to where we want them. I think Huygens is a great success, and so is Deep Impact. =D>

In effect, it cleans up our engineering task for space flight with a better physics. It's not that we fail to get there, since using gravitational assist trajectories are of necessity self correcting (G * M as a product value is still the same, even if G and M are wrong), but that we may be handicapped with a constant G. A better way may be to use local kilograms (as opposed to Earth kilograms) to work out the dynamics of how a spaceprobe will behave in the vicinity of another planet, hypothetically.
Dynamics does not need "local" kilograms, because the kilogram is the same.
Ah, "kilograms are the same", so we're back to square one! "Who was right?" The Xians or the Earthians?
Consider that your questions are all based on your misunderstanding of unit of kilograms for mass and for weight.

So, can a (variable) measure of mass size and density improve on Newton's physics for a variable G?
You assume that Newtonian physics relies on a universal G.
I already explained to you several times that taking into account a variable G is mathematically trivial, and it would be obvious from the orbits of objects like comets.
You have to invoke a magical re-adjustment of mass to keep the orbits consistent with a constant G, in which the change of mass would be obvious from measurements of the inertial mass and moment of inertia (in objects like the probes we sent out).
All the observations disprove your idea of a "variable G + mass compensation"
This revives the "why are comets not showing orbital anomalies" question. I don't know if they are or not, since we don't have two way communications with them. By all "appearances" they seem to be where they're supposed to be, around the Sun in a very large elliptical orbit. Remember, comets are only specs of dust on the Sun's back.
So, you have no idea why we get it right.
Maybe because we are right?

That ultimately is the question. Can we better understand very distant bodies, such as neutron stars with variable kg, adjusted for local G? Would our overall understanding of cosmology be improved, if G is found to be a variable (something we still do not know)?
None seem to have given us a groundbreaking new understanding of the Universe.
What are you refering to? MOND? Dark Matter? Testing for Einstein's Relativity from Jupiter using the Sun's gravitational lensing? Like I said before, there is no "hard" evidence for a variable G, only anecdotal soft evidence, and even that is contested. Ibid. We don't know.
Actually, you do not know.

These were my reasons for bringing up this question. (At this point, however, I don't even want to get close to what this means for Einstein's General Relativity theory.) For this reason, I titled this question as "hypothetical" only, until such time that we find G to be otherwise than now postulated.
The postulate is based on a wealth of experimental evidence, which you simply ignored.
No, not ignored, but only considering as possible theoretical explanations without final verdict.
Where did you consider it?

Tassel
2005-Jul-06, 01:18 PM
Heres my two cents, mass is the amount of stuff in an object, weight is the effect that two objects have on one another say the earth and a person which is F in both equations.
f=m*a or sometimes written f=m*g is the force of acceraleration in gravitional feild of the the earth and can change if the mass of the larger object changed to like a star or blackhole or another planet and the f= m*m*G/(r^2) is just two particles in isolated system like deep
space but this G is a constant of the universe. At least that my version of whats going on i could be could wrong I am often.
Particularly because this thread is in the "General Astronomy" forum, I think that it should be pointed out that you're right on the money, David. All evidence points to a constant G and it seems to me that you correctly understand the difference between weight and mass.

nutant gene 71
2005-Jul-06, 11:05 PM
Heres my two cents, mass is the amount of stuff in an object, weight is the effect that two objects have on one another say the earth and a person which is F in both equations.
f=m*a or sometimes written f=m*g is the force of acceraleration in gravitional feild of the the earth and can change if the mass of the larger object changed to like a star or blackhole or another planet and the f= m*m*G/(r^2) is just two particles in isolated system like deep
space but this G is a constant of the universe. At least that my version of whats going on i could be could wrong I am often. In metric or SI the unit of weight is the newton in imperial system it is the lb so if you did change the G the weight of the objects may change but the amount of material inside the objects has not changed. and by the way since I am in canada my 2 cents is worth about a penny us currency. sorry for any grammer or spelling mistakes and if I had scientific I could spit the numbers for up above I am at work so I dont have the calculator and hopefully i do not have re edit again. (bold mine)
Thanks for your input davidlpf, 2 cents is good, as indeed I should earnestly thank ALL who had contributed to this debate on hypothetical variable mass. Turning over these ideas like this is an important form of conceptual examination. It is not my wish to be proven right on this issue, but rather to find a right conceptual way to cope with a universe if our concept of G should prove wrong.

Yes, we know this is true if "G is a constant in the universe", and if this is so, then there is no debate here. This hypothetical question stems from a possibility that at some future point we may discover than the 1 G we know is not everywhere the same, and that would change things. In fact, I suspect it might even change how we model cosmology in general, which could be a significant change. But not there yet, since at present there is no firm reason to doubt the 1 G universe.

nutant gene 71
2005-Jul-06, 11:18 PM
A 10 kg mass weighs 98 Newtons on Earth. It weighs about 16 Newtons on the Moon, but still has a mass of 10 kg.

If somehow G were to be made 10 times bigger than it is now and if the Earth does not compress and get smaller then that 10 kg mass will still be 10 kg but it will weigh 980 Newtons.

This is elementary physics. Please review a good elementary physics text on the distinction between mass and weight.
Thanks C.M., that is how I see it too, in a 1 G universe scenario. I will show below the fundamental difference between that and a variable G universe scenario, should there be a different G elsewhere. It may have a profound effect on lots of things we now cannot seem to explain. Remember that the "kilogram" is dimensionless, a scalar, unlike G, which is a cubic meter per kilogram per second squared (m^3 kg^-1 s^-2) dimensional. Take out the kilograms, and you still have the cubic meters per second squared, so this is something not contingent on the kilogram itself, but a "proportional" force in its own right which affects the mass. The mass is in kilograms. Now, the current existing physics books do not address the possibility that these kilograms, really an arbitrary unit developed from the (metric system) cubic decimeter of water, which we use to measure mass are capable of being anything else. That's what's being explored here.

nutant gene 71
2005-Jul-06, 11:55 PM
I don't think I've seen the actual value of G used anywhere in this thread, only the symbol. Doesn't that suggest to you, mutant gene, that the actual value is irrelevant to your argument, depsite your argument being about its value changing?
Worzel, I treat this question as "hypothetical" because to date no hard evidence had been found to doubt Newton's G, which is G = 6.67E-11 Nm^2 kg^-2 (or m^3 kg^-1 s^-2). As far as we know, that is what G is, so the question of measurement in any kind of variable kilograms had not come up before, to my knowledge.

What is being explored is that IF Earth's G were instead tenfold, i.e., G' = 66.7E-11 Nm^2 kg^2,
Well simply quoting the number over and over now doesn't change my point. You haven't used the value in any of your arguments.

what would be the equivalence (inertial) mass of a cubic decimeter of water in kilograms? Would that same cube, which would not change in size, still be the same kilogram as before
Depends on what you mean by kilogram. If you mean the amount of mass then yes, if you mean how much does it weigh then no. The former is the correct usage of the term kilogram, the latter is a convenience because we live in an almost constant gravitational field where a 1kg "weight" is the amount of mass that happens to weigh 9.8N.

"Mass (http://simple.wikipedia.org/wiki/Mass) is the amount of matter in a body. The mass of an object is the same everywhere."

That's the long and short of it, as papageno and others have explained here. This is how we now see it, period. The question I am raising is whether or not this is still true if G is different elsewhere.

and still maintain the equivalence principle?
In Newtonian mechanics it was a bit of mystery why the inertial mass equals the gravitional mass - but given that they're equal, I don't see any reason to suppose they wouldn't be if G changed because that would require a big coincidence: that they are equal now!
Exactly, there seems no apparent reason to doubt the Equivalence Principle if G changed, but that would be in a 1 G universe. What if, let's say where exists Dark Matter, G were different? If inertial mass and gravitational mass are the same (http://www.npl.washington.edu/eotwash/equiv.html), what are they in a different G?

Einstein's resolution was that spacetime is warped so that free falling masses are following Newton's first law (they keep going in a straight line) as best they can in warped spacetime (they follow geodesics). For the equivalence to break there would have to be a different warping of spacetime for different masses starting out on the same geodesic.

As I see it, a more interesting question would be: if intertia is resistance to the gravitional field (http://chaos.fullerton.edu/~jimw/general/inertia/) of the universe as a whole (Mach, Lense Thirring), then is there even any meaning to the quesion "what if G changed?" ?
That depends on whether Machian background gravity is 1 G or X G. If the universal space vacuum, at say past the Oort Cloud, is some G much greater than here, then the Machian principle would not apply directly, unless it was figured out in a different G. (What would the Machian gravity background be in a Dark Matter Galaxy, for example?) The same would be true of the geodesic, where its curvature would be different. Again, we can't calculate any of this because we have no reason to assume G is different anywhere else, and thus we operate conceptually in a 1 G universe. But if gravitational mass and inertial mass are the same, the resulting values for a different G should be different.

nutant gene 71
2005-Jul-07, 02:48 AM
Yes, I agree with you and Tobin Dax, that it doesn't matter what units of measuring mass are used (it could be stones), since mass is mass, always the same.
So what is the problem?
See my response to worzel above.

Nothing changed as far as the physical properties of the object of mass is concerned, whether weighed or accelerated. What I had been trying to show, why this is a problem for me, is that a unit of measure for mass derived in one G is going to be different for the same unit of measure in another G. In other words, 1 kg doesn not equal 10 kg for the same mass. So this unit of measure chosen is dependent on G where it is derived.
It is not.
As long as the Equivalence Principle holds, one can use purely dynamical measurements to determine a unit of mass, which is valid for gravitational measurements as well.
It has nothing to do with the value of G.
What is the Equivalence Principle for G' = 66.7E-11 Nm^2 kg^-2?
What part of the Equivalence Principle does not depend on the value of G is not clear?
To the best of our knowledge, the actual value of G, in any units, has no effect whatsoever on the equivalence of inertail and gravitational mass:
m(inertial) = m(gravit.), whatever the units and whatever the value of G.

Per the Equivalence Principle, gravitational mass equals inertial mass, always. "Gravitational" mass is a function of G, as my basket of apples illustration shows above, which is equivalent to its "inertial" mass. The same basket, or cubic decimeter of water, can be either one kilograms (in 1 G), or ten kilograms (in 10 G). If we want the whole universe to be figured in Earth's arbitrary measure for mass, our kilograms, then 10 G means the inertial-gravitational masses are always 10 kilograms. But this causes a problem with how masses interact locally, because if kilograms for their local measure of G are different from ours, then each kilogram will need to be "locally" different from ours. The ramifications of this is that local mass interacts differently than here on Earth. If so, then using Earth's kilograms becomes a poor, and ingenuous choice, since it fails to explain how in a different gravitational G "proportional" masses may hold together in ways that our 1 G (1 kg) cannot explain. Let me explain this further in your next.

Let me show it another way. As shown above, G can be defined as G = r^2 * a/ m.
Not defined, but measured.
The formula you just wrote is a way to measure G.
Right. Once measured, that had been its "definition" as Newton's G, as a universal constant.
Wrong.
The definition of G as physical quantity is in Newton's law for gravitation.
The formula above is a way to measure the value of G.
Why would you say G is a "physical quantity"? Is it not merely a "proportional" quantity between gravitationally attracted masses? If G attracts at 1 G, that is the proportional attraction between the masses, which are measured in kilograms. If G's proportional attracts at ten times our 1 G, then the proportional attraction between masses increases by tenfold. But if each mass is now tenfold in terms of its (equivalent) inertial mass, then the attraction is ten times ten (two bodies interacting), so the interaction between them is what? If each side has an "eqivalent" mass that is ten times greater, and the G "proportional" between them is ten times greater, saying merely that the mass is now 10 kilograms is not enough, because they attract by a larger proportional. And that, really, is why it is important to redefine our (arbitrary) kilogram in a different G scenario.

This is also why I brought up the question in the first place, because I don't know if the answer is tenfold or one hundredfold. This question cannot be raised in a 1 G universe, but it can be raised in a (hypothetical) variable G universe. So, yes, I understand what you are saying, if the universe is only 1 G throughout, but I am forced to disagree with you, because it does not apply in a variable G universe.

But then it must also be: 10G = r^2 * 10a/ m, so mass is still the same, only G' is ten times what we know.
You measure 10G because the acceleration is 10a.
And the acceleration is 10a because the gravitational force is ten times stronger.
And the gravitational force is ten times stronger because G' = 10 G.
Do you now understand that what you wrote enables to measure G from the acceleration of the mass? And that this relies on the Equivalence Principle, and does not affect the unit for mass?
Yes, I understand that, but there is a difference. What you had not understood, it seems, is that if G' were different, the inertial mass is affected, per Equivalence.
No.
You still do not understand that the Equivalence Principle is independent of the value of G.
How many times do I have to refer you to the equations?
No. The Equivalence Principle is NOT independent of the value of G. We know the inertial mass and gravitational mass are the same. Gravity acceleration is a function of its "proportional" G. Acceleration of mass is inertially "proportional" to G. Therefore, they are BOTH a function of the G proportional. Change G and you change the "proportional" to how mass interacts. Why is this so difficult to understand? Or is it because you think ONLY of G as a "universal constant", and cannot imagine it being something different? Okay, for now, that is how the world of physics sees it. But change G, hypothetically, and what have you got? The same "proportinal"? No!

Newtons' formula for gravitation:
F = G * (m*M) / r^2 (1)

Newton's second law:
F = M * a (2)

Equivalence principle: M in (1) is the same as M in (2).

M(1) = M(2): where does G enter?
a = G * m/ r^2.

Since this is such a challenging question, let's spend some time here, and compare apples to apples. Let's say I have a basket of apples which weigh about 2.2 pounds (US) and I put them on a (European) scale in kilograms. What I find is this same basked of apples now weighs 1 kilogram. (I will address the difference between "weight" and "mass" in a minute.) Now put this 1 Kg. on a balance scale where on the other side is a cubic decimeter of water (or one kilogram platinum-irdidum rod) and what do we have? They balance, at least in Earth's gravity they balance. Now suppose that G' were different, let's say tenfold, for Earth. Would the scale still balance? Yes, the cubic decimeter of water would balance against the 1 Kg. apples (and the same for the scale using pounds), but with one difference: if you "weighed" the cubic decimeter of water in G' = 66.7E-11 Nm^2 kg^2, (meaning the Earth's gravitational acceleration was pulling on this cube with ten times the force), would it still be one kilogram compared to G where it was only a tenth, viz. G = 6.67E-11 Nm^2 kg^-2? If you put the basket of apples in 10 G (were it possible) and the cubic decimeter of water in 1 G, which would "weigh" more? Same mass.
Weight is a force!
Changing G changes the force, hence the weight. It does not affect the mass.
You're really stubbornly holding on to this notion that our 1 G is it. Let's go back to this:

F = Ma

If the gravity F is ten times, then (as an either or case) either 10 F = M * 10 a; or, 10 F = 10 M * a. Which would you choose?

They are not the same: If you choose the prior, mass is calculated in Earth's 1 G kilograms, and acceleration is tenfold (gravity acts ten times on mass). This has been your argument all along, I believe. On Jupiter, a much greater mass than Earth's, the acceleration is increased by its greater (1 G) gravity.

But if you choose the latter, you're in a 10 G universe, then mass is calculated in 10 G "kilograms" (where each kilogram is tenfold ours, same cubic decimeter of water but "weighs" ten times ours, and ten times per equivalence), but acceleration remains the "same". Is this the same acceleration we had in our 1 G universe? I don't think so, since it is already tenfold ( a = 10 G * m/ r^2 ), so that it pulls ten times as hard on the (tenfold kilograms) of mass. The end result is that in 10 G universe, tenfold acceleration pulls on tenfold mass.

Whether we are pushing or pulling on this mass, it should remain equivalent. If 10 F = 10 M * a, and the "a" is already tenfold because G is tenfold, then gravity acts ten times on a mass that is ten times greater. Therefore, in 10 G universe, the 10 F (gravitational equivalence) acting on mass is tremendous (a square of 10), and that means matter interacts there differently from our 1 G universe. Conversely, per equivalence, the nertial mass will now take a much greater (1 G) force to move the 10 G mass (10 squared). If the inertial mass is now ten times (10 kg) what it was in our 1 G universe, the force needed to move it will be 100 times our 1 G force. but only tenfold in local 10 G "kilograms". And THAT is why a variable G universe is different from our known 1 G universe.

Are we conceptually prepared to think this way? In my opinion, we are not. Equivalence is still preserved, but it takes a different set of rules for a (hypothetical) universe where G is variable:

Mass has not changed, only how we measure it changed.

To "weigh" mass is merely to subject it to Earth's gravitational acceleration, a = 9.8 m/s^2, so its "weight" could be said to be 9.8 kg m s^-2, but it's still the same kilogram. The apples still "weigh" one kilogram on the balance scale, their mass had not changed. If per equivalence you accelerated the basket of apples by the same rate, they would show the same "weight", but they are still one kilogram of mass, mass had not changed.

The question then remains, that if this one "kilogram" of apples were accelerated at 98 m/s^2 (where Earth's G' is tenfold), would it still be the same "kilogram"? No, the "weight" would change to 98 kg m/s^2, but the mass is still the same (1 kg) basket of apples, but now they weight 10 kgs.
kilogram is a unit of mass, not weight.
The weight you get is (1 kg)*(local g) N, where "local g" is the local gravitational acceleration on the surface of the planet; on Earth local g = 9.8 m/s^2, so 1 kg -> 9.8 N of weight.
On your planet, local g = 98 m/s^2, so 1 kg -> 98 N of weight.

Same 1 kg. basket of apples, same mass, but Earth changed its G. What happened? Does the balance scale tip towards the apples rather than towards the one cubic decimeter of water? No, it does not. Now the cubic water mass is 10 kg. Does 1 kg = 10 kg? No sir, it does not.
I see that you are confusing weight and mass.
1 kg(weight on Earth) = 1 kg(mass)*g = 9.8 N.
On your planet, g is different: g' = 10 g, hence
1 kg(weight on planet) = 1 kg(mass)*g' = 98 N.
ibid.

Now can you see why I find this hypothetical question so challenging? I realize this is merely a "what if" question, but what if we find that other worlds or regions of space have a different G? If we find this, then it is truly exciting!
I see only a confusion about the unit kilogram.
It is not for weight, but for mass.
ibid.

Now, if the Xians (per illustration above) think their 10G is merely Gx (one unit of G'), then of necessity their equation would be: Gx = r^2 * 10a/ (?m). This is the problem I'm trying to show. Should (?m) now not be, in Xian kilograms, 10m? So per "their" equivalence, kgx = 10kg in ours.
And how is that different from using another system of units?
Who is right, the Xians or the Earthians?
They are both right: they are just using different units.
Europeans are right when they say that I am 180 cm tall;
US citizens are right when they say that I am 71 inches tall.
Yes! It's all relative to where you measure.

...snip...

In fact, if Gx is ten times G, the density of the planet need not be affected, only the results of what things would weigh there, and by equivalence, how things would respond to acceleration (and perhaps also affect their centripetal force, so affect their planetary spin).
Wrong.
The dynamical measurements would not be affected by a different value of G.
Only the gravitational force would be different, not the mass.
Wrong. Yes, only "the gravitational force would be different", but so also would the effective "kilograms" for mass.
In this case, there is no such thing as "effective kilograms".
The kg unit for mass odes not change.
What changed is the gravitational acceleration at the surface, hence the gravitational force is difeerent = different weight.
Mass does not change, same basket of apples. But the inertial mass changed.

Same mass, but in 10 G, equivalence is now 10 kg (for the 1 kg basket of apples). Remember that it is G we're talking about for a planet that had not changed in size or volume, only the G changed. The cubic decimeter of water had not changed, only its effective "weight" had changed. Where m = 1 kg before (at 1 G), the mass (same mass, still cubic decimeter) is now m' = 10 kg (at 10 G).
Only if you use kg for weight, which is wrong.
You should use N, the unit for force.
Either or. Kilograms are derived from Earth's gravitational force on one cubic centimeter of water (plantinum-irridium artefact) and also a standard of measure for weight through most of the world.

I see that as a change in the mass's "dynamic measurement", where a new "kilogram" defines (measures) the cubic decimeter of water. (Remember, 1 kg does not equal 10 kg for the same mass.)
And you see wrong.
Becuase the inertial mass has not changed.
In your 1 G universe, inertial mass has not changed. In a variable G universe, it has changed, as per above.

Can you see where this is taking me?
You start from a very serious and deep misunderstanding of measurement unit, force, mass and acceleration.
This won't take you anywhere.
I am beginning to see you have a serious conceptual disconnect with what is being discussed here.
You do not grasp the difference between weight and mass, confusing the unit kg(mass) with the unit kg(weight on Earth) = 1 kg(mass)*9.8 m/s^2.
ibid.

It looks to me like you're still thinking 1 G, but ten times more powerful, in a 1 G universe. The conceptual adjustment necessary is to think in terms of a 10 G (or any variable G) universe, and then measure the new "kilograms" from there.
Wrong.
I already explaiend to you uncountable times that a variable G is not an exotic concept, but unsupported by observations.
Correct, a variable G is unsupported by current observations. That's why this exercise in reason is only hypothetical.

If, for example, a neutron star (so called) has a great mass equivalent (in our G terms) to several solar masses, we of necessity must think it is very dense for the amount of gravity it displays (hence the resultant spin is very great), that its composition must of necessity be something we do not have here in our vicinity of space, though we know it is very small by comparrison to our Sun. But if it is the G of the neutron star that is so great that it "appears" as if it were several solar masses (I'll skip for now the reasoning why this might be), then its size may be even larger than we estimate (as if made of only neutrons), and density not necessarily so compact. In fact, it may be not too much denser than our own Sun, and perhaps only comparatively smaller if its internal radiation pressure is less, meaning it was a small star to start with.

Take another example. I read somewhere that Jupiter may have a rocky core about two or three Earth masses. Whether or not this is true, I can't confirm since I never saw how this was arrived at, whether through radar probing of Jupiter's interior, or derived from atmospheric occultation, or from ephemeris spin data(?). But if true, given a constant G, how could a small rocky core hold such a vast atmosphere?
The atmosphere has amss of its own, and it interacts gravitationally.
How do you think the Sun formed: it does not even have a rocky core!
I am refering to the current need to invent "exotic matter" to explain the very high gravity in neutron stars, namely a neutron only composition.
See this Giant Planets article (http://www.space.com/scienceastronomy/mystery_monday_050307.html)

..snip...

How would such a buble of gas exist otherwise? We're not talking about soap bubbles here (where the outer surface is held together by water tension), but vast collections of molecules put into immense spin, around what? More gas? Since I obviously don't know, I'd be curious as to your idea of what's inside Jupiter, really! :lol:
I see no numbers to support your idea.
Can you show that the mass of Jupiter cannot be held together by gravity?

...snip...

These were my reasons for bringing up this question. (At this point, however, I don't even want to get close to what this means for Einstein's General Relativity theory.) For this reason, I titled this question as "hypothetical" only, until such time that we find G to be otherwise than now postulated.
The postulate is based on a wealth of experimental evidence, which you simply ignored.
No, not ignored, but only considering as possible theoretical explanations without final verdict.
Where did you consider it?
Experiments are not final judgements forever. They need to be periodically and critically reviewed, so new models emerge. Otherwise, your skating dangerously close to the thin edges of dogma.

nutant gene 71
2005-Jul-07, 04:02 AM
Heres my two cents, mass is the amount of stuff in an object, weight is the effect that two objects have on one another say the earth and a person which is F in both equations.
f=m*a or sometimes written f=m*g is the force of acceraleration in gravitional feild of the the earth and can change if the mass of the larger object changed to like a star or blackhole or another planet and the f= m*m*G/(r^2) is just two particles in isolated system like deep
space but this G is a constant of the universe. At least that my version of whats going on i could be could wrong I am often.
Particularly because this thread is in the "General Astronomy" forum, I think that it should be pointed out that you're right on the money, David. All evidence points to a constant G and it seems to me that you correctly understand the difference between weight and mass.
This is all we know for now. If we had reason to suspect otherwise, I am sure we'd be looking into it. And looking is exactly what we are about to do.

Also:

So, why do you think anomalies like the Pioneer anomalies, have anything to do with G?
Let's leave that to ESA's future probe (http://physicsweb.org/articles/world/17/9/3), where they will have a separated "flywheel" registering inertial mass, and the equivalence principle. But what if they find that inertial mass and spin are different there? Would "equivalence" not dictate a different gravity at work in that part of space, say at 10-30 AU from the Sun, for example? And if that gravity is different, would it not impact how G is measured out there? And if it shows to be a variable, are we prepared to cope with that conceptually?

But I personally think the Pioneers Anomaly is the least of our problems,that there are enough holes out there in our cosmological theories to allow for new thinking. For example:

[1.] How are stars born from hydrogen clouds, since there is insufficient gravity to make the clouds come together with enough force to start nuclear fusion, a mystery.

[2.] "Dark matter" foils our understanding of how galaxies rotate, in violation of the inverse square law for gravity.

[3.] Why do the gas giants sport such massive atmospheres, especially if the inner core planet is small? Why does our local moon have no atmosphere, but Pluto a small planet a fraction of our moon, or Saturn's Arizon sized Enceladus, does have an atmosphere?

[4.] How can a "dark matter galaxy" (http://www.space.com/scienceastronomy/050223_dark_galaxy.html) exist, when it has no visible stars?

[5.] Cosmic light redshift, though now ascribed to a Doppler effect of expanding universe, may find a different interpretation if gravity is different for 99.99% of the spacevacuum between galaxies.

[6.] Why are there unmodeled gravitational anomalies on Mars, or Venus, for their prominent features? Or why are the giant's rings dirtier near the planet than further out, where they're almost pure water ice?

[7.] Why did astronomers find "clumps of dark matter" (http://www.spaceflightnow.com/news/n0501/08clumps/) in space?
I think the Pioneers Anomaly is the smallest of our mysteries, judging from the list above. Explaining these things takes incredible contortions of reason, if not outright fancy, such as the Big Bang; including theorized gravity waves never detected, or more "elementary" particles yet to be discovered, like the Higgs boson. There's lots of room for an improved theory on how gravity works, or how it interacts closer in to star generated energies. Can a variable gravity, a variable G as discussed here, be the answer? I don't know. But we should try to see if that may not work better than what we had been working with so far, assuming a universal Newton's 1 G proportional for the whole cosmos.

Celestial Mechanic
2005-Jul-07, 04:19 AM
A 10 kg mass weighs 98 Newtons on Earth. It weighs about 16 Newtons on the Moon, but still has a mass of 10 kg.

If somehow G were to be made 10 times bigger than it is now and if the Earth does not compress and get smaller then that 10 kg mass will still be 10 kg but it will weigh 980 Newtons.

This is elementary physics. Please review a good elementary physics text on the distinction between mass and weight.
Thanks C.M., that is how I see it too, in a 1 G universe scenario. I will show below the fundamental difference between that and a variable G universe scenario, should there be a different G elsewhere. It may have a profound effect on lots of things we now cannot seem to explain. Remember that the "kilogram" is dimensionless, a scalar, unlike G, which is a cubic meter per kilogram per second squared (m^3 kg^-1 s^-2) dimensional.
Wrong, the kilogram is a unit, coequal with the meter and the second.

Take out the kilograms, and you still have the cubic meters per second squared, so this is something not contingent on the kilogram itself, but a "proportional" force in its own right which affects the mass. The mass is in kilograms.
Again, the kilogram is a unit and cannot be dispensed with. There is an exception to this, though. It is a common practice to set c=1 (eliminating seconds in favor of meters) and h-bar=1 (eliminating meters in favor of kilograms^-1). This simplifies computations considerably. However, it is understood that when it is time to give real results in SI units the answers must be multiplied by the appropriate powers of c and h-bar so that the answer is in SI units. We haven't really dispensed with any of these units at all.

Now, the current existing physics books do not address the possibility that these kilograms, really an arbitrary unit developed from the (metric system) cubic decimeter of water, which we use to measure mass are capable of being anything else. That's what's being explored here.
You (and your various previous incarnations/identities) have been pointed to numerous references in the literature where variable G and variable masses have been considered.

I happen to believe in variable G and variable masses myself, but not in anything as drastic and noticeable as what you and the now-banned Jerry have been promoting. In my theory (which is a long way from ready for prime time) the masses of leptons and quarks depend on the expectation values of at least two scalar fields, and ...

Excuse me, there's a knock at the door.

[A little later]

It's OK, it was a razor salesman at the door. I assured him that I was not multiplying entities endlessly, in fact, I use those two scalar fields to provide the value for my varying G. The curious thing is that while the lepton and quark masses vary, the nucleon masses do not vary as much because chromodynamic fields are the main source of nucleon mass, not the quark content. In my theory, if the scalar fields are smaller by a factor of x, lepton and quark masses are smaller by a factor of x, nucleon masses are maybe (1-x/20) times as large (can't calculate this yet), spectral lines are longer (redder) by a factor of 1/x and G is stronger by a factor of x^-2.

I am not optimistic of this variability existing on the galactic scale where it could explain the velocity curves, but there may be a small variation on cosmological scales. More research is needed.

papageno
2005-Jul-07, 10:15 AM
Per the Equivalence Principle, gravitational mass equals inertial mass, always. "Gravitational" mass is a function of G, as my basket of apples illustration shows above, which is equivalent to its "inertial" mass. The same basket, or cubic decimeter of water, can be either one kilograms (in 1 G), or ten kilograms (in 10 G).
:roll:
I see that you still do not grasp the difference between kilogram-mass and kilogram-weight.

You basket has a weight of 1 kg-weight = 1 kg-mass * g on Earth, and of 10 kg-weight = 1 kg-mass * (10*g) on your planet.
The inertial mass of 1 kg-mass has not changed.

If we want the whole universe to be figured in Earth's arbitrary measure for mass, our kilograms, then 10 G means the inertial-gravitational masses are always 10 kilograms.
Wrong.
The force on them is ten time higher: their masses are not.

But this causes a problem with how masses interact locally, because if kilograms for their local measure of G are different from ours, then each kilogram will need to be "locally" different from ours.
Of course you are referring to the weight, measured in kg-weight, and not to the mass, measured in kg-mass.

The ramifications of this is that local mass interacts differently than here on Earth. If so, then using Earth's kilograms becomes a poor, and ingenuous choice, since it fails to explain how in a different gravitational G "proportional" masses may hold together in ways that our 1 G (1 kg) cannot explain. Let me explain this further in your next.
The masses are not proportional to G.

Why would you say G is a "physical quantity"?
Because it can be measured.
Newton's law for gravitation gives at least an operative definition of G, which allows researchers to measure it.

Is it not merely a "proportional" quantity between gravitationally attracted masses?
Is it not a quantity that can be measured?

If G attracts at 1 G, that is the proportional attraction between the masses, which are measured in kilograms. If G's proportional attracts at ten times our 1 G, then the proportional attraction between masses increases by tenfold.
The force increases, but not the masses.

But if each mass is now tenfold in terms of its (equivalent) inertial mass,....
It is not: the gravitational force increased because you increased G not the masses.

... then the attraction is ten times ten (two bodies interacting), so the interaction between them is what?
Why?
You already increased G: G' = 10 G.
Then you assume out of the blue an increase in mass: m' = 10 m, M' = 10 M.
That would make the force: F' = 10*10*10*F = 1000 F.

If each side has an "eqivalent" mass that is ten times greater, and the G "proportional" between them is ten times greater, saying merely that the mass is now 10 kilograms is not enough, because they attract by a larger proportional. And that, really, is why it is important to redefine our (arbitrary) kilogram in a different G scenario.
It is not, because the definition of kilogram-mass does not depend on the value of G.

This is also why I brought up the question in the first place, because I don't know if the answer is tenfold or one hundredfold. This question cannot be raised in a 1 G universe,...
Your question can be raised with G as universal constant.
Your problem is the distinction between kg-mass and kg-weight: you just need to go to the Moon to raise that question, and find the answer.

....but it can be raised in a (hypothetical) variable G universe. So, yes, I understand what you are saying, if the universe is only 1 G throughout, but I am forced to disagree with you, because it does not apply in a variable G universe.
No, you do not understand.
You still confuse weight with mass.

No. The Equivalence Principle is NOT independent of the value of G.
So, how come it does not show up in the relevant equation?
M(grav) = M(inertial).

We know the inertial mass and gravitational mass are the same. Gravity acceleration is a function of its "proportional" G.
And of the distance and mass of the other mass (for example, Earth).

Acceleration of mass is inertially "proportional" to G.
Because the gravitational force is proportional to G.

Therefore, they are BOTH a function of the G proportional.
:roll:
If the acceleration of a mass is due to gravity, then your "gravity acceleration" and "acceleration of mass" are one and the same!
And sinc this acceleration depends on the garvitational force exerted on the mass, of course it depends on G.
But this does not imply that the mass that is accelerated depends on G.

Change G and you change the "proportional" to how mass interacts.
Because the force changes.

Why is this so difficult to understand? Or is it because you think ONLY of G as a "universal constant", and cannot imagine it being something different? Okay, for now, that is how the world of physics sees it. But change G, hypothetically, and what have you got? The same "proportinal"? No!
Change G and you change the force.
Change the force and you change the acceleration.

G mM/r^2 = F = a M, and
G' mM/r^2 = F' = a' M
If G' = 10 G,
G mM/r^2 = F = a M, and
(10 G) mM/r^2 = (10 F) = (10 a) M.

You see? M is subjected to ten times the acceleration, becasue the force is ten times stronger.
But you, with no justification whatsoever, assume a change in M.

Newtons' formula for gravitation:
F = G * (m*M) / r^2 (1)

Newton's second law:
F = M * a (2)

Equivalence principle: M in (1) is the same as M in (2).

M(1) = M(2): where does G enter?
a = G * m/ r^2.
So, you admit that G, even through a, does not enter M(1) = M(2).

Weight is a force!
Changing G changes the force, hence the weight. It does not affect the mass.
You're really stubbornly holding on to this notion that our 1 G is it. Let's go back to this:

F = Ma

If the gravity F is ten times, then (as an either or case) either 10 F = M * 10 a; or, 10 F = 10 M * a. Which would you choose?
Experimental results show that it is (10 a).
Why are you attributing the change to M?

They are not the same: If you choose the prior, mass is calculated in Earth's 1 G kilograms, and acceleration is tenfold (gravity acts ten times on mass). This has been your argument all along, I believe. On Jupiter, a much greater mass than Earth's, the acceleration is increased by its greater (1 G) gravity.
And on Jupiter the weight of 1 kg-mass is higher!

But if you choose the latter, you're in a 10 G universe, then mass is calculated in 10 G "kilograms" (where each kilogram is tenfold ours, same cubic decimeter of water but "weighs" ten times ours, and ten times per equivalence), but acceleration remains the "same".
Unfortunately for you, we observe the first case, where the acceleration is larger.

Is this the same acceleration we had in our 1 G universe? I don't think so, since it is already tenfold ( a = 10 G * m/ r^2 ), so that it pulls ten times as hard on the (tenfold kilograms) of mass. The end result is that in 10 G universe, tenfold acceleration pulls on tenfold mass.
You forgot to justify where the change in mass comes from.

Whether we are pushing or pulling on this mass, it should remain equivalent. If 10 F = 10 M * a, and the "a" is already tenfold because G is tenfold, then gravity acts ten times on a mass that is ten times greater.
You cannot even be consistent.
You had mutually exclusive options: either 10 M or 10 a.
You chose 10 M, so you cannot change a: a stay the same (not the "same").

Therefore, in 10 G universe, the 10 F (gravitational equivalence) acting on mass is tremendous (a square of 10), and that means matter interacts there differently from our 1 G universe. Conversely, per equivalence, the nertial mass will now take a much greater (1 G) force to move the 10 G mass (10 squared). If the inertial mass is now ten times (10 kg) what it was in our 1 G universe, the force needed to move it will be 100 times our 1 G force. but only tenfold in local 10 G "kilograms". And THAT is why a variable G universe is different from our known 1 G universe.
You forgot that the inertial mass can be measure using non-gravitational forces.
And by doing such experiments, you would see that the inertial mass has not changed, even if G changes.

Are we conceptually prepared to think this way? In my opinion, we are not.
Your opinion on serious misconceptions and deep misunderstandings.
I explained to you uncountable times that a variable G is not an exotic concept, nor mathematically challenging.
It would exactly like a variable dielectric constant in electrostatics, which is commonplace, and easily dealt with.

Equivalence is still preserved, but it takes a different set of rules for a (hypothetical) universe where G is variable:

Mass has not changed, only how we measure it changed.
:roll:
You still do not understand how mass is measured.

To "weigh" mass is merely to subject it to Earth's gravitational acceleration, a = 9.8 m/s^2, so its "weight" could be said to be 9.8 kg m s^-2, but it's still the same kilogram. The apples still "weigh" one kilogram on the balance scale, their mass had not changed. If per equivalence you accelerated the basket of apples by the same rate, they would show the same "weight", but they are still one kilogram of mass, mass had not changed.

The question then remains, that if this one "kilogram" of apples were accelerated at 98 m/s^2 (where Earth's G' is tenfold), would it still be the same "kilogram"? No, the "weight" would change to 98 kg m/s^2, but the mass is still the same (1 kg) basket of apples, but now they weight 10 kgs.
kilogram is a unit of mass, not weight.
The weight you get is (1 kg)*(local g) N, where "local g" is the local gravitational acceleration on the surface of the planet; on Earth local g = 9.8 m/s^2, so 1 kg -> 9.8 N of weight.
On your planet, local g = 98 m/s^2, so 1 kg -> 98 N of weight.

Same 1 kg. basket of apples, same mass, but Earth changed its G. What happened? Does the balance scale tip towards the apples rather than towards the one cubic decimeter of water? No, it does not. Now the cubic water mass is 10 kg. Does 1 kg = 10 kg? No sir, it does not.
I see that you are confusing weight and mass.
1 kg(weight on Earth) = 1 kg(mass)*g = 9.8 N.
On your planet, g is different: g' = 10 g, hence
1 kg(weight on planet) = 1 kg(mass)*g' = 98 N.
ibid.
So, you still do not understand.

Now can you see why I find this hypothetical question so challenging? I realize this is merely a "what if" question, but what if we find that other worlds or regions of space have a different G? If we find this, then it is truly exciting!
I see only a confusion about the unit kilogram.
It is not for weight, but for mass.
ibid.

Now, if the Xians (per illustration above) think their 10G is merely Gx (one unit of G'), then of necessity their equation would be: Gx = r^2 * 10a/ (?m). This is the problem I'm trying to show. Should (?m) now not be, in Xian kilograms, 10m? So per "their" equivalence, kgx = 10kg in ours.
And how is that different from using another system of units?
Who is right, the Xians or the Earthians?
They are both right: they are just using different units.
Europeans are right when they say that I am 180 cm tall;
US citizens are right when they say that I am 71 inches tall.
Yes! It's all relative to where you measure.

...snip...
The numbers attributed to physical quantities depends on the system of units.
You have shown that this concept is beyond your grasp.

In fact, if Gx is ten times G, the density of the planet need not be affected, only the results of what things would weigh there, and by equivalence, how things would respond to acceleration (and perhaps also affect their centripetal force, so affect their planetary spin).
Wrong.
The dynamical measurements would not be affected by a different value of G.
Only the gravitational force would be different, not the mass.
Wrong. Yes, only "the gravitational force would be different", but so also would the effective "kilograms" for mass.
In this case, there is no such thing as "effective kilograms".
The kg unit for mass odes not change.
What changed is the gravitational acceleration at the surface, hence the gravitational force is difeerent = different weight.
Mass does not change, same basket of apples. But the inertial mass changed.
You are contradicting yourself: either mass does not change, or it does.

Same mass, but in 10 G, equivalence is now 10 kg (for the 1 kg basket of apples). Remember that it is G we're talking about for a planet that had not changed in size or volume, only the G changed. The cubic decimeter of water had not changed, only its effective "weight" had changed. Where m = 1 kg before (at 1 G), the mass (same mass, still cubic decimeter) is now m' = 10 kg (at 10 G).
Only if you use kg for weight, which is wrong.
You should use N, the unit for force.
Either or. Kilograms are derived from Earth's gravitational force on one cubic centimeter of water (plantinum-irridium artefact) and also a standard of measure for weight through most of the world.
:roll:
Why can't you understand the difference between kg-mass and kg-weight?
Why is it so hard for you?

I see that as a change in the mass's "dynamic measurement", where a new "kilogram" defines (measures) the cubic decimeter of water. (Remember, 1 kg does not equal 10 kg for the same mass.)
And you see wrong.
Becuase the inertial mass has not changed.
In your 1 G universe, inertial mass has not changed. In a variable G universe, it has changed, as per above.
The "above" is misguided and wrong.

Can you see where this is taking me?
You start from a very serious and deep misunderstanding of measurement unit, force, mass and acceleration.
This won't take you anywhere.
I am beginning to see you have a serious conceptual disconnect with what is being discussed here.
You do not grasp the difference between weight and mass, confusing the unit kg(mass) with the unit kg(weight on Earth) = 1 kg(mass)*9.8 m/s^2.
ibid.
Translated: "I have no clue." Is this what you mean?

It looks to me like you're still thinking 1 G, but ten times more powerful, in a 1 G universe. The conceptual adjustment necessary is to think in terms of a 10 G (or any variable G) universe, and then measure the new "kilograms" from there.
Wrong.
I already explaiend to you uncountable times that a variable G is not an exotic concept, but unsupported by observations.
Correct, a variable G is unsupported by current observations. That's why this exercise in reason is only hypothetical.
But you clearly do not grasp the basics for this kind of speculations.
Hence you are reaching wrong conclusions.

[snip!]

How would such a buble of gas exist otherwise? We're not talking about soap bubbles here (where the outer surface is held together by water tension), but vast collections of molecules put into immense spin, around what? More gas? Since I obviously don't know, I'd be curious as to your idea of what's inside Jupiter, really! :lol:
I see no numbers to support your idea.
Can you show that the mass of Jupiter cannot be held together by gravity?
You made claim: the burden of proof is yours.

...snip...

These were my reasons for bringing up this question. (At this point, however, I don't even want to get close to what this means for Einstein's General Relativity theory.) For this reason, I titled this question as "hypothetical" only, until such time that we find G to be otherwise than now postulated.
The postulate is based on a wealth of experimental evidence, which you simply ignored.
No, not ignored, but only considering as possible theoretical explanations without final verdict.
Where did you consider it?
Experiments are not final judgements forever. They need to be periodically and critically reviewed, so new models emerge. Otherwise, your skating dangerously close to the thin edges of dogma.
:roll:
Galileo's experimental results are still valid today.

Maksutov
2005-Jul-07, 12:32 PM
Jerry may be gone, but his "spirit" lives on.

You have a lot more patience than I, papageno! That's praiseworthy. But after all these corrections with no meaningful responses, it must be wearing at least a little thin by now.

Meanwhile, good work re your replies. A shame your target audience doesn't seem to comprehend physics fundamentals. :-?

papageno
2005-Jul-07, 12:42 PM
A shame your target audience doesn't seem to comprehend physics fundamentals. :-?
My jaw dropped to the ground when I realized nutant gene 71's misconception about the kilogram. :o
It's a pity that he persists in it.

nutant gene 71
2005-Jul-07, 11:12 PM
Jerry may be gone, but his "spirit" lives on.

You have a lot more patience than I, papageno! That's praiseworthy. But after all these corrections with no meaningful responses, it must be wearing at least a little thin by now.

Meanwhile, good work re your replies. A shame your target audience doesn't seem to comprehend physics fundamentals. :-?

Maksutov, I think all here have had a lot of patience with my attempt to communicate something I see as a conceptual conundrum, but it seems no one picked up on it. So I take it as my personal failure to communicate something I see, but I don't know how else to present it. I've run out of ideas!

In papageno's last big post I counted at least 10 times where he says I'm "wrong" or "don't understand", so must accept this as my failure. In my previous incarnation (as "Lunatik" now in permanent 'safing') I had shown a paper on Atomic Mass with some mathematical scratchings on how G is different for each orbital region (which happens to unexpectedly grow at the rate of one G per AU), but then too I was told how wrong I was, or didn't understand, by the same parties, so pulled it from its site for review (it's up for peer review at the moment). I don't have the answers, and would not have them unless further astrophysical research measured a different Newton's G, which has not happened. So I remain isolated with my conceptual conundrum, which is okay with me, as it is something on which I can further meditate.

I would like to leave off here with this article at Space.com: First Invisible Galaxy Discovered in Cosmology Breakthrough (http://www.space.com/scienceastronomy/050223_dark_galaxy.html), which too me is worth more than "a picture's worth a thousand words", since no picture appears.

I sincerely hope we can discover a variable G in the future, because if we cannot, then God help us, for this universe, in the present form of our understanding, makes absolutely no sense. I see our present cosmology as pure fantasy.

In my vision of the universe, gravity is a variable determined by the G proportional where it is measured. That is not the same as to say that greater gravity is merely greater acceleration between masses, but it means that the actual mass itself has changed, per equivalence, and thus the resulting interactions of these masses is between the greater masses. Obviously, not one else thinks so, at least not at present.

No fault. My next project is to work out how much gravity we would need in the 99.99% of the space "vacuum" to account for redshift z = 1, but haven't worked it out yet. Thanks for your inputs, and to all who had their say, appreciate it.

papageno
2005-Jul-08, 09:49 AM
In papageno's last big post I counted at least 10 times where he says I'm "wrong" or "don't understand", so must accept this as my failure. In my previous incarnation (as "Lunatik" now in permanent 'safing') I had shown a paper on Atomic Mass with some mathematical scratchings on how G is different for each orbital region (which happens to unexpectedly grow at the rate of one G per AU), but then too I was told how wrong I was, or didn't understand, by the same parties, so pulled it from its site for review (it's up for peer review at the moment).
Let's be clear: it has been shown and explained where you are wrong.

I don't have the answers, and would not have them unless further astrophysical research measured a different Newton's G, which has not happened. So I remain isolated with my conceptual conundrum, which is okay with me, as it is something on which I can further meditate.
We tried to get you out of your coneptual conudrum, but you simply dismiss our explanations by saying that we just would not understand.

Now, do you understand that kilogram-mass is not the same as kilogram-weight?

nutant gene 71
2005-Jul-08, 11:42 PM
A shame your target audience doesn't seem to comprehend physics fundamentals. :-?
My jaw dropped to the ground when I realized nutant gene 71's misconception about the kilogram. :o
It's a pity that he persists in it.
Sure hope you didn't break a tooth! :)

You forget I wrote in my above, July 5th:

To "weigh" mass is merely to subject it to Earth's gravitational acceleration, a = 9.8 m/s^2, so its "weight" could be said to be 9.8 kg m s^-2, but it's still the same kilogram. The apples still "weigh" one kilogram on the balance scale, their mass had not changed. If per equivalence you accelerated the basket of apples by the same rate, they would show the same "weight", but they are still one kilogram of mass, mass had not changed.

The question then remains, that if this one "kilogram" of apples were accelerated at 98 m/s^2 (where Earth's G' is tenfold), would it still be the same "kilogram"? No, the "weight" would change to 98 kg m/s^2, but the mass is still the same (1 kg) basket of apples, but now they weight 10 kgs. Same 1 kg. basket of apples, same mass, but Earth changed its G. What happened? Does the balance scale tip towards the apples rather than towards the one cubic decimeter of water? No, it does not. Now the cubic water mass is 10 kg. Does 1 kg = 10 kg? No sir, it does not.

Mass does not change. But it's measured inertial mass, per equivalence, just like its weight, does change. Place 1 kilogram cubic decimeter of water on a balance scale in a 10 G universe, it would still balance with 1 kg of platinum-iridium rod on the other side; but this rod/cubic water is now on a scale in 10G; so it registers (each side independently) as 10 kg. That's the equivalence of gravitational mass (weight/gravity related) and inertial mass (acceleration related/equivalence) that makes a difference in the new measure of mass, where the same 1 kg of mass registered differently, now as m' = 10 kg for G' = 10 G. And that dear sir, is the conundrum.

It's not that the Earth is suddenly ten times larger in mass, same planet same mass, but the G changed, and that is what changes the (measured) mass, not the acceleration, but the mass. Remember my original question:

Hypothetically, per Equivalence Principle, what would kilograms be for any given mass in a variable G?

You answer it with 'acceleration' changes in terms of G; I answer 'mass' changes in terms of G.

Your answer makes sense in a 1 G universe, forever universally the same. My answer makes sense in a universe where G is variable. But I know you cannot (or will not?) see this. My conclusion is that if you did see it, you'd have to admit to the possibility our current take on the universe is possibly wrong, and variable gravity is a real possibility. By rejecting my idea, you're safe, nothing need change, and G is a universal constant.

I do not wish to offend you Sir, but that is how I read your responses, though I do appreciate them all the same, really. ;)

nutant gene 71
2005-Jul-09, 01:56 AM
Eureka! I think I figured out a way to rephrase this question in such a way that your answer will have to make you come clear.

Okay, let's assume Earth, same planet, same mass, is now located in a 10 G universe. Now take one cubic decimeter of water, which is 1 kilogram of mass (on our Earth it weighs 9.8 kg m s^-2), where it now is equal in weight to 98 kg m s^-2, what would its mass be in 10 G?

(an open question to everyone) :)

Celestial Mechanic
2005-Jul-09, 03:44 AM
[Snip!]Okay, let's assume Earth, same planet, same mass, is now located in a 10 G universe. Now take one cubic decimeter of water, which is 1 kilogram of mass (on our Earth it weighs 9.8 kg m s^-2), where it now is equal in weight to 98 kg m s^-2, what would its mass be in 10 G?[Snip!]
The mass is still 1 kilogram. If my theory has any validity, that 1 kg mass might have a smaller mass, say 950 grams. I can't give any better answer than that in my theory, so until I can calculate this and find evidence of variable mass on cosmological scales, I will have to go with the earlier answer of 1 kilogram, no more, no less.

papageno
2005-Jul-09, 11:54 AM
A shame your target audience doesn't seem to comprehend physics fundamentals. :-?
My jaw dropped to the ground when I realized nutant gene 71's misconception about the kilogram. :o
It's a pity that he persists in it.
Sure hope you didn't break a tooth! :)

You forget I wrote in my above, July 5th:

To "weigh" mass is merely to subject it to Earth's gravitational acceleration, a = 9.8 m/s^2, so its "weight" could be said to be 9.8 kg m s^-2, but it's still the same kilogram. The apples still "weigh" one kilogram on the balance scale, their mass had not changed. If per equivalence you accelerated the basket of apples by the same rate, they would show the same "weight", but they are still one kilogram of mass, mass had not changed.

The question then remains, that if this one "kilogram" of apples were accelerated at 98 m/s^2 (where Earth's G' is tenfold), would it still be the same "kilogram"? No, the "weight" would change to 98 kg m/s^2, but the mass is still the same (1 kg) basket of apples, but now they weight 10 kgs. Same 1 kg. basket of apples, same mass, but Earth changed its G. What happened? Does the balance scale tip towards the apples rather than towards the one cubic decimeter of water? No, it does not. Now the cubic water mass is 10 kg. Does 1 kg = 10 kg? No sir, it does not.

Mass does not change. But it's measured inertial mass, per equivalence, just like its weight, does change.
No, because inertial mass can be measured using forces that have nothing to do with gravity.
And experiments show that inertial mass does not change.

Place 1 kilogram cubic decimeter of water on a balance scale in a 10 G universe, it would still balance with 1 kg of platinum-iridium rod on the other side; but this rod/cubic water is now on a scale in 10G; so it registers (each side independently) as 10 kg.
It registers as a weight of (1 kg)*(10 g).
If you measure the inertial mass, you would find 1 kg.

That's the equivalence of gravitational mass (weight/gravity related) and inertial mass (acceleration related/equivalence) that makes a difference in the new measure of mass, where the same 1 kg of mass registered differently, now as m' = 10 kg for G' = 10 G. And that dear sir, is the conundrum.
The conundrum is that you still cannot distinguish kg-mass from kg-weight.
Why don't you write all weights in N, as you should since weight is a force?

It's not that the Earth is suddenly ten times larger in mass, same planet same mass, but the G changed, and that is what changes the (measured) mass, not the acceleration, but the mass.
By increasing G, you increased the force, hence the weight increased.
Your attribution of the change to the mass is completely unjustified, and disproven by experiments.

Remember my original question:

Hypothetically, per Equivalence Principle, what would kilograms be for any given mass in a variable G?

You answer it with 'acceleration' changes in terms of G; I answer 'mass' changes in terms of G.
There is no theoretical reason for it, and the experiemtns show that it does not happen.

Your answer makes sense in a 1 G universe, forever universally the same. My answer makes sense in a universe where G is variable.
My answer does not depend on the value of G, since we were examining the case where G changes value.

But I know you cannot (or will not?) see this. My conclusion is that if you did see it, you'd have to admit to the possibility our current take on the universe is possibly wrong, and variable gravity is a real possibility. By rejecting my idea, you're safe, nothing need change, and G is a universal constant.
Experiments put very tight constraints on a variable G.
So tight that assuming it is a universal constant, is justified.
You, on the other hand, cannot justify why a variable G would affect the inertial mass, and do not accept that the observations disproven your pet idea.
Instead of accepting the experimental results, you accuse your critics of ignoring or not understanding your idea on ideological basis.
I have given you enough explanations and references to show you that a variable G is not beyond our grasp or outside our safety zone.

I do not wish to offend you Sir, but that is how I read your responses, though I do appreciate them all the same, really. ;)
Since you won't accept the idea that you might not understand what you are talking about, it is clear that you read my responses incorrectly.

nutant gene 71
2005-Jul-09, 03:41 PM
No, because inertial mass can be measured using forces that have nothing to do with gravity.
And experiments show that inertial mass does not change.
Equivalence Principle has to do with gravity and inertial mass. Aren't you jumping the gun on ESA's future probe's results for inertial mass away from our inner solar system?

Place 1 kilogram cubic decimeter of water on a balance scale in a 10 G universe, it would still balance with 1 kg of platinum-iridium rod on the other side; but this rod/cubic water is now on a scale in 10G; so it registers (each side independently) as 10 kg.
It registers as a weight of (1 kg)*(10 g).
If you measure the inertial mass, you would find 1 kg. -bold mine
...
The conundrum is that you still cannot distinguish kg-mass from kg-weight.
Why don't you write all weights in N, as you should since weight is a force?
...
By increasing G, you increased the force, hence the weight increased.
Your attribution of the change to the mass is completely unjustified, and disproven by experiments.
...
My answer does not depend on the value of G, since we were examining the case where G changes value.
...
Experiments put very tight constraints on a variable G.
So tight that assuming it is a universal constant, is justified.
You, on the other hand, cannot justify why a variable G would affect the inertial mass, and do not accept that the observations disproven your pet idea.
Instead of accepting the experimental results, you accuse your critics of ignoring or not understanding your idea on ideological basis.

Okay, per the fundamental issue in my bold above, I can see where this has gone, where rather than being a scientific argument, based on experimentation and results away from Earth's 1 G, we delve into the realms of ideology and philosophy. Granted this is a hypothetical question on an unobserved phenomenon, a G that is variable not within very narrow constraints, but wide open to orders of magnitude. Consequently, the exploration of this idea is more philosophical in nature, since I, nor we, have any evidence to the contrary. As far as we know, G is a constant.

[Snip!]Okay, let's assume Earth, same planet, same mass, is now located in a 10 G universe. Now take one cubic decimeter of water, which is 1 kilogram of mass (on our Earth it weighs 9.8 kg m s^-2), where it now is equal in weight to 98 kg m s^-2, what would its mass be in 10 G?[Snip!]
The mass is still 1 kilogram. If my theory has any validity, that 1 kg mass might have a smaller mass, say 950 grams. I can't give any better answer than that in my theory, so until I can calculate this and find evidence of variable mass on cosmological scales, I will have to go with the earlier answer of 1 kilogram, no more, no less.
Thanks C.M. for your answer. Indeed if I were to look at this with eyes other than my own, I would see it a 1 kg also. Being I see it differently, the answer can be also 10 kg for this (1 kg) cube of water in a 10 G universe. The conceptual difference (as my eyes see it) is that in a 1 G universe, you merely apply the increased 10 G acceleration to the same mass; while in a 10 G universe, the mass itself is now measured/redefined within the parameters of this increased gravity state, which changes the mass. So it truly is becoming a philosophical question centering on how do we measure gravity in a different gravitational environment, than that known at 1 G where we live. Since we never had reason to doubt the 1 G universe before, it is understandable why we have no other way to see it.

Now, let me ask another question, as an extension of the one above:

If Earth has a known mass of M = 5.974E +24 kg., what would Earth's mass be in 10 G ?

Remember it is the same Earth, same size and shape. If Earth were suddenly ten times larger, then the answer is one way; but since Earth has not changed, only G changed, then will the answer be another way? I am curious to see how others understand this, because I think how we answer this tells how we think of gravity and mass.

I should add that philosophical questions often preceed scientific inquiry, where the posited question is answered with theory and verified with scientific observation. This hypothetical question on variable mass in a hypothetical variable G is such a question, to my mind.

papageno
2005-Jul-09, 03:55 PM
No, because inertial mass can be measured using forces that have nothing to do with gravity.
And experiments show that inertial mass does not change.
Equivalence Principle has to do with gravity and inertial mass.
The Equivalence Principle does not depend on the value of G.
You claim that the value of G affects the mass.
Measurements using non-gravitational forces show that the inertial mass is not affected.
Following the Equivalence Principle, the gravitational mass is also not affected.
Conclusion: your claim is disproven by experiments.

Place 1 kilogram cubic decimeter of water on a balance scale in a 10 G universe, it would still balance with 1 kg of platinum-iridium rod on the other side; but this rod/cubic water is now on a scale in 10G; so it registers (each side independently) as 10 kg.
It registers as a weight of (1 kg)*(10 g).
If you measure the inertial mass, you would find 1 kg. -bold mine
...
The conundrum is that you still cannot distinguish kg-mass from kg-weight.
Why don't you write all weights in N, as you should since weight is a force?
...
By increasing G, you increased the force, hence the weight increased.
Your attribution of the change to the mass is completely unjustified, and disproven by experiments.
...
My answer does not depend on the value of G, since we were examining the case where G changes value.
...
Experiments put very tight constraints on a variable G.
So tight that assuming it is a universal constant, is justified.
You, on the other hand, cannot justify why a variable G would affect the inertial mass, and do not accept that the observations disproven your pet idea.
Instead of accepting the experimental results, you accuse your critics of ignoring or not understanding your idea on ideological basis.

Okay, per the fundamental issue in my bold above, I can see where this has gone, where rather than being a scientific argument, based on experimentation and results away from Earth's 1 G, we delve into the realms of ideology and philosophy.
I showed that your claim is disproven by experiments.
And the experiments are not all based on Earth, nor do they all depend on the value of G.

Granted this is a hypothetical question on an unobserved phenomenon, a G that is variable not within very narrow constraints, but wide open to orders of magnitude. Consequently, the exploration of this idea is more philosophical in nature, since I, nor we, have any evidence to the contrary. As far as we know, G is a constant.
And I explained to you that a variable G is no more exotic than refraction in optics.
If you want to see analogies of the effect of a variable G, study electromagnetism.

[Snip!]Okay, let's assume Earth, same planet, same mass, is now located in a 10 G universe. Now take one cubic decimeter of water, which is 1 kilogram of mass (on our Earth it weighs 9.8 kg m s^-2), where it now is equal in weight to 98 kg m s^-2, what would its mass be in 10 G?[Snip!]
The mass is still 1 kilogram. If my theory has any validity, that 1 kg mass might have a smaller mass, say 950 grams. I can't give any better answer than that in my theory, so until I can calculate this and find evidence of variable mass on cosmological scales, I will have to go with the earlier answer of 1 kilogram, no more, no less.
Thanks C.M. for your answer. Indeed if I were to look at this with eyes other than my own, I would see it a 1 kg also. Being I see it differently, the answer can be also 10 kg for this (1 kg) cube of water in a 10 G universe. The conceptual difference (as my eyes see it) is that in a 1 G universe, you merely apply the increased 10 G acceleration to the same mass; while in a 10 G universe, the mass itself is now measured/redefined within the parameters of this increased gravity state, which changes the mass. So it truly is becoming a philosophical question centering on how do we measure gravity in a different gravitational environment, than that known at 1 G where we live. Since we never had reason to doubt the 1 G universe before, it is understandable why we have no other way to see it.
There is no qualitative difference between G and 10 G.

Now, let me ask another question, as an extension of the one above:

If Earth has a known mass of M = 5.974E +24 kg., what would Earth's mass be in 10 G ?
Exactly the same, because the mass is not affected by G.

Remember it is the same Earth, same size and shape. If Earth were suddenly ten times larger, then the answer is one way; but since Earth has not changed, only G changed, then will the answer be another way? I am curious to see how others understand this, because I think how we answer this tells how we think of gravity and mass.
:roll:

nutant gene 71
2005-Jul-09, 04:13 PM
Thanks papageno for your speedy response. You wrote:

The Equivalence Principle does not depend on the value of G.
You claim that the value of G affects the mass.
Measurements using non-gravitational forces show that the inertial mass is not affected.
Following the Equivalence Principle, the gravitational mass is also not affected.
Conclusion: your claim is disproven by experiments.
But is this conclusion not based on experiments performed only within Earth's region of our solar system? As to why you cannot see how gravity and inertial mass are equivalent is beyond me. Why would they show independently different measurements if G were different? That would violate the "equivalence" of the principle. We are not talking about equivalence for a larger Earth, for example, same mass size of matter as before, but about equivalence within a different G. If Earth were suddenly half the G we now know, wouldn't its mass also be adjusted by half? Same size, but the linkage between gravity and inertial mass (now half) would have to be conserved.

Quite frankly, I am genuinely puzzled why this is such a difficult issue to understand. If "equivalence" is truly equivalent between gravity and inerital mass, then any change in G (gravity) there must be a comensurate change in m (inertial mass), for equivalence to be conserved. :-?

papageno
2005-Jul-09, 04:29 PM
Thanks papageno for your speedy response. You wrote:

The Equivalence Principle does not depend on the value of G.
You claim that the value of G affects the mass.
Measurements using non-gravitational forces show that the inertial mass is not affected.
Following the Equivalence Principle, the gravitational mass is also not affected.
Conclusion: your claim is disproven by experiments.
But is this conclusion not based on experiments performed only within Earth's region of our solar system?
General Relativity works fine outside the Solar System, as far as we know.
And we have plenty of probes throughout the Solar System, outside the Earth's region.

As to why you cannot see how gravity and inertial mass are equivalent is beyond me.
If you are saying "gravitational mass = inertial mass", this is exactly the Equivalence Principle.
If you are saying "gravitational mass depends on G", then you are wrong.

Why would they show independently different measurements if G were different?
They do not.
The difference you measure is in the acceleration, which is due to a change in force.
There is no change in the mass.

That would violate the "equivalence" of the principle. We are not talking about equivalence for a larger Earth, for example, same mass size of matter as before, but about equivalence within a different G.
The Equivalence Principle (gravitational mass = inertial mass) does not depend on the value of G.

If Earth were suddenly half the G we now know, wouldn't it's mass also be adjusted by half?
It would not.
The force between the Earth and other objects would change (we would lose half of the weight), but the mass would not change.

Same size, but the linkage between gravity and inertial mass (now half) would have to be conserved.
Why do you assume such linkage between G and mass?

nutant gene 71
2005-Jul-09, 04:49 PM
Okay, I see the crux of the problem, though I may not know how to address it right now. You said:

If you are saying "gravitational mass = inertial mass", this is exactly the Equivalence Principle.
If you are saying "gravitational mass depends on G", then you are wrong.
But is this not denying gravity as a function of G? If we assume G to be always the same, then gravity affects the amount of mass in the ways we know, and per equivalence the inertial mass is the same. But if G is altered, is gravity not altered? After all, G is merely a proportional ratio of how masses attract. So if gravity is now in "proportion" to this altered G, and the size of the matter of mass has not changed, then where does that leave its gravitational mass? And per equivalence, what happens to the commensurate inertial mass? This is the real issue, isn't it?

nutant gene 71
2005-Jul-09, 06:14 PM
If you are saying "gravitational mass = inertial mass", this is exactly the Equivalence Principle.
If you are saying "gravitational mass depends on G", then you are wrong.

Let's focus on this one for a moment, I think it is a threshold point.

We know from Newton's orbital equation that GM = rv^2. Now if G is different, and rv^2 is as we had always observed, what would happen to M if G were different? (I'll need another cappuccino for this one. :) )

We know that G is 6.67E-11 m^3 kg^-1 s^-2, and we know mass for let's say Jupiter, M = ~19E+26 kg (vs. Earth's ~5.97E+24 kg), and we know Jupiter's rv^2 from astronomical observations, which is a fixed value. Now for the same rv^2 fixed value, what if ESA's 'gravitational' probe* (http://physicsweb.org/articles/world/17/9/3) found that at 5.2 AU, where is Jupiter's orbit, that G was greater than 1 G? For the sake of example, hypothetically, let's say they measure it's 5 times, or about Gj = 33.35E-11 m^3 kg^-1 s^-2. In this hypothetical scenario, what would happen to M?

Could M now be only Mj = 3.9E+26 kg? Not to any known physics it can't. The rv^2 had not changed in G * M = rv^2, so to accommodate 5xG we are forced to "shrink" Jupiter to 1/5th its mass M, dimensionally speaking, in terms of kilograms. But this cannot be! It's still the same Jupiter. To make Jupiter NOT fly out of its rv^2 orbit (Jupiter's orbit is what it is), it would have to mean that Jupiter's kilograms are different, by five times different, meaning that each "kilogram" there is five times what we had figured before discovering Jupiter's (new) 5 G = 33.35E-11 N... Per force, that means that Jupiter's mass, which is now Mj = 3.9E+26 kg must be multiplied by its own kilograms, which are five times greater than ours, to arrive at our estimated mass of Mj = 19E+26 kg. Now, this is only hypothetical since we have not found this fivefold increased G to be true. But if it were... :-?

Now, for our ealier argument about how G affects gravitational mass, I think I am allowed here to call "check". Gravitational mass is a function of G. And if we did find Jupiter's G to be five times, I would be forced to call "checkmate". The coffee helped, though I personally find chess boring. :)

(*)Quote from the article: "The European Space Agency is considering a unique experiment that could explain strange gravitational phenomena in the outer solar system." - bold mine

I wish I had a "time machine" to take me to that gravitational experiment now!

Tassel
2005-Jul-10, 04:41 PM
We know that G is 6.67E-11 m^3 kg^-1 s^-2, and we know mass for let's say Jupiter, M = ~19E+26 kg (vs. Earth's ~5.97E+24 kg), and we know Jupiter's rv^2 from astronomical observations, which is a fixed value. Now for the same rv^2 fixed value, what if ESA's 'gravitational' probe* found that at 5.2 AU, where is Jupiter's orbit, that G was greater than 1 G? For the sake of example, hypothetically, let's say they measure it's 5 times, or about Gj = 33.35E-11 m^3 kg^-1 s^-2. In this hypothetical scenario, what would happen to M?

Could M now be only Mj = 3.9E+26 kg? Not to any known physics it can't. The rv^2 had not changed in G * M = rv^2, so to accommodate 5xG we are forced to "shrink" Jupiter to 1/5th its mass M, dimensionally speaking, in terms of kilograms. But this cannot be! It's still the same Jupiter.
In the case that you describe, Jupiter's mass would not be ~19E+26 kg as you claim. How could it be? In your imagination, we arrived at that number by using the "wrong" value for G. As you've said seemingly hundreds of times in these forums, GM stays the same. If we were to learn that G is higher than we think at Jupiter, we'd have to recalculate Jupiter's mass based on the new G. Using the "new" mass along with the "new" G, everything would work just fine. You're trying to use the "old" mass with the "new" G to claim we need to change the definition of kilograms which is of course, incorrect.

The kilogram problem you're trying to manufacture doesn't exist.

To make Jupiter NOT fly out of its rv^2 orbit (Jupiter's orbit is what it is), it would have to mean that Jupiter's kilograms are different
FYI, Jupiter's mass is irrelevant when determining Jupiter's orbit. If Jupiter's mass were to suddenly change, Jupiter's moons would "fly out of their orbits". Jupiter's orbit would be essentially unchanged.

Aren't you jumping the gun on ESA's future probe's results for inertial mass away from our inner solar system?

Now for the same rv^2 fixed value, what if ESA's 'gravitational' probe* found that at 5.2 AU, where is Jupiter's orbit, that G was greater than 1 G?
The article you linked does not talk about any proposed experiments. Can you provide references for your claims that the ESA is going to measure G or inertial mass away from the inner solar system?

nutant gene 71
2005-Jul-10, 10:52 PM
n the case that you describe, Jupiter's mass would not be ~19E+26 kg as you claim. How could it be? In your imagination, we arrived at that number by using the "wrong" value for G. As you've said seemingly hundreds of times in these forums, GM stays the same. If we were to learn that G is higher than we think at Jupiter, we'd have to recalculate Jupiter's mass based on the new G. Using the "new" mass along with the "new" G, everything would work just fine. You're trying to use the "old" mass with the "new" G to claim we need to change the definition of kilograms which is of course, incorrect.
Yes, the product G*M remains the same, so if one changes, the other is adjusted. And no, Jupiter would not stay in orbit if its mass changed, relative to the Sun's mass (as two body system where they spin around each other on a common center, which I believe is outside the Sun) it would have a different relationship. If M changed (in same G) per GM = rv^2, either r or v^2 would be adjusted. And yes, the G used for Jupiter is a cooked up number, hypothetical only, to illustrate a point.

See Future Test paper (http://arXiv.org/abs/gr-qc/0411077), on page 12, sect. 3.4, you'll see a description of the "formation flying" sub-satellite, where they want to put an inertial reference mass separated from the main craft.

Tassel
2005-Jul-11, 01:27 AM
Yes, G*M remains the same.
Right, so we don't need to change the definition of a kilogram. Everything works just fine, even for a variable G, if you are consistent with the masses you use.

And no, Jupiter would not stay in orbit if its mass changed, because relative to the Sun's mass it would have a different relationship, hence orbit (per GM = rv^2).
You're wrong. The mass of any planet could change drastically and there would be essentially no change in the orbit, since the mass of any planet in our solar system is very small compared to that of the sun. If an astronaut steps out of the space shuttle while orbiting Earth, he doesn't go "flying" out of the shuttle's orbit. The shuttle and the astronaut effectively share an orbit in spite of the large difference in their masses because both masses are so small compared to Earth's mass. The same applies to Jupiter and the sun. This is pretty basic stuff.

And yes, the G I used is a cooked up number, hypothetical only, to illustrate a point.
The value you chose for G is irrelevant and I didn't comment on it at all. I believe your point was that if you used Jupiter's real mass and combined it with your invented G, it has "to mean that Jupiter's kilograms are different". And I believe I showed you that this was incorrect by showing that if G were different than we think, Jupiter's mass would be different than we think and everything would work just fine.

See Future Test paper, on page 12, sect. 3.4, you'll see a description of the "formation flying" sub-satellite, where they want to put an inertial reference mass separated from the main craft.
Really a very interesting paper. But I could not find anything about testing for changes in inertial mass. It seems to me that the "sub-satellite" would be used so that they could more accurately track any unexpected trajectory changes, since the "sub-satellite" would be isolated from any interference from systems on the main spacecraft. The main spacecraft would track the distance between itself and the "sub-satellite", and the distance between Earth and the main spacecraft would be tracked as normal. See:

The idea is to avoid the inherent problems of self disturbance of an inertial sensor on board the primary spacecraft by placing the inertial reference mass(es) (i.e. subsatellite(s)) outside the craft at a sufficient, but not too large distance. A laser ranging sensor that employs a mWlaser monitors the 3-dimensional vector of mutual separation between the spacecraft and subsatellite. Any subsatellite is covered with corner-cube retro-reflectors that enable precise laser ranging similar to that currently used for satellite and lunar laser ranging.
Very clever idea, I think, if you're looking for un-modeled accelerations. But not an experiment to test for changes in inertial mass, as far as I can tell.

And nothing on the ESA testing for variation in G, then? I would have thought that since this isn't the first time you've made that claim, that you would have had something to back it up this time. I guess not.

Gsquare
2005-Jul-11, 03:27 AM
...Yes, the product G*M remains the same, so if one changes, the other is adjusted. And no, Jupiter would not stay in orbit if its mass changed, relative to the Sun's mass ...

Why do you think the orbit must change? If GM remains constant the orbit must remain the same:

Indeed Kepler's 3rd law states that (for any planet) the ratio R^3/T^2 is constant and equal to GM/4pi^2.
IOW, the orbital period T, vs. radius R is constant because GM is invariant. (Here M is solar mass; Jupiter's mass doesn't even enter in). (This of course works because the mass of a planet is rather negligible compared to that of the sun, as is any change in mass, as mentioned by Tessel.)

However, the same equation allows you to speculate that G and M for Jupiter could each change in an offsetting way without it being detectable, at least, not by orbital variations. :wink:
The question then becomes one of whether the equivalence principle still holds so as to make it (un)recognizable with an inertial test.

G^2

Celestial Mechanic
2005-Jul-11, 03:40 AM
And for a different perspective, I will now channel Gertrude Stein. Cue the cheesy theremin music!

Woo-OOO, woo-OOO-oo!

A gram is a gram is a gram.
:lol:

nutant gene 71
2005-Jul-11, 03:40 AM
If an astronaut steps out of the space shuttle while orbiting Earth, he doesn't go "flying" out of the shuttle's orbit. The shuttle and the astronaut effectively share an orbit in spite of the large difference in their masses because both masses are so small compared to Earth's mass. The same applies to Jupiter and the sun. This is pretty basic stuff.
This is a nonsense example, since where does the astronaut's G change? You made a foolish statement which has no application to what is being discussed. I suggest you go back and reread what I wrote if you wish to discuss this with some modicum of understanding.

And nothing on the ESA testing for variation in G, then? I would have thought that since this isn't the first time you've made that claim, that you would have had something to back it up this time. I guess not.

And if you read the ESA (future test) paper with no understanding, than you will not understand how inertial mass and G may be different out there. So in your case, it is best to merely wait until the results are in, and somebody else can tell you. ... later...

papageno
2005-Jul-11, 10:05 AM
If you are saying "gravitational mass = inertial mass", this is exactly the Equivalence Principle.
If you are saying "gravitational mass depends on G", then you are wrong.

Let's focus on this one for a moment, I think it is a threshold point.

We know from Newton's orbital equation that GM = rv^2.
You already used the Equivalence Principle, by dropping the other mass.

Now if G is different, and rv^2 is as we had always observed, what would happen to M if G were different? (I'll need another cappuccino for this one. :) )
Nothing, unless you assume (and justify) that rv^2 has not changed.
If you assume that M changes because of the change in G, you have to explain what happens to the other mass (the one you dropped from the equation).
Then you have to explain why no change in the inertial mass is observed.

We know that G is 6.67E-11 m^3 kg^-1 s^-2, and we know mass for let's say Jupiter, M = ~19E+26 kg (vs. Earth's ~5.97E+24 kg), and we know Jupiter's rv^2 from astronomical observations, which is a fixed value. Now for the same rv^2 fixed value, what if ESA's 'gravitational' probe* (http://physicsweb.org/articles/world/17/9/3) found that at 5.2 AU, where is Jupiter's orbit, that G was greater than 1 G?
If any mass changed because of G, you have to explain why the probe has not shown a change in its mass and its moment of inertia.
Don't forget that there is no evidence of violations of the Equivalence Principle.

For the sake of example, hypothetically, let's say they measure it's 5 times, or about Gj = 33.35E-11 m^3 kg^-1 s^-2. In this hypothetical scenario, what would happen to M?
M would not change.

Could M now be only Mj = 3.9E+26 kg? Not to any known physics it can't.
And that includes all the experimental result.

The rv^2 had not changed in G * M = rv^2, so to accommodate 5xG we are forced to "shrink" Jupiter to 1/5th its mass M, ...
M is not the mass of Jupiter, but the mass of the Sun.
Jupiter's mass is the one you dropped from the equation.

...dimensionally speaking, in terms of kilograms. But this cannot be! It's still the same Jupiter. To make Jupiter NOT fly out of its rv^2 orbit (Jupiter's orbit is what it is), it would have to mean that Jupiter's kilograms are different, by five times different, meaning that each "kilogram" there is five times what we had figured before discovering Jupiter's (new) 5 G = 33.35E-11 N... Per force, that means that Jupiter's mass, which is now Mj = 3.9E+26 kg must be multiplied by its own kilograms, which are five times greater than ours, to arrive at our estimated mass of Mj = 19E+26 kg. Now, this is only hypothetical since we have not found this fivefold increased G to be true. But if it were... :-?
This whole "reasoning" is based on the error I pointed out above: in all the equations you wrote, the mass of Jupiter is absent.

Now, for our ealier argument about how G affects gravitational mass, I think I am allowed here to call "check". Gravitational mass is a function of G. And if we did find Jupiter's G to be five times, I would be forced to call "checkmate". The coffee helped, though I personally find chess boring. :)
The coffee was not enough.
Jupiter's mass was never in the formulae you wrote in this post: you dropped it when you applied the Equivalence Principle to write the first equation.

Tassel
2005-Jul-11, 01:58 PM
This is a nonsense example, since where does the astronaut's G change? You made a foolish statement which has no application to what is being discussed. I suggest you go back and reread what I wrote if you wish to discuss this with some modicum of understanding.
I was giving you an example to show you that you were wrong to say that Jupiter would "fly out of its orbit" if its mass were different. You were wrong to say that, and this was a perfectly valid example that shows you were wrong. I'm sorry if you don't understand how it is applicable.

And if you read the ESA (future test) paper with no understanding, than you will not understand how inertial mass and G may be different out there. So in your case, it is best to merely wait until the results are in, and somebody else can tell you. ... later...
So, in other words, you can't back up your statements that the ESA will be testing for changes in inertial mass or variations in G.

Continuing to say that the ESA is going to test for a variable G is not going to make it true. I'm sorry that it upsets you when I point out that you post false information from time to time.

nutant gene 71
2005-Jul-11, 05:14 PM
...Yes, the product G*M remains the same, so if one changes, the other is adjusted. And no, Jupiter would not stay in orbit if its mass changed, relative to the Sun's mass ...

Why do you think the orbit must change? If GM remains constant the orbit must remain the same:

Indeed Kepler's 3rd law states that (for any planet) the ratio R^3/T^2 is constant and equal to GM/4pi^2.
IOW, the orbital period T, vs. radius R is constant because GM is invariant. (Here M is solar mass; Jupiter's mass doesn't even enter in). (This of course works because the mass of a planet is rather negligible compared to that of the sun, as is any change in mass, as mentioned by Tessel.)

However, the same equation allows you to speculate that G and M for Jupiter could each change in an offsetting way without it being detectable, at least, not by orbital variations. :wink:
The question then becomes one of whether the equivalence principle still holds so as to make it (un)recognizable with an inertial test.

G^2
Okay, the example I gave is still not good enough, as Tassel, papageno, and you point out. So let me go back and figure out another way, stay tuned. ;)

I'm not eager to argue that G is different, only that if it were different how things change, namely how we measure mass changes. That's been my argument all along here. If the universe sings in the key of G flat, and we modeled it that way, everything works and all is hunky dory. But if we should find in the future that perhaps it's also (elsewhere) singing at G sharp, then we have to know how to model that too. This is the point I want to make, though perhaps I had not yet succeeded in doing so. So stay tuned.

pghnative
2005-Jul-11, 08:58 PM
Okay, the example I gave is still not good enough, as Tassel, papageno, and you point out. So let me go back and figure out another way, stay tuned. ;)

I'm not eager to argue that G is different, only that if it were different how things change, namely how we measure mass changes. That's been my argument all along here. If the universe sings in the key of G flat, and we modeled it that way, everything works and all is hunky dory. But if we should find in the future that perhaps it's also (elsewhere) singing at G sharp, then we have to know how to model that too. This is the point I want to make, though perhaps I had not yet succeeded in doing so. So stay tuned.
I'd suggest going back to your "1 kg of water" thought experiment. If we all can't agree on the mass of 1 kg of water in a 10G environment, then any discussion of planets, orbits, etcetera will go nowhere.

Perhaps the best place to start is to decide how to define a kilogram. You seem to be under the impression that if an object weighs 9.8 Newtons on earth, then it is a kilogram. From that you conclude that the object was transfered to a 10G world where that same object weighed 98 Newtons, then therefore the object would be 10 kg. There is no basis for this conclusion. There is no reason why that object cannot still be 1 kg of mass and 98 Newtons in weight.

By the way --- you do know what a Newton is, don't you???????

nutant gene 71
2005-Jul-12, 02:05 AM
By the way --- you do know what a Newton is, don't you???????
Yesssirree! A Newton is a measure of force, with SI units of "m kg s^2".

Okay, let's back to the beginning, by the numbers.

We know GM = rv^2 is a short form, a reduced simplified equation. We also know that what's on the left has to be equivalent to what's on the right, which is what an equation equals. One way to see this is to reconstruct from whence this simplified equation came from. We know the following:

F = GMm/ r^2, and we know F = ma, which is also F = M*(Gm/r^2), where a = (Gm)/ r^2, but we can also see Newton's orbital equation one more way, as:

F = M*(Gm)/r^2 = mv^2/ r.

I showed the M separate, since it represents the Sun, a fixed value of mass, unaffected by G. The (Gm) is together because these are the variables, hypothetically, that need to be addressed. And on the right (mv^2/r) is a version of the same equation. When M*(Gm)/r^2 = mv^2/r is netted out, it gives us:

GM = rv^2

But there something wrong here, since we don't see the G on the right side. And we won't because it is "hidden" within the (Keplerian-Newton?) parameters of rv^2. So any change in G on the left will be reflected on the right side in rv^2, with the mass intact. This is how it's supposed to work in a universe where G is always constant, where the right side is a function of the G on the left. But now it gets interesting, because what if G is not the same?

If we say, hypothetically, that G changes, then what happens to M*(Gm)/r^2 = m v^2/r ? Something in the rest of the equation has to change. We can't change the v^2 nor r because these are fixed values, as astronomical observation shows. Nor does m on the right side change, since it is equal to the left side in toto of the equation. We know M is the Sun, which does not change, so the only m left to adjust is within (G*m). Therefore, in the full equation: M*(Gm)/r^2 = mv^2/r, we know observationally that neither side can change, as a total equation. There is only (Gm) on the left, where G changes. Soooh... What is m's mass within (G*m)?

Going back to the Jupiter example, if G changes for Jupiter, it's m has to do what? Now before you all raise a howl in protest and start banging the table with your shoe, let's plug in some numbers:

Hypothetically, Jupiter's G is 5 times Earth's, so it reads as: G' = 33.35E-11 N..., which if we put that into the left side, we get:

M * (33.35E-11 N..) * m'/ r^2 = m v^2/r. Remember the m, r, and v^2 are known values as measured astronomically, Jupiter is where it is in orbit traveling at its velocity, same as M is the Sun in it's place, so only G' and m' are affected by the changed gravity "proportional" G. Thus, per (G'*m'), any change in G' has to be reflected in the m' if the equation's equality is to be conserved. And we're talking hypothetically here, but if G is increased five fold, then m' has to decrease one fifth.

Remember (critical point) neither Jupiter's orbit nor its mass are changed, only how we measure the mass in terms of an increased G. So let's plug in some numbers for Jupiter as we know them in (Earth's) "universal constant" G, where G = 6.67E-11 m^3 kg^-1 s^-2 and Jupiter's m = 19E+26 kg:

M * (Gm)/r^2 = mv^2/r, so that M*(6.67E-11 m^3 kg^-1 s^-2)(19E+26 kg)/ r^2 = m v^2/r, which is M*(126.73E+15 m^3 s^-2)/r^2 = m v^2/r

(Remember the right side is in Earth units, so does not change.)

Now let's plug in the new Jupiter G', fivefold:

M*[(33.35E-11 m^3 kg^-1 s^-2) * m'] /r^2 = m v^2/r, and to conserve the product (G'm') =126.73E+15 m^3 s^-2, divided by 33.35E-11 kg, m' becomes = 3.8E+26 kg (which is one fifth the mass of what Jupiter was in a G = 6.67E-11 N.. equation).

Now I ask you, where did I go wrong? I'm once again faced with the same challenge, where Jupiter is still the same Jupiter in terms of mass and location and orbital velocity, but if it should be in a different G, its mass is different. To adjust for the mass in Jupiter's G' mass back to Earth's G mass, the kilograms have to increase fivefold. This means m' is in new kilograms!

Back to a = (Gm)/ r^2, if G changed, m had to change too as shown above, and r^2 stayed the same (Jupiter did not change its orbital distance from the Sun), which means that "a" stays the same. Taking that to the Equivalence Principle:

F = ma = m (G'm')/ r^2, if G' is changed, and commensurately m' is changed, does "a" change? Think about it... No! Remember the product of (G*m) is conserved, so the acceleration (a = Gm/r^2) is still the same acceleration, but the parameters of G and m are different in relation to each other. Now, if G changed, but "a" stayed the same, what happens to "m" in F = ma, if equivalence is conserved? Is the Equivalence Principle broken? I'll leave that one for you to figure out...

So to recap: GM = rv^2 is really the (pre-reduced) equation GMm/r^2 = mv^2/r. The Sun's M is unvariable, it is what it is within the parameters of whatever G the Sun is (figured same as Earth's G), and we know Jupiter's mass and orbital configurations are unvariable. So if G changes, we are left with only one possibility, that m changes within the (G*m) product. And if m is different, we cannot use Earth's kilograms, since now we need to "re-normalize" them (a term I just re-invented) for Jupiter's measured mass in its own "kilograms".

Well, that's all folks. I'm sure you'll find a way to disagree with this, but I will then find another way to explain it. To me, it's simple, as plain as the nose on your face, except I suspect for you it's invisible. Cheers. :)

...rook takes queen... "check?"

papageno
2005-Jul-12, 10:13 AM
By the way --- you do know what a Newton is, don't you???????
Yesssirree! A Newton is a measure of force, with SI units of "m kg s^2".

Okay, let's back to the beginning, by the numbers.

We know GM = rv^2 is a short form, a reduced simplified equation. We also know that what's on the left has to be equivalent to what's on the right, which is what an equation equals. One way to see this is to reconstruct from whence this simplified equation came from.
Instead of using the simplified form, why not use the equations where it comes from?

We know the following:

F = GMm/ r^2, and we know F = ma, which is also F = M*(Gm/r^2), where a = (Gm)/ r^2, but we can also see Newton's orbital equation one more way, as:

F = M*(Gm)/r^2 = mv^2/ r.
Why not use this:
GMm/r^2 = F = ma. (1)

Applying to Equivalence Principle to m, we get:
a = GM/r^2. (2)

I showed the M separate, since it represents the Sun, a fixed value of mass, unaffected by G. The (Gm) is together because these are the variables, hypothetically, that need to be addressed. And on the right (mv^2/r) is a version of the same equation. When M*(Gm)/r^2 = mv^2/r is netted out, it gives us:

GM = rv^2
Note that m no longer appears explicitly in the equation, and that you just applied the Equivalence Principle, without worrying about the actual value of G.

But there something wrong here, since we don't see the G on the right side. And we won't because it is "hidden" within the (Keplerian-Newton?) parameters of rv^2.
There is nothing wrong.
G is a parameter determining the force on m.
The force on m determines its acceleration.

So any change in G on the left will be reflected on the right side in rv^2, with the mass intact.
That is: if you change the force, you change the acceleration of m.
And the same happens if you change the distance between m and M, or the mass M.

This is how it's supposed to work in a universe where G is always constant, where the right side is a function of the G on the left. But now it gets interesting, because what if G is not the same?
If G changed with the distance between the masses (in which case we would wirte it G(r), where (r) means "function of r"), there is no guarantee that gravity is a conservative force and the planets, satellites, asteroids, comets, meteoroids, artificial probes/satellites, space junk and dust would no longer follow conical sections as orbits.
Unless you assume, against all the experimental evidence, that the Equivalence Principle is violated.

If we say, hypothetically, that G changes, then what happens to M*(Gm)/r^2 = m v^2/r ? Something in the rest of the equation has to change. We can't change the v^2 nor r because these are fixed values, as astronomical observation shows. Nor does m on the right side change, since it is equal to the left side in toto of the equation. We know M is the Sun, which does not change, so the only m left to adjust is within (G*m). Therefore, in the full equation: M*(Gm)/r^2 = mv^2/r, we know observationally that neither side can change, as a total equation. There is only (Gm) on the left, where G changes. Soooh... What is m's mass within (G*m)?
It is the gravitational mass.
On the right-hand side you have the inertial mass.
If you assume that the Equivalence Priniple is still valid, the two masses are one and the same, and you are left with a contradiction in your reasoning.
The only way out for you is a violation of the Equivalence Principle, which is disproven by observations.
But the Equivalence Principle is anyway unrelated to the actual value of G.

Going back to the Jupiter example, if G changes for Jupiter, it's m has to do what?
You can change the gravitational mass.
But then you have to explain why the probes we sent out there did not show any violation of the Equivalence Principle.

Now before you all raise a howl in protest and start banging the table with your shoe, let's plug in some numbers:

Hypothetically, Jupiter's G is 5 times Earth's, so it reads as: G' = 33.35E-11 N..., which if we put that into the left side, we get:
G has not the dimensions of a force.
And do you think that a change in the gravitational mass by 5 times would not be observable?

M * (33.35E-11 N..) * m'/ r^2 = m v^2/r. Remember the m, r, and v^2 are known values as measured astronomically, Jupiter is where it is in orbit traveling at its velocity, same as M is the Sun in it's place, so only G' and m' are affected by the changed gravity "proportional" G. Thus, per (G'*m'), any change in G' has to be reflected in the m' if the equation's equality is to be conserved. And we're talking hypothetically here, but if G is increased five fold, then m' has to decrease one fifth.
Which means that the Equivalence Principle would be glaringly violated.

Remember (critical point) neither Jupiter's orbit nor its mass are changed, only how we measure the mass in terms of an increased G.
No.
The inertial mass has not changed.
You assume that the gravitaional mass has changed.
This violates the Equivalence Principle, and you have to explain why General Relativity works.

So let's plug in some numbers for Jupiter as we know them in (Earth's) "universal constant" G, where G = 6.67E-11 m^3 kg^-1 s^-2 and Jupiter's m = 19E+26 kg:

M * (Gm)/r^2 = mv^2/r, so that M*(6.67E-11 m^3 kg^-1 s^-2)(19E+26 kg)/ r^2 = m v^2/r, which is M*(126.73E+15 m^3 s^-2)/r^2 = m v^2/r

(Remember the right side is in Earth units, so does not change.)

Now let's plug in the new Jupiter G', fivefold:

M*[(33.35E-11 m^3 kg^-1 s^-2) * m'] /r^2 = m v^2/r, and to conserve the product (G'm') =126.73E+15 m^3 s^-2, divided by 33.35E-11 kg, m' becomes = 3.8E+26 kg (which is one fifth the mass of what Jupiter was in a G = 6.67E-11 N.. equation).

Now I ask you, where did I go wrong?
You assumed that a change in G would not affect the orbit.
In order to do so, the conclusion is that the Equivalence Principle must be violated.

I'm once again faced with the same challenge, where Jupiter is still the same Jupiter in terms of mass and location and orbital velocity, but if it should be in a different G, its mass is different. To adjust for the mass in Jupiter's G' mass back to Earth's G mass, the kilograms have to increase fivefold. This means m' is in new kilograms!
Since our probes out there worked as expected without "changing" kilograms, your conclusion is obviously incorrect.

Back to a = (Gm)/ r^2, if G changed, m had to change too as shown above, and r^2 stayed the same (Jupiter did not change its orbital distance from the Sun), which means that "a" stays the same. Taking that to the Equivalence Principle:
I already explained that following your reasoning, you have to change the gravitational mass without changing the inertial mass.

F = ma = m (G'm')/ r^2, if G' is changed, and commensurately m' is changed, does "a" change? Think about it... No!
You assumed that a does not change!
This is circular reasoning.

Remember the product of (G*m) is conserved, so the acceleration (a = Gm/r^2) is still the same acceleration, but the parameters of G and m are different in relation to each other. Now, if G changed, but "a" stayed the same, what happens to "m" in F = ma, if equivalence is conserved? Is the Equivalence Principle broken? I'll leave that one for you to figure out...
In your reasoning you adjusted G and m(gravitaional) so that the force would not change.

So to recap: GM = rv^2 is really the (pre-reduced) equation GMm/r^2 = mv^2/r.
On the left m(gravitational).
On the right m(inertial).

The Sun's M is unvariable, it is what it is within the parameters of whatever G the Sun is (figured same as Earth's G), and we know Jupiter's mass and orbital configurations are unvariable. So if G changes, we are left with only one possibility, that m changes within the (G*m) product. And if m is different, we cannot use Earth's kilograms, since now we need to "re-normalize" them (a term I just re-invented) for Jupiter's measured mass in its own "kilograms".
We do not.
Kilogram is the unit of mass.
Your reasoning concludes with the violation of the Equivalence Principle, which is disproven by experiments.

Well, that's all folks. I'm sure you'll find a way to disagree with this, but I will then find another way to explain it. To me, it's simple, as plain as the nose on your face, except I suspect for you it's invisible.
I see: you are convinced that you are right, and nothing we show or explain will change your opinion.

Tassel
2005-Jul-12, 02:34 PM
M*[(33.35E-11 m^3 kg^-1 s^-2) * m'] /r^2 = m v^2/r, and to conserve the product (G'm') =126.73E+15 m^3 s^-2, divided by 33.35E-11 kg, m' becomes = 3.8E+26 kg (which is one fifth the mass of what Jupiter was in a G = 6.67E-11 N.. equation).

Now I ask you, where did I go wrong? I'm once again faced with the same challenge, where Jupiter is still the same Jupiter in terms of mass and location and orbital velocity, but if it should be in a different G, its mass is different. To adjust for the mass in Jupiter's G' mass back to Earth's G mass, the kilograms have to increase fivefold. This means m' is in new kilograms!
This is the problem right here. There's no justification for "new kilograms". What would be the problem with simply using 3.8E+26 kg for the mass of Jupiter in all the familiar equations?

nutant gene 71
2005-Jul-12, 04:30 PM
M*[(33.35E-11 m^3 kg^-1 s^-2) * m'] /r^2 = m v^2/r, and to conserve the product (G'm') =126.73E+15 m^3 s^-2, divided by 33.35E-11 kg, m' becomes = 3.8E+26 kg (which is one fifth the mass of what Jupiter was in a G = 6.67E-11 N.. equation).

Now I ask you, where did I go wrong? I'm once again faced with the same challenge, where Jupiter is still the same Jupiter in terms of mass and location and orbital velocity, but if it should be in a different G, its mass is different. To adjust for the mass in Jupiter's G' mass back to Earth's G mass, the kilograms have to increase fivefold. This means m' is in new kilograms!
This is the problem right here. There's no justification for "new kilograms". What would be the problem with simply using 3.8E+26 kg for the mass of Jupiter in all the familiar equations?
Did Jupiter just shrink in size per its "shrunk" mass?

It is the gravitational mass.
On the right-hand side you have the inertial mass.
If you assume that the Equivalence Priniple is still valid, the two masses are one and the same, and you are left with a contradiction in your reasoning.
The only way out for you is a violation of the Equivalence Principle, which is disproven by observations.
But the Equivalence Principle is anyway unrelated to the actual value of G.
Are you saying (Gm) = m, or not ? :-) ...Show me.

G has not the dimensions of a force.
And do you think that a change in the gravitational mass by 5 times would not be observable?
G is in SI units: m^3/ kg/ s^2 or Nm^2/kg^2.

The second part question belongs on ATM, since it had not been observed. Any variance in G could be observable, but ask ESA if that's what they're after, in situ.

Tassel
2005-Jul-12, 04:34 PM
Did Jupiter just shrink in size per its "shrunk" mass?
No, we used a constant G to measure its mass so therefore, if G were different, we would have measured it's mass wrong. As long as we use the correct G, we get the correct mass.

What would be the problem with using that mass in all our familiar fomulas?

papageno
2005-Jul-12, 04:42 PM
It is the gravitational mass.
On the right-hand side you have the inertial mass.
If you assume that the Equivalence Priniple is still valid, the two masses are one and the same, and you are left with a contradiction in your reasoning.
The only way out for you is a violation of the Equivalence Principle, which is disproven by observations.
But the Equivalence Principle is anyway unrelated to the actual value of G.
Are you saying (Gm) = m, or not ? :-) ...Show me.

Where did you get this?
Did you actually read my post, or are you just trying to evade the point?

Tassel
2005-Jul-12, 04:53 PM
Perhaps the best place to start is to decide how to define a kilogram.
This comment seems to have been glossed over, probably because it cuts to the heart of the issue and is difficult for "nutant" to respond to.

nutant, what is your proposed definition for the kilogram?

nutant gene 71
2005-Jul-12, 05:45 PM
If G changed with the distance between the masses (in which case we would wirte it G(r), where (r) means "function of r"), there is no guarantee that gravity is a conservative force and the planets, satellites, asteroids, comets, meteoroids, artificial probes/satellites, space junk and dust would no longer follow conical sections as orbits.
Unless you assume, against all the experimental evidence, that the Equivalence Principle is violated.
Are your refering again to (Gm) = m, as per above? If the product of G*m remains the same no matter where G is measured, why would it affect gravity at a distance? If in proportion to the Sun's gravity, (Gm) will still respond the same whether G is changed or not (provided m is adjusted), so the same inverse square law is not violated, and all things in orbit would still follow conical sections as before. If in proportion to any other heavenly body, the same applies. The (Gm) product hasn't changed anywhere (only its internal components of G and m are mutally adjusted).

Equivalence is conserved, but that requires somebody answer my question above (re what happens to m in F = ma ?), which nobody has, or just ignored it.

nutant gene 71
2005-Jul-12, 05:59 PM
Perhaps the best place to start is to decide how to define a kilogram.
This comment seems to have been glossed over, probably because it cuts to the heart of the issue and is difficult for "nutant" to respond to.

nutant, what is your proposed definition for the kilogram?
You may want to re-read this post July 5 th (http://www.badastronomy.com/phpBB/viewtopic.php?p=498202#498202) where I discuss this issue. Pay special attention to the BALANCE between kilograms (equal weights both sides) in response to G and 10G example:
Viz. Let's say I have a basket of apples which weigh about 2.2 pounds (US) and I put them on a (European) scale in kilograms. What I find is this same basked of apples now weighs 1 kilogram. (I will address the difference between "weight" and "mass" in a minute.) Now put this 1 Kg. on a balance scale where on the other side is a cubic decimeter of water (or one kilogram platinum-irdidum rod) and what do we have? They balance, at least in Earth's gravity they balance. Now suppose that G' were different, let's say tenfold, for Earth. Would the scale still balance? Yes, the cubic decimeter of water would balance against the 1 Kg. apples (and the same for the scale using pounds), but with one difference: if you "weighed" the cubic decimeter of water in G' = 66.7E-11 Nm^2 kg^2, (meaning the Earth's gravitational acceleration was pulling on this cube with ten times the force), would it still be one kilogram compared to G where it was only a tenth, viz. G = 6.67E-11 Nm^2 kg^-2? If you put the basket of apples in 10 G (were it possible) and the cubic decimeter of water in 1 G, which would "weigh" more? Same mass.

I'll see if I can better explain this above later, when have some free time. The kilogram is always referenced back to one cubic decimeter of water, so the mass is always the same, only how we measure this mass in terms of the G where it is. More later...

Tassel
2005-Jul-12, 07:13 PM
Continuing to assume for the sake of example that G = 5G at Jupiter (or G=10G...whatever you want), and assuming a radius of approximately 68,000km for Jupiter (http://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html), can you show how you would calculate the acceleration due to gravity at 68,000km from the center of Jupiter (ie: surface gravity)? I'm interested in the steps you would take to measure the mass of Jupiter (and what value you think that should be), and then the steps you would take to use that mass to determine the acceleration due to gravity.

Perhaps working through an example would help us understand what you are trying to say.

nutant gene 71
2005-Jul-12, 09:14 PM
Continuing to assume for the sake of example that G = 5G at Jupiter (or G=10G...whatever you want), and assuming a radius of approximately 68,000km for Jupiter (http://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html), can you show how you would calculate the acceleration due to gravity at 68,000km from the center of Jupiter (ie: surface gravity)? I'm interested in the steps you would take to measure the mass of Jupiter (and what value you think that should be), and then the steps you would take to use that mass to determine the acceleration due to gravity.

Perhaps working through an example would help us understand what you are trying to say.
Comparing apples to oranges. Earth's surface gravity is 9.8 m/s^2, while Jupiter's "surface" gravity is 23 m/s^2, but the two are not comparable directly, since we don't know Jupiter's real surface (use top of atmosphere, not same on Earth where we use planet's crust surface).

Per the above listed examples, at Jupiter (hypothetical) 5 G, the force would not affect surface (whatever that is) acceleration since the mass of the planet is adjusted for G (in local terms): i.e., F = 5(0.2 m)a

papageno[/b] is referring to in his: [i]"In your reasoning you adjusted G and m(gravitational) so that the force would not change."
This is how I see it, that the force does not change, nor how the planet interacts gravitationally per (G*m) per orbital equation, even if G and m are different (adjusted separately) so product remains same.]

Jupiter has not changed, same planet, so its gravity acceleration force has not changed. The mass of Jupiter, hypothetically, is adjusted for local G only, but it is still the same planet we figured in Earth's 1 G. It's just that Jupiter's "kilograms" are different from Earth derived kilograms. This has been my point for all of the above posts, though I am sure there would be some (all?) who would disagree.

[Edited for error in prior, viz. 5F = (0.2m)a, which was wrong. I sometimes think too fast for my fingers.] :oops:

Tassel
2005-Jul-12, 09:30 PM
Comparing apples to oranges.
I didn't ask you to compare anything.

Will you work through the calculations with numbers or not? I believe it should make it easier for you to explain your point.

nutant gene 71
2005-Jul-12, 09:58 PM
Comparing apples to oranges.
I didn't ask you to compare anything.

Will you work through the calculations with numbers or not? I believe it should make it easier for you to explain your point.
Can you give me an example of what you have in mind? Show any math, for example?

Can you give me an equation representing what you're asking?

Gotta run, talk later. 8)

Tassel
2005-Jul-12, 10:00 PM
Continuing to assume for the sake of example that G = 5G at Jupiter (or G=10G...whatever you want), and assuming a radius of approximately 68,000km for Jupiter (http://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html), can you show how you would calculate the acceleration due to gravity at 68,000km from the center of Jupiter (ie: surface gravity)? I'm interested in the steps you would take to measure the mass of Jupiter (and what value you think that should be), and then the steps you would take to use that mass to determine the acceleration due to gravity.

Perhaps working through an example would help us understand what you are trying to say.
You keep saying we need "new" kilograms to support (hypothetical, of course...) variable G. So, I'd like to see the practical application of the "variable kilogram" idea.

I'm also still waiting on your definition of "kilogram".

frogesque
2005-Jul-13, 12:36 AM
nutant gene 71 wrote:

...

I'll see if I can better explain this above later, when have some free time. The kilogram is always referenced back to one cubic decimeter of water, so the mass is always the same, only how we measure this mass in terms of the G where it is. More later...

I beg to differ. NPL: Frequently asked questions - mass and density (http://www.npl.co.uk/mass/faqs/kilogram.html)

Finally...
Any better ideas on a postcard please. :lol:

nutant gene 71
2005-Jul-13, 01:41 AM
Continuing to assume for the sake of example that G = 5G at Jupiter (or G=10G...whatever you want), and assuming a radius of approximately 68,000km for Jupiter (http://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html), can you show how you would calculate the acceleration due to gravity at 68,000km from the center of Jupiter (ie: surface gravity)? I'm interested in the steps you would take to measure the mass of Jupiter (and what value you think that should be), and then the steps you would take to use that mass to determine the acceleration due to gravity.

Perhaps working through an example would help us understand what you are trying to say.
You keep saying we need "new" kilograms to support (hypothetical, of course...) variable G. So, I'd like to see the practical application of the "variable kilogram" idea.

I'm also still waiting on your definition of "kilogram".
Sorry I couldn't answer you earlier, but I was catching your posts on the fly in between other projects. Can you be more specific? Please give me an equation to which I can answer, since your question is vague.

The bottom line, if I understand what you're asking, is that 23 meters per second squared is still Jupiter's gravity (at its atmospheric surface), same as now have it. Remember the planet did not change, it's still the same planet, even if its G is different affecting how its mass is measured. Measured in our 1 G postulate for the universe, or measured for its local G, the end result is the same. Please provide me with an equation to address this better. Thanks.

nutant gene 71
2005-Jul-13, 02:06 AM
nutant gene 71 wrote:

...

I'll see if I can better explain this above later, when have some free time. The kilogram is always referenced back to one cubic decimeter of water, so the mass is always the same, only how we measure this mass in terms of the G where it is. More later...

I beg to differ. NPL: Frequently asked questions - mass and density (http://www.npl.co.uk/mass/faqs/kilogram.html)

Finally...
Any better ideas on a postcard please. :lol:
Correct, cubic decimeter of water is not longer used, which has been replaced by a high controlled environment platinum-irridium rod, where the accumulated molecules are carefully cleaned off periodically. My reference was to the original idea behind what represented the mass of one kilogram. If you look up historical references to the kilogram, you'll see what I mean.

A "kilogram" is a scalar, arbitrarily chosen unit of mass. Here's a reference: http://physics.nist.gov/cuu/Units/kilogram.html

nutant gene 71
2005-Jul-13, 04:07 AM
WHAT HAPPENS TO THE KILOGRAM IN A VARIABLE G?

I'd like to add this, how I see it. I don't know if any of this is true or not, and will not know for certain until such time that we find G varies, so for now this is all hypothetical.

I suspect, per my above reasonings, that the kilogram, which is an arbitrary unit of mass (and with which we also calculate weight), is a defined unit in its own right. Taking the same representative mass and putting it elsewhere where G might be different does not change the mass itself, except in how it interacts with other mass at that locale. This means the same gravitational parameters exist as before, but the size of the material mass may be different. If we use 5G as an example, in that locale, the material size of a representative unit of kilogram would be five times smaller, but still exhibit the same characteristics we know for one kilogram (on Earth). The effect should be that smaller size bonds with other equivalent smaller size, what we can also call "weight" for gravitational, or molecular attraction for chemical, as if the smaller mass weighed more. Not knowing how this will affect chemical bonding, I am forced to leave that to some future study. And ditto for how it affects metal springs, or rocket propulsion, both unknowns. I assume that "heavier" gravitational mass will somehow affect these, but can't guess how. Obviously the Cassini-Huygens springs worked, though operating in extreme cold space (which might make them brittle?) and launched the separated craft away from the mother craft without mishap. But I do suspect that this different "kilogram", which is heavier than our kilogram per volume, does affect gases and matter. There should be more compactness to the planet cores for the gas giants, if such a core exists, and there should be more abundant gas retention per mass. This means that even if Jupiter's core planet is only two Earth masses (in size and volume) it may retain an atmosphere that is substantially greater than if Earth was twice its own size here. I can't calculate this, since I don't know how, so treat this only as a conceptual idea worth examining. When I learn how to calculate it, I will. Separately, though I am not yet ready to release it, I did calculate what the gravitational mass is for the hydrogen atoms in 99.9% of deep space (I used one atom of hydrogen per cubic centimeter of space), away from those tiny galaxy islands of electromagnetic radiant energy, and came up with a startling number, where G is 100,000 times what it is on Earth. When I applied this greater G against a gravitational-redshift value (measured on Earth), I came up with close to Hubble's constant. But that's all I can say for now, not ready to show this yet.

Overall, I am intrigued by the idea that maybe we don't have the G "constant" right, and that perhaps its variance had been (very well) hidden from us all this time. Do I know this for sure? Hardly! I'm fascinated by it. And if it's self delusion, then it's self delusion, and no loss to anybody. But if it's right, how exciting it will be to know that the isotropic universe is still homogenous, but at a substantially higher G, and that the galaxies and stars within them are but low G islands within the cosmic scale of things. If we find that G is variable, our universe is a very different place from what we had thought before, and very exciting too.

Well, gotta walk the dogs on the beach and as usual look up at the sky, except today it's more of the same, cloudy. I hope I added something of value to you all in considering these ideas, as I know for sure that you had added something of value to me by questioning them. Cheers.

papageno
2005-Jul-13, 09:37 AM
WHAT HAPPENS TO THE KILOGRAM IN A VARIABLE G?

I'd like to add this, how I see it. I don't know if any of this is true or not, and will not know for certain until such time that we find G varies, so for now this is all hypothetical.

I suspect, per my above reasonings, that the kilogram, which is an arbitrary unit of mass (and with which we also calculate weight), is a defined unit in its own right. Taking the same representative mass and putting it elsewhere where G might be different does not change the mass itself, except in how it interacts with other mass at that locale.
You mean: the weight changes.

This means the same gravitational parameters exist as before, but the size of the material mass may be different.
No.
The kg-mass is the same on Earth, on Mars, on Jupiter or on your 5G-Earth.
1 kg-mass will still be accelerated by 1 m/s^2 if 1 N of force is applied, whether this force is gravitational or not.

The difference between the Earth and your 5G-Earth is the gravtitational force at the surface, hence the wieght is different.
And since the force on the 1 kg-mass at the surface, changes, the gravitational acceleration (downwards) changes.
But if you go ice-skating on 5G-Earth, you still need the same force as on Earth to accelerate horizontally, even if your weight is different.

This is what happened to the Apollo astronauts on the Moon.

If we use 5G as an example, in that locale, the material size of a representative unit of kilogram would be five times smaller, but still exhibit the same characteristics we know for one kilogram (on Earth).
No need to change the size, because the (inertial) mass does not change.
You can take take a sample kilogram from and use it on 5G-Earth.

The effect should be that smaller size bonds with other equivalent smaller size, what we can also call "weight" for gravitational, or molecular attraction for chemical, as if the smaller mass weighed more.
That would work at most in the vertical direction, but not horizontally.
Since the inertial mass of atoms does not change, I do not expect significant changes in molecular bonding.
If chemical bonds were affected by the weight of atmos and molecules, we would not be able to recognize spectra from the Sun or the gas giants.
All the spectroscopy in astrophysics would be useless, because we would not be able to compare spectra taken in labs on Earth with spectra from stars.
The electronics of the probes we sent throughout the Solar System and of the artificial satellites around Earth would not work as designed, becuase chemical bonds determine the band-structure of the semiconductors used (which includes CCDs), and hence it determines their electric properties.

Not knowing how this will affect chemical bonding, I am forced to leave that to some future study. And ditto for how it affects metal springs, or rocket propulsion, both unknowns.
Since spectroscopy works fine, and the probes we sent out work fine, it is safe to say that weight does not significantly affect chemical bonding.

I assume that "heavier" gravitational mass will somehow affect these, but can't guess how. Obviously the Cassini-Huygens springs worked, though operating in extreme cold space (which might make them brittle?) and launched the separated craft away from the mother craft without mishap. But I do suspect that this different "kilogram", which is heavier than our kilogram per volume, does affect gases and matter.
Why?
Spectra from the Sun look like spectra taken on Earth, hence atoms in gases have still the same electronic structure, which determines chemical bonds.

There should be more compactness to the planet cores for the gas giants, if such a core exists, and there should be more abundant gas retention per mass. This means that even if Jupiter's core planet is only two Earth masses (in size and volume) it may retain an atmosphere that is substantially greater than if Earth was twice its own size here.
Do you still think that the mass of gas of Jupiter does not exert gravitational force?
WHy do you treat gases as passive objects in gravitational interaction?

I can't calculate this, since I don't know how, so treat this only as a conceptual idea worth examining. When I learn how to calculate it, I will. Separately, though I am not yet ready to release it, I did calculate what the gravitational mass is for the hydrogen atoms in 99.9% of deep space (I used one atom of hydrogen per cubic centimeter of space), away from those tiny galaxy islands of electromagnetic radiant energy, and came up with a startling number, where G is 100,000 times what it is on Earth. When I applied this greater G against a gravitational-redshift value (measured on Earth), I came up with close to Hubble's constant. But that's all I can say for now, not ready to show this yet.
You should show how yopu performed these calculations.

Overall, I am intrigued by the idea that maybe we don't have the G "constant" right, and that perhaps its variance had been (very well) hidden from us all this time.
Do you think you can "hide" a 100000G?

Do I know this for sure? Hardly! I'm fascinated by it. And if it's self delusion, then it's self delusion, and no loss to anybody. But if it's right, how exciting it will be to know that the isotropic universe is still homogenous, but at a substantially higher G, and that the galaxies and stars within them are but low G islands within the cosmic scale of things. If we find that G is variable, our universe is a very different place from what we had thought before, and very exciting too.
You should take into account what we already know.

Metricyard
2005-Jul-13, 12:52 PM
Cool, who knew that BABB's came with re-runs?

Let's take one of the experiments I proposed in the now defunct Huygens thread and modify it a bit. No math or physics should be needed for these experiments.

Let's pretend that an alien race from the planet Fisiks comes to visit us and we find that indeed that G is 10 times more at Fisiks then around Earth.

To test our theory, we're going to build a cube using carbon-14 atoms.
The measurement will be 1 billion atoms length, 1 billion atoms width, 1 billion atoms height.

The scientist from the planet Fisiks (we'll call him Dr. Kalkulous) also will make a cube of carbon-14 with the same amount of atoms and dimensions on his home planet.

Dr. Kalkulous takes you and your cube to the planet Fisiks to see if the cubes mass are the same. He places his cube on a balance scale, you place your cube on the opposite side of the scale. Hmm, they balance.

Next both cubes will be dropped in a vacuum chamber at the same height. Yep, they drop at the same rate. Of course with the higher G, they both drop extremely fast.

Now, Dr. Kalkulous takes us back to Earth to preform the same experiments. Same results. The cubes balance on the scale, and drop at the same speed.

Now without using any mathematics, explain why the cubes shouldn't balance or fall at the same rate. Maybe this will help shed some light on where you're heading.

edit: spelling and clarity.

Tassel
2005-Jul-13, 01:58 PM
Sorry I couldn't answer you earlier, but I was catching your posts on the fly in between other projects. Can you be more specific? Please give me an equation to which I can answer, since your question is vague.
Moi? According to you, I don't even have a "modicum of understanding" while you, on the other hand, have a new theory of gravity up for peer review. How could I possibly provide you with equations?

My question is not vague. A high school physics student could produce the calculations I'm asking for using Newtonian physics. You've been claiming that the need for a "variable kilogram" to support a "hypothetical" variable G is as plain as the nose on your face. You've been taking queens and calling "check". You have your variable G "hypothesis". So...use it.

Pretend your fantasy has come true and you're right: we just discovered G is different in the outer solar system. Use your variable G "hypothesis" to model our universe, starting with Jupiter. Since the need for "new kilograms" is as plain as the nose on your face, we should all be able to see the truth of it. All you need to do is use your physics to model what we observe. Then we could all see "new kilograms" in action.

The bottom line, if I understand what you're asking, is that 23 meters per second squared is still Jupiter's gravity (at its atmospheric surface), same as now have it. Remember the planet did not change, it's still the same planet, even if its G is different affecting how its mass is measured.
Great! See, this is what I'm talking about. Nothing has changed, we just wrongly assumed G was constant. Now pretend you've been proven right. G varies. Tell us what the value of G is at Jupiter, and then walk us through the calculations used to model what we observe. Since "new kilograms" are critical, and plain as the nose on your face, they should just fall right out of the model and be obvious to all.

Bear in mind, I'm not asking for vague "well, it's the same". The results may be the same, as you claim, but with variable G and "new kilograms" certainly how we arrive at the results will be different. So we need to start from scratch and be able to model interactions at Jupiter that match what we observe. Calculating the acceleration due to gravity, step by step, would be an excellent start, I believe.

nutant gene 71
2005-Jul-13, 04:06 PM
No.
The kg-mass is the same on Earth, on Mars, on Jupiter or on your 5G-Earth.
1 kg-mass will still be accelerated by 1 m/s^2 if 1 N of force is applied, whether this force is gravitational or not.

The difference between the Earth and your 5G-Earth is the gravitational force at the surface, hence the weight is different.
And since the force on the 1 kg-mass at the surface, changes, the gravitational acceleration (downwards) changes.
But if you go ice-skating on 5G-Earth, you still need the same force as on Earth to accelerate horizontally, even if your weight is different.

This is what happened to the Apollo astronauts on the Moon.
Not exactly same comparison because our Moon is still in a 1G-Earth region.

Remember the Apollo astronauts had not shrunk in size, which they would have to do in a 5G environment to maintain the same inertial mass. Remember F = 5(0.2m)a, if they don't shrink, their gravitational mass is now five times their former inertial mass, which per equivalence would take five times the force to move, which is not so. This is why the comparison is not accurate, in my view.

RE: Do you think you can "hide" a 100000G?
This is the most interesting thing, that we figured out physics within a 1G-Earth environment, then projected this 1G to the rest of the cosmos, including how spectra redshifts in the rest of the cosmos, and we can see no evidence of a variable G! That's quite a puzzle, though we never measured the same star spectra from Saturn for example, and we never went outside the solar system to check on G out there. We only know things from within our inner solar system region, and everything looks great within the parameters we set for the rest of the universe. That's why I think any future studies of gravity or inertial mass in the outer solar system should be so interesting. If the product of (G*m) can hide gravitational anomalies so well within our solar system, it could as well hide these for the rest of the cosmos. But if we have 100000G-Earth in most of intergalactic deep space, hydrogen clouds can more easily come together to ignite stars, for example, because each hydrogen molecule acts as if it were five orders of magnitude "heavier" than on Earth, and perhaps that's why fusion can ignite. I'm sorry I can't show you my work on that, but I am not ready at this point to do so, still work in progress.

To test our theory, we're going to build a cube using carbon-14 atoms.
The measurement will be 1 billion atoms length, 1 billion atoms width, 1 billion atoms height.

The scientist from the planet Fisiks (we'll call him Dr. Kalkulous) also will make a cube of carbon-14 with the same amount of atoms and dimensions on his home planet.

Dr. Kalkulous takes you and your cube to the planet Fisiks to see if the cubes mass are the same. He places his cube on a balance scale, you place your cube on the opposite side of the scale. Hmm, they balance.

Next both cubes will be dropped in a vacuum chamber at the same height. Yep, they drop at the same rate. Of course with the higher G, they both drop extremely fast.

Now, Dr. Kalkulous takes us back to Earth to perform the same experiments. Same results. The cubes balance on the scale, and drop at the same speed.
Great thought experiment, could've written it myself, though Dr. Kalkulous's explanation is all Greek to me! :)

Yes, this is the conundrum in a nutshell, that a different G is not easily spotted. I suspect that the planet Fisiks has a thicker and higher atmosphere per its planet's size and volume compared to Earth. At 10G, it would be more like Saturn, my guess.

A high school physics student could produce the calculations I'm asking for using Newtonian physics.
Produce the (high school student) equation you're looking for and I will try to answer it. For now I have given you all the equations I worked with, except for how I calculated G in 99.9% deep space, which is not yet ready for viewing. Your turn... ;)

papageno
2005-Jul-13, 04:25 PM
No.
The kg-mass is the same on Earth, on Mars, on Jupiter or on your 5G-Earth.
1 kg-mass will still be accelerated by 1 m/s^2 if 1 N of force is applied, whether this force is gravitational or not.

The difference between the Earth and your 5G-Earth is the gravitational force at the surface, hence the weight is different.
And since the force on the 1 kg-mass at the surface, changes, the gravitational acceleration (downwards) changes.
But if you go ice-skating on 5G-Earth, you still need the same force as on Earth to accelerate horizontally, even if your weight is different.

This is what happened to the Apollo astronauts on the Moon.
Not exactly same comparison because our Moon is still in a 1G-Earth region.

Remember the Apollo astronauts had not shrunk in size, which they would have to do in a 5G environment to maintain the same inertial mass.
Why? Because you say so?
Or do you have actual evidence that the Equivalence Principle is violated?

Remember F = 5(0.2m)a, if they don't shrink, their gravitational mass is now five times their former inertial mass, which per equivalence would take five times the force to move, which is not so. This is why the comparison is not accurate, in my view.
Where did you provide evidence that gravity affects molecular bonding?
Where did you show that this "shrinking" actualy changes the mass, instead of the volume?

RE: Do you think you can "hide" a 100000G?
This is the most interesting thing, that we figured out physics within a 1G-Earth environment, then projected this 1G to the rest of the cosmos, including how spectra redshifts in the rest of the cosmos, and we can see no evidence of a variable G! That's quite a puzzle, though we never measured the same star spectra from Saturn for example, and we never went outside the solar system to check on G out there.
We identify elements in stars using spectroscopy, comparing the stellar spectra with spectra obtained on Earth.
Cassini's electronics worked fine, which mean that the band-structure of the semiconductors composing the electronic devices worke as designed on Earth.

We only know things from within our inner solar system region,...
And here I thought that the spectra of star have been measured!

...and everything looks great within the parameters we set for the rest of the universe. That's why I think any future studies of gravity or inertial mass in the outer solar system should be so interesting.
It has been thoroughly explained to you that the probes we already sent out are, give a lot of information about you "hypthetical" variable G.

If the product of (G*m) can hide gravitational anomalies so well within our solar system, it could as well hide these for the rest of the cosmos.
But you have nothing to hide with.
A violation of the Equivalence Principle would result in a experimental failure of General Relativity.
And a variable G would result in orbits that do not follow conical sections.

But if we have 100000G-Earth in most of intergalactic deep space, hydrogen clouds can more easily come together to ignite stars, for example, because each hydrogen molecule acts as if it were five orders of magnitude "heavier" than on Earth, and perhaps that's why fusion can ignite. I'm sorry I can show you my work on that, but I am not ready at this point to do so, still work in progress.
But we would not be able to identify the emission as coming from Hydrogen, if it even started nuclear fusion.

To test our theory, we're going to build a cube using carbon-14 atoms.
The measurement will be 1 billion atoms length, 1 billion atoms width, 1 billion atoms height.

The scientist from the planet Fisiks (we'll call him Dr. Kalkulous) also will make a cube of carbon-14 with the same amount of atoms and dimensions on his home planet.

Dr. Kalkulous takes you and your cube to the planet Fisiks to see if the cubes mass are the same. He places his cube on a balance scale, you place your cube on the opposite side of the scale. Hmm, they balance.

Next both cubes will be dropped in a vacuum chamber at the same height. Yep, they drop at the same rate. Of course with the higher G, they both drop extremely fast.

Now, Dr. Kalkulous takes us back to Earth to perform the same experiments. Same results. The cubes balance on the scale, and drop at the same speed.
Great thought experiment, could've written it myself, though Dr. Kalkulous's explanation is all Greek to me! :)

Yes, this is the conundrum in a nutshell, that a different G is not easily spotted.
Which requires a violation of the Equivalence Principle.
Or a variation of G so small that we still cannot detect it (and nowhere near 5G or 100000G; more like G/100000).

I suspect that the planet Fisiks has a thicker and higher atmosphere per its planet's size and volume compared to Earth. At 10G, it would be more like Saturn, my guess.

A high school physics student could produce the calculations I'm asking for using Newtonian physics.
Produce the (high school student) equation you're looking for and I will try to answer it. For now I have given you all the equations I worked with, except for how I calculated G in 99.9% deep space, which is not yet ready for viewing. Your turn... ;)
:roll:

Again, where is the evidence that Equivalence Principle is violated?

Tassel
2005-Jul-13, 05:37 PM
A high school physics student could produce the calculations I'm asking for using Newtonian physics.
Produce the (high school student) equation you're looking for and I will try to answer it. For now I have given you all the equations I worked with, except for how I calculated G in 99.9% deep space, which is not yet ready for viewing. Your turn... ;)
I'm not looking for an equation (how do you "answer" an equation anyway?). I'm looking for you to demonstrate the usage of "new kilograms" using your variable G "hypothesis". I suggested calculating the acceleration due to gravity at Jupiter, which you should easily be able to do. You gave me the answer in the back of the book...but I'm looking for you to show your work, so we can see precisely where "new kilograms" come into play.

I guess you are unwilling or unable to do this.

nutant gene 71
2005-Jul-13, 06:29 PM
Which requires a violation of the Equivalence Principle.
Or a variation of G so small that we still cannot detect it (and nowhere near 5G or 100000G; more like G/100000).

Again, where is the evidence that Equivalence Principle is violated?

Its not. [-X

frogesque
2005-Jul-13, 08:54 PM
nutant gene 71 wrote:

...

I'll see if I can better explain this above later, when have some free time. The kilogram is always referenced back to one cubic decimeter of water, so the mass is always the same, only how we measure this mass in terms of the G where it is. More later...

I beg to differ. NPL: Frequently asked questions - mass and density (http://www.npl.co.uk/mass/faqs/kilogram.html)

Finally...
Any better ideas on a postcard please. :lol:
Correct, cubic decimeter of water is not longer used, which has been replaced by a high controlled environment platinum-irridium rod, where the accumulated molecules are carefully cleaned off periodically. My reference was to the original idea behind what represented the mass of one kilogram. If you look up historical references to the kilogram, you'll see what I mean.

A "kilogram" is a scalar, arbitrarily chosen unit of mass. Here's a reference: http://physics.nist.gov/cuu/Units/kilogram.html

But this is the crux of the matter. A kilogram is currently defined as a particular quantity of matter. It has absolutely nothing to do with local g or the universal gravitational constant G.

Yes, we can use scales to compare one mass against another but a kilogram is a kilogram is a kilogram whether it is on Earth, Saturn or the micro-g of interplanetary space. The force it exerts under different accelerations will be different (according to Newton's tried and trusted formula; Force = Mass x Acceleration) but the mass cannot change by definition

nutant gene 71
2005-Jul-14, 03:43 AM
A high school physics student could produce the calculations I'm asking for using Newtonian physics.
Produce the (high school student) equation you're looking for and I will try to answer it. For now I have given you all the equations I worked with, except for how I calculated G in 99.9% deep space, which is not yet ready for viewing. Your turn... ;)
I'm not looking for an equation (how do you "answer" an equation anyway?). I'm looking for you to demonstrate the usage of "new kilograms" using your variable G "hypothesis". I suggested calculating the acceleration due to gravity at Jupiter, which you should easily be able to do. You gave me the answer in the back of the book...but I'm looking for you to show your work, so we can see precisely where "new kilograms" come into play.

I guess you are unwilling or unable to do this.
Tassel, I'm beginning to believe that you don't really know what your question is, and you want me to answer it so you can formulate in your own mind what your question is. Alas, unless you work harder to ask a specific question, I will not work hard to answer it. So far, you said to me in effect: you worked out different kilograms for Jupiter per its G, which has a diameter of about 68,000 km, now show me. But show you what? :-?

Show me some math, then we'll talk.

Look at frogesque's sub-signature: "By asking questions we sometimes get the wrong answers, from wrong answers we learn to ask the right questions." :)

Very nice!

nutant gene 71
2005-Jul-14, 04:00 AM
But this is the crux of the matter. A kilogram is currently defined as a particular quantity of matter. It has absolutely nothing to do with local g or the universal gravitational constant G.
You're absolutely right. For the postulated 1 G for our whole universe, that is how a kilogram works, and it has nothing to do with any local G. But it does get tricky if the local G is different from our postulated 1 G (which we applied to the whole cosmos), because then the same kilogram of mass acts differently, not as if 1 G times mass, but as variable G times mass, so the mass itself is affected. This is something "papageno" cannot see, though I've explained it over and over again, so he keeps arguing his case in a 1 G universe. All well and good, but it's not good enough for me. One kilogram of mass is always the same mass, only it registers differently if G (not variable G times 1 G mass) is different, since then it is "variable G times its own G mass".

My hunch is that the greater intergalactic cosmos operates at a different G, about 100,000 times greater than ours, and the mass out there measured in that G acts differently from how it acts on Earth. Of course, once more everybody, "it's only hypothetical!". :lol:

Tassel
2005-Jul-14, 04:21 AM
But show you what? :-?
Ok, so you can't do it.

Show me some math, then we'll talk.
If I were the one making claims about "new kilograms" and "peer reviewed" papers, I'd show math.

Tensor
2005-Jul-14, 04:24 AM
But show you what? :-?

Let's make this simple. What equation, using your new variable G, are you using to come up with a seemingly match to the H0?

Show me some math, then we'll talk.

Well, see that's what Tassel's asking for, and I'd be interested in seeing also, your math equations. Of course, you stated you can't calculate it or aren't ready to release it. Wondering why you would use the Earth measured gravitational redshift value in an equation with an extra-galactic large variable G. You should use the large variable G redshift value. It all come down to this: without producing the equations showing how you got the number for H0, all you have is a bit of numerology.

papageno
2005-Jul-14, 09:01 AM
Which requires a violation of the Equivalence Principle.
Or a variation of G so small that we still cannot detect it (and nowhere near 5G or 100000G; more like G/100000).

Again, where is the evidence that Equivalence Principle is violated?

Its not. [-X
So, we would expect to see non-elliptical orbits for planets, satellites, asteroids, satellites and space junk.

What good is your speculation, if you cannot work within the constraints given by the experimental results?

papageno
2005-Jul-14, 09:10 AM
But this is the crux of the matter. A kilogram is currently defined as a particular quantity of matter. It has absolutely nothing to do with local g or the universal gravitational constant G.
You're absolutely right. For the postulated 1 G for our whole universe, that is how a kilogram works, and it has nothing to do with any local G. But it does get tricky if the local G is different from our postulated 1 G (which we applied to the whole cosmos), because then the same kilogram of mass acts differently,...
The weight is different.

... not as if 1 G times mass, but as variable G times mass, so the mass itself is affected.
As I said uncountable times, the gravitational force is different.

This is something "papageno" cannot see, though I've explained it over and over again, so he keeps arguing his case in a 1 G universe.
Wrong.
I showed that your reasoning has nothing to do with the particular value of G.
I showed the consequences of a variable G and argued that, given the constraints well established by experiments, G is not variable by the amount you claim.

All well and good, but it's not good enough for me. One kilogram of mass is always the same mass, only it registers differently if G (not variable G times 1 G mass) is different, since then it is "variable G times its own G mass".
So, you are arguing that the gravitational mass changes, but the inertial mass does not.
You are claiming that the Equivalence Principle must be violated in your variable G "speculation".

My hunch is that the greater intergalactic cosmos operates at a different G, about 100,000 times greater than ours, and the mass out there measured in that G acts differently from how it acts on Earth. Of course, once more everybody, "it's only hypothetical!". :lol:
You "hypothesis" is disproven by observations.
A violation of the Equivalence Principle implies that General Relativity is wrong, hence its predictions should not match the observations.

nutant gene 71
2005-Jul-14, 07:17 PM
But show you what? :-?

Let's make this simple. What equation, using your new variable G, are you using to come up with a seemingly match to the H0?

Show me some math, then we'll talk.

Well, see that's what Tassel's asking for, and I'd be interested in seeing also, your math equations. Of course, you stated you can't calculate it or aren't ready to release it. Wondering why you would use the Earth measured gravitational redshift value in an equation with an extra-galactic large variable G. You should use the large variable G redshift value. It all come down to this: without producing the equations showing how you got the number for H0, all you have is a bit of numerology.

Lets answer this request first with a look at this equation (http://www.badastronomy.com/phpBB/viewtopic.php?p=501100#501100) on Redshift in Solar System. It says:

What is the distance of (delta) 1 z?

I figured it out as 129.2 million light-years, which at ~9.46E+6 meters per light-year, I figured it as delta z = 1 (1% of lightspeed), the total distance for light to travel (ignoring expanding space for now) is about D_1z = ~1.222E+24 meters. Did anyone else ever figure this out? Is this a good number?

If so, then the light redshift for our solar system, assuming it ends somewhere at 70 AU, is rather small, something to the effect of z = 0.54E-6, if I take 70 AU and divide it by 129.2E+6 AU. if so, we're talking about a half of a millionth of light speed for redshift at the edge of our solar system, if 70 AU is really it's edge. I think this works out for a value of lightshift z = ~v/c = 0.54E-6 = v/3E+8 m/s, which gives us a (Doppler equivalent) v = ~162 m/s, at 70 AU.
In fact, I am wrong on where our heliosphere ends, that number should be more like 85-90 AU, but the question is still the same for distance of 1z . (Note, the Doppler-equivalent velocity is irrelevant here, other forces may be at work at 70 AU, like the Suns gravitational pull, which negate this Doppler velocity, so there should be no space-expansion within our solar system.)

Now, if this distance for 1 z, viz. D_1z = ~1.222E+24 m, is correct, then we can proceed to the next step. (No one answered my question, so dont know if right or not.) I believe we now assume that empty intergalactic space has about one atom per cubic centimeter, most of which (92%) are hydrogen atoms. If we take that volume of atoms per one linear meter, were talking about 100 atoms (within a 1 cm square-tube one meter long), and if multiplied by the distance for 1z, then we arrive at how many atoms lie in a continuous line, which works out to be = 1.222E+26 atoms of (mostly) hydrogen per the distance of light traveling delta 1 z.

And if this is correct, what is the total (linear) mass of all those atoms?

Ill let you ponder this for now. Any comments are welcome, or math assistance.
Lets take this one step at a time, so will check back later. Thanks.

nutant gene 71
2005-Jul-15, 12:48 AM
All well and good, but it's not good enough for me. One kilogram of mass is always the same mass, only it registers differently if G (not variable G times 1 G mass) is different, since then it is "variable G times its own G mass".
So, you are arguing that the gravitational mass changes, but the inertial mass does not.
You are claiming that the Equivalence Principle must be violated in your variable G "speculation".
Here is once again the kernel of the argument, why you and I see it differently.

Inertial-mass is always equivalent to its gravitational-mass. Let me illustrated it thus:

Take a one kilogram mass (same cube or rod or ball or whatever, did not change its composition in any way) from Earth's 1 G and move it to Jupiter's (hypo) 5 G. Place this kilogram of mass on a balance scale, and it balances against another kilogram mass. Now try lifting this kilogram off Jupiter's surface: it will take 5 times the force to do so. I interpret this as 5F = 5m*a; while you interpret this as 5F = m*5a; this is our fundamental difference. But Jupiter is not 5 times the mass it was before, same mass. If Jupiter were 5 times larger mass, than we operate in 1G, where 5F = m* 5a; but if Jupiter is still the same mass, we operate in 5G, where 5F = 5m*a, which means Jupiter's "kilogram" is one-fifth Earth's kilogram.. Now take a one "kilogram" mass from Jupiter, which is one-fifth the size of our previous mass, and take it to Earth's 1 G. Place that 0.2 mass on a balance scale with Earth's one kilogram mass, and the Earthian kilogram wins hands down. This is why Jupiter (if it really is in 5 G) is a far smaller planet core than we realize. Because this core is hidden by its incredibly vast atmosphere, we can't see it, though Voyager reported that it is about two or three Earth masses, which is still very small. Yet Jupiter's small core is operating in 5 G, so it balances on a scale with Earth 1 G mass, but it would take ten to fifteen Earths to do so. If you tried to move 10 to 15 Earths rather than two to three Earths, you would need a great deal more force, and that is why the inertial mass and gravitational mass are equivalent.

Here is something by way of illustration that maybe will give you a better sense of what I am talking about: Enceladus (http://news.bbc.co.uk/1/hi/sci/tech/4676069.stm), especially where they say:

"Likewise, with the atmosphere, Enceladus does not possess the gravitational attraction necessary to hold on to a cloud of water ions, so this must be being replenished also."

Only problem, Enceladus being about the size of Arizona, is too small gravitationally in 1 G; but it may not be too small in 5-10 G (I think Saturn has ~10 G). The scientists who are puzzling over this (they know nothing of a greater G) are saying that there must be a source replenishing these water ions, and atmosphere. But there is no evidence of this since we began looking, so they're mystified, naturally.

This is why mass behaves differently in 1 G or 5 G: in 5 G it can do things that in 1 G it cannot, like hold onto more ambient mass, gases, water ions, etc. BTW, the current thought is that Saturn's core planet must be much larger than Jupiter's small core, but that again is a function of not knowing that G on Saturn may be twice that of Jupiter, so to compensate theory they give Saturn a bigger core, which has not been found to date due to Saturn's thicker (bigger G, hypothetically) atmosphere.

I think where you run into trouble here, conceptually (and why this whole idea rubs you the wrong way), is that by referring back to all our spaceprobes that successfully maneuvered their way around Jupiter and Saturn, there was no variable G ever detected. And this is positively true, but we may have the planet's mass-volume figured in 1 G, when it fact it should be substantially smaller in 5 G. We use gravity assist, which is a function of (G*m) to navigate the spacecrafts, and the spacecraft mass is infinitesimal in comparison to the mass of the planet, so everything more or less works fine (after a few inflight adjustments). You see, it is well hidden, even if wrong, because we modeled the universe after our planet's 1 G environment, so even if the G is different elsewhere, we can't tell... well, except for all the puzzling things like an atmosphere on Enceladus or Pluto, where we explain it away saying that they are "gassing out". End result? We get a very confused view of cosmological events from our solar system out to deep space. Is the Pioneers Anomaly giving us a clue? Most people, including yourself, don't think any of these puzzling phenomena are gravitational. I do...* But I do not wish to convince, only to illustrate how I see it, and remember I may be wrong. :)

*(I think it is at least possible, but we can't know until we find a way to measure for this, so only hypothetical for now.)

nutant gene 71
2005-Jul-15, 01:20 AM
I'd like to make an editorial comment here.

I know all you guys, and gals, are smart here. Yes, Tassel, I think you're smart too. I wouldn't be going through the trouble to explain this if I didn't think so. I had been on other boards where I did not find this the case (won't mention names) and left after some time. But in the same vein I do not want you to accept my word on this, because until we have evidence to bear this out, we cannot assume any of it is right. My goal is for you to see that there is another way to understand what gravity is doing out there, and to file it away into your little grey cells until such time that something gets revealed that maybe it is so. Until then, what you learn in school, or had always accepted professionally, will have to serve. Three hundred years of science cannot be dismissed as wrong so easily, and it should not be, but if there should be evidence in the future that we did not get gravity right, I ask only this, at least be prepared conceptually to deal with it. But for now, all we know is what we know.

Metricyard
2005-Jul-15, 02:50 AM
Inertial-mass is always equivalent to its gravitational-mass. Let me illustrated it thus:

Take a one kilogram mass (same cube or rod or ball or whatever, did not change its composition in any way) from Earth's 1 G and move it to Jupiter's (hypo) 5 G. Place this kilogram of mass on a balance scale, and it balances against another kilogram mass. Now try lifting this kilogram off Jupiter's surface: it will take 5 times the force to do so. I interpret this as 5F = 5m*a; while you interpret this as 5F = m*5a; this is our fundamental difference. But Jupiter is not 5 times the mass it was before, same mass. If Jupiter were 5 times larger mass, than we operate in 1G, where 5F = m* 5a; but if Jupiter is still the same mass, we operate in 5G, where 5F = 5m*a, which means Jupiter's "kilogram" is one-fifth Earth's kilogram.. Now take a one "kilogram" mass from Jupiter, which is one-fifth the size of our previous mass, and take it to Earth's 1 G. Place that 0.2 mass on a balance scale with Earth's one kilogram mass, and the Earthian kilogram wins hands down. This is why Jupiter (if it really is in 5 G) is a far smaller planet core than we realize. Because this core is hidden by its incredibly vast atmosphere, we can't see it, though Voyager reported that it is about two or three Earth masses, which is still very small. Yet Jupiter's small core is operating in 5 G, so it balances on a scale with Earth 1 G mass, but it would take ten to fifteen Earths to do so. If you tried to move 10 to 15 Earths rather than two to three Earths, you would need a great deal more force, and that is why the inertial mass and gravitational mass are equivalent.

Ah, were back to changing variable mass.
When does the variable light speed come back to your theory? :lol:

I interpret this as 5F = 5m*a; while you interpret this as 5F = m*5a; this is our fundamental difference.

What are you showing us here??? 5F=5ma -> 5*F=5*m*a #-o
This is basic algebra. Of course they're going to balance. Moving an '*' isn't going to revolutionize physics anytime soon.

Edit to add. Sorry, it must be way past my bed time. Actually, your formula will never balance. Why do you have F multiplied by 5?

If I make m=2 and a=3 (no units needed here) then F should equal 30 using you formula.

But in your formula, F would equal 150, hence, they would never balance.

Sleep on this I must.

papageno
2005-Jul-15, 11:23 AM
All well and good, but it's not good enough for me. One kilogram of mass is always the same mass, only it registers differently if G (not variable G times 1 G mass) is different, since then it is "variable G times its own G mass".
So, you are arguing that the gravitational mass changes, but the inertial mass does not.
You are claiming that the Equivalence Principle must be violated in your variable G "speculation".
Here is once again the kernel of the argument, why you and I see it differently.

Inertial-mass is always equivalent to its gravitational-mass.
So, since you claim that the mass changes, it wouuld observable using non-gravitational forces.
These experiments show that the inertial mass does not change.

Let me illustrated it thus:

Take a one kilogram mass (same cube or rod or ball or whatever, did not change its composition in any way) from Earth's 1 G and move it to Jupiter's (hypo) 5 G. Place this kilogram of mass on a balance scale, and it balances against another kilogram mass. Now try lifting this kilogram off Jupiter's surface: it will take 5 times the force to do so. I interpret this as 5F = 5m*a; while you interpret this as 5F = m*5a; this is our fundamental difference.
And the "5a" interpretation is consistent with the observations.
And we know that, because we have observe known masses in situations where the gravitational force is different.
And because the spectra from other planets and stars are recognizable from spectra taken on Earth, and because the electonics of our porbes worked fine (if it did not, we would not have any image taken with CCD-based cameras).

But Jupiter is not 5 times the mass it was before, same mass.
But the gravitationally force Jupiter exerts is 5 times larger.
This would be observable, unless you invoke a violation of the Equivalence Principle.

If Jupiter were 5 times larger mass, than we operate in 1G, where 5F = m* 5a; but if Jupiter is still the same mass, we operate in 5G, where 5F = 5m*a, which means Jupiter's "kilogram" is one-fifth Earth's kilogram..
The weight of 1 kg-mass would be 5 times larger on Jupiter.
But Jupiter's kg-mass would be exactly the same as Earth's.

Now take a one "kilogram" mass from Jupiter, which is one-fifth the size of our previous mass, and take it to Earth's 1 G.

Place that 0.2 mass on a balance scale with Earth's one kilogram mass, and the Earthian kilogram wins hands down.
Of course, because 1 kg > 0.2 kg.
But 0.2 kg is not Juptier's kg, it is 0.2 Jupiter's kg.

You are still confused about weight and mass.
Why don't you express the wieght in Newtons?

This is why Jupiter (if it really is in 5 G) is a far smaller planet core than we realize. Because this core is hidden by its incredibly vast atmosphere, we can't see it, though Voyager reported that it is about two or three Earth masses, which is still very small. Yet Jupiter's small core is operating in 5 G, so it balances on a scale with Earth 1 G mass, but it would take ten to fifteen Earths to do so. If you tried to move 10 to 15 Earths rather than two to three Earths, you would need a great deal more force, and that is why the inertial mass and gravitational mass are equivalent.
You still forget that Jupiter's atmosphere itself has a significant mass.

Here is something by way of illustration that maybe will give you a better sense of what I am talking about: Enceladus (http://news.bbc.co.uk/1/hi/sci/tech/4676069.stm), especially where they say:

"Likewise, with the atmosphere, Enceladus does not possess the gravitational attraction necessary to hold on to a cloud of water ions, so this must be being replenished also."

Only problem, Enceladus being about the size of Arizona, is too small gravitationally in 1 G; but it may not be too small in 5-10 G (I think Saturn has ~10 G). The scientists who are puzzling over this (they know nothing of a greater G) are saying that there must be a source replenishing these water ions, and atmosphere. But there is no evidence of this since we began looking, so they're mystified, naturally.

This is why mass behaves differently in 1 G or 5 G: in 5 G it can do things that in 1 G it cannot, like hold onto more ambient mass, gases, water ions, etc. BTW, the current thought is that Saturn's core planet must be much larger than Jupiter's small core, but that again is a function of not knowing that G on Saturn may be twice that of Jupiter, so to compensate theory they give Saturn a bigger core, which has not been found to date due to Saturn's thicker (bigger G, hypothetically) atmosphere.

I think where you run into trouble here, conceptually (and why this whole idea rubs you the wrong way), is that by referring back to all our spaceprobes that successfully maneuvered their way around Jupiter and Saturn, there was no variable G ever detected. And this is positively true, but we may have the planet's mass-volume figured in 1 G, when it fact it should be substantially smaller in 5 G. We use gravity assist, which is a function of (G*m) to navigate the spacecrafts, and the spacecraft mass is infinitesimal in comparison to the mass of the planet, so everything more or less works fine (after a few inflight adjustments).
5 times the expected force is not small.

You see, it is well hidden,...
To hide it, you need a violation of the Equivalence Principle.

...even if wrong, because we modeled the universe after our planet's 1 G environment, so even if the G is different elsewhere, we can't tell... well, except for all the puzzling things like an atmosphere on Enceladus or Pluto, where we explain it away saying that they are "gassing out". End result? We get a very confused view of cosmological events from our solar system out to deep space. Is the Pioneers Anomaly giving us a clue? Most people, including yourself, don't think any of these puzzling phenomena are gravitational. I do...* But I do not wish to convince, only to illustrate how I see it, and remember I may be wrong. :)
You just claimed that the variable G is well hidden.
Why would it show up only in the Pionerr probes?

*(I think it is at least possible, but we can't know until we find a way to measure for this, so only hypothetical for now.)

nutant gene 71
2005-Jul-15, 06:36 PM
Now take a one "kilogram" mass from Jupiter, which is one-fifth the size of our previous mass, and take it to Earth's 1 G.

Place that 0.2 mass on a balance scale with Earth's one kilogram mass, and the Earthian kilogram wins hands down.

Of course, because 1 kg > 0.2 kg.
But 0.2 kg is not Juptier's kg, it is 0.2 Jupiter's kg.

You are still confused about weight and mass.
Why don't you express the wieght in Newtons?

Ive avoided the weight discussion because it may confuse. Your weight on Jupiter will be the same as we now estimated, since your material mass had not changed when you got there.

So if you weigh on earth 100 kg (Earth scale), your weight on Earth (at 9.8 m/s^2) is 980 N. If G is 5 times greater on Jupiter, your Jupiter weight woud be five times Earths N (times 23 m/s^2), so you would weigh 23 * 4900 N, which now makes you in Newtons force = 112,700 N. However, that is incorrect. Your weight on Jupiter is the same as it was before we (hypothetically) discovered it had a different G, so the correct weight on Jupiter is only 100 kg * 23 m/s^2, so you would weigh 2300 N. So you see, it doesnt work, which is why I avoided this analogy.

How would you calculate it to show your analogy better?

Tassel
2005-Jul-15, 06:41 PM
The definition of Kilogram (http://en.wikipedia.org/wiki/Kilogram):

The kilogram or kilogramme, (symbol: kg) is the SI base unit of mass. A gram is defined as one thousandth of a kilogram. Conversion of units describes equivalent units of mass in other systems.

Since 1889, the SI system defines the unit to be equal to the mass of the international prototype of the kilogram, which is made from an alloy of platinum and iridium of 39 mm height and diameter, and kept at the Bureau International des Poids et Mesures (International Bureau of Weights and Measures).

The definition of Mass (http://en.wikipedia.org/wiki/Mass):

Mass is a property of physical objects that, roughly speaking, measures the amount of matter they contain. It is a central concept of classical mechanics and related subjects.

Based on the definitions above, kilograms are a numeric representation of the amount of matter that an object is composed of. 1 kilogram is defined as the amount of matter that composes a specific platinum and iridium rod.

We could take that rod anywhere in the universe, even in a universe where G varies greatly (which we know isn't the case in our universe). Since the amount of matter in the rod does not change, then by definition the mass of the rod is still one kilogram.

The amount of matter that makes up an object is a physical property that is measured in kilograms. What 1 kilogram means is not up for debate. It is a defined value. That definition does not include the value of G. Don't bother: it doesn't include the value of G not because "we've always assumed constant G", but because G has no bearing on how much matter an object is composed of.

Kilograms are a measure of mass. Mass is a physical property of an object that represents the amount of matter that makes up the object. The amount of matter in an object does not depend on the value G. Therefore, kilograms, which are the unit we use to represent quantities of matter, would not be different if G were different.

If you want to debate what happens to the unit "kilogram" if G were somehow different, you are not debating anyone in this forum. You are debating the definition of "kilogram".

Yes, Tassel, I think you're smart too.
You launched an ad hominem attack on my intelligence while simultaneously demonstrating that you do not have any kind of intuitive understanding about how orbits are determined. With that, and your history of intellectual dishonesty in mind, I couldn't possibly care any less if you think I'm smart or not.

nutant gene 71
2005-Jul-15, 06:59 PM
...snip...

If you want to debate what happens to the unit "kilogram" if G were somehow different, you are not debating anyone in this forum. You are debating the definition of "kilogram".

Yes, Tassel, I think you're smart too.
You launched an ad hominem attack on my intelligence while simultaneously demonstrating that you do not have any kind of intuitive understanding about how orbits are determined. With that, and your history of intellectual dishonesty in mind, I couldn't possibly care any less if you think I'm smart or not.
You're absolutely right, it is not a debate against anyone's person but a debate against understanding, or lack of it. Since this is becoming nasty, I'm going to move on from this "1 kilogram" discussion and turn my attention to an earlier question raised about the mass of atoms in space at 1 z distance. Later.

Tassel
2005-Jul-15, 07:40 PM
So if you weigh on earth 100 kg (Earth scale), your weight on Earth (at 9.8 m/s^2) is 980 N. If G is 5 times greater on Jupiter, your Jupiter weight woud be five times Earths N (times 23 m/s^2), so you would weigh 23 * 4900 N, which now makes you in Newtons force = 112,700 N. However, that is incorrect. Your weight on Jupiter is the same as it was before we (hypothetically) discovered it had a different G, so the correct weight on Jupiter is only 100 kg * 23 m/s^2, so you would weigh 2300 N. So you see, it doesnt work, which is why I avoided this analogy.
You are so confused it is becoming painful to read this thread.

W = mg

The mass of an object is 100kg.
The weight on Earth is: 100kg * 9.8m/s^2 = 980N
The weight on Jupiter is: 100kg * 23m/s^2 = 2300N

G is required to calculate the mass of Jupiter. If we discovered G were different at Jupiter, then we'd realize we had always had Jupiter's mass wrong. We'd recalculate Jupiter's mass and find it was much lower than we thought. You've pointed this out seemingly hundreds of times: GM remains constant. Therefore, acceleration due to gravity remains constant. The mass of the object in question remains constant. If m (mass of the object) and g (acceleration due to gravity) are the same before and after the "discovery" that we were wrong about G, guess what? W is the same before and after the "discovery". If you'd just take 30 seconds and walk through the calculations step-by-step instead of running around randomly multiplying everything by 5, you'd see this.

nutant gene 71
2005-Jul-15, 09:45 PM
So if you weigh on earth 100 kg (Earth scale), your weight on Earth (at 9.8 m/s^2) is 980 N. If G is 5 times greater on Jupiter, your Jupiter weight woud be five times Earths N (times 23 m/s^2), so you would weigh 23 * 4900 N, which now makes you in Newtons force = 112,700 N. However, that is incorrect. Your weight on Jupiter is the same as it was before we (hypothetically) discovered it had a different G, so the correct weight on Jupiter is only 100 kg * 23 m/s^2, so you would weigh 2300 N. So you see, it doesnt work, which is why I avoided this analogy.
You are so confused it is becoming painful to read this thread.

W = mg

The mass of an object is 100kg.
The weight on Earth is: 100kg * 9.8m/s^2 = 980N
The weight on Jupiter is: 100kg * 23m/s^2 = 2300N

G is required to calculate the mass of Jupiter. If we discovered G were different at Jupiter, then we'd realize we had always had Jupiter's mass wrong. We'd recalculate Jupiter's mass and find it was much lower than we thought. You've pointed this out seemingly hundreds of times: GM remains constant. Therefore, acceleration due to gravity remains constant. The mass of the object in question remains constant. If m (mass of the object) and g (acceleration due to gravity) are the same before and after the "discovery" that we were wrong about G, guess what? W is the same before and after the "discovery". If you'd just take 30 seconds and walk through the calculations step-by-step instead of running around randomly multiplying everything by 5, you'd see this. -- bold mine

Exactly! So what is Jupiter's "kilograms" per my bold? Is 100 kg on Earth still 100 kg on Jupiter, in (assumed) 5 G? Or is it this same mass now 20 "kg" per Jupiter's 5X (Earth) G? Don't you see that in Jupiter's greater graivtational G "proportional" each kilogram is now five times greater? Of course the end result is the same mass with 100 kg, but can you see the qualitative difference here, or not?

Don't get frustrated about it, but think of what is happpening if a body's gravitational G is different: to arrive at the same kilograms of mass we use on Earth, the "kilograms" of a different G world have to be adjusted for the higher G. There is no getting around this, and this is why the mass acts differently there, because (per above example) each kg.-Earth is now equivalent to 0.2 kg-Jupiter. This is not so confusing, and I'm sure if you can understand Relativity, you can understand a variable G, in fact it's easier!

Metricyard
2005-Jul-15, 10:20 PM
Exactly! So what is Jupiter's "kilograms" per my bold? Is 100 kg on Earth still 100 kg on Jupiter, in (assumed) 5 G? Or is it this same mass now 20 "kg" per Jupiter's 5X (Earth) G? Don't you see that in Jupiter's greater graivtational G "proportional" each kilogram is now five times greater? Of course the end result is the same mass with 100 kg, but can you see the qualitative difference here, or not?

What happens to the other 80kgs? And how does having a larger G replace the missing mass? This is why I think you didn't want to do my little thought experiment. To prove you theory without using any math, you're going to have to remove mass. All you have to do is explain why a larger G would have anything to do with replacing mass.

Don't get frustrated about it, but think of what is happpening if a body's gravitational G is different: to arrive at the same kilograms of mass we use on Earth, the "kilograms" of a different G world have to be adjusted for the higher G. There is no getting around this, and this is why the mass acts differently there, because (per above example) each kg.-Earth is now equivalent to 0.2 kg-Jupiter. This is not so confusing, and I'm sure if you can understand Relativity, you can understand a variable G, in fact it's easier!

Why is this easier? You're creating a formula that is way overcomplicated. Why spend so much time to try and prove a variable G and m that's trying to give the same results as a standard formula would?

Tassel
2005-Jul-15, 10:39 PM
Exactly! So what is Jupiter's "kilograms" per my bold? Is 100 kg on Earth still 100 kg on Jupiter, in (assumed) 5 G?
Yes. Kilograms are a unit of measure indicating the amount of matter that an object is composed of. The amount of matter making up the object has not changed.

Or is it this same mass now 20 "kg" per Jupiter's 5X (Earth) G?
No.

W = 20kg * 23m/s^2 = 460N which is NOT what we would observe, if we were wrong about G at Jupiter.

Do the math, and post the calculations step-by-step (you won't do this). Start with Jupiter's mass based on the currently accepted value for G and calculate "g" for Jupiter. Then, replace the currently accepted value of G with any value you choose. Recalculate Jupiter's mass utilizing this new value of G and then recalculate "g". Since GM remains the same, you will find that "g" is the same.

"g" and the mass of the object you are weighing are all you need to determine weight, since "g" is calculated using G.

You do not need to change the definition of kilograms to support variable G. You will never realize this until you actually sit down, take a deep breath, and work through the calculations step by step. You are trying to take shortcuts, and you are making mistakes. For example:

So if you weigh on earth 100 kg (Earth scale), your weight on Earth (at 9.8 m/s^2) is 980 N. If G is 5 times greater on Jupiter, your Jupiter weight woud be five times Earths N (times 23 m/s^2), so you would weigh 23 * 4900 N, which now makes you in Newtons force = 112,700 N.
You are saying that W(Jupiter) is W(Earth) * 5 * g(Jupiter). This is just horrendously wrong. In the real, constant G universe, you don't multiply W(earth) by g(Jupiter) to get W(Jupiter). You multiply the mass of the object by the "g" of the planet you are weighing it on. You further mangle the issue by multiplying by the increase in G, which also makes no sense, since g(Jupiter) already includes G (old or "new").

Again, I suggest you work out the calculations with real numbers, starting with the currently accepted values for G and Jupiter's mass. If you just do this, you will see why variable G works just fine (mathematically...) in all regards with the current definition of "kilogram".

Edit to add: if you do the calculations the value for "g" at Jupiter should be closer to 25m/s^2, which is Jupiter's "surface gravity". On the fact sheet (http://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html), "surface acceleration" (which for Jupiter is listed at around 23m/s^2) includes the effects of the rotation of the planet.

nutant gene 71
2005-Jul-16, 12:33 AM
No.

W = 20kg * 23m/s^2 = 460N which is NOT what we would observe, if we were wrong about G at Jupiter.
What happened to the 5 G? You're missing it.

(1G*m) = (5G*0.2m)

W = 5(20kg) * 23m/s^2 = 2300 N.

So if you weigh on earth 100 kg (Earth scale), your weight on Earth (at 9.8 m/s^2) is 980 N. If G is 5 times greater on Jupiter, your Jupiter weight woud be five times Earths N (times 23 m/s^2), so you would weigh 23 * 4900 N, which now makes you in Newtons force = 112,700 N.
You are saying that W(Jupiter) is W(Earth) * 5 * g(Jupiter). This is just horrendously wrong. In the real, constant G universe, you don't multiply W(earth) by g(Jupiter) to get W(Jupiter). You multiply the mass of the object by the "g" of the planet you are weighing it on. You further mangle the issue by multiplying by the increase in G, which also makes no sense, since g(Jupiter) already includes G (old or "new").
You are correct, my mistake. I had double multiplied Earth's N. The correct version should have been:

W = 100 kg * 9.8 m/s^2 = 980 N for Earth
W = 100 kg * 23 m/s^2 = 2300 N for Jupiter.

That is how we know it in one G. Now, if (1G*m) = (5G*0.2m), for Jupiter's five times higher G, then it should read:

W" = 5(20 "kg" * 23 m/^2) = 2300 N for Jupiter, if you factor in the 5 G.

The way you see it, the 5 G doesn't exist anywhere, and if I did it your way, if I understand what you're saying, factoring in the 5 G, it would look like this:

W'" = 5(100 kg * 23.m^2) = 11,500 N , which is wrong.

Metricyard
2005-Jul-16, 12:55 AM
You are correct, I had double multiplied Earth's N. The correct version should have been:

W = 100 kg * 9.8 m/s^2 = 980 N for Earth
W = 100 kg * 23 m/s^2 = 2300 N for Jupiter.

That is how we know it in one G. Now, if (1G*m) = (5G*0.2m), for Jupiter's five times higher G, then it should read:

W" = 5(20 kg * 23 m/^2) = 2300 N for Jupiter, if you factor in the 5 G.

The way you see it, the 5 G doesn't exist anywhere, and if I did it your way, if I understand what you're saying, factoring in the 5 G, it would look like this:

W'" = 5(100 kg * 23.m^2) = 11,500 N , which is wrong.

But where is the mass going? In the above example, you've lost 80kgs of mass. Why and where??

nutant gene 71
2005-Jul-16, 03:17 AM
But where is the mass going? In the above example, you've lost 80kgs of mass. Why and where??
80 kg is not lost: 20 "kg" of Jupiter = 100 kg of Earth. The "loss" is that they are NOT the same kilograms, but different units. Viz., each Jupiterian "kilogram", per this 5 G scenario, is worth 5 Earthian kilograms.

This is why I said "hypothetical variable mass" in my original question, because the mass varies in a "hypo variable G" scenario. It's been my whole point all along, though it seems I had not communicated this to anybody, at least not so far... :-?

I'm now working on the other puzzle, about how much gravity would deep space need to do a "gravitational redshift" of 1 z, within the distance light travels in 1 z. Still working on it conceptually, since the SI units also need to balance. Stay tuned.

Tensor
2005-Jul-16, 04:58 AM
But where is the mass going? In the above example, you've lost 80kgs of mass. Why and where??
80 kg is not lost: 20 "kg" of Jupiter = 100 kg of Earth. The "loss" is that they are NOT the same kilograms, but different units. Viz., each Jupiterian "kilogram", per this 5 G scenario, is worth 5 Earthian kilograms.

You have a problem then. The thrust of the engines of probes around Jupiter would produce too much thrust and the probes would not follow the plotted trajectory. And, even if you get around this, you have the problem of reactions wheels that are set with the measured G near earth in mind. These would also not move the spacecraft as planned .

You really also need to come up the amount of change around the different planets.

And once you get past that, you have to explain why the tests of GR outside our solar system (such as the amount of observed periastron change around PSR 1913 + 16 (http://nobelprize.org/physics/laureates/1993/press.html) and theoretical calculations concerning that binary pairmatch (http://pdg.ift.unesp.br/2005/reviews/gravrpp.pdf) so closely (see figure 18.1) with a non variable G assumption and explain how well your idea matches the observations just as well.

nutant gene 71
2005-Jul-16, 05:14 AM
COSMIC LIGHT REDSHIFT FROM GRAVITATIONAL FACTORS ONLY, part 1.

...
Lets answer this request first with a look at this equation (http://www.badastronomy.com/phpBB/viewtopic.php?p=501100#501100) on Redshift in Solar System. It says:

What is the distance of (delta) 1 z?

I figured it out as 129.2 million light-years, which at ~9.46E+6 meters per light-year, I figured it as delta z = 1 (1% of lightspeed), the total distance for light to travel (ignoring expanding space for now) is about D_1z = ~1.222E+24 meters. Did anyone else ever figure this out? Is this a good number?

If so, then the light redshift for our solar system, assuming it ends somewhere at 70 AU, is rather small, something to the effect of z = 0.54E-6, if I take 70 AU and divide it by 129.2E+6 AU. if so, we're talking about a half of a millionth of light speed for redshift at the edge of our solar system, if 70 AU is really it's edge. I think this works out for a value of lightshift z = ~v/c = 0.54E-6 = v/3E+8 m/s, which gives us a (Doppler equivalent) v = ~162 m/s, at 70 AU.
In fact, I am wrong on where our heliosphere ends, that number should be more like 85-90 AU, but the question is still the same for distance of 1z . (Note, the Doppler-equivalent velocity is irrelevant here, other forces may be at work at 70 AU, like the Suns gravitational pull, which negate this Doppler velocity, so there should be no space-expansion within our solar system.)

Now, if this distance for 1 z, viz. D_1z = ~1.222E+24 m, is correct, then we can proceed to the next step. (No one answered my question, so dont know if right or not.) I believe we now assume that empty intergalactic space has about one atom per cubic centimeter, most of which (92%) are hydrogen atoms. If we take that volume of atoms per one linear meter, were talking about 100 atoms (within a 1 cm square-tube one meter long), and if multiplied by the distance for 1z, then we arrive at how many atoms lie in a continuous line, which works out to be = 1.222E+26 atoms of (mostly) hydrogen per the distance of light traveling delta 1 z.

And if this is correct, what is the total (linear) mass of all those atoms?

Ill let you ponder this for now. Any comments are welcome, or math assistance.
Lets take this one step at a time, so will check back later. Thanks.
The answer to this question, as to what is the linear mass of all hydrogen-atoms in the distance light traveled at 1 z is pretty much straightforward. Multiply the atomic mass of hydrogen by the number of atoms in that distance:

hydrogen mass: m= 1.67E-27 kg
atoms in 1 z (if correct) = 1.222E+26 h atoms

1) M_1z = (1.67E-27 kg)*(1.222E+26) = 2.04E-1 kg (less than one kilogram of mass for all that distance of 1z space!)

Now take a ratio of this mass for hydrogen over the distance of 1z, and multiply it by the distance of redshift 1z (per kilogram), so you again have 0.204 kg. Gravitationally, if we take the same 1z in its gravitational per meter value, per Wiki's Gravitational Redshift (http://en.wikipedia.org/wiki/Gravitational_redshift), by dividing the delta z value at 22 meters, 2.5E-15 / 22 m, we arrive at 1.136E-16 redshift (per kilogram), in Earth's 1 G for one meter.

[Note (this is new): I say atoms "per meter" and "per kilogram" because we are converting distance meters into kilograms equivalent of space-mass "kilograms" per meter, as it applies to gravitational lightshift per meter. So per a linear meter of mass, the gravitational lightshift z is also per the kilograms in that linear meter. If we multiply this linear meter cum kilograms by the basic gravitational lightshift measured (in Earth's 1 G), we get: delta G-z = 1.136E-16 kg^-1. ]

We can now figure what is the G needed for deep intergalactic space to accommodate lightshift of 1Z over the distance 1z_D = 129 million light years (which is D = 1.222E+24 m). The equation works out to be:

[(delta G-z) * (h M)] / G-Earth = G-space

where: delta G-z = 1.136E-16 kg^-1
hydgrogen M-space = 2.04E-1 kg (which is: h mass = 1.67E-27 kg times 1 z distance 1.222E+24 m, times 100 atoms in space)
G-Earth = 6.67E-11 m^3 kg^-1 s^-2

Working it out, it becomes:

(1.136E-16 kg^-1)*(2.04E-1 kg) / 6.67E-11 m^3 kg^-1 s^-2 = 0.347E-6 m^3 kg^-1 s^-2 = 0.347E-6 m^3 kg^-1 s^-2 = deep space-G

In words: We are taking the gravitational-lighshift (per kilogram) as we find it in 1 G, times the distance of 1z in meters, times the product of hydrogen mass and number of atoms per linear meter (per kilogram), times the total mass of space hydrogen atoms in 1z, all divided by Earth's G; all to equal deep space gravity in space-G.

This deep space-G = ~0.347E-6 Nm^2 kg^2 is what would be needed to redshift light at 1z per the atomic hydrogen mass in that distance, in terms of the gravitational-lightshift on Earth, as adjusted per Earth's G. I believe we have to allow for an error here, since deep intergalactic space is denser than figured here, since "empty" space is not merely hydrogen atoms but has other gases as well, which are heavier than hydrogen. I suspect this so derived value for deep space-G is actually higher.

But it gets more interesting.

We can figure now what is the atomic mass for hydrogen in deep space-G by the following:

(h mass) * (G-space) / G-Earth = proton Mass-space

this works out as:

(1.67E-27 kg) * (0.347E-6 Nm^ kg^2) / 6.67E-11 Nm^2 kg^2 = 0.087E-22 kg, or

deep space hydrogen Mass = ~0.87E-21 kg (six orders of magnitude Earth's hydrogen mass)

But this is a fooler! If variable mass works the way I think it does, then in deep space G, if it is over 1,000,000 times greater than Earth's G, also means that hydrogen atoms "kilograms" in space act "is if" they are over 1,000,000 times our kilograms, which means gravitational attraction for space gases is immense. And that is why hydrogen clouds in deep intergalactic space can come together, because of this greater atomic mass, and merge into a mass where if enough is present, fusion results, and it combusts into new stars.

But this still leaves the conundrum. Are these deep space "kilograms" Earth kilograms? I think the answer is yes, they are. Of course, all hypothetical until we can find a variable G out there.

I'm going to take a break here (sleep), but when I return I will show how this nearly same space-G was arrived at using the Axiomatic Equation, which is much simpler.

[Edited for calculation errors, add title.]

Tensor
2005-Jul-16, 05:51 AM
...
Lets answer this request first with a look at this equation (http://www.badastronomy.com/phpBB/viewtopic.php?p=501100#501100) on Redshift in Solar System. It says:
What is the distance of (delta) 1 z?

I figured it out as 129.2 million light-years, which at ~9.46E+6 meters per light-year, I figured it as delta z = 1 (1% of lightspeed), the total distance for light to travel (ignoring expanding space for now) is about D_1z = ~1.222E+24 meters. Did anyone else ever figure this out? Is this a good number?

...snip

If we multiply this linear meter cum kilograms by the basic gravitational lightshift measured (in Earth's 1 G), we get: delta G-z = 1.136E-16 kg^-1. ]

We can now figure what is the G needed for deep intergalactic space to accommodate lightshift of 1Z over the distance 1z_D = 129 million light years (which is D = 1.222E+24 m). The equation works out to be:

snip again...

I'm going to take a break here, but when I return I will show how this nearly same space-G was arrived at using the Axiomatic Equation, which is much simpler .

To quote EtaC's signature block "This isn't right, it isn't even wrong".

First off, how can you use a simplified GR equation (which uses a constant G) in your idea, when you haven't shown the equation applies in the case of a variable G or variable mass?

Second, the equation, using z, in the Wikipedia article is Use to determine gravitational redshft, not the cosmological and can be used to determine the schwartzchild radius of the mass in question, not the cosmological redshift.

Third, the this site (http://iapetus.phy.umist.ac.uk/Teaching/Cosmology/ScaleFactor.html) gives the equations (for determining z for the doppler and cosmological redshift. Notice the cosmological redshift, near the bottom(which is what your trying to find) does not include G or M. It uses the scale factor. Which is simply a calculation of the difference in curvature of the universe when the light was emitted and when it was recieved. But, again, these are from GR, which you haven't shown to work with a variable G or variable mass.

You're not even using the correct equations.

edited to correct the use of the equation in the wikipedia article. Changes have been italizied. I shouldn't post when it's late and I'm tired.

papageno
2005-Jul-16, 10:16 AM
Now take a one "kilogram" mass from Jupiter, which is one-fifth the size of our previous mass, and take it to Earth's 1 G.

Place that 0.2 mass on a balance scale with Earth's one kilogram mass, and the Earthian kilogram wins hands down.

Of course, because 1 kg > 0.2 kg.
But 0.2 kg is not Juptier's kg, it is 0.2 Jupiter's kg.

You are still confused about weight and mass.
Why don't you express the wieght in Newtons?

Ive avoided the weight discussion because it may confuse.
Yes, you are confused.

Your weight on Jupiter will be the same as we now estimated, ...
Wrong.
The weight = the gravitational force pulling downwards would be greater, because you increased G.

You are still confusing mass and weight.

So if you weigh on earth 100 kg (Earth scale), your weight on Earth (at 9.8 m/s^2) is 980 N. If G is 5 times greater on Jupiter, your Jupiter weight woud be five times Earths N (times 23 m/s^2), so you would weigh 23 * 4900 N, which now makes you in Newtons force = 112,700 N.
Why don't we orbit at a distance from Jupiter, so that the gravitational acceleration is the same as on the Earth's surface: it would simplify the calculations.

Let's try again:

Before increasing G
Earth: m = 100 kg, g = 9.8 m/s^2, w = m*g = 980 N.
Jupiter: m = 100 kg, g' = 9.8 m/s^2, w' = m*g' = 980 N.

Set G' = 5*G on Jupiter
Earth: m = 100 kg, g = 9.8 m/s^2, w = m*g = 980 N.
Jupiter: m = 100 kg, g'' = 5*9.8 m/s^2, w'' = m*g'' = 4900 N.

The weight at Jupiter is 5 times the weight on Earth.

Let's interpret the weight at Jupiter as if G' = G.
We perform experiments at Jupiter: using non-gravititational forces we measure the inertial mass of m, and using an Eotvos-type experiemant we test the Equivalence Principle.
We already measured the weight using springs.
We find that the inertial mass is the same as on Earth, and that the Equivalence still works.
Let's assume that we forgot to do an Cavendish-like experiment to measure G'.

What would be our conclusion?
Jupiter's mass is 5 times larger than we thought.

However, that is incorrect. Your weight on Jupiter is the same as it was before we (hypothetically) discovered it had a different G,...
Wrong, because you increased the gravitational force.
In order to have the same weight, you have to assume, without any justification whatsoever, that the gravitational mass of the probe mass has changed, without affecting the inertial mass.

...so the correct weight on Jupiter is only 100 kg * 23 m/s^2, so you would weigh 2300 N. So you see, it doesnt work, which is why I avoided this analogy.
I think you miscalculated, based on your misunderstandings.

How would you calculate it to show your analogy better?
See above.

You want the Equivalence Principle to hold, but the only way for your hypothesis to be constsnet is to violate it.

nutant gene 71
2005-Jul-16, 06:34 PM
HYPOTHETIC DEEP SPACE GRAVITY, part 2 (continued from above).

This is a continuation of the above post (http://www.badastronomy.com/phpBB/viewtopic.php?p=504217&amp;sid=38402e500523916325de0d2 75a7920d3#504217) where I showed how cosmic light redshift may be a function of intergalactic deep space gravity G, which hypothetically may be substantially greater than Earth's 1 G environment. Here will be shown that a similar result can be found using the cut-off wavelength of the photo-electric effect on deep space.

Per the modified Planck-Einstein-deBroglie equation, what I call the Axiomatic Equation (http://www.humancafe.com/cgi-bin/discus/show.cgi?70/108.html), I found that there may be a different gravitational G in deep space from what it is commonly assumed to be a universal constant at 1 G. The Axiomatic says, in its simplest stripped down form:

E = hc / em lambda*proton mass = ~(1-g)c^2,

where the right side is E = mc^2, for m =1 kg/kg, and g as the proton-to-proton gravitational "constant".

I used the "cut-off" wavelength of the photo-electric effect, with lambda as ~3.97E-7 m (orange light, just under 400 nanometers) to figure total Energy, leaving all other factors unchanged.

h = Planck's constant: 6.626E-34 m^2 kg s^-1
c = lightspeed: 3E+8 m s^-1
proton-hydrogen mass (~hydrogen mass equivalent) = 1.67E-27 kg
lambda (cut-off) = 3.97E-7 m
E = Energy

When I solved for E, I got:

(6.626E-34 m^2 kg s^-1)*(3E+8 m s^-1) / (3.97E-7 m)*(1.67E-27 kg kg^-1) = E, which becomes:

E = 3E+8 m^2 kg s^-2 or Joules. This is the square root of the usual 90 petajoules for this same equation, when em lambda is 1.32E-15 m (deBroglie wavelength for proton?); and note the kilograms for proton-hydrogen mass was adjusted to 1 kg/kg, which should raise eyebrows, but it is a function of "kilograms of mass per kilograms of space" (new idea). I also could have written E as J/kg, but more on that later.

Now, taking this Energy value back into the above Axiomatic equation, but re-inserting the standard em lambda, we can solve for the proton-hydrogen mass:

3E+8 J = (6.626E-34 m^2 kg s^-1)*(3E+8 m s^-1) / (1.32E-15 m)*(prot-h mass), which gives us a value for proton-hydrogen mass = 5.02E-19 kg kg^-1.

What this mass value represents is the kilograms (Earthian) per the "kilograms" of the space vacuum, which is a new factor not presently used in physics, since we now assume space has no kilograms function. This kilograms can be moved over to the Energy side instead, so that E = 3E+8 J/kg, as a new convention of how this system works.

Also per the Axiomatic equation, there is a method to convert the proton-hydrogen gravitational "constant" into Newton's G equivalent. This is achieved through the conversion equation (discussed in the link above), which give us:

G^2 = gc^2 * pi^2

If we take Earthian proton gravitational constant, for proton-hydrogen mass, as = ~5.9E-39, which is proportional to m = 1.67E-27 kg, then we find that for m' = 5.02E-19 kg, the equivalent is g' = 1.777E-30. Plugging that into the gravitation conversion equation, we get:

G^2 = (1.777E-30)*(9E+16 J)*(9.89) = 158.2E-14, taking the square root gives us:

G' = 12.58E-7 Nm^2 kg^2, which is:

G' = 1.258E-6 Nm^2 kg^2 (see above link for Si conversion)

Note this is not too far removed from the above gravitational-redshift derived value of G = 0.347E-6 Nm^2 kg^2, similar values using two different methodologies.

So this represents the "cut-off" gravitational G for deep space at the photo-electric lambda for em energy. For reasons I cannot explain and only speculate, this same effect has a "cut-off" value for gravity, where at this low level of Energy in deep space, light em energy "fails" to modify gravity, and thus it remains very great. As shown here, it is nearly a million times greater than Earth's Newton G "constant". In my interpretation of this phenomenal gravity, it should flatten at this rate for all of intergalactic space where Energy is low, and now change substantially until the space-vacuum gravity (space-kilograms) encounters the vicinity of a hot star. I also entertain the idea that different stars and galaxies have different Energy outputs, so that this G "constant" may be different for other solar systems, and possibly very different for (low Energy) neutron and brown dwarfs, where G would be very great.

This is of interest to me, though it may be and should be viewed critically by others, because it these two comparisons of deep space gravity dovetail correctly, hypothetically, then the universe is a very different place from what we had imagined thus far. If we should find in the future that Newton's G "proportional" is not constant but a variable instead, then everything in our cosmology thinking must change, including why cosmic light redshfits, what is the isotropic 99.99% space function for our universe, why light bends around stars and galaxies, why the gas giant planets are where they are, and why the Pioneers deep space probes are slowing. I guess all that's missing is finding the true orbits for highly elliptical comets and why Mercury precesses. On the latter, I suspect it is merely a moment of inertia transfer from our hot star's rotation to its nearest neighbor, which should respond to this as if it had a very light mass, only a speculation.

These ideas for now remain "work in progress", really not more than mental musings, but with the hope that in time, with more information, I can write an eventual science paper to show this in a more formal presentation.

Thanks for your comments, and especially for your kind patience, with these purely hypothetical ideas. They are not theories, merely an exploration into what may be very far away from us, and different.

Okay, now you can bring out the peanuts and cat calls. Any comments? :)

nutant gene 71
2005-Jul-16, 07:11 PM
Thanks Tensor, but I believe we are comparing apples and oranges.

Re yours:
First off, how can you use a simplified GR equation (which uses a constant G) in your idea, when you haven't shown the equation applies in the case of a variable G or variable mass?

Second, the equation, using z, in the Wikipedia article is Use to determine gravitational redshft, not the cosmological and can be used to determine the schwartzchild radius of the mass in question, not the cosmological redshift.

Third, the this site gives the equations (for determining z for the doppler and cosmological redshift. Notice the cosmological redshift, near the bottom(which is what your trying to find) does not include G or M. It uses the scale factor. Which is simply a calculation of the difference in curvature of the universe when the light was emitted and when it was recieved. But, again, these are from GR, which you haven't shown to work with a variable G or variable mass.

You're not even using the correct equations.
In fact the two systems are very different, so totally unrelated. In mine I looked for a measured value of gravitational redshift in a 1 G setting, on Earth, so the value I found fits my need showing how it may be a purely gravitational phenomenon. In GR delta z is assumed to be a Doppler related effect in an expanding universe. If the measured redshift on Earth has Swartzchild applications, it is a different matter, inaplicable to what mine shows. Also you forgot to mention that in my original (http://www.badastronomy.com/phpBB/viewtopic.php?p=501100#501100) I said:

"I think this works out for a value of lightshift z = ~v/c = 0.54E-6 = v/3E+8 m/s"... which is a reference to a non-relativistic z.

In your referenced Redshift and the Scale Factor (http://iapetus.phy.umist.ac.uk/Teaching/Cosmology/ScaleFactor.html), it says:

Here lambda(t) is the wavelength at time of emission, i.e. the rest wavelength. If Dt is small:

z >> *v / c = H(t) r / c = H(t) c Dt / c (18 )
This is same how I treated this above light redshift, as if there were a very small Dt, so that the wavelength at the time of emmission shifted by only a small amount in 1 z (1% of lighspeed), in effect a non-relativistic treatment, where z = v/c is shown purely within a gravitational context. Expansion is disregarded.

Think about it: if light redshifts over great cosmic distances is due to gravitational factors, postulating a very great G in intergalactic space, then relativistic expansion is no longer a factor, since there is no longer the need for a Doppler expansion to make this light redshift. So what you are referencing and what I had posited are two very different ideas. If G were found to be very great in deep space, would GR in its current form survive? I suspect it would not, since it is postulated on a universal constant G, at 1 G. If it is not merely 1 G throughout the universe, then current thinking on GR would have to be reconsidered. So you see, two very different approaches to cosmic light redshift, and in mine Relativity plays no part except as an observational science within the limits of maximum electromagnetic energy at v = c.

:)
To quote EtaC's signature block "This isn't right, it isn't even wrong".
I believe it was Richard Feynman who said this? Did he also not say, regarding Quantum Uncertainty Principles: "You're not supposed to understand it"? ;)

Of course, this is only hypothetical, so I really don't know. I could be totally wrong.

nutant gene 71
2005-Jul-16, 08:03 PM
Your weight on Jupiter will be the same as we now estimated, ...
Wrong.
The weight = the gravitational force pulling downwards would be greater, because you increased G.
We're talking about different Gs. In yours, you're referring to Jupiter's greater gravitational mass. In mine, I'm referring to Jupiter's mass same, but it is in a higher G region, hypothetically, five times what we have on Earth.

Why don't we orbit at a distance from Jupiter, so that the gravitational acceleration is the same as on the Earth's surface: it would simplify the calculations.

Let's try again:

Before increasing G
Earth: m = 100 kg, g = 9.8 m/s^2, w = m*g = 980 N.
Jupiter: m = 100 kg, g' = 9.8 m/s^2, w' = m*g' = 980 N.

Set G' = 5*G on Jupiter
Earth: m = 100 kg, g = 9.8 m/s^2, w = m*g = 980 N.
Jupiter: m = 100 kg, g'' = 5*9.8 m/s^2, w'' = m*g'' = 4900 N.

The weight at Jupiter is 5 times the weight on Earth.
I don't understand your comparisons here. Isn't Jupiter's "surface" acceleration 23 m/s^2 ? How did you arrive at 5*9.8 m/s^2? I don't get it, I don't.

Let's interpret the weight at Jupiter as if G' = G.
We perform experiments at Jupiter: using non-gravititational forces we measure the inertial mass of m, and using an Eotvos-type experiemant we test the Equivalence Principle.
We already measured the weight using springs.
We find that the inertial mass is the same as on Earth, and that the Equivalence still works.
Let's assume that we forgot to do an Cavendish-like experiment to measure G'.

What would be our conclusion?
Jupiter's mass is 5 times larger than we thought.
First of all, we never performed this experiment for G' on Jupiter. Second, Jupiter's mass is what it is, not five times larger. The adjustment to a 5 G region is that our mass reading (from our 1 G postulate) is correct; what changes is that the SIZE of the mass is SMALLER in 5 G, i.e., Jupiter's core is only about two or three Earth masses. But this smaller size "acts as if" it were larger, which it would be if Jupiter's G' is only 1 G. The massive gaseous atmosphere is a separate issue.

However, that is incorrect. Your weight on Jupiter is the same as it was before we (hypothetically) discovered it had a different G,...
Wrong, because you increased the gravitational force.
In order to have the same weight, you have to assume, without any justification whatsoever, that the gravitational mass of the probe mass has changed, without affecting the inertial mass.
Again I don't get it, I just don't. If 100 kg on Earth are 980 N, and your same 100 kg (same person) is 2300 N on Jupiter, then weight has changed per Jupiter's greater gravity, but if you went to Jupiter in 2006, and came back again after finding out that Jupiter's G is five times, in 2012, you still weighed the same. It's just that your concept of Jupiterian "kilograms" would be five times smaller than before, same mass, same weight. Only a new "variable mass" idea would kick in, that's all.

Look at it this way: Jupiter is Jupiter, same mass all the time. A visitor is the same visitor, same mass all the time. His/her weight on Jupiter would be higher than on Earth because Jupiter is a more massive planet. But the G factor is a separate issue, which is why I never used "weight" to explain this. A separate issue is Newton's G as a universal constant, at our 1 G, as opposed to a variable G away from Earth. Mass as measured in 1 G is a separate "kilogram" from same mass measured in a higher or lower G. Mass merely "acts" differently in a different G setting. This may be why Mercury is "dragged" along by the Sun's rotational spin, because so close to this hot radiant body its G is low, at 0.4 AU it's about 2.5 times as light in terms of its own "kilograms" as here, so it has less inertial resistance to the momentum transfer from the Sun's rotation; Saturn's G may, at 9.5 AU, may be nearly ten times ours, so it's mass "kilograms" act as if they're heavier than our kilograms, so it holds a massive atmosphere, but it's still the same mass. But here I do not wish to argue for a variable G being fact, because we don't know that it is, something to be studied in the future. What I want to show is how a variable G leads to a variable "kilogram", which per the Equivalence Principle acts differently than what we had assumed in a flat G universe. But you don't get it, do you? You just don't get it. :)

Anybody get it? I'm stumped, running out of ways of explaining this very simple phenomenon. It's a piece of cake, easy as pie. Why am I the only one who sees this? Is everyone else like this? [-(

I can imagine why. If mass "acts" differently in a different G setting, all cosmology is affected. So... resist! It has to be only hypothetical, until we find evidence that G is not universally the same. Until then, it's pie in the sky. :)

Tensor
2005-Jul-16, 08:48 PM
Thanks Tensor, but I believe we are comparing apples and oranges.

Re yours:
First off, how can you use a simplified GR equation (which uses a constant G) in your idea, when you haven't shown

Snip...

It uses the scale factor. Which is simply a calculation of the difference in curvature of the universe when the light was emitted and when it was recieved. But, again, these are from GR, which you haven't shown to work with a variable G or variable mass.

You're not even using the correct equations.

In fact the two systems are very different, so totally unrelated.

Which is the point. You used that equation, and you can't, until you show how you derived it from your math. not from GR.

In mine I looked for a measured value of gravitational redshift in a 1 G setting, on Earth, so the value I found fits my need showing how it may be a purely gravitational phenomenon.

Fits your need? So all you are then doing is playing with numbers until you find something that fits.

In GR delta z is assumed to be a Doppler related effect in an expanding universe.

Ah....no. The doppler is a SR effect, expanding universe based on the scale factor and not doppler.

If the measured redshift on Earth has Swartzchild applications, it is a different matter, inaplicable to what mine shows. Also you forgot to mention that in my original (http://www.badastronomy.com/phpBB/viewtopic.php?p=501100#501100) I said:

"I think this works out for a value of lightshift z = ~v/c = 0.54E-6 = v/3E+8 m/s"... which is a reference to a non-relativistic z.

Well, if you want to talk about forgetting, this (http://badastronomy.com/phpBB/viewtopic.php?p=504212&amp;sid=90419d91d3f28dd71251bb7 6b594039a#504212) posts asks how you account for reaction wheels, thrust differential and binary neutron stars in your model, since the data matches GR.

In your referenced Redshift and the Scale Factor (http://iapetus.phy.umist.ac.uk/Teaching/Cosmology/ScaleFactor.html), it says:
Here lambda(t) is the wavelength at time of emission, i.e. the rest wavelength. If Dt is small:

z >> *v / c = H(t) r / c = H(t) c Dt / c (18 )
This is same how I treated this above light redshift, as if there were a very small Dt, so that the wavelength at the time of emmission shifted by only a small amount in 1 z (1% of lighspeed), in effect a non-relativistic treatment, where z = v/c is shown purely within a gravitational context. Expansion is disregarded.

But the scale factor is not a gravitational effect. Look at the equations, for the scale factor, there is no G and no mass.

Think about it: if light redshifts over great cosmic distances is due to gravitational factors, postulating a very great G in intergalactic space, then relativistic expansion is no longer a factor, since there is no longer the need for a Doppler expansion to make this light redshift.

You are confused here. The cosmological redshift is not a gravitational or doppler effect. You are using gravitational or doppler equations where they can't be used.

So what you are referencing and what I had posited are two very different ideas.

Then why are you using GR eqations? They are not valid using your idea. What you also don't seem to understand it that these are simplified versions of the full GR equations (Looks like a PPN formulization to me). And you have yet to show how

If it is not merely 1 G throughout the universe, then current thinking on GR would have to be reconsidered.

Again, you continue to blow off the binary neutron star example. That seems to be pretty good evidence that our Earth G here is the same there, so why do you keep arguing for a difference in G away from the Earth? The is one example where the observations make it appear it is the same.

To quote EtaC's signature block "This isn't right, it isn't even wrong".
I believe it was Richard Feynman who said this? Did he also not say, regarding Quantum Uncertainty Principles: "You're not supposed to understand it"? ;)

Wolfgan Pauli.

Of course, this is only hypothetical, so I really don't know. I could be totally wrong.

Well, with the lack of evidence you've provided, the interchange of equations from two different systems and misunderstanding how the expansion is calculated mathematically, your last sentence is pretty accurate.

Metricyard
2005-Jul-16, 08:49 PM
Anybody get it? I'm stumped, running out of ways of explaining this very simple phenomenon. It's a piece of cake, easy as pie. Why am I the only one who sees this? Is everyone else like this? Not talking

Because your theory cannot exist, at least in the form your trying to explain.

4. Physics. An observable event.

The reason you can't explain this "very simple phenomenon" is because it has not been observed. Anywhere. After all this time, you're still trying to create a formula to match your theory, which has never been observed. What's the point?

nutant gene 71
2005-Jul-17, 05:11 PM
And once you get past that, you have to explain why the tests of GR outside our solar system (such as the amount of observed periastron change around PSR 1913 + 16 and theoretical calculations concerning that binary pairmatch so closely (see figure 18.1) with a non variable G assumption and explain how well your idea matches the observations just as well.
In your PSR 1913+16 reference (http://nobelprize.org/physics/laureates/1993/press.html), it says:
The discovery of the binary pulsar Hulse's and Taylor's discovery in 1974 of the first binary pulsar, called PSR 1913 + 16 (PSR stands for pulsar, and 1913 + 16 specifies the pulsar's position in the sky) thus brought about a revolution in the field. We have here two very small astronomical bodies, each with a radius of some ten kilometres but with a mass comparable with that of the sun, and at a short distance from each other, only several times the moon's distance from the earth. Here the deviations from Newton's gravitational physics are large.
Note how small the diameters but how incredibly great the gravitational masses. This is exactly what I'm talking about. In a hypothetical variable mass cum variable G, this is how it's supposed to look. I made a similar case for Jupiter, where its inner core planet is only two or three Earth masses, but this small mass can hold an incredibly large atmosphere, the largest in our solar system. The binary small mass in a very high G setting would have the same effect, their mass size and volume is tiny for the tremendous gravity they exhibit. Indirectly, this is an example of the variable mass idea, but our interpretation of this observable phenomenon are different.

What I had done in comparing the two sets of systems, part 1 &amp; 2, of my above, was to show how two different approaches mathematically yield a similar result. I did this to show how using a Quantum equation in the second part (http://www.badastronomy.com/phpBB/viewtopic.php?p=504364#504364), where the Planck's-deBroglie equation is matched against Einstein's famous equation (E=mc^2 is not SR/GR, and in fact predates Einstein), so using the physical phenomenon cut-off wavelength of the photoelectric effect (something done in a lab, and which earned Einstein a Nobel) that Quantum equation yields a very low Energy result. When I then reversed the wavelength back to normal, per this low Energy result, I got the proton-hydrogen mass, which was much higher. This is representative of a "variable mass" for the atom. In the first part (http://www.badastronomy.com/phpBB/viewtopic.php?p=504217&amp;sid=38402e500523916325de0d2 75a7920d3#504217) I used a gravitational-redshift "sample", the only one I could find, and then applied that against the computed mass over the distance of 1z redshift in space, for all the hydrogen atoms there (~92% of space is believed to be hydrogen) to arrive at how much (Earth kilograms) mass there is out there in intergalactic space for that distance. When I applied this mass (linear mass the light traveled through) to the sample grav-redshift, I came up with a number, which in proportion to our Newton's G turns out to be orders of magnitude greater than here. That's all I did, was to take the sample grav-redshift times mass to equal Earth's G times deep space G, and the value I got was similar to the Quantum equation solution. The fact they yielded similar results made me pause, and wonder why are they so close. (The other 8% of non-hydrogen gas and dust may fill the gap between the two results, since they would have greater mass, thus raise the value in the first example.)

In neither case did I use Relativistic equations, nor did I need the Doppler effect expansion of space. The gravitational-redshift does all the work, at a much higher G, to give us the impression that space is expanding. But it may not be. What's wrong with this picture is that we have found no other evidence to support a variable G idea. That's where we're at today, and as you correctly pointed out (and others too) is that all our spacecraft behave normally as if 1 G is it, except for minor inflight adjustments which are not outside expectations. This is quite a conundrum, since how can it be? Surely G must be a universal constant. (I actually wake up in the middle of the night thinking about it, that it cannot be anything else, and that 300 years of science must be correct.) So there is this nagging thought that maybe, just maybe, we've overlooked something, because the variable G seems to fit astronomical observations, as I just pointed out for the exceptionally high gravity binaries. How can a variable G hide within a constant G universe? My hypothetical answer is: the masses are different per local G, which still mimics the results of a constant universal G. And that is not an easy thing to spot, unless you actually measure for it in situ. Remember that all our cosmology is predicated on how we measured mass based on a universal constant G. So if this G were different, the mass we figured with 1 G would still be the same mass, but in its own G locale, adjusted for its own "kilograms" it would act differently only in how matter binds gravitationally, but all else would remain equal. Thus, orbital velocities, gravity assist trajectories, attitude control reaction wheels, rocket propulsion, all look normal in relation to G adjusted mass, because figuring it all out with either a constant G or variable G gives us similar results. Also, the difference in G in our solar system, hypothetically, is still very small, and does not get very large until we exit the galaxy. If G is inversely proportional to em energy, as the Axiomatic equation predicts, then that's how it would work... hypothetically, of course.

Here is a potential test for how I see this (hypothetical) variable G: Test the binaries for chemical content. If they show only neutron atoms (of what element?), then they truly are neutron stars. But if they show a typical range of elements of other stars, then perhaps they are nothing more than "cold" stars, with very high G, but they behave as if (in 1 G interpretation) they are neutron-only composition. Only difference is that the normal composition "acts as if" it were very compact neutrons, so that the gravity registered there is exceptionally great, as exhibited by fast spin, very high rotation velocities, etc. I am not aware this kind of study had been done, however, but it could be revealing to understanding what we're looking at.

So that's why I keep thinking to myself, "my idea has to be wrong", because how could a variable G have escaped our notice for the past three hundred years? I see myself as a mere "court jester" compared to all the great minds of science, so I don't take my idea too seriously. But that's how the numbers work out, which then explains a lot of unexplained phenomena in astronomy. The only thing this idea does not explain directly are black-holes, because I think they're a special case. The Axiomatic Equation points to their being where all em lambda cancels on a point (Swartzchild radius?) so that the greatest possible space-gravity G is reaffirmed. All the other stuff fits in very lovely with a very high space G, or slightly higher G within our solar system. This is what we should look for.

nutant gene 71
2005-Jul-17, 05:27 PM
Anybody get it? I'm stumped, running out of ways of explaining this very simple phenomenon. It's a piece of cake, easy as pie. Why am I the only one who sees this? Is everyone else like this? Not talking

Because your theory cannot exist, at least in the form your trying to explain.

4. Physics. An observable event.

The reason you can't explain this "very simple phenomenon" is because it has not been observed. Anywhere. After all this time, you're still trying to create a formula to match your theory, which has never been observed. What's the point?

The definition of phenomenon : an observable event
Did you ever look at something and not see it? Or if you saw it, like in a magic trick, it was not what you "observed"? It's not the observable phenomenon that is in question regarding gravity physics, but our interpretation of it.

"What's point?" the point is we should not stop looking, because we may not yet have it right. In other words, the universe tricked us into thinking G is a universal constant, like a magician, where there may be cause to "uncloak" the truth behind what gravity is all about.

Tassel
2005-Jul-17, 05:27 PM
W" = 5(20 "kg" * 23 m/^2) = 2300 N for Jupiter, if you factor in the 5 G.
This equation is not correct. W does not equal G*m*g. You're making up equations in an attempt to get the answer you're looking for.

The way you see it, the 5 G doesn't exist anywhere, and if I did it your way, if I understand what you're saying, factoring in the 5 G, it would look like this:

W'" = 5(100 kg * 23.m^2) = 11,500 N , which is wrong.
You're still terminally confused.

The "5G" exists in the calculation of "g". W does not equal G*m*g. There's no justification for multiplying by "5" as the "new G" is already included (and you can be doubly sure since the units wouldn't balance). See:

W = mg

g = GM/r^2, therefore:

W = m * (GM/R^2)

m is the mass of the object you are weighing.
G is whatever G your fevered imagination thinks it should be at Jupiter
M is the mass of Jupiter, calculated based on whatever G you used
R is the distance from the center of Jupiter

As you can see, "W" already includes "new G", whatever value that may be.

As predicted, you won't just do the calculations step-by-step. Instead you make incorrect assumptions. You're never going to learn anything until you stop trying to get the answer your looking for.

nutant gene 71
2005-Jul-17, 05:45 PM
W = mg

g = GM/r^2, therefore:

W = m * (GM/R^2)

m is the mass of the object you are weighing.
G is whatever G your fevered imagination thinks it should be at Jupiter
M is the mass of Jupiter, calculated based on whatever G you used
R is the distance from the center of Jupiter
I believe the correct equation is: g = GMm/ r^2, where in your above example M is the Sun's mass, while m is Jupiter's mass.

You're never going to learn anything until you stop trying to get the answer your looking for.
Looking for answers is exactly what I plan to do. Whether or not you learn anything is a separate issue. :)

Tassel
2005-Jul-17, 06:06 PM
I believe the correct equation is: g = GMm/ r^2, where in your above example M is the Sun's mass, while m is Jupiter's mass.
Well, what you believe is wrong.

F = GMm/ r^2

g = GM/R^2, which is what I said.

Now that you know the correct equation to use, please go back to my last post and address it.

Looking for answers is exactly what I plan to do.
Then why don't you work through the calculations step-by-step?

nutant gene 71
2005-Jul-17, 07:34 PM
I believe the correct equation is: g = GMm/ r^2, where in your above example M is the Sun's mass, while m is Jupiter's mass.
Well, what you believe is wrong.

F = GMm/ r^2

g = GM/R^2, which is what I said.

Now that you know the correct equation to use, please go back to my last post and address it.

Looking for answers is exactly what I plan to do.
Then why don't you work through the calculations step-by-step?
If you go back to this post (http://www.badastronomy.com/phpBB/viewtopic.php?p=501297&amp;sid=eddb5168c33a43225e6ec67 7cad4a915#501297) you'll see how I was corrected of a similar error, where I had M for Jupiter instead of the Sun. For a better explanation, go to Wiki's Gravity (http://en.wikipedia.org/wiki/Gravity) explanation.

About yours above that the units would not balance in W = 5G(100kg * 23 m/s^2) = 11,500 N. Other than the fact that the outcome is simply wrong, the units work out like this:

W = 5G'/G (100 kg * 23 m/s^2) = 11,500 kg m/s^2, which is in Newtons

The G'/G cancel out dimensionally, so you are left with a 5 dimensionless multiplier.

About me showing step by step calculations, you are ignorning what I did in my two posts above, part 1 and 2. Which one don't you like?

Tassel
2005-Jul-17, 07:50 PM
If you go back to this post (http://www.badastronomy.com/phpBB/viewtopic.php?p=501297&amp;sid=eddb5168c33a43225e6ec67 7cad4a915#501297) you'll see how I was corrected of a similar error, where I had M for Jupiter instead of the Sun. For a better explanation, go to Wiki's Gravity (http://en.wikipedia.org/wiki/Gravity) explanation.

Ok, so you keep making the same error...?

g = GM/R^2 is the correct formula for determining the acceleration due to gravity at Jupiter.

About yours above that the units would not balance in W = 5G(100kg * 23 m/s^2) = 11,500 N. Other than the fact that the outcome is simply wrong, the units work out like this:

W = 5G'/G (100 kg * 23 m/s^2) = 11,500 kg m/s^2, which is in Newtons

The G'/G cancel out dimensionally, so you are left with a 5 dimensionless multiplier.
So in response to being shown you were using an incorrect formula to calculate weight, you just invent a new term to add onto the correct formula, for no apparent reason. Now the units balance which is good, but there is still no justification for adding that term. Why does "Earth G" matter at Jupiter in a variable G universe?

About me showing step by step calculations, you are ignorning what I did in my two posts above, part 1 and 2. Which one don't you like?
I've asked you to show the calculations step-by-step starting with calculating Jupiter's mass. You won't do this because if you do, you will demonstrate that there is no need to change the definition (to a definition you've never provided) of "kilogram" for a variable G universe.

nutant gene 71
2005-Jul-18, 04:04 AM
g = GM/R^2 is the correct formula for determining the acceleration due to gravity at Jupiter.

About yours above that the units would not balance in W = 5G(100kg * 23 m/s^2) = 11,500 N. Other than the fact that the outcome is simply wrong, the units work out like this:

W = 5G'/G (100 kg * 23 m/s^2) = 11,500 kg m/s^2, which is in Newtons

The G'/G cancel out dimensionally, so you are left with a 5 dimensionless multiplier.
So in response to being shown you were using an incorrect formula to calculate weight, you just invent a new term to add onto the correct formula, for no apparent reason. Now the units balance which is good, but there is still no justification for adding that term. Why does "Earth G" matter at Jupiter in a variable G universe?

About me showing step by step calculations, you are ignorning what I did in my two posts above, part 1 and 2. Which one don't you like?
I've asked you to show the calculations step-by-step starting with calculating Jupiter's mass. You won't do this because if you do, you will demonstrate that there is no need to change the definition (to a definition you've never provided) of "kilogram" for a variable G universe.
Since your question is not clear, I'll answer it to what I think you're asking. I am assuming you want me to give an example using both Earth 1G kilograms, and Jupiter's (hypothetical 5G) "kilograms".

g = GM/R^2, since I don't know what you have in mind for R, I'll figure it with R^2 unresolved, something you can fill in yourself, if that fits your question.

Earth's G = 6.67E-11
Jupiter's (Earth kg) mass = 1,898.6E+24 kg

1) Earth kilograms (at 1G):

g = (6.67E-11 m^3 kg^-1 s^-2)*(1,898.6E+24 kg)/ R^2

g R^2 = 12,663.7E+13 m^3 s^-2, which is 126.6E+15 m^3 s^-2

2) For Jupiter's (hypothetical 5G) "kilograms" (1/5th of Earth kg):

g' = (33.35E-11 m^3 kg^-1 s^-2)*(379E+24 kg)/ R^2

g' R^2 = 12663.7E+13 m^3 s^-2, which is same, 126.6E+15 m^3 s^-2.

I don't know what you're looking for, so this is the best I can do for you. Otherwise, I'd rather move on from this silly exercise and move on to more important stuff, like whether gravitational-redshift can be a purely gravitational phenomenon, or why hydrogen clouds in deep space do not have enough gravity to combust into fusion, thus giving birth to a star, but with G orders of magnitude 5 or 6 times that of our G, perhaps it could. Or what is so called "dark matter". That's much more interesting than flogging the same dead horse. I hope you understand, but if I haven't answered your question, it is because I have no idea what you're getting at. Hopefully what I showed here will help you resolve your problem, since I'm still left guessing what is your question. Good luck.

papageno
2005-Jul-18, 09:22 AM
Your weight on Jupiter will be the same as we now estimated, ...
Wrong.
The weight = the gravitational force pulling downwards would be greater, because you increased G.
We're talking about different Gs. In yours, you're referring to Jupiter's greater gravitational mass. In mine, I'm referring to Jupiter's mass same, but it is in a higher G region, hypothetically, five times what we have on Earth.
By increasing G, you increased the gravitational force exerted by Jupiter.
I have not done anything to Jupiter's mass. I explained what happens if you increased G: Your weight would be greater than before.

Why don't we orbit at a distance from Jupiter, so that the gravitational acceleration is the same as on the Earth's surface: it would simplify the calculations.

Let's try again:

Before increasing G
Earth: m = 100 kg, g = 9.8 m/s^2, w = m*g = 980 N.
Jupiter: m = 100 kg, g' = 9.8 m/s^2, w' = m*g' = 980 N.

Set G' = 5*G on Jupiter
Earth: m = 100 kg, g = 9.8 m/s^2, w = m*g = 980 N.
Jupiter: m = 100 kg, g'' = 5*9.8 m/s^2, w'' = m*g'' = 4900 N.

The weight at Jupiter is 5 times the weight on Earth.
I don't understand your comparisons here. Isn't Jupiter's "surface" acceleration 23 m/s^2 ? How did you arrive at 5*9.8 m/s^2? I don't get it, I don't.

Why don't we orbit at a distance from Jupiter, so that the gravitational acceleration is the same as on the Earth's surface: it would simplify the calculations.

Let's interpret the weight at Jupiter as if G' = G.
We perform experiments at Jupiter: using non-gravititational forces we measure the inertial mass of m, and using an Eotvos-type experiemant we test the Equivalence Principle.
We already measured the weight using springs.
We find that the inertial mass is the same as on Earth, and that the Equivalence still works.
Let's assume that we forgot to do an Cavendish-like experiment to measure G'.

What would be our conclusion?
Jupiter's mass is 5 times larger than we thought.
First of all, we never performed this experiment for G' on Jupiter.
Weird, I thought we were talking hypothetically: we can well do thought experiments, can't we?

Second, Jupiter's mass is what it is, not five times larger.

Let's interpret the weight at Jupiter as if G' = G.

The adjustment to a 5 G region is that our mass reading (from our 1 G postulate) is correct; what changes is that the SIZE of the mass is SMALLER in 5 G, i.e., Jupiter's core is only about two or three Earth masses.
That would happen to all other masses, and it would be measureble.
A change like that would affect everything, including chemical bonds and bandstructures of semiconductors.
Electronincs and CCDs on space probes would not work, and the spectra would not be recognizable.

But this smaller size "acts as if" it were larger, which it would be if Jupiter's G' is only 1 G. The massive gaseous atmosphere is a separate issue.
It is not, because gas has mass.
Then, why would a change in size have any effect on the gravitational force? It depends only on the masses involved and their distances.
The size affects tidal effects, and how objects behave on rotations.
Since the moment of inertia of our probes behave as expected, there is no reason to assume a change in size.

However, that is incorrect. Your weight on Jupiter is the same as it was before we (hypothetically) discovered it had a different G,...
Wrong, because you increased the gravitational force.
In order to have the same weight, you have to assume, without any justification whatsoever, that the gravitational mass of the probe mass has changed, without affecting the inertial mass.
Again I don't get it, I just don't. If 100 kg on Earth are 980 N, and your same 100 kg (same person) is 2300 N on Jupiter, then weight has changed per Jupiter's greater gravity, but if you went to Jupiter in 2006, and came back again after finding out that Jupiter's G is five times, in 2012, you still weighed the same. It's just that your concept of Jupiterian "kilograms" would be five times smaller than before, same mass, same weight. Only a new "variable mass" idea would kick in, that's all.
kilogram is for inertial mass.
|Since inertial mass is the same here and on Jupiter, there is no change in kilograms.
A variable mass would "kick in" only if the Equivalence Principle is violated, but it is not.

Look at it this way: Jupiter is Jupiter, same mass all the time. A visitor is the same visitor, same mass all the time. His/her weight on Jupiter would be higher than on Earth because Jupiter is a more massive planet. But the G factor is a separate issue, which is why I never used "weight" to explain this.
If you change G, you change the weight without changing the (gravitational) mass.

A separate issue is Newton's G as a universal constant, at our 1 G, as opposed to a variable G away from Earth. Mass as measured in 1 G is a separate "kilogram" from same mass measured in a higher or lower G.
It is only if you assume that the value of G affects the mass.

Mass merely "acts" differently in a different G setting.
Different G, different force, different acceleration.
No need to change the mass.

This may be why Mercury is "dragged" along by the Sun's rotational spin, because so close to this hot radiant body its G is low, at 0.4 AU it's about 2.5 times as light in terms of its own "kilograms" as here, so it has less inertial resistance to the momentum transfer from the Sun's rotation;...
General Relativity accounts quantitatively for this effect.
Your speculative handwaving requires a violation of the Equivalence Principle, one of GR's postulate.
Conclusion: your hypothesis fails the experimetnal test.

Saturn's G may, at 9.5 AU, may be nearly ten times ours, so it's mass "kilograms" act as if they're heavier than our kilograms, so it holds a massive atmosphere, but it's still the same mass.
Yet Huygens landed as predicted and Cassini is working fine.
Again, your hypothesis fails the experimental test.

But here I do not wish to argue for a variable G being fact, because we don't know that it is, something to be studied in the future.
We have very tight experimental constraints on a varibale G.
Your hypothesis assume a variable G beyond thos constraints, therefore it fails the experimental test.

What I want to show is how a variable G leads to a variable "kilogram", which per the Equivalence Principle acts differently than what we had assumed in a flat G universe. But you don't get it, do you? You just don't get it. :)

You are the one confusing gravitational mass, inertial mass, and weight.
You apparently that your handwaving could "work" if you assume that the Equivalence Principle is violated.

Anybody get it? I'm stumped, running out of ways of explaining this very simple phenomenon. It's a piece of cake, easy as pie. Why am I the only one who sees this? Is everyone else like this? [-(
Did it occur to you, that you are the one not understanding?

I can imagine why. If mass "acts" differently in a different G setting, all cosmology is affected. So... resist! It has to be only hypothetical, until we find evidence that G is not universally the same. Until then, it's pie in the sky. :)
I gave you plenty of references about reserachers working on variable G scenarios.
Yet again, your conlclusions fail the experimetnal test.

nutant gene 71
2005-Jul-19, 03:01 AM
Your weight on Jupiter will be the same as we now estimated, ...
Wrong.
The weight = the gravitational force pulling downwards would be greater, because you increased G.
We're talking about different Gs. In yours, you're referring to Jupiter's greater gravitational mass. In mine, I'm referring to Jupiter's mass same, but it is in a higher G region, hypothetically, five times what we have on Earth.
By increasing G, you increased the gravitational force exerted by Jupiter.
I have not done anything to Jupiter's mass. I explained what happens if you increased G: Your weight would be greater than before.
Yes, increased G increases gravitational force exerted by Jupiter, but for a smaller "equivalent" mass, i.e., lower "kilograms". Weight stays the same as before.

Why don't we orbit at a distance from Jupiter, so that the gravitational acceleration is the same as on the Earth's surface: it would simplify the calculations.

Let's try again:

Before increasing G
Earth: m = 100 kg, g = 9.8 m/s^2, w = m*g = 980 N.
Jupiter: m = 100 kg, g' = 9.8 m/s^2, w' = m*g' = 980 N.

Set G' = 5*G on Jupiter
Earth: m = 100 kg, g = 9.8 m/s^2, w = m*g = 980 N.
Jupiter: m = 100 kg, g'' = 5*9.8 m/s^2, w'' = m*g'' = 4900 N.

The weight at Jupiter is 5 times the weight on Earth.
I don't understand your comparisons here. Isn't Jupiter's "surface" acceleration 23 m/s^2 ? How did you arrive at 5*9.8 m/s^2? I don't get it, I don't.

Why don't we orbit at a distance from Jupiter, so that the gravitational acceleration is the same as on the Earth's surface: it would simplify the calculations.
Okay, I understand what you're saying. But I don't know how to do this orbital equation. Example?

Let's interpret the weight at Jupiter as if G' = G.
We perform experiments at Jupiter: using non-gravititational forces we measure the inertial mass of m, and using an Eotvos-type experiemant we test the Equivalence Principle.
We already measured the weight using springs.
We find that the inertial mass is the same as on Earth, and that the Equivalence still works.
Let's assume that we forgot to do an Cavendish-like experiment to measure G'.

What would be our conclusion?
Jupiter's mass is 5 times larger than we thought.
First of all, we never performed this experiment for G' on Jupiter.
Weird, I thought we were talking hypothetically: we can well do thought experiments, can't we?
Yes, hypothetically a 'thought experiment'. Actual measurement would be better.

Second, Jupiter's mass is what it is, not five times larger.

Let's interpret the weight at Jupiter as if G' = G.

The adjustment to a 5 G region is that our mass reading (from our 1 G postulate) is correct; what changes is that the SIZE of the mass is SMALLER in 5 G, i.e., Jupiter's core is only about two or three Earth masses.
That would happen to all other masses, and it would be measureble.
No. Orbital gravitational effect on all ambient masses, i.e., moons and spaceprobes, remains same.

A change like that would affect everything, including chemical bonds and bandstructures of semiconductors.
Electronincs and CCDs on space probes would not work, and the spectra would not be recognizable.
Don't know this, an assumption only that it would change. We had not measured this effect on chemical-electrical forces in non-Earth locations. Would need to specifically test for it in outer solar system, for example.

But this smaller size "acts as if" it were larger, which it would be if Jupiter's G' is only 1 G. The massive gaseous atmosphere is a separate issue.
It is not, because gas has mass.
Why wouldn't this gas keep accumulating indefinitely? What is the cut-off for how gas interacts with itself? Hypothetically, if your assessment that gas alone is sufficient to cause atmospheres to accumulate, Jupiter could grow to infinity in time. On what do you base this assumption, that the atmosphere can exist without a central solid planet core? Do we have evidence to this effect?

Then, why would a change in size have any effect on the gravitational force? It depends only on the masses involved and their distances.
Mostly yes. However, if G is greater, then the "equivalent" mass to what we had measured in 1G is smaller, lower volume, same effect.

The size affects tidal effects, and how objects behave on rotations.
Since the moment of inertia of our probes behave as expected, there is no reason to assume a change in size.
No change in size to probes, but our "assumed" mass size for outer planets may be wrong. See this post on Uranus &amp; Neptune (http://www.badastronomy.com/phbBB/viewtopic.php?t=22972) for similar discussion. Planet rotations are a separate issue. I.e., Why does Sun have rotation, for example?

However, that is incorrect. Your weight on Jupiter is the same as it was before we (hypothetically) discovered it had a different G,...
Wrong, because you increased the gravitational force.
In order to have the same weight, you have to assume, without any justification whatsoever, that the gravitational mass of the probe mass has changed, without affecting the inertial mass.
Again I don't get it, I just don't. If 100 kg on Earth are 980 N, and your same 100 kg (same person) is 2300 N on Jupiter, then weight has changed per Jupiter's greater gravity, but if you went to Jupiter in 2006, and came back again after finding out that Jupiter's G is five times, in 2012, you still weighed the same. It's just that your concept of Jupiterian "kilograms" would be five times smaller than before, same mass, same weight. Only a new "variable mass" idea would kick in, that's all.
kilogram is for inertial mass.
|Since inertial mass is the same here and on Jupiter, there is no change in kilograms.
A variable mass would "kick in" only if the Equivalence Principle is violated, but it is not.
"kilogram is for inertial mass", for same mass, but at 1/5th Jupiter "kilograms" in (hypo) 5G: each new "kg" is gravitationally 5 times as powerfull as for our Earth's 1G kg, for same mass. Ditto for every molecule on Jupiter, it is 5 times as gravitationally "active" as on Earth, so total size of Jupiter's core is five times smaller than it would be in Earth's 1G estimation of mass. Yet, same mass!

Look at it this way: Jupiter is Jupiter, same mass all the time. A visitor is the same visitor, same mass all the time. His/her weight on Jupiter would be higher than on Earth because Jupiter is a more massive planet. But the G factor is a separate issue, which is why I never used "weight" to explain this.
If you change G, you change the weight without changing the (gravitational) mass.
Other way around: When you change G, you change the gravitational mass without changing the weight, for same mass, only size and volume change. Each atom and molecule acts with more force, but there are fewer of them. I think this is the part you haven't seen yet in your mind's eye. It is elemental to how variable G works: (hypothetically) Mercury has 2 1/2 times more molecules per same mass; (hypothetically) Jupiter has 5 times fewer molecules for same mass per this system. And that affects kilograms.

Viz., if G' = 5G, and a = g = 23 ms^-2 for Jup, then either F = 5(20 kg * 23 ms^-2) or F = 5(100 kg * 23 ms^-2)... which would you choose?

A separate issue is Newton's G as a universal constant, at our 1 G, as opposed to a variable G away from Earth. Mass as measured in 1 G is a separate "kilogram" from same mass measured in a higher or lower G.
It is only if you assume that the value of G affects the mass.
Yes.

Mass merely "acts" differently in a different G setting.
Different G, different force, different acceleration.
No need to change the mass.
Yes, if different force, then difference acceleration for same mass: this is the Equivalence Principle.
No, if "kilograms" are adjusted for G, same acceleration for same mass: EP preserved. Each atom and molecule "acts as if" it had more "force" intrinsically, so it takes more force to move a smaller volume of mass equivalent to a larger volume of mass on Earth.

This may be why Mercury is "dragged" along by the Sun's rotational spin, because so close to this hot radiant body its G is low, at 0.4 AU it's about 2.5 times as light in terms of its own "kilograms" as here, so it has less inertial resistance to the momentum transfer from the Sun's rotation;...
General Relativity accounts quantitatively for this effect.
Correct. Einstein's GR found a way to explain this. I think it might be a "square peg in a round hole" however, and a better explanation may exist.

Your speculative handwaving requires a violation of the Equivalence Principle, one of GR's postulate.
Conclusion: your hypothesis fails the experimetnal test.
No, not "handwaving" but reasoning, something that seems to be lost on you. Yes, it contradicts GR in its current form, since a much greater G in 99.99% of space means it is isotropic and homogeneous at a very different level from now assumed. Hypothesis has not failed any tests because they had not yet been performed. ESA is closest to test that idea, and still years away. (Still, I must admire how you're still standing on your feet slugging, not taking it lying down. :) )

Saturn's G may, at 9.5 AU, may be nearly ten times ours, so it's mass "kilograms" act as if they're heavier than our kilograms, so it holds a massive atmosphere, but it's still the same mass.
Yet Huygens landed as predicted and Cassini is working fine.
Again, your hypothesis fails the experimental test.
Effects of variable G had been well hidden from us, to date, except for "a few extra molecules" in Titan's upper atmosphere. Future tests may show different results, however. I am saying only this: it is hypothetically possible that 1G is not universal, but a large variable.

But here I do not wish to argue for a variable G being fact, because we don't know that it is, something to be studied in the future.
We have very tight experimental constraints on a varibale G.
Your hypothesis assume a variable G beyond thos constraints, therefore it fails the experimental test.
Yes, very tight constraints within 1 AU, which is 1G. We haven't checked for these same constraints at Jupiter's 5.2 AU, Saturn's 9.5 AU, Neptune's 30 AU, etc., not even at ~0.4 AU where is Mercury's orbit. Do we really know?

What I want to show is how a variable G leads to a variable "kilogram", which per the Equivalence Principle acts differently than what we had assumed in a flat G universe. But you don't get it, do you? You just don't get it. :)

You are the one confusing gravitational mass, inertial mass, and weight.
You apparently that your handwaving could "work" if you assume that the Equivalence Principle is violated.
No. Very clear in my reasoning, not "handwaving" as you claim, nor confusing gravitational mass with weight. Inertial mass and gravitational mass are same, as I've explained over and over again: if each "kilogram" is gravitationally more potent for the molecules it represents, smaller in size, but it still takes the same amount of force to move an equivalent inertial mass. When we test for this, we should find that it takes the same force to move the same "kilogram" in space in the outersolar system than on Earth, but we will find that one kilogram is smaller in size for same matter than we expected.
I should add here for clarification: we are talking about propulsion, kinetic push energy. Gravitational assist energy is something entirely different, since the greater inertial "kilograms" mass in outersolar system is still pulled along by gravity in the ways we now expect it to, as mass attracts mass. To illustrate how different these two forces on mass are: Why do we not "feel" the barycenter wobble of the Earth? Do you ever find yourself being pulled one way or another as the planet wobbles in its barrycenter in relation to the Moon? (This may be one reason why tides bulge on both sides, BTW, so the Earth's acceleration "cancels" out.) But if you were pushed by a giant rocket ship moving you out of your orbital position, you would notice your coffee slosh. You see, they're different.

Anybody get it? I'm stumped, running out of ways of explaining this very simple phenomenon. It's a piece of cake, easy as pie. Why am I the only one who sees this? Is everyone else like this? [-(
Did it occur to you, that you are the one not understanding?
In a word? No. I am reasoning my way through a hypothetical situation. My conclusion is that we may not yet have gravity right, GR notwithstanding.

I can imagine why. If mass "acts" differently in a different G setting, all cosmology is affected. So... resist! It has to be only hypothetical, until we find evidence that G is not universally the same. Until then, it's pie in the sky. :)
I gave you plenty of references about reserachers working on variable G scenarios.
Yet again, your conlclusions fail the experimetnal test.

Let's look at it this way, and let me take inventory here:

1) Your understanding is thus: Greater G' increases gravitational force, therefore it increases acceleration, but violates Equivalence Principle. Kilograms remain the same. Greater G' is impossible.

2) My understanding is thus: Greater G' increases gravitational mass, therefore it leaves acceleration unchanged for same mass, but per Equivalence Principle the mass's volume size is smaller. "Kilograms" are adjusted in inverse proportion to G'. Greater G' is a hidden phenomenon.

Does this pretty much sum up our conceptual differences? You say that G' can vary only slightly within very tight parameters. I say G' can vary to orders of magnitude. My hypothetical vision of the universe makes more sense now, why galaxy rotation "acts as if" there was Dark Matter, or why stars can form from hydrogen nebulea, or why 'neutron stars' can have immense gravity and spin: their G is massively different.

Net net: Astronomical observation interpreted within the parameters of a constant universal G leaves too many unanswered questions. The only way to resolve this astronomy conundrum is to go out there into space far from Earth's 1 G and test for it.

Can you agree on any of my above points? Of course, it's okay if not, and you're still standing and slugging, which is cool. 8)

papageno
2005-Jul-19, 09:32 AM
Your weight on Jupiter will be the same as we now estimated, ...
Wrong.
The weight = the gravitational force pulling downwards would be greater, because you increased G.
We're talking about different Gs. In yours, you're referring to Jupiter's greater gravitational mass. In mine, I'm referring to Jupiter's mass same, but it is in a higher G region, hypothetically, five times what we have on Earth.
By increasing G, you increased the gravitational force exerted by Jupiter.
I have not done anything to Jupiter's mass. I explained what happens if you increased G: Your weight would be greater than before.
Yes, increased G increases gravitational force exerted by Jupiter, but for a smaller "equivalent" mass, i.e., lower "kilograms". Weight stays the same as before.
So, you are changing to parameters at the same time.
You do not simply increase G, but you also chnage the mass of Jupiter, and you assume that the weight is the same.
If you change the mass of Juptier, you have to admit that all the other masses must be changed, because gravity is symmetric with respect to the masses involved.

At this point, you are left with either an observable change of the inertial mass of our probe, or an observable violation of the Equivalence Principle.

Why don't we orbit at a distance from Jupiter, so that the gravitational acceleration is the same as on the Earth's surface: it would simplify the calculations.

Let's try again:

Before increasing G
Earth: m = 100 kg, g = 9.8 m/s^2, w = m*g = 980 N.
Jupiter: m = 100 kg, g' = 9.8 m/s^2, w' = m*g' = 980 N.

Set G' = 5*G on Jupiter
Earth: m = 100 kg, g = 9.8 m/s^2, w = m*g = 980 N.
Jupiter: m = 100 kg, g'' = 5*9.8 m/s^2, w'' = m*g'' = 4900 N.

The weight at Jupiter is 5 times the weight on Earth.
I don't understand your comparisons here. Isn't Jupiter's "surface" acceleration 23 m/s^2 ? How did you arrive at 5*9.8 m/s^2? I don't get it, I don't.

Why don't we orbit at a distance from Jupiter, so that the gravitational acceleration is the same as on the Earth's surface: it would simplify the calculations.
Okay, I understand what you're saying. But I don't know how to do this orbital equation. Example?
The same as for Earth: you just need to use Jupiter's mass and find the right distance.
By the way, you do not need to find the distance.
My point was to simplify the reasoning.

Let's interpret the weight at Jupiter as if G' = G.
We perform experiments at Jupiter: using non-gravititational forces we measure the inertial mass of m, and using an Eotvos-type experiemant we test the Equivalence Principle.
We already measured the weight using springs.
We find that the inertial mass is the same as on Earth, and that the Equivalence still works.
Let's assume that we forgot to do an Cavendish-like experiment to measure G'.

What would be our conclusion?
Jupiter's mass is 5 times larger than we thought.
First of all, we never performed this experiment for G' on Jupiter.
Weird, I thought we were talking hypothetically: we can well do thought experiments, can't we?
Yes, hypothetically a 'thought experiment'. Actual measurement would be better.
Show us how you "actually" change G.

Second, Jupiter's mass is what it is, not five times larger.

Let's interpret the weight at Jupiter as if G' = G.

The adjustment to a 5 G region is that our mass reading (from our 1 G postulate) is correct; what changes is that the SIZE of the mass is SMALLER in 5 G, i.e., Jupiter's core is only about two or three Earth masses.
That would happen to all other masses, and it would be measureble.
No. Orbital gravitational effect on all ambient masses, i.e., moons and spaceprobes, remains same.
Why? Because you say so?
Newton's law is perfectly symmetric with repect to the masses.
You cannot assume that only Jupiter's mass is affected, while all the other masses in the same region remain the same.

A change like that would affect everything, including chemical bonds and bandstructures of semiconductors.
Electronincs and CCDs on space probes would not work, and the spectra would not be recognizable.
Don't know this, an assumption only that it would change. We had not measured this effect on chemical-electrical forces in non-Earth locations. Would need to specifically test for it in outer solar system, for example.
Bandstructure of semiconductors and electronic structure of atoms and moclecules depend on the masses of electrons and nuclei.
By changing the masses, you change the electronic structure, hence all the semiconductor electronics and the spectra are affected.
If the electronic structure changed outside the Earth, astronomers would not be able to recognize spectra from stars and planets, by comparison to spectra taken on Earth.
The electronics on our probes would not work as designed, CCDs would not give pictures.

But this smaller size "acts as if" it were larger, which it would be if Jupiter's G' is only 1 G. The massive gaseous atmosphere is a separate issue.
It is not, because gas has mass.
Why wouldn't this gas keep accumulating indefinitely? What is the cut-off for how gas interacts with itself? Hypothetically, if your assessment that gas alone is sufficient to cause atmospheres to accumulate, Jupiter could grow to infinity in time. On what do you base this assumption, that the atmosphere can exist without a central solid planet core? Do we have evidence to this effect?
Gas have a pressure, which counteracts the effect of gravity.
On the Sun, the pressure was not enough, and the gas started fusion.

On what do you base your assumptions?

Then, why would a change in size have any effect on the gravitational force? It depends only on the masses involved and their distances.
Mostly yes. However, if G is greater, then the "equivalent" mass to what we had measured in 1G is smaller, lower volume, same effect.
Why would the volume change?
It would be observable.

The size affects tidal effects, and how objects behave on rotations.
Since the moment of inertia of our probes behave as expected, there is no reason to assume a change in size.
No change in size to probes, but our "assumed" mass size for outer planets may be wrong.
Why?
Our probes worked out there, and Huygens landed as planned.

See this post on Uranus &amp; Neptune (http://www.badastronomy.com/phbBB/viewtopic.php?t=22972) for similar discussion. Planet rotations are a separate issue. I.e., Why does Sun have rotation, for example?
You do not understand my point.
A change in mass would affect the moment of inertia, which tells us how an object moves under rotations.
This is observable in our probes.

However, that is incorrect. Your weight on Jupiter is the same as it was before we (hypothetically) discovered it had a different G,...
Wrong, because you increased the gravitational force.
In order to have the same weight, you have to assume, without any justification whatsoever, that the gravitational mass of the probe mass has changed, without affecting the inertial mass.
Again I don't get it, I just don't. If 100 kg on Earth are 980 N, and your same 100 kg (same person) is 2300 N on Jupiter, then weight has changed per Jupiter's greater gravity, but if you went to Jupiter in 2006, and came back again after finding out that Jupiter's G is five times, in 2012, you still weighed the same. It's just that your concept of Jupiterian "kilograms" would be five times smaller than before, same mass, same weight. Only a new "variable mass" idea would kick in, that's all.
kilogram is for inertial mass.
|Since inertial mass is the same here and on Jupiter, there is no change in kilograms.
A variable mass would "kick in" only if the Equivalence Principle is violated, but it is not.
"kilogram is for inertial mass", for same mass, but at 1/5th Jupiter "kilograms" in (hypo) 5G: each new "kg" is gravitationally 5 times as powerfull as for our Earth's 1G kg, for same mass.
The weight (= force) is five times larger, but the inertial mass does not change.

Ditto for every molecule on Jupiter, it is 5 times as gravitationally "active" as on Earth, so total size of Jupiter's core is five times smaller than it would be in Earth's 1G estimation of mass. Yet, same mass!
So, Jupiter's core is denser than Earth.
I don't think that astronomer are surprised by this.
Water at the bottom of the oceans is denser than at the surface, and air at sea level is denser than air at the top of mountains.

Look at it this way: Jupiter is Jupiter, same mass all the time. A visitor is the same visitor, same mass all the time. His/her weight on Jupiter would be higher than on Earth because Jupiter is a more massive planet. But the G factor is a separate issue, which is why I never used "weight" to explain this.
If you change G, you change the weight without changing the (gravitational) mass.
Other way around: When you change G, you change the gravitational mass without changing the weight, for same mass, only size and volume change.
That is your assumption: you arbitrarily change both G and the mass.

Each atom and molecule acts with more force, but there are fewer of them. I think this is the part you haven't seen yet in your mind's eye. It is elemental to how variable G works: (hypothetically) Mercury has 2 1/2 times more molecules per same mass; (hypothetically) Jupiter has 5 times fewer molecules for same mass per this system. And that affects kilograms.
It is even mor elementary if you do not assume ad hoc that the masses change.

Viz., if G' = 5G, and a = g = 23 ms^-2 for Jup, then either F = 5(20 kg * 23 ms^-2) or F = 5(100 kg * 23 ms^-2)... which would you choose?
I choose what the experiments tell me.

A separate issue is Newton's G as a universal constant, at our 1 G, as opposed to a variable G away from Earth. Mass as measured in 1 G is a separate "kilogram" from same mass measured in a higher or lower G.
It is only if you assume that the value of G affects the mass.
Yes.
And your assumptions fail the experimental tests.

Mass merely "acts" differently in a different G setting.
Different G, different force, different acceleration.
No need to change the mass.
Yes, if different force, then difference acceleration for same mass: this is the Equivalence Principle.
No, if "kilograms" are adjusted for G, same acceleration for same mass: EP preserved.
The is not the same, because gravity is symmetric.
The mass you changed, has a different acceleration, unless you violated the EP.

Each atom and molecule "acts as if" it had more "force" intrinsically, so it takes more force to move a smaller volume of mass equivalent to a larger volume of mass on Earth.
You do you keep changing the volume?

This may be why Mercury is "dragged" along by the Sun's rotational spin, because so close to this hot radiant body its G is low, at 0.4 AU it's about 2.5 times as light in terms of its own "kilograms" as here, so it has less inertial resistance to the momentum transfer from the Sun's rotation;...
General Relativity accounts quantitatively for this effect.
Correct. Einstein's GR found a way to explain this. I think it might be a "square peg in a round hole" however, and a better explanation may exist.
And a better explanation than your variable G does exist.

Your speculative handwaving requires a violation of the Equivalence Principle, one of GR's postulate.
Conclusion: your hypothesis fails the experimetnal test.
No, not "handwaving" but reasoning, something that seems to be lost on you.
You keep confusing weight and mass, you do not realize that your variable G affects both masses involved and result in a violation of the Equivalence Principle, and you have trouble with the maths of Newton's laws.
Where is the quantitative analysis?

Yes, it contradicts GR in its current form, since a much greater G in 99.99% of space means it is isotropic and homogeneous at a very different level from now assumed. Hypothesis has not failed any tests because they had not yet been performed.
Your hypothesis requires a violation of the Equivalence Principle.
GR requires the validity of the EP, and has been succesfully tested in a wide range of situations.
You lose.

ESA is closest to test that idea, and still years away. (Still, I must admire how you're still standing on your feet slugging, not taking it lying down. :) )
You apparently do not understand that experiments have already constrained a potentially variable G within boundaries that are much tighter than in your hypothesis.

Saturn's G may, at 9.5 AU, may be nearly ten times ours, so it's mass "kilograms" act as if they're heavier than our kilograms, so it holds a massive atmosphere, but it's still the same mass.
Yet Huygens landed as predicted and Cassini is working fine.
Again, your hypothesis fails the experimental test.
Effects of variable G had been well hidden from us, to date, except for "a few extra molecules" in Titan's upper atmosphere. Future tests may show different results, however. I am saying only this: it is hypothetically possible that 1G is not universal, but a large variable.

But here I do not wish to argue for a variable G being fact, because we don't know that it is, something to be studied in the future.
We have very tight experimental constraints on a varibale G.
Your hypothesis assume a variable G beyond thos constraints, therefore it fails the experimental test.
Yes, very tight constraints within 1 AU, which is 1G. We haven't checked for these same constraints at Jupiter's 5.2 AU, Saturn's 9.5 AU, Neptune's 30 AU, etc., not even at ~0.4 AU where is Mercury's orbit. Do we really know?
I gave you references about tests well beyond our Solar System.

What I want to show is how a variable G leads to a variable "kilogram", which per the Equivalence Principle acts differently than what we had assumed in a flat G universe. But you don't get it, do you? You just don't get it. :)

You are the one confusing gravitational mass, inertial mass, and weight.
You apparently that your handwaving could "work" if you assume that the Equivalence Principle is violated.
No. Very clear in my reasoning, not "handwaving" as you claim, nor confusing gravitational mass with weight. Inertial mass and gravitational mass are same, as I've explained over and over again: if each "kilogram" is gravitationally more potent for the molecules it represents, smaller in size, but it still takes the same amount of force to move an equivalent inertial mass.
Then we would see it the motion of any kind of object in the Solar System.
To hide this effect you assume a change in mass, with the result of violating the equivalence.

When we test for this, we should find that it takes the same force to move the same "kilogram" in space in the outersolar system than on Earth, but we will find that one kilogram is smaller in size for same matter than we expected.
Probes are not smaller, and we tested it.
Your hypothesis fails the experimental test, again.

I should add here for clarification: we are talking about propulsion, kinetic push energy. Gravitational assist energy is something entirely different, since the greater inertial "kilograms" mass in outersolar system is still pulled along by gravity in the ways we now expect it to, as mass attracts mass. To illustrate how different these two forces on mass are: Why do we not "feel" the barycenter wobble of the Earth? Do you ever find yourself being pulled one way or another as the planet wobbles in its barrycenter in relation to the Moon? (This may be one reason why tides bulge on both sides, BTW, so the Earth's acceleration "cancels" out.) But if you were pushed by a giant rocket ship moving you out of your orbital position, you would notice your coffee slosh. You see, they're different.
:roll:

Anybody get it? I'm stumped, running out of ways of explaining this very simple phenomenon. It's a piece of cake, easy as pie. Why am I the only one who sees this? Is everyone else like this? [-(
Did it occur to you, that you are the one not understanding?
In a word? No. I am reasoning my way through a hypothetical situation. My conclusion is that we may not yet have gravity right, GR notwithstanding.
I explained you the flaws of your reasoning, and showed that your conclusions are disproven by experiments.
You simply don't accept it.

I can imagine why. If mass "acts" differently in a different G setting, all cosmology is affected. So... resist! It has to be only hypothetical, until we find evidence that G is not universally the same. Until then, it's pie in the sky. :)
I gave you plenty of references about reserachers working on variable G scenarios.
Yet again, your conlclusions fail the experimetnal test.

Let's look at it this way, and let me take inventory here:

1) Your understanding is thus: Greater G' increases gravitational force, therefore it increases acceleration, but violates Equivalence Principle. Kilograms remain the same. Greater G' is impossible.
Wrong. The acceleration is greater, because the Equivalence Principle is valid.

2) My understanding is thus: Greater G' increases gravitational mass, therefore it leaves acceleration unchanged for same mass, but per Equivalence Principle the mass's volume size is smaller. "Kilograms" are adjusted in inverse proportion to G'. Greater G' is a hidden phenomenon.
The Equivalence Princple has nothing to do with volume.
A greater mass result in greater force: it is the other mass that must violate the equivalence to have the same acceleration.

Does this pretty much sum up our conceptual differences?
No, you misunderstood my points.

You say that G' can vary only slightly within very tight parameters.
Set by experiments.

I say G' can vary to orders of magnitude. My hypothetical vision of the universe makes more sense now,...
Your vision, but the Universe does not share that vision with you.

... why galaxy rotation "acts as if" there was Dark Matter, or why stars can form from hydrogen nebulea, or why 'neutron stars' can have immense gravity and spin: their G is massively different.

Net net: Astronomical observation interpreted within the parameters of a constant universal G leaves too many unanswered questions. The only way to resolve this astronomy conundrum is to go out there into space far from Earth's 1 G and test for it.
Maybe you should do more research.

Can you agree on any of my above points? Of course, it's okay if not, and you're still standing and slugging, which is cool. 8)
What's to agree?
There are unanswered questions: that's why research goes on.
But the questions are not the ones you think.

Tassel
2005-Jul-19, 02:30 PM
Since your question is not clear, I'll answer it to what I think you're asking. I am assuming you want me to give an example using both Earth 1G kilograms, and Jupiter's (hypothetical 5G) "kilograms".
As it has been from the beginning, my question was very clear.

g = GM/R^2
So you now accept that this is the correct equation. I see you've also dropped your unjustified term as well. When having a discussion like this, it's customary to acknowledge one's errors. It's also a part of suggestion #13 in the Advice for ATM theory supporters (http://www.badastronomy.com/phpBB/viewtopic.php?t=19638).

since I don't know what you have in mind for R, I'll figure it with R^2 unresolved, something you can fill in yourself, if that fits your question.
We'll use Jupiter's equitorial radius, so we can check the numbers. I also asked a couple of times for you to compute Jupiter's mass:

I've asked you to show the calculations step-by-step starting with calculating Jupiter's mass.
Since mass is the heart of the debate, it would be helpful if you showed how you arrive at the numbers you show below for Jupiter's mass. But anyway....

1) Earth kilograms (at 1G):

g = (6.67E-11 m^3 kg^-1 s^-2)*(1,898.6E+24 kg)/ R^2

Ok, so completing the calculations for R = 71,492km:

g = (6.67E-11 * 1.9E+27) / 5.11E+15m = 24.8m/s^2

2) For Jupiter's (hypothetical 5G) "kilograms" (1/5th of Earth kg):

g' = (33.35E-11 m^3 kg^-1 s^-2)*(379E+24 kg)/ R^2

g' R^2 = 12663.7E+13 m^3 s^-2, which is same, 126.6E+15 m^3 s^-2.

Again, you didn't show how you calculated Jupiter's mass, but you used the correct number that we would calculate if we suddenly realized that G at Jupiter was 5G. In any case, GM was the same, R was the same, so naturally, "g'" is the same. g = g'

So now you've demonstrated that "g" for Jupiter in a "constant G" universe and g' for Jupiter in a "variable G universe" where G=5G at Jupiter are the same. As you've shown above, our estimation of Jupiter's mass (which I made bold for emphasis) is all that changes if we are wrong about what G is at Jupiter.

Now let's return to the example we were discussing a few days ago. If we were to take a 100kg mass and weigh it on Earth and on Jupiter, what would we expect it to weigh? Here's what you correctly said (http://www.badastronomy.com/phpBB/viewtopic.php?p=504124&amp;#504124) we should expect to observe:

You are correct, my mistake. I had double multiplied Earth's N. The correct version should have been:

W = 100 kg * 9.8 m/s^2 = 980 N for Earth
W = 100 kg * 23 m/s^2 = 2300 N for Jupiter.

That is how we know it in one G. Now, if (1G*m) = (5G*0.2m), for Jupiter's five times higher G, then it should read:

W" = 5(20 "kg" * 23 m/^2) = 2300 N for Jupiter, if you factor in the 5 G.

Emphasis mine. So, even though you used an incorrect formula for determining weight, it is clear that you (correctly) expect our 100kg mass to weigh 980N on Earth and 2300N on Jupiter (assuming g=23m/s^2, of course). We know the formula for weight is W = mg, so let's plug in the numbers we just generated above:

For Earth:
W = mg
W = 100kg * 9.8m/s^2
W = 980N

For Jupiter (5G):
W = mg'
W = 100kg * 24.8m/s^2
W = 2480N

Which is right about what we expected to find.

So using the value of g' that you calculated for Jupiter in a 5G environment, with no modifications to our test mass or the units used to measure that mass, we modeled the weight we would observe on Earth (in a 1G environment) and on Jupiter (in a 5G environment).

This example clearly shows that neither mass, nor the definition of "kilogram" change in a variable G universe.

nutant gene 71
2005-Jul-19, 07:31 PM
...snip...

Now let's return to the example we were discussing a few days ago. If we were to take a 100kg mass and weigh it on Earth and on Jupiter, what would we expect it to weigh? Here's what you correctly said (http://www.badastronomy.com/phpBB/viewtopic.php?p=504124&amp;#504124) we should expect to observe:

You are correct, my mistake. I had double multiplied Earth's N. The correct version should have been:

W = 100 kg * 9.8 m/s^2 = 980 N for Earth
W = 100 kg * 23 m/s^2 = 2300 N for Jupiter.

That is how we know it in one G. Now, if (1G*m) = (5G*0.2m), for Jupiter's five times higher G, then it should read:

W" = 5(20 "kg" * 23 m/^2) = 2300 N for Jupiter, if you factor in the 5 G.

Emphasis mine. So, even though you used an incorrect formula for determining weight, it is clear that you (correctly) expect our 100kg mass to weigh 980N on Earth and 2300N on Jupiter (assuming g=23m/s^2, of course). We know the formula for weight is W = mg, so let's plug in the numbers we just generated above:

For Earth:
W = mg
W = 100kg * 9.8m/s^2
W = 980N

For Jupiter (5G):
W = mg'
W = 100kg * 24.8m/s^2
W = 2480N

Which is right about what we expected to find.

So using the value of g' that you calculated for Jupiter in a 5G environment, with no modifications to our test mass or the units used to measure that mass, we modeled the weight we would observe on Earth (in a 1G environment) and on Jupiter (in a 5G environment).

This example clearly shows that neither mass, nor the definition of "kilogram" change in a variable G universe.

:-? How did you figure that for the same mass in a variable G the "kilograms" are the same? :-?

In your example above, G is the same 1G, I don't see 5G anywhere.

BTW, I never figured Jupiter's mass, took it off NASA data page, but for 5G, I changed the "kilograms" inversely, so each "kg" is 5 times larger than our Earth based kg.

An example: If there are 100 million molecules in a piece of mass on Earth, the same mass on Jupiter would only have 20 million molecules, smaller piece of matter but same mass. So if 5G (were real), then the 20 million Jupiter molecules would have same mass (as 100 million Earth molecule), and weighing the same on Jupiter as would 100 million molecules weigh on Jupiter (figured in Earth's 1G).

Is this any clearer than mud? :)

nutant gene 71
2005-Jul-19, 07:45 PM
Your weight on Jupiter will be the same as we now estimated, ...
Wrong.
The weight = the gravitational force pulling downwards would be greater, because you increased G.
We're talking about different Gs. In yours, you're referring to Jupiter's greater gravitational mass. In mine, I'm referring to Jupiter's mass same, but it is in a higher G region, hypothetically, five times what we have on Earth.
By increasing G, you increased the gravitational force exerted by Jupiter.
I have not done anything to Jupiter's mass. I explained what happens if you increased G: Your weight would be greater than before.
Yes, increased G increases gravitational force exerted by Jupiter, but for a smaller "equivalent" mass, i.e., lower "kilograms". Weight stays the same as before.
So, you are changing to parameters at the same time.
You do not simply increase G, but you also chnage the mass of Jupiter, and you assume that the weight is the same.
If you change the mass of Juptier, you have to admit that all the other masses must be changed, because gravity is symmetric with respect to the masses involved.

At this point, you are left with either an observable change of the inertial mass of our probe, or an observable violation of the Equivalence Principle.
...SNIP...

Can you agree on any of my above points? Of course, it's okay if not, and you're still standing and slugging, which is cool. 8)
What's to agree?
There are unanswered questions: that's why research goes on.
But the questions are not the ones you think.
Yes, the unanswered questions remain, and research is paramount.

So... net-net-net, what have we got? The way I read yours, it looks to me like this: Only 1G has been observed, calculated, understood, and except for very tight parameters, a large variable G is impossible, so kilograms are always kilograms, and any change in G is merely a change in weight-force. It has nothing to do with any change in mass.

If this is correct? If so, then there is absolutely nothing I can say or do, mor even stand on my head in the sand, to make you see otherwise.
:-? [-( :( [-X :roll: :lol: :o 8) :^o =D> :P

Tensor
2005-Jul-19, 08:11 PM
I hope pap don't mind me jumping in here.

If this is correct? If so, then there is absolutely nothing I can say or do, mor even stand on my head in the sand, to make you see otherwise.

Well, lets go at it this way. As has pointed out, for your idea to work,

you are left with either an observable change of the inertial mass of our probe, or an observable violation of the Equivalence Principle.

Now, the reaction wheels on the probes that use them base their movment on their inertial mass of the probe with a constant g. They have been observed to drive them as expected, so the first one is out. That leaves a violation of the Equivalence Principle. But,you claim (http://www.badastronomy.com/phpBB/viewtopic.php?p=502730#502730) that there is no violation of the Equivalence Principle . So either way, your idea doesn't work, unless you are retracting your claim and are now throwing out the Equivalence Principle and GR (along with the matching GR observations) . And the only way to convince us otherwise is to now show how a violation of the Equivalence Principle actually works, which you haven't done.

nutant gene 71
2005-Jul-19, 08:41 PM
I hope pap don't mind me jumping in here.

If this is correct? If so, then there is absolutely nothing I can say or do, mor even stand on my head in the sand, to make you see otherwise.

Well, lets go at it this way. As has pointed out, for your idea to work,

you are left with either an observable change of the inertial mass of our probe, or an observable violation of the Equivalence Principle.

Now, the reaction wheels on the probes that use them base their movment on their inertial mass of the probe with a constant g. They have been observed to drive them as expected, so the first one is out. That leaves a violation of the Equivalence Principle. But,you claim (http://www.badastronomy.com/phpBB/viewtopic.php?p=502730#502730) that there is no violation of the Equivalence Principle . So either way, your idea doesn't work, unless you are retracting your claim and are now throwing out the Equivalence Principle and GR (along with the matching GR observations) . And the only way to convince us otherwise is to now show how a violation of the Equivalence Principle actually works, which you haven't done. --bold mine--

So you are basing your whole conclusion on reaction wheels observed to control attitude as expected? The ill-operating wheels are discarded? So no more tests, please, we're not interested any further. The very question of variable G is absurd...

How scientific is that? :)

Metricyard
2005-Jul-19, 08:58 PM
An example: If there are 100 million molecules in a piece of mass on Earth, the same mass on Jupiter would only have 20 million molecules, smaller piece of matter but same mass. So if 5G (were real), then the 20 million Jupiter molecules would have same mass (as 100 million Earth molecule), and weighing the same on Jupiter as would 100 million molecules weigh on Jupiter (figured in Earth's 1G).

Is this any clearer than mud?

Nope.. How can removing 80% of the mass still equal the same mass?

100 million != 20 million. irregardless of the value of G.

And of course they weigh would way the same. All you've done here is remove 80% of the mass, and multilpied it by 5. Your formula doesn't replace the mass, so you're mass can not be the same.

substance: that which has mass and occupies space; "an atom is the smallest indivisible unit of matter"

How can you have smaller matter with the same mass(assuming the same material)? You're removing large junks of mass from your matter.

W = 100 kg * 9.8 m/s^2 = 980 N for Earth
W = 100 kg * 23 m/s^2 = 2300 N for Jupiter.

That is how we know it in one G. Now, if (1G*m) = (5G*0.2m), for Jupiter's five times higher G, then it should read:

W" = 5(20 "kg" * 23 m/^2) = 2300 N for Jupiter, if you factor in the 5 G.

Your own formula shows that the matter is not equal. Look closely. 100kg !=20kg

pghnative
2005-Jul-19, 09:57 PM
...snip...

Now let's return to the example we were discussing a few days ago. If we were to take a 100kg mass and weigh it on Earth and on Jupiter, what would we expect it to weigh? Here's what you correctly said (http://www.badastronomy.com/phpBB/viewtopic.php?p=504124&amp;#504124) we should expect to observe:

You are correct, my mistake. I had double multiplied Earth's N. The correct version should have been:

W = 100 kg * 9.8 m/s^2 = 980 N for Earth
W = 100 kg * 23 m/s^2 = 2300 N for Jupiter.

That is how we know it in one G. Now, if (1G*m) = (5G*0.2m), for Jupiter's five times higher G, then it should read:

W" = 5(20 "kg" * 23 m/^2) = 2300 N for Jupiter, if you factor in the 5 G.

Emphasis mine. So, even though you used an incorrect formula for determining weight, it is clear that you (correctly) expect our 100kg mass to weigh 980N on Earth and 2300N on Jupiter (assuming g=23m/s^2, of course). We know the formula for weight is W = mg, so let's plug in the numbers we just generated above:

For Earth:
W = mg
W = 100kg * 9.8m/s^2
W = 980N

For Jupiter (5G):
W = mg'
W = 100kg * 24.8m/s^2
W = 2480N

Which is right about what we expected to find.

So using the value of g' that you calculated for Jupiter in a 5G environment, with no modifications to our test mass or the units used to measure that mass, we modeled the weight we would observe on Earth (in a 1G environment) and on Jupiter (in a 5G environment).

This example clearly shows that neither mass, nor the definition of "kilogram" change in a variable G universe.

:-? How did you figure that for the same mass in a variable G the "kilograms" are the same? :-?

Tassel's point is that if one speculates that G at Jupiter is five times the G at earth, and if one further speculates that 1 kg at Earth is five times a kg at Jupiter, then a 100 kg object would weight 980N on Earth and 2480N on Jupiter.

Also, if one speculates that G is the same at Jupiter as at Earth, and if one further speculates that 1 kg at Earth is 1 kg at Jupiter, then a 100 kg object would weigh 980N on Earth and 2480N on Jupiter.

So if all calculations arrive at the same conclusion, what's the flippin' point of babbling about a variable G??? Another way of stating this is that if one is to speculate that G at Jupiter is different than G at earth, then one is REQUIRED to speculate that kg's are different at Jupiter than on Earth --- otherwise Jupiter's gravity would be different, and the variable G theory would not match observations.

So this explains why nutant is arguing so vociferously in favor of variable mass --- without it, he cannot have his variable G theory.

So I see at two major flaws in nutant's variable G, variable mass theory. (There are lot's of minor ones, but let's stick to the major conceptual flaws)

1) Since the theory relies on M varying exactly opposite to G, such that G*M remains a constant, the theory does not predict ANYTHING that current "constant G" theory predicts. So it is therefore useless

2) Since the theory requires changes to a host of other theories, it is unnecessarily complicated. To postulate a variable G, and yet still predict planet and moon orbits, one needs to then postulate a variable mass. To then predict how reaction wheels and springs affect space probes (whose mass presumably changes in route), one needs to handwave some arguments. And if planet's masses are different that we think, then theories about material properties and chemical properties need to be altered.

In short, lots of added complications with no improvements in predictions. Utterly useless.

Tensor
2005-Jul-19, 10:52 PM
So you are basing your whole conclusion on reaction wheels observed to control attitude as expected? The ill-operating wheels are discarded?

In Cassini's case,there are four wheels on Cassini, one had a problem. Let's see, we should believe that the variable g is causing only one of the wheels to act strangely for a while, then work normally. So, to your way of thinking, I'm ignoring one wheel that has an anamolly for a while, then works as advertised. And you're ignoring three wheels that worked normally all the time and one wheel that worked normally except for a short time. As to the other probes, do you have any evidence that all the wheels on a specific probe failed at the same time? It seems you are basing your whole conclusion on several temporarily balky wheels, ignoring the rest that worked with no problems. I could have pointed out the propellant reactions in driving the probes (which also matches current thought on g, observations of planetary motions that match current thought etc.

So no more tests, please, we're not interested any further. The very question of variable G is absurd...

And I said no more tests where? Come on, point us to any solid evidence, that's all anyone is asking. As for the idea of a variable g, it has been pointed out to you several times that it has been looked into, and is stll being looked into and the limits of variability have been found to be much less than the amount of variability that you have been proposing. It's not that a variable g is absurd, just your version of it.

How scientific is that? :)

It's a lot more scientific than basing it on a temporarily balky reaction wheels (while ignoring the rest of the wheels that worked), ignoring good matches between current theory and observation, and providing equations with no derivations for the math from your own idea (using GR or Newonian Gravitational redshift equations in a variable g environment sin't allowed) etc, which is what you have been doing.

nutant gene 71
2005-Jul-19, 11:04 PM
;)

Gentlemen, choose your "counter-intuitives". At twenty paces turn, take aim, and fire.

You may choose your weapons from "variable lengths and time", or from "variable G and mass".

Your seconds will attend silently, and mark the time. May G*d have mercy on your souls, for only one of you will remain standing... Gentlemen... take positions.

:lol: |_*"_l+> :lol: ... I wish there were an emoticon for rolling-on-the-floor- laughing!.

pghnative
2005-Jul-19, 11:42 PM
Can you translate that last post into English?

Tassel
2005-Jul-19, 11:55 PM
In your example above, G is the same 1G, I don't see 5G anywhere.
"5G" was used to calculate g'. You used 5G right here:

g' = (33.35E-11 m^3 kg^-1 s^-2)*(379E+24 kg)/ R^2

BTW, I never figured Jupiter's mass, took it off NASA data page,
No kidding. I've been asking you to calculate Jupiter's mass for a week now and you won't do it. You do know how we determine the masses of the planets, right?

but for 5G, I changed the "kilograms" inversely, so each "kg" is 5 times larger than our Earth based kg.
No, you divided Jupiter's mass by 5, which is actually the right thing to do. We observe Jupiter and its moons to behave a certain way, and calculate Jupiter's mass based on that and the known value of G. If we were to learn that G is different than we expect, we would have to recalculate Jupiter's mass. You did that (see bold):

2) For Jupiter's (hypothetical 5G) "kilograms" (1/5th of Earth kg):

g' = (33.35E-11 m^3 kg^-1 s^-2)*(379E+24 kg)/ R^2

g' R^2 = 12663.7E+13 m^3 s^-2, which is same, 126.6E+15 m^3 s^-2.

We went from calculating Jupiter's mass (sort of...) right down to figuring what a 100kg mass would weigh on Earth and on Jupiter in a "hypothetical" 5G environment and never once had to change the definition of kilogram. You did most of the work yourself, and don't really seem to know what you did.

If you think something was done incorrectly, point out the specific problem you have with those calculations, or alternatively, show where the concept of "new kilograms" came into play.

nutant gene 71
2005-Jul-20, 06:06 AM
A change like that would affect everything, including chemical bonds and bandstructures of semiconductors.
Electronincs and CCDs on space probes would not work, and the spectra would not be recognizable.
Don't know this, an assumption only that it would change. We had not measured this effect on chemical-electrical forces in non-Earth locations. Would need to specifically test for it in outer solar system, for example.
Bandstructure of semiconductors and electronic structure of atoms and moclecules depend on the masses of electrons and nuclei.
By changing the masses, you change the electronic structure, hence all the semiconductor electronics and the spectra are affected.
If the electronic structure changed outside the Earth, astronomers would not be able to recognize spectra from stars and planets, by comparison to spectra taken on Earth.
The electronics on our probes would not work as designed, CCDs would not give pictures.
This is a most interesting question, as to whether or not electronics, chemical reactions, spectrography, would all work the same in a hypothetical variable G. I would like to answer this after I've had better familiarization with Coulomb's Law, electric force, the Coulomb's constant. It may also entail hydrogen energy levels, and possibly the deBroglie's v=h/[(lambda)*(mass)], where v = c. So will need some time to do this work before I can get back on that... later. Carry on.

nutant gene 71
2005-Jul-20, 06:28 AM
We went from calculating Jupiter's mass (sort of...) right down to figuring what a 100kg mass would weigh on Earth and on Jupiter in a "hypothetical" 5G environment and never once had to change the definition of kilogram. You did most of the work yourself, and don't really seem to know what you did.
The definition, in this hypothetical case of Jupiter's 5G, for kilograms is:

For equal mass, (Earth)1 kilogram = (Jupiter) 5 "kilograms".

The mass on Jupiter, for equivalent mass on Earth, would have for each molecule 5 times the gravitational force, so it would take 1/5 th as many molecules on Jupiter to equal one Earth kilogram. Earth is in a very light gravity region, only 1G, so each molecule is only 1/5 th as gravitationally strong as at Jupiter's 5G. Take in toto as a product of G*m, the masses are equivalent for each, though kilograms-"kilograms" differ. Therefore, one kilogram of Earth mass would register as 5 "kilograms" on Jupiter. Equivalence is conserved.

Earth, as is our inner solar system, resides in a very light gravity region. Our solar system is a relatively light gravity region within the overall galaxy. The galaxy is a fairly light gravity region within the great stretches of deep space gravity, which is very great, perhaps 5 or 6 orders of mangitude of the G we know here. This is the model I'm working with, that we are light gravity "islands" within a very strong gravity universe, which is isotropic and homogeneous for itself, except for the little scattered "oasis" of galaxies and stars. Star energy, especially hot stars like our Sun, moderate gravity, to where it is very light, what we know of as 1G at 1 AU. Nobody has to like it, nor accept it, nor agree with it. But this is where I'm at. What you do with it, whether you take an interest in exploring this possibility, or reject if off hand, is your choice. Either way for me.

Nevertheless, I truly appreciate your questions.., err.. mostly challenges, because they also stimulate my interest in this subject. But think about it! Cheers. :)

papageno
2005-Jul-20, 09:52 AM
[SNIP!]

So... net-net-net, what have we got? The way I read yours, it looks to me like this: Only 1G has been observed, calculated, understood, and except for very tight parameters, a large variable G is impossible, so kilograms are always kilograms, and any change in G is merely a change in weight-force. It has nothing to do with any change in mass.
The units kilogram does not depend on the value of G, nor does it depend on the Equivalence Principle.
Whether G is 1 G or 5 G, iit is inconsequential to the kilogram.

Experiments suggest that the Equivalence Principle holds exactly, and put very tight constraints on a potentially variable G.
A variable G would not affect anyway gravitational or inertial masses.

If this is correct? If so, then there is absolutely nothing I can say or do, mor even stand on my head in the sand, to make you see otherwise.
You could not show that the value of G has any effect on the mass, whether gravitational or inertial.
In your posts you displayed serious misconceptions about gravity, mass, weight and motion, and a deep misunderstanding of Newton's laws for gravity and motion.

papageno
2005-Jul-20, 09:59 AM
A change like that would affect everything, including chemical bonds and bandstructures of semiconductors.
Electronincs and CCDs on space probes would not work, and the spectra would not be recognizable.
Don't know this, an assumption only that it would change. We had not measured this effect on chemical-electrical forces in non-Earth locations. Would need to specifically test for it in outer solar system, for example.
Bandstructure of semiconductors and electronic structure of atoms and moclecules depend on the masses of electrons and nuclei.
By changing the masses, you change the electronic structure, hence all the semiconductor electronics and the spectra are affected.
If the electronic structure changed outside the Earth, astronomers would not be able to recognize spectra from stars and planets, by comparison to spectra taken on Earth.
The electronics on our probes would not work as designed, CCDs would not give pictures.
This is a most interesting question, as to whether or not electronics, chemical reactions, spectrography, would all work the same in a hypothetical variable G. I would like to answer this after I've had better familiarization with Coulomb's Law, electric force, the Coulomb's constant. It may also entail hydrogen energy levels, and possibly the deBroglie's v=h/[(lambda)*(mass)], where v = c. So will need some time to do this work before I can get back on that... later. Carry on.
It was not a question. It is a consequence of changing mass.

If you relly have more familiarity with Coulomb's law, it cannot have escaped you that a G dependent on position in Newton's law is mathematically the same a position-dependent dielectric constant in Coulomb's law.
Nothing exotic.

Also, I would be careful in extending De Broglie's formula to relativistic particles. Those are best left to Quantum Electrodynamics.

Celestial Mechanic
2005-Jul-20, 12:43 PM
In case you missed it the first time, I'll post this again. For a different perspective, I will now channel Gertrude Stein. Cue the cheesy theremin music!

Woo-OOO, woo-OOO-oo!

A gram is a gram is a gram.
What an honour! We have Winston Churchill on the line. Tell me sir, you've seen Lunatik/Luna2Uno/nutant gene 71's "Axiomatic Equation" and the website around it, what is your opinion of it?

It is a host of errors wrapped in a fallacy inside a misconception. Speaking hypothetically, of course.
:lol:
All right, post number 1400!

pghnative
2005-Jul-20, 01:21 PM
We went from calculating Jupiter's mass (sort of...) right down to figuring what a 100kg mass would weigh on Earth and on Jupiter in a "hypothetical" 5G environment and never once had to change the definition of kilogram. You did most of the work yourself, and don't really seem to know what you did.
The definition, in this hypothetical case of Jupiter's 5G, for kilograms is:

For equal mass, (Earth)1 kilogram = (Jupiter) 5 "kilograms".

The mass on Jupiter, for equivalent mass on Earth, would have for each molecule 5 times the gravitational force, so it would take 1/5 th as many molecules on Jupiter to equal one Earth kilogram. Earth is in a very light gravity region, only 1G, so each molecule is only 1/5 th as gravitationally strong as at Jupiter's 5G. Take in toto as a product of G*m, the masses are equivalent for each, though kilograms-"kilograms" differ.
I refer again to what I posted above:

Tassel's point is that if one speculates that G at Jupiter is five times the G at earth, and if one further speculates that 1 kg at Earth is five times a kg at Jupiter, then a 100 kg object would weigh 980N on Earth and 2480N on Jupiter.

Also, if one speculates that G is the same at Jupiter as at Earth, and if one further speculates that 1 kg at Earth is 1 kg at Jupiter, then a 100 kg object would weigh 980N on Earth and 2480N on Jupiter.

So if all calculations arrive at the same conclusion, what's the flippin' point of babbling about a variable G???

This is the model I'm working with, that we are light gravity "islands" within a very strong gravity universe, which is isotropic and homogeneous for itself, except for the little scattered "oasis" of galaxies and stars. Star energy, especially hot stars like our Sun, moderate gravity, to where it is very light, what we know of as 1G at 1 AU. Nobody has to like it, nor accept it, nor agree with it. But this is where I'm at. What you do with it, whether you take an interest in exploring this possibility, or reject if off hand, is your choice.
There is a third choice: that we explored it, quickly saw that it showed no improvement in predicting anything and introduced needless complications.

I noticed you haven't responded to the points in my previous post.

Tassel
2005-Jul-20, 01:26 PM
The mass on Jupiter, for equivalent mass on Earth, would have for each molecule 5 times the gravitational force, so it would take 1/5 th as many molecules on Jupiter to equal one Earth kilogram.
If there is 5 times the gravitational force then things weigh 5 times as much. That's it. By definition, mass is a property of matter that is independent of gravitational force. This isn't meant to be an attack, but a simple truth: you do not have an intuitive understanding of the difference between weight and mass. Your statement above is all the evidence that is needed. "Gravitational force" is weight. So take what you said above and replace "gravitational force" with "weight", "Jupiter" with "5G" and "Earth" with "1G" and you get a true statement:

"The mass [in 5G], for equivalent mass [in 1G], would have for each molecule 5 times the weight, so it would take 1/5 th as many molecules [in 5G] to equal one Newton."

We worked through some calculations that show we can model what would be observed if G = 5G at Jupiter and we never needed to change the definition of mass or kilograms. In fact it showed that we can't use "different kilograms" and still get the expected answer. I asked you to point out the flaw in the calculations. You can't.

I asked you to calculate the mass of Jupiter based on observation. You won't, because like the calculations we just went through (which you cannot find fault with), it will show how everything works just fine in a variable G universe with our current definition of mass and "kilogram".

Tensor
2005-Jul-20, 02:14 PM
Gentlemen, choose your "counter-intuitives". At twenty paces turn, take aim, and fire.

Let's see,

You may choose your weapons from "variable lengths and time",

Experimental match (http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html) and Observation match (http://www-pdg.lbl.gov/2004/reviews/gravrpp.pdf)

or from "variable G and mass".

No presented experimental match or observation match.

Obviously, from your comment in a previous post, I should choose from "no presented experiment or observation", because that is so much more scientific than basing my choice on experiment and observations.

nutant gene 71
2005-Jul-20, 05:42 PM
Tassel's point is that if one speculates that G at Jupiter is five times the G at earth, and if one further speculates that 1 kg at Earth is five times a kg at Jupiter, then a 100 kg object would weigh 980N on Earth and 2480N on Jupiter.

Also, if one speculates that G is the same at Jupiter as at Earth, and if one further speculates that 1 kg at Earth is 1 kg at Jupiter, then a 100 kg object would weigh 980N on Earth and 2480N on Jupiter.

So if all calculations arrive at the same conclusion, what's the flippin' point of babbling about a variable G??? bold mine

The mass on Jupiter, for equivalent mass on Earth, would have for each molecule 5 times the gravitational force, so it would take 1/5 th as many molecules on Jupiter to equal one Earth kilogram.
If there is 5 times the gravitational force then things weigh 5 times as much. That's it. By definition, mass is a property of matter that is independent of gravitational force. This isn't meant to be an attack, but a simple truth: you do not have an intuitive understanding of the difference between weight and mass. Your statement above is all the evidence that is needed. "Gravitational force" is weight. So take what you said above and replace "gravitational force" with "weight", "Jupiter" with "5G" and "Earth" with "1G" and you get a true statement:

"The mass [in 5G], for equivalent mass [in 1G], would have for each molecule 5 times the weight, so it would take 1/5 th as many molecules [in 5G] to equal one Newton."

We worked through some calculations that show we can model what would be observed if G = 5G at Jupiter and we never needed to change the definition of mass or kilograms. In fact it showed that we can't use "different kilograms" and still get the expected answer. I asked you to point out the flaw in the calculations. You can't.

I asked you to calculate the mass of Jupiter based on observation. You won't, because like the calculations we just went through (which you cannot find fault with), it will show how everything works just fine in a variable G universe with our current definition of mass and "kilogram".

The flippin point is that if youre a Jupiterian, and you know only your own kilogram, these are 5 times Earths kilograms, and your kg volume wise is 1/5th the size. They would be quick to tell you that Earths mass (http://www.nyu.edu/classes/gspscience/hou/html/vlab01.html) is calculated (http://www.wonderquest.com/calculation-mass-of-earth.htm) at 1.195E+24 kg in Jewtons G of 33.34E-11 Nm^2 kg^2, so each cubic decimeter of Jupiterian aqua is one kilogram.

This is the re-definition of the kilogram in variable G and variable mass.

The fact that I used Earth kilograms to explain this seems to leave with the impression that nothing has changed, weight same, etc. Youre missing the flippin point. And if you got it you would instantly see the ramifications for cosmology are immense. Take note that if the Jupiterians were as stuck on their 5G as you appear to be on 1G, they would think Earthian kilogram are the feverish rants of a lunatic..

Geez Louise, Gertrude, and Winston Churchill too aargh whats the flippin point.

[Edit for spell]

pghnative
2005-Jul-20, 06:04 PM
The flippin point is that if youre a Jupiterian, and you know only your won kilogram, these are 5 times Earths kilograms, and your kg volume wise is 1/5th the size. They would be quick to tell you that Earths mass (http://www.nyu.edu/classes/gspscience/hou/html/vlab01.html) is calculated (http://www.wonderquest.com/calculation-mass-of-earth.htm) at 1.195E+24 kg in Jewtons G of 33.34E-11 Nm^2 kg^2, so each cubic decimeter of Jupiterian aqua is one kilogram.
In this scenario, an Earthling would weigh a 100 kg object as being 2480N on Jupiter*. The Earthling would give the object to the Jupiterian, who would weigh it as 2480N*. They would log on to the Jupiternet (BadJupstronomy.com) and argue for years about whether it is a 100 kg object or a 500 kg object and whether G was the same or 5X different, but every other calculation would be the same.

So again -- no flippin point. In some ways, this is just unit conversion.

*side issue -- this number doesn't seem right. Shouldn't it be more??
**edited to add that the 2480 number appears about right based on a quick websearch

Tassel
2005-Jul-20, 06:41 PM
The flippin point is that if youre a Jupiterian, and you know only your won kilogram, these are 5 times Earths kilograms, and your kg volume wise is 1/5th the size.
Why, because you say so?

They would be quick to tell you that Earths mass (http://www.nyu.edu/classes/gspscience/hou/html/vlab01.html) is calculated (http://www.wonderquest.com/calculation-mass-of-earth.htm) at 1.195E+24 kg in Jewtons G of 33.34E-11 Nm^2 kg^2
And we would be quick to retort, "You used the wrong value of G to calculate the mass of our planet. G by Earth is 6.67E10-11 Nm^2 kg^2".

To which they would respond, "Oh, how about that. We'll need to recalculate the mass of your planet then since we assumed G was 33.34E-11 Nm^2 kg^2 everywhere. One second...that's 6.67...carry the 1......here it is, now that we know the correct value of G at Earth, we calculate the mass of your planet to be 5.97E+24kg. And when we use this correct value for the mass of your planet, along with the correct value for G, all our equations still work just fine."

I see what the problem is now: You are under the impression that, in a fictitious variable G universe, we should calculate the masses of the planets using a constant G and then correct the resulting errors by changing the definition of a kilogram. This is patently absurd.

In a variable G universe, you would calculate the masses of the planets using the value for G at that planet. You would then get the correct mass for that planet. Use these correct numbers in any calculation you like, and everything works fine.

Again, if you would just start from scratch, observe Jupiter, calculate it's mass using whatever value of G you want, and then use the resulting number for mass in any equation you like, you will find that it all still works. We did most of this already and you saw that there were no problem using the current definition of mass and kilogram. But like every other result that conflicts with your belief system, you just ignore it and carry on.

I have no idea why you won't just do the calculations from start to finish (actually, I have some idea...). If you're so sure "variable mass" is required, just do the math starting with calculating Jupiter's mass and show how nothing works unless we modify kilograms.

nutant gene 71
2005-Jul-20, 08:32 PM
The flippin point is that if youre a Jupiterian, and you know only your won kilogram, these are 5 times Earths kilograms, and your kg volume wise is 1/5th the size.
Why, because you say so?

They would be quick to tell you that Earths mass (http://www.nyu.edu/classes/gspscience/hou/html/vlab01.html) is calculated (http://www.wonderquest.com/calculation-mass-of-earth.htm) at 1.195E+24 kg in Jewtons G of 33.34E-11 Nm^2 kg^2
And we would be quick to retort, "You used the wrong value of G to calculate the mass of our planet. G by Earth is 6.67E10-11 Nm^2 kg^2".

To which they would respond, "Oh, how about that. We'll need to recalculate the mass of your planet then since we assumed G was 33.34E-11 Nm^2 kg^2 everywhere. One second...that's 6.67...carry the 1......here it is, now that we know the correct value of G at Earth, we calculate the mass of your planet to be 5.97E+24kg. And when we use this correct value for the mass of your planet, along with the correct value for G, all our equations still work just fine."

I see what the problem is now: You are under the impression that, in a fictitious variable G universe, we should calculate the masses of the planets using a constant G and then correct the resulting errors by changing the definition of a kilogram. This is patently absurd.

In a variable G universe, you would calculate the masses of the planets using the value for G at that planet. You would then get the correct mass for that planet. Use these correct numbers in any calculation you like, and everything works fine.

Again, if you would just start from scratch, observe Jupiter, calculate it's mass using whatever value of G you want, and then use the resulting number for mass in any equation you like, you will find that it all still works. We did most of this already and you saw that there were no problem using the current definition of mass and kilogram. But like every other result that conflicts with your belief system, you just ignore it and carry on.

I have no idea why you won't just do the calculations from start to finish (actually, I have some idea...). If you're so sure "variable mass" is required, just do the math starting with calculating Jupiter's mass and show how nothing works unless we modify kilograms.
Spoken like a true "Earthian"! Remember this post on Xians (http://www.badastronomy.com/phpBB/viewtopic.php?p=495886#495886)? I guess your retort is that of an Earthian's "true blue" kilograms? \:D/

BTW, did you ever answer the question, of who's right? :)

Celestial Mechanic
2005-Jul-20, 09:06 PM
BTW, it's "Jovian", not "Jupiterian". [-X

Tassel
2005-Jul-20, 09:08 PM
So no calculations then? No pointing out the flaws in the calculations we already did? No pointing out where "variable mass" entered the picture? No explanation as to why you would measure the mass of a planet using the wrong value for G...and then suggest we change the definition of a kilogram to correct the mistake?

I'm shocked.

You have your "variable mass" idea. You have your variable G "hypothesis" complete with a "hypothetical" value for G at Jupiter. Instead of posting essays, run the numbers starting with calculating Jupiter's mass based on observation. Show us precisely where we need "variable kilograms". I'll be sure to hold my breath waiting.

nutant gene 71
2005-Jul-20, 11:49 PM
So no calculations then? No pointing out the flaws in the calculations we already did? No pointing out where "variable mass" entered the picture? No explanation as to why you would measure the mass of a planet using the wrong value for G...and then suggest we change the definition of a kilogram to correct the mistake?

Dear Tassel,

I can appreciate what you are suggesting, that I formally mathematize this discussion on this hypothetical issue of variable mass in a variable G scenario, and if I were to write a more formal paper on it, I would do so. However, regrettably, nay humbly, I must decline your fine invitation to do so. Since though this subject had been well received, as six page testify, many of which already have math in them, it has nevertheless been less than enthusiastically endorsed, so perhaps my motivations for writing such a formal paper are not yet present. For every hard effort there should be some reward, and in this case, my regrets, I feel that such work might be in vain, or perceived within the context of my conceit. Still, thank you for your suggestion, and be assured that should we take it from mere essay to more substantive mathematics, as you have demonstrated a certain skill in these matters, I will solicit your help.

Kindly excuse my demur response to your offer for writing a more formal treatise. For now, I am turning my attention to the Coulomb-Lorentz Force idea to address papageno's concerns with a hypothetical electronic-chemical effect of a variable G, which may yet take some time. I am a mere layman in these matters, so it will take some effort on my part to make a sensible study and response. I appreciate your sincere interest. More later.

With regards,

nutant.

nutant gene 71
2005-Jul-20, 11:56 PM
BTW, it's "Jovian", not "Jupiterian". [-X
Well, "we" say Jovian, while "they" say Jupiterian, though they're not yet convinced that our kilograms are the best. ;)

nutant gene 71
2005-Jul-21, 04:59 PM
This PhysicsReview article The Weight of Light (http://focus.aps.org/story/v16/st1) says light gravitationally redshifts, as measured in 1960 Harvard experiment by Rebka and Pound, at the rate of 10^-15, which is the pure gravitational effect, and which matches Einstein's prediction with 10% accuracy.

Does this mean this "classic test of relativity" that photons of light or gamma rays have "mass" to cause gravitational interaction? Or does it mean something different, that gravitational forces cause redshift, a lengthening in wavelength, which then mimmics light having mass? In this part 1 post (http://www.badastronomy.com/phpBB/viewtopic.php?p=504217#504217), I show an example of pure gravitational redshift, without relativistic considerations, while in part 2 post (http://www.badastronomy.com/phpBB/viewtopic.php?p=504364#504364) I show how light "cut-off" wavelength may account for this greater gravitational effect through most of intergalactic deep space.

If light, or any em energy, has no "mass" as we know it, then can the natural gravitational redshift of such energy account for what "mimmics" mass in the gravitational bending of light? And if this is so, then is this gravitational bending of light a valid test of Relativity? This of necessity raises the question as to what exactly is "mass".

pghnative
2005-Jul-21, 05:33 PM
... This of necessity raises the question as to what exactly is "mass".Keep in mind that just because a question is raised doesn't mean that you are the one with the answer.

See, you make the classic mistake that you are the only one who is open-minded. Nothing could be further from the truth. I think I can speak for everyone here when I state that we would be enthralled with a modified theory of gravity. If someone showed that G was variable and that observation X, Y, and Z were now better explained by their new theory, we'd be overjoyed.

But your theory doesn't explain anything. You are unable to provide calculations that predict anything of any usefulness. Your theory predicts that objects of known mass weigh the same on Jupiter as our theory predicts they would. Your theory predicts the same orbits for Jupiters satellites as our theory does. YOU are unable to present any calculations. How can anyone claim to have a new theory of gravity without being able to perform the most basic physics calculations???

Here's my suggestions: when you can use your theory to calculate anything, then come back and let us know. Otherwise, we're all just babbling, and this is the wrong forum for that.

Tensor
2005-Jul-21, 05:46 PM
This PhysicsReview article The Weight of Light (http://focus.aps.org/story/v16/st1) says light gravitationally redshifts, as measured in 1960 Harvard experiment by Rebka and Pound, at the rate of 10^-15, which is the pure gravitational effect, and which matches Einstein's prediction with 10% accuracy.

Does this mean this "classic test of relativity" that photons of light or gamma rays have "mass" to cause gravitational interaction? Or does it mean something different, that gravitational forces cause redshift, a lengthening in wavelength, which then mimmics light having mass?

Nutant, in GR gravity doesn't couple with "mass" (note that "mass" in this post is some solid object, the way most people think of it), gravity couples to the stress-energy tensor. It's the amount of rest energy and stress (think pressure in this instance) that causes the curvature of space. Photons have energy (heck, that's what their definition is: energy quanta), but only when they are moving, so they have no rest energy. "Massive" particles (like planets, particles etc) do have a rest energy and thus can cause a curvature of space.

If light, or any em energy, has no "mass" as we know it, then can the natural gravitational redshift of such energy account for what "mimmics" mass in the gravitational bending of light?

In GR zero rest mass particles paths are on null geodesics, the shortest path, timewise, through space. The bending of that path is what we observe as the bending of space-time.

And if this is so, then is this gravitational bending of light a valid test of Relativity?

Yes, it is a valid test.

This of necessity raises the question as to what exactly is "mass".

That is the tricky part. Which is why GR uses the rest energy of the object. You can figure out the rest energy of a "mass" by multiplying the "mass" by c^2.

nutant gene 71
2005-Jul-21, 08:29 PM
...
Here's my suggestions: when you can use your theory to calculate anything, then come back and let us know. Otherwise, we're all just babbling, and this is the wrong forum for that.

Did you understand what this sentence, my earlier, actually meant?

"The mass on Jupiter, for equivalent mass on Earth, would have for each molecule 5 times the gravitational force, so it would take 1/5 th as many molecules on Jupiter to equal one Earth kilogram."

pghnative
2005-Jul-21, 08:36 PM
...
Here's my suggestions: when you can use your theory to calculate anything, then come back and let us know. Otherwise, we're all just babbling, and this is the wrong forum for that.

Did you understand what this sentence, my earlier, actually meant?

"The mass on Jupiter, for equivalent mass on Earth, would have for each molecule 5 times the gravitational force, so it would take 1/5 th as many molecules on Jupiter to equal one Earth kilogram."
I understand it completely. Once you postulate that G is 5 times greater at Jupiter, you must adjust the mass by 5 or else you can't explain the orbits of Jovian satellites. Once you adjust both (such that G*M remains constant) then calculations of weight, orbits, etc, give the same results in your theory as they do ours. You seem to be fooling yourself that the results are different at some level (such as at very near the surface of Titan), but in reality the results are the same.

If the results ARE different, I'm eager for you to show me how. But given your reluctance to perform simple math for us, I think I'll be waiting a long time.

nutant gene 71
2005-Jul-21, 09:16 PM
...
Here's my suggestions: when you can use your theory to calculate anything, then come back and let us know. Otherwise, we're all just babbling, and this is the wrong forum for that.

Did you understand what this sentence, my earlier, actually meant?

"The mass on Jupiter, for equivalent mass on Earth, would have for each molecule 5 times the gravitational force, so it would take 1/5 th as many molecules on Jupiter to equal one Earth kilogram."
I understand it completely. Once you postulate that G is 5 times greater at Jupiter, you must adjust the mass by 5 or else you can't explain the orbits of Jovian satellites. Once you adjust both (such that G*M remains constant) then calculations of weight, orbits, etc, give the same results in your theory as they do ours. You seem to be fooling yourself that the results are different at some level (such as at very near the surface of Titan), but in reality the results are the same.

If the results ARE different, I'm eager for you to show me how. But given your reluctance to perform simple math for us, I think I'll be waiting a long time.
I'm cutting it down some more, so focus on this part:

".. for each molecule 5 times the gravitational force, so it would take 1/5 th as many molecules.."

The catch word is "molecules", which is more than what you refer to in yours. "Each molecule" has greater gravitational mass. Yes, the weight results are the same, but for "fewer molecules" on Jupiter than Earth. All the other stuff on math etc, is superfluous if you can't get this one point, and no amount of math would make a lick of difference. :x

TravisM
2005-Jul-21, 09:33 PM
Jerry may be gone, but his "spirit" lives on.

You have a lot more patience than I, papageno! That's praiseworthy. But after all these corrections with no meaningful responses, it must be wearing at least a little thin by now.

Meanwhile, good work re your replies. A shame your target audience doesn't seem to comprehend physics fundamentals. :-?

Wow... this was from page 2. I can't wait to catch up to this post of mine...

Tassel
2005-Jul-21, 10:42 PM
"Each molecule" has greater gravitational mass.
This is wrong. Each molecule weighs more. Weight is a function of mass (and g). Mass is not a function of weight. This is intuitive, as the weight of objects can change, even in our real constant G universe, while their mass (by definition) remains the same.

Perhaps talking about Jupiter is overcomplicating the issue. Instead of assuming we got G wrong, what would happen to mass and weight if G were to suddenly change?

W = mg
W = m * (GM/R^2)

One day the flying spaghetti monster, ruler of the universe, suddenly changes G to 5G, leaving everything else the same. What do we expect would happen? The force of gravity becomes stronger throughout the universe. Orbits change. Everything becomes heavier.

Nothing happens to mass. Looking at the equations above, if G increases W increases, as we would expect. There is no reason or justification to change m or M.

nutant gene 71
2005-Jul-22, 02:10 AM
"Each molecule" has greater gravitational mass.
This is wrong. Each molecule weighs more. Weight is a function of mass (and g). Mass is not a function of weight. This is intuitive, as the weight of objects can change, even in our real constant G universe, while their mass (by definition) remains the same.

Perhaps talking about Jupiter is overcomplicating the issue. Instead of assuming we got G wrong, what would happen to mass and weight if G were to suddenly change?

W = mg
W = m * (GM/R^2)

One day the flying spaghetti monster, ruler of the universe, suddenly changes G to 5G, leaving everything else the same. What do we expect would happen? The force of gravity becomes stronger throughout the universe. Orbits change. Everything becomes heavier.

Nothing happens to mass. Looking at the equations above, if G increases W increases, as we would expect. There is no reason or justification to change m or M.
Is this really "wrong"? Think about it. Isn't this the real issue, where you say "tomayto" and I say "tomahto"?

You say the molecules don't change, while I say there are fewer molecules for the same mass of one kilogram. In suddenly changing the G to 5G, you immediately assume that weight is changed. I say not. The weight is the same, but there are five times fewer molecules to account for it. This is where the math is deceiving, because in:

W = mg
W = m * (GM/R^2)

the change in G*M appears normal, though it should not be. If G is greater, then M is lesser, meaning "fewer molecules", and that changes everything. It's not that the whole universe went to 5G, but that specifically Jupiter resides in 5G, while we're in 1G. So it is not "overcomplicating the issue", but rather stating the issue as it is being presented, that G varies with distance from the Sun, hypothetically.

There is a very simple way to resolve this, in my mind. Take a piece of matter one kilogram to Jupiter and weight it per its gravitational weight. It should weigh 5 times as much as here. If it does, then there's our answer to this whole conundrum. Or take a kilogram of mass from Jupiter and bring it here, and it should weigh 1/5 th of what we thought we had. This is the most direct way, forget reaction wheels, etc., to find out if G out there is really different from here. And if it's the same? Then you guys were right and I was way off, totally wrong!

If I'm wrong, nothing changes in cosmology, so we carry on as before. It also means Einstein had it right. But if I'm right, then we have to start thinking seriously of what it means to most of our astrophysical theory, and astronomy would be revolutionized by it.

Tensor
2005-Jul-22, 03:35 AM
There is a very simple way to resolve this, in my mind. Take a piece of matter one kilogram to Jupiter and weight it per its gravitational weight. It should weigh 5 times as much as here. If it does, then there's our answer to this whole conundrum. Or take a kilogram of mass from Jupiter and bring it here, and it should weigh 1/5 th of what we thought we had. This is the most direct way, forget reaction wheels, etc., to find out if G out there is really different from here.

Why forget reaction wheels etc? Those are direct measurements whether you want to acknowledge them as such or not. You may want to forget the reaction wheels, but you can't ignore the impulse supplied by the rockets to put them in orbit around Jupiter and Saturn. The length of the rocket firings to put those probes in orbit where calculated on the mass of the probes based on Earth G. If it is 5G at Jupiter, length of those firings would have been too short and the probes would never had made it into orbit. If you want to argue that the smaller mass at Jupiter allowed them to get into orbit, then you have to explain where 4/5 of the molecules went and also why the probes are still working when missing 4/5 of their molecules.

papageno
2005-Jul-22, 09:31 AM
I'm cutting it down some more, so focus on this part:

".. for each molecule 5 times the gravitational force, so it would take 1/5 th as many molecules.."
Our probes out there have not changed their inertial mass (tested with non-gravitational forces and through the moment of inertia), and it is clear they have not lost 4/5th of their molecules (otherwise they would not work).
Since they move and work as planned, your hypothesis fails the experimental test.

The catch word is "molecules", which is more than what you refer to in yours. "Each molecule" has greater gravitational mass. Yes, the weight results are the same, but for "fewer molecules" on Jupiter than Earth. All the other stuff on math etc, is superfluous if you can't get this one point, and no amount of math would make a lick of difference. :x
You have only vague hand-waving based on misunderstandings of gravity and dynamics, and misconceptions about wieght and mass.
If you propose a scientific hypothesis, you are supposed to provide some quantitative estimates.
That "other stuff on math" is what allows scientists to tell correct theories from incorrect ones.

papageno
2005-Jul-22, 09:45 AM
Is this really "wrong"? Think about it. Isn't this the real issue, where you say "tomayto" and I say "tomahto"?
No, you are simply wrong.

You say the molecules don't change, while I say there are fewer molecules for the same mass of one kilogram. In suddenly changing the G to 5G, you immediately assume that weight is changed. I say not. The weight is the same, but there are five times fewer molecules to account for it.
So, now you assume without any justification whatsoever that the amount of matter changes.
Do you realize that we can measure the amount of matter? (SI unit mole.)

This is where the math is deceiving, because in:

W = mg
W = m * (GM/R^2)

the change in G*M appears normal, though it should not be. If G is greater, then M is lesser,...
Not then.
You can say "then M is lesser" only if you assume that W does not change.
And you can do that only if your hypothesis worked like this: "Let's assume that G is greater, but the weight does not change. Then we have to assume that M changes...", at which point you have to justify why M changes, and not m, since the formulae are symmetric for the masses:
W = M*(Gm/R^2).

.... meaning "fewer molecules", and that changes everything.
And now you even change the amount of matter, without justification.
Why don't you consider non-material objects, such as photons?

It's not that the whole universe went to 5G, but that specifically Jupiter resides in 5G, while we're in 1G.
And you keep coming up with ad hoc assumptions, that are incompatible with well-established observations.

So it is not "overcomplicating the issue",...
Introducing assumptions without any justification is complicating the issue.

... but rather stating the issue as it is being presented, that G varies with distance from the Sun, hypothetically.
ANd we explained what would happen, hypothetically.

There is a very simple way to resolve this, in my mind. Take a piece of matter one kilogram to Jupiter and weight it per its gravitational weight.
Have you forgotten that we have probes in the outer Solar System?
Do you think that an increase in gravitational mass wouldn't have been notice in "sling-shots" and orbits?

It should weigh 5 times as much as here. If it does, then there's our answer to this whole conundrum.
We gave reference that answered the questions.
You chose to ignore them.

Or take a kilogram of mass from Jupiter and bring it here, and it should weigh 1/5 th of what we thought we had. This is the most direct way, forget reaction wheels, etc.,...
Let's not forget them, because they are valid methods to test your hyptothesis, whether you like it or not.

... to find out if G out there is really different from here. And if it's the same? Then you guys were right and I was way off, totally wrong!
We already have shown that you are wrong.
But you cannot accept it.

If I'm wrong, nothing changes in cosmology, so we carry on as before. It also means Einstein had it right. But if I'm right, then we have to start thinking seriously of what it means to most of our astrophysical theory, and astronomy would be revolutionized by it.
You are wrong, and research goes on without you.

pghnative
2005-Jul-22, 12:21 PM
I'm cutting it down some more, so focus on this part:
".. for each molecule 5 times the gravitational force, so it would take 1/5 th as many molecules.."

Trust me, I'm jiggy wit your theory. The product G*M remains the same. So a 5X increase in G means a corresponding 5X decrease in M.

All the other stuff on math etc, is superfluous if you can't get this one point, and no amount of math would make a lick of difference

sorry dude -- physics IS math, and no amount of essay on your part changes that. If you can't do math, your theory ain't worth the ones and zero's it's posted with.

Tassel
2005-Jul-22, 02:28 PM
You say the molecules don't change, while I say there are fewer molecules for the same mass of one kilogram. In suddenly changing the G to 5G, you immediately assume that weight is changed. I say not.
Pardon me? Weight wouldn't change? What exactly would be the observable change if G suddenly changed?

The weight is the same, but there are five times fewer molecules to account for it.
So are you saying that if we woke up one day and G here on Earth were 5 times greater, we would find the platinum and iridium bar (the very definition of 1 kilogram) in France would weigh the same because it magically lost molecules? Is this what you are saying?

This is where the math is deceiving
Math is never deceiving.

the change in G*M appears normal, though it should not be. If G is greater, then M is lesser, meaning "fewer molecules", and that changes everything.
So the Earth lost molecules? Where did they go?

It's not that the whole universe went to 5G, but that specifically Jupiter resides in 5G, while we're in 1G.
We're dealing with another, simpler example now. I'm sorry that you don't like the one I brought up, but you have to deal with it.

forget reaction wheels
I'm sure you'd like to, but as others have pointed out, you can't. They are evidence. You're supposed to be on a search for the "truth". Why would you want to "forget" perfectly valid evidence in your search?

If I'm wrong, nothing changes in cosmology, so we carry on as before. It also means Einstein had it right. But if I'm right, then we have to start thinking seriously of what it means to most of our astrophysical theory, and astronomy would be revolutionized by it.
False dichotomy. You are wrong, and scientists will continue to "think seriously" about and explore gravity and cosmology.

At this point it will be interesting to see if you are willing to be honest and acknowledge that you are wrong about "variable mass" in a "hypothetical" variable G universe.

nutant gene 71
2005-Jul-22, 09:01 PM
You say the molecules don't change, while I say there are fewer molecules for the same mass of one kilogram. In suddenly changing the G to 5G, you immediately assume that weight is changed. I say not.

Pardon me? Weight wouldn't change? What exactly would be the observable change if G suddenly changed?

I'm cutting it down some more, so focus on this part:
".. for each molecule 5 times the gravitational force, so it would take 1/5 th as many molecules.."

Trust me, I'm jiggy wit your theory. The product G*M remains the same. So a 5X increase in G means a corresponding 5X decrease in M.

This is where the math is deceiving, because in:

W = mg
W = m * (GM/R^2)

the change in G*M appears normal, though it should not be. If G is greater, then M is lesser,

Not then.
You can say "then M is lesser" only if you assume that W does not change.
And you can do that only if your hypothesis worked like this: "Let's assume that G is greater, but the weight does not change. Then we have to assume that M changes...", at which point you have to justify why M changes, and not m, since the formulae are symmetric for the masses:
W = M*(Gm/R^2).
These are all the same issue. If you could "weigh" Huygens sitting on Titan's surface, adjusted for the gravitational difference between Earth and Titan, it should weigh more than its kilograms on Earth, perhaps 10 times. Likewise, if you could weigh the rovers on Mars, they should weigh about 1.5 times their kilogram weight on Earth. If you could go and scoop from Jupiter's atmosphere 1 kilogram (Jupiter's) of gas and bring it back to Earth, it would only weigh about 1/5 th of what we'd expect when weighed on Earth. If you could bring back a scoop of soil from Mars and weigh it here on Earth, it would weigh about 2/3 of what it weighed on Mars. That's what I'm talking about.

Not that molecules somehow magically changed, but the same molecules would register different weights in different places. It's the mass that is variable in a variable G, then to have equal weights, you have to adjust the number of molecules. That's all I said.

Finally, regarding papageno's and others that the gravity assist and orbitals showed no difference for hypothetical different G, keep in mind what (I think it was Tassel?) had been said before, that an astronaut can fly in space next to the ISS though they have great difference in mass because gravity acts on mass equally. So if it is "mass" that is adjusted for variable G, as I had been arguing all along, then the weight differences are immaterial when it comes to orbital and gravity assist trajectories. It doesn't matter if the mass of Huygens is suddenly 10 times what it was on Earth, if in Saturn's domain of G the pull of gravity acts equally on mass. So if Huygens were the size of a bread box, or the size of a rail car, the gravity would still act the same. That's why we haven't seen appreciable differences in our orbital gravity assist trajectories.

The key to this understanding, essay words only, is that if mass changes with G, weight for the same mass is the same because the mass is adjusted. And if mass is adjusted, it doesn't matter what it is for orbital and orbit assist behaviors, they stay the same. This is per F = GMm/R^2, where for different G you get different M, keeping GM constant. For W = m*(GM/R^2), the GM is constant, so W is constant. For W =mg, and g = (GM/R^2), then any change in G will not affect W. Easy as pie! :)

The real question is: How can we find if G is different elsewhere? Should we bring back a weighed sample of soil from Mars and put it on an Earth scale to find out? Or should we send a spring scale to Mars and weigh what the landed probe weighs when there? If vairable mass in a hypothetical variable G works as I describe, then these experiments should yield positive proof. You may not agree with the expected results, but you must agree that this would be a valid test. If not, why not?

Last but not least:
Why forget reaction wheels etc? Those are direct measurements whether you want to acknowledge them as such or not. You may want to forget the reaction wheels, but you can't ignore the impulse supplied by the rockets to put them in orbit around Jupiter and Saturn. The length of the rocket firings to put those probes in orbit where calculated on the mass of the probes based on Earth G. If it is 5G at Jupiter, length of those firings would have been too short and the probes would never had made it into orbit. If you want to argue that the smaller mass at Jupiter allowed them to get into orbit, then you have to explain where 4/5 of the molecules went and also why the probes are still working when missing 4/5 of their molecules.

Readction wheels should exhibit anomalous acceleration in higher G domains, but my point was that other tests may be better, since only some of the wheels reacted funny. Rocket firings would appear to be affected by variable mass in variable G, or so it would seem intuitively. But we don't know this, since there could be factors of which we are still ignorant: rocket fuel mass density in higher G may give different burn, so reaction force is greater than in Earth's 1G domain, for example.

Regarding where did the 4/5 of molecules go? This is such a gross misunderstanding of what I said that I really don't feel any priority in answering it.

Tassel
2005-Jul-22, 09:35 PM
Not that molecules somehow magically changed, but the same molecules would register different weights in different places. It's the mass that is variable in a variable G, then to have equal weights, you have to adjust the number of molecules. That's all I said.

So are you saying that if we woke up one day and G here on Earth were 5 times greater, we would find the platinum and iridium bar (the very definition of 1 kilogram) in France would weigh the same because it magically lost molecules?

nutant gene 71
2005-Jul-22, 09:50 PM
Not that molecules somehow magically changed, but the same molecules would register different weights in different places. It's the mass that is variable in a variable G, then to have equal weights, you have to adjust the number of molecules. That's all I said.

So are you saying that if we woke up one day and G here on Earth were 5 times greater, we would find the platinum and iridium bar (the very definition of 1 kilogram) in France would weigh the same because it magically lost molecules?
I love this example! Actually, it wouldn't quite work out that way, hypothetically of course.

Taking the bar from one G to another changes its weight, as I showed above. To have same weight in different G domains, you need to adjust the molecules for same mass. So in your case, if G suddently changed on Earth from 1G to 5G, the same mass would weigh 5 times more. To have it weigh what it used to say, back to 1G from 5G (5 kg=> 1 kg), you'd have to take the bar and knock off 4/5 ths of its molecules, or cut it down to size.

Metricyard
2005-Jul-22, 09:56 PM
Not that molecules somehow magically changed, but the same molecules would register different weights in different places. It's the mass that is variable in a variable G, then to have equal weights, you have to adjust the number of molecules. That's all I said.

So are you saying that if we woke up one day and G here on Earth were 5 times greater, we would find the platinum and iridium bar (the very definition of 1 kilogram) in France would weigh the same because it magically lost molecules?
I love this example! Actually, it wouldn't quite work out that way, hypothetically of course.

Taking the bar from one G to another changes its weight, as I showed above. To have same weight in different G domains, you need to adjust the molecules for same mass. So in your case, if G suddently changed on Earth from 1G to 5G, the same mass would weigh 5 times more. To have it weigh what it used to way, back to 1G from 5G, you'd have to take the bar and knock off 4/5 ths of its molecules, or cut it down to size.

So did the Huygens probe have a built in cutting torch to remove parts of itself on it's descent to Titan?

Taking the bar from one G to another changes its weight, as I showed above. To have same weight in different G domains, you need to adjust the molecules for same mass. So in your case, if G suddently changed on Earth from 1G to 5G, the same mass would weigh 5 times more. To have it weigh what it used to way, back to 1G from 5G, you'd have to take the bar and knock off 4/5 ths of its molecules, or cut it down to size.

This is the first thing you've said that's made any sense. Read what you wrote carefully. You actually got it right.

nutant gene 71
2005-Jul-22, 10:03 PM
Not that molecules somehow magically changed, but the same molecules would register different weights in different places. It's the mass that is variable in a variable G, then to have equal weights, you have to adjust the number of molecules. That's all I said.

So are you saying that if we woke up one day and G here on Earth were 5 times greater, we would find the platinum and iridium bar (the very definition of 1 kilogram) in France would weigh the same because it magically lost molecules?
I love this example! Actually, it wouldn't quite work out that way, hypothetically of course.

Taking the bar from one G to another changes its weight, as I showed above. To have same weight in different G domains, you need to adjust the molecules for same mass. So in your case, if G suddently changed on Earth from 1G to 5G, the same mass would weigh 5 times more. To have it weigh what it used to way, back to 1G from 5G, you'd have to take the bar and knock off 4/5 ths of its molecules, or cut it down to size.

So did the Huygens probe have a built in cutting torch to remove parts of itself on it's descent to Titan?

To assume that you are not an embarrassment to your intelligence, I must conclude that you just missed this paragraph:

"Finally, regarding papageno's and others that the gravity assist and orbitals showed no difference for hypothetical different G, keep in mind what (I think it was Tassel?) had been said before, that an astronaut can fly in space next to the ISS though they have great difference in mass because gravity acts on mass equally. So if it is "mass" that is adjusted for variable G, as I had been arguing all along, then the weight differences are immaterial when it comes to orbital and gravity assist trajectories. It doesn't matter if the mass of Huygens is suddenly 10 times what it was on Earth, if in Saturn's domain of G the pull of gravity acts equally on mass. So if Huygens were the size of a bread box, or the size of a rail car, the gravity would still act the same. That's why we haven't seen appreciable differences in our orbital gravity assist trajectories."

Huygens landed with all its parts intact, as far as I know.

pghnative
2005-Jul-22, 10:03 PM
If you could "weigh" Huygens sitting on Titan's surface, adjusted for the gravitational difference between Earth and Titan, it should weigh more than its kilograms on Earth, perhaps 10 times.

Oh goody, some predictions.

OK, nutant, let's try this one more time. I would calculate the following:

1) mass of Huygens: 350 kg
2) mass of Titan: 1.35 e23 kg
3) radius of Titan: 2,580,000 m
4) G at Titan: 6.67e-11 m3 kg-1 s-2
5) "weight" (that is, "gravity force") of Huygens on Titan: 475 N

Can you tell us, if G is variable (let's say that at Titan it is 10 times it's value on Earth), what are the following values for:

1) mass of Huygens
2) mass of Titan
4) G at Titan
5) "weight" ("gravity force") of Huygens on Titan

nutant gene 71
2005-Jul-22, 10:23 PM
If you could "weigh" Huygens sitting on Titan's surface, adjusted for the gravitational difference between Earth and Titan, it should weigh more than its kilograms on Earth, perhaps 10 times.

Oh goody, some predictions.

OK, nutant, let's try this one more time. I would calculate the following:

1) mass of Huygens: 350 kg
2) mass of Titan: 1.35 e23 kg
3) radius of Titan: 2,580,000 m
4) G at Titan: 6.67e-11 m3 kg-1 s-2
5) "weight" (that is, "gravity force") of Huygens on Titan: 475 N

Can you tell us, if G is variable (let's say that at Titan it is 10 times it's value on Earth), what are the following values for:

1) mass of Huygens
2) mass of Titan
4) G at Titan
5) "weight" ("gravity force") of Huygens on Titan

Oh, it's been a long day, a long week, but here goes... yawhnnn...

Assuming Titan is (hypo) in 10 G:

1) mass of Hygens = 350 kg (no molecules chopped off)
2) mass Titan = 0.135E+23 kg
3) radius of Titan = 2.58E+6 m
4) G of Titan = 66.7E-11 Nm^2 kg^-2
5) "Weight" of Hyugens probe on Titan = 473.5 N

Here's the equation, with something lost in rounding off:

F = GMm/R^2

W = (66.7E-11 Nm^2 kg^-2) (0.135E+23 kg) (350 kg) divided by/(6.656E+12 m^2)

W = 473.5 N ... snorrre...zzzzz.

What, did someone wake me? I know I did say before:

"If you could "weigh" Huygens sitting on Titan's surface, adjusted for the gravitational difference between Earth and Titan, it should weigh more than its kilograms on Earth, perhaps 10 times."

But are these Earth kilograms, or Titan's "kilograms"? Back to the conundrum again, deja vu all over again. ;)

Tassel
2005-Jul-22, 11:20 PM
Taking the bar from one G to another changes its weight, as I showed above. To have same weight in different G domains, you need to adjust the molecules for same mass. So in your case, if G suddently changed on Earth from 1G to 5G, the same mass would weigh 5 times more. To have it weigh what it used to say, back to 1G from 5G (5 kg=> 1 kg), you'd have to take the bar and knock off 4/5 ths of its molecules, or cut it down to size.
Great, so the unmodified rod does weigh 5 times as much. G changes from G to 5G, and the rod goes from weighing 9.8N to 49N, which is what we'd expect.

Before G change:

W = m * (GM/R^2)
W = 1kg * (6.67E-11 * 5.97E+24 / 6.38E+06) = 9.8N

After G Change:

W = m * (GM/R^2)
W = 1kg * (3.34E-10 * 5.97E+24 / 6.38E+06) = 49N

Therefore, 1kg in 1G equals 1kg in 5G. Mass and kilograms do not vary.

nutant gene 71
2005-Jul-23, 01:40 AM
QUESTION OF MASS-KG?

You'all been asking me lots of questions, so now it's my turn.

For Titan's mass (1G) = 1.35E+23 kg = (10G) 0.135E+23 "kg", as per above:

(1) Remembering that Titan did not change shape or size, if Titan's mass is 1.35E+23 kg in Earth's 1G measure, what is the value of Titan's "kilograms" in its 10G equivalent 0.135E+23 "kg"?

(2) Imagine we sent miners to 10G Titan and they discover the planet is made of platinum and iridium. If they manufactured a pure bar on Titan (titanium2) that has same size and volume as an equivalent French prototype kilogram on Earth (titanium2 "kilogram"), what would be this bar's weight and mass on 1G Earth?

Hypothetically, of course.

pghnative
2005-Jul-23, 02:09 AM
[quote=pghnative]OK, nutant, let's try this one more time. I would calculate the following:

1) mass of Huygens: 350 kg
2) mass of Titan: 1.35 e23 kg
3) radius of Titan: 2,580,000 m
4) G at Titan: 6.67e-11 m3 kg-1 s-2
5) "weight" (that is, "gravity force") of Huygens on Titan: 475 N

Assuming Titan is (hypo) in 10 G:

1) mass of Hygens = 350 kg (no molecules chopped off)
2) mass Titan = 0.135E+23 kg
3) radius of Titan = 2.58E+6 m
4) G of Titan = 66.7E-11 Nm^2 kg^-2
5) "Weight" of Hyugens probe on Titan = 473.5 N

OK, so mass of Huygens is the same in our theory and yours.

And gravity force is the same in our theroy and yours.

That means that the acceleration of Huygens toward Titan is the same in our theory and yours.

So no difference in predictions between your theory and ours.

I think it should be clear to all what this thread is about. If G were discovered to be different at Saturn, then the mass of Saturn would have to be different than we think in order to explain how objects orbit Saturn at the speed we observe*.

Of course, if my brain is discovered to be a tuna sandwich, then biologists need to adjust their theories to explain why my heart is still beating.

---------------------------------------------------------------------------------
*Of course, there are subtle problems which nutant has not bothered to suss out. Specifically, since he agrees that G = 6.67e-11 m kg-1 s-2 at Earth, then surely he agrees that the mass of the Sun is 1.99e30 kg --- otherwise, how can he explain Earth's orbit?

But if the Sun is 1.99e30 kg, and G = 6.67e-10 at Saturn, shouldn't Saturn be traveling sqrt(10) times faster than it is?

How do you account for that in your theory, Nutant?

pghnative
2005-Jul-23, 02:52 AM
QUESTION OF MASS-KG?

You'all been asking me lots of questions, so now it's my turn.

For Titan's mass (1G) = 1.35E+23 kg = (10G) 0.135E+23 "kg", as per above:

(1) Remembering that Titan did not change shape or size, if Titan's mass is 1.35E+23 kg in Earth's 1G measure, what is the value of Titan's "kilograms" in its 10G equivalent 0.135E+23 "kg"?

(2) Imagine we sent miners to 10G Titan and they discover the planet is made of platinum and iridium. If they manufactured a pure bar on Titan (titanium2) that has same size and volume as an equivalent French prototype kilogram on Earth (titanium2 "kilogram"), what would be this bar's weight and mass on 1G Earth?

Hypothetically, of course.

Just for giggles, I'll give it a try. Easier question first.

(2) Imagine we sent miners to 10G Titan and they discover the planet is made of platinum and iridium. If they manufactured a pure bar on Titan (titanium2) that has same size and volume as an equivalent French prototype kilogram on Earth (titanium2 "kilogram"), what would be this bar's weight and mass on 1G Earth?
Mass would be 1 kg; Weight would be 9.8N.

(1) Remembering that Titan did not change shape or size, if Titan's mass is 1.35E+23 kg in Earth's 1G measure, what is the value of Titan's "kilograms" in its 10G equivalent 0.135E+23 "kg"?
If Titan's mass is 1.35e23 in Earth's 1G measure, then it's mass is 1.35e23 in its 10G environment. Titan would be travelling around Saturn at 3.16X it's current velocity. Titan and Saturn together would be travelling around the Sun at 3.16X their current velocity. If G really did vary with distance from the sun, then the orbit of Saturn around the sun would not be an ellipse, nor would any of Kepler's laws be valid.

I would ask that you answer my question about Saturn's orbit, from my previous post -- I really think that we are close to seeing each other eye-to-eye, even if we don't choose to agree with each other.

nutant gene 71
2005-Jul-23, 08:31 PM
Thanks for taking a stab at this question, pgh. It seems the others who had been posting are shy at the moment. Here is how I see it, and my answer to yours is in (3) below.
__________________________________________________ _______________________
RULE OF THUMB FOR VARIABLE MASS IN HYPO VARIABLE G, and why it does not show up in our space trajectories.

"The rule of thumb for variable mass in hypo variable G is that the kilograms for mass are taken from the G where it is measured."

(1) This is the only solution possible to resolve the variable mass cum variable G conundrum. Let's use the above example of Titan, which near Saturn has a hypothetical 10G. Knowing Titan's mass, as calculated in Earth's 1G, m = 1.35E+23 kg. And as figured in 10G, "m" = 0.135E+23 "kg", per the above F = 10G"/G (0.135E+23 kg)m/R^2. Taking a ratio of the two:

1.35E+23 kg/ kg = 0.135E+23 "kg"/ "kg" so that

kg = ("kg" * 1.35E+23 kg )/ 0.135E+23 "kg", which is kg = 10 "kg", or "kg" = 1/10 kg.

Remember that the two masses, per volume and size, are identical, so that

"m" = m, though they register different kilograms.

Therefore, if measuring mass in Earth's 1G, the kilogram used is Earth's kg; but if measuring mass from a hypothetical variable G, say 10G, then the kilograms used are that G's domain "kg", as shown above.

(2) The second question about a bar of platinum-iridium was more of a trick question, because the size and volume for the two bars may be different.

If I take 1 kg of platinum-iridium of the size and volume of the French prototype, I know it is one kilogram of mass. But I took the same size and volume, still the same mass, to a hypo 10G domain, it would be 10 times heavier within the gravity of that domain than in Earth's 1G.

Conversely, if I took a one "kilogram" of platinum-iridium from Titan, in a hypo 10G, it would be 10 times lower volume and size than what we had in Earth's 1G. Therefore, a 1 "kg" on Titan is smaller by volume and size than 1 kg on Earth. But if brought to Earth's 1G, that 1 "kg" from Titan would weigh 1/10 th of Earth's prototype bar. So it depends upon where the material is gathered. If gathered on Earth, 1 kg is of one size and volume; if gathered from Titan, then 1 "kg" is of a different size and volume. Given that we hypothetically gave Titan 10G, the volume and size is 1/10 th of Earth's kilogram. In effect, Titan's "kilogram" has only 10% of the molecules necessary to make one kilogram on Earth.

The weight of either kilogram or "kilogram" does not change. And this is why this is such a conundrum: If you measure the weight of a Titanian "kilogram" it will register the same Newtons as if weighed on Earth. Why not, if it is same mass? And if you take an Earthian kilogram to Titan, it will still weight the same Newtons. But they will be different in size and volume, and thus the kilogram mass from Earth will have a different weight from the "kilogram" mass from Titan. As we saw in the example above, once the kg-"kg" are adjusted for G, the Newtons remain the same. But this is contingent upon which measure of mass you used, whether kg or "kg" for the given volume and size of the block of material involved. The difference between the two is that a kilogram on Earth (low 1G) will be 10 times the size and volume of a "kilogram" on Titan (high 10G).

So the Huygens probe leaving Earth and arriving on Titan is still the same mass, and same weight as before when measured in Newtons. But conversely, if the "Huygens" were assembled on Titan, its "kilogram" mass measured in 10G would have made it 10 times heavier, using local "kilograms". So now instead of 350 kg, it would register on Titan as 3500 "kg" because the gravitational force is 10 times that of Earth's; remember that Titan's "kg" is 1/10 th of Earth's kg, so 350 kg is not 35 "kg" (as one would suppose), but 10 times kg: i.e., you need 10 times as many "kg" to arrive at 350 kg. Consequently, when this 3500 "kg" probe landed on Earth, it would show as 350 kg in our 1G domain, as gravity went from 10G to 1G. However, the Newton force had never changed, only the gravitational effect of where the probe was, (as it went in reverse from Titanian "1G" to Earthian 1/10 th "G"). Whether measured in Earth's kilograms or Titan's "kilograms" we will have an equivalent weight in their native measures of mass. And!.. (papageno pay attention here) it is for this reason that the Equivalence Principle is never violated, regardless of G. It is always in relation to the G where it is located.

(3) So why doesn't this show up in our orbitals or trajectories? Does it even matter as some suggested, that it is a null sum game? I think it does not show up, but it does matter.

G*M is what always rules, that the product of any (hypo) G adjusted for its native kilograms will always be the same. The Sun has an Earth derived mass of M = ~2E+30 kg (assuming Sun has 1G, which may not be true), so Saturn's mass (in hypo 10G) still answers to the inverse square law 1/R^2, because the product of G*M is still the same. And the same for the Sun responding to Saturn's gravitational attraction, where (10G*0.10m) is still the same product, so nothing changed with regards to their orbital relationship, the barycenter is unaffected, and neither would be their LaGrange points. I realize this is a pernicious problem, but a variable G is a very well hidden phenomenon, hypothetically. Trajectories, especially those affected by gravitation assist maneuvers, would act in the same fashion, virtually unaffected.

The only way this hypo variable G would show up would be in two ways: 1) in how mass sticks together in a higher or lower G: Mercury has to be heavy metal, because lighter substances in its very light G (close to the Sun gives it ~0.4G) so more volatile materials are blown off by the solar wind. And the very large gas giants may be very small in terms of size and volume, but because of their very great G, they act as if they were very large, and thus retain vast atmospheres. 2) The other effect would be on an "escape velocity" spaceprobe, since as it travels from light G to progressively higher G on its way out of the solar system, it is behaving as if it were gaining "kg" mass. In fact, nothing changed in its mass, except how it is affected by the Sun's very distant but massive gravity: it's progressively pulling it back into the Sun as its "equivalent" inertial mass is growing with G.

All else remains equal. I hope this meets with your satisfaction, but if not, sound off. :)

BTW, this above reasoning implies that the reason Kepler's law of elliptical orbits works is because the inertial mass varies with G, so that it is light and faster nearer the Sun, but inertially "heavier" and slower away from it. The fact that some moons also display some elliptical orbits, especially if close to the planet, then also implies that the planet is "hot" in its radiation of electromagnetic energy, if G and em energy are inversely proportional as I suspect, hypothetically. But that is another subject all together, and regrettably I have not worked out the math.

BTW2, pgh, I think you have a typo in yours:

"If Titan's mass is 1.35e23 in Earth's 1G measure, then it's mass is 1.35e23 in its 10G environment."

Did you not mean to say that it's mass in 10G environment its mass "m" = 0.135E+23 "kg"?

Tensor
2005-Jul-24, 04:19 AM
Last but not least:
Why forget reaction wheels etc? Those are direct measurements whether you want to acknowledge them as such or not. You may want to forget the reaction wheels, but you can't ignore the impulse supplied by the rockets to put them in orbit around Jupiter and Saturn. The length of the rocket firings to put those probes in orbit where calculated on the mass of the probes based on Earth G. If it is 5G at Jupiter, length of those firings would have been too short and the probes would never had made it into orbit. If you want to argue that the smaller mass at Jupiter allowed them to get into orbit, then you have to explain where 4/5 of the molecules went and also why the probes are still working when missing 4/5 of their molecules.

Readction wheels should exhibit anomalous acceleration in higher G domains, but my point was that other tests may be better, since only some of the wheels reacted funny.

For Cassini, it malfunctioned only for a short time and only for low speeds. It was suspected that the low speed did not provide enough lubricant for the wheel. After the wheel was run at high speed, the problem went away and has worked fine since. The performace of those wheels, before and after the one apparently mechanical problem, are ample evidence that the mass of Cassini has not changed at a distance away from G measured at Earth.

Rocket firings would appear to be affected by variable mass in variable G, or so it would seem intuitively. But we don't know this, since there could be factors of which we are still ignorant: rocket fuel mass density in higher G may give different burn, so reaction force is greater than in Earth's 1G domain, for example.

So now you have to have another unobserved postulated effect, just to get your idea to match observations.

Regarding where did the 4/5 of molecules go? This is such a gross misunderstanding of what I said that I really don't feel any priority in answering it.

Well, I apologize for misunderstanding, but this quote by you seems to claim that there is mass missing. If the mass is missing, you have to have 4/5 of it gone to have a G*M match in a 5G environment. So, exactly how should the following be understood?

You say the molecules don't change, while I say there are fewer molecules for the same mass of one kilogram.

Fewer molecules sure sounds like some are missing. Either the molecules change or mass is disappearing. Which is it? Not to mention, if they are disappearing, when Cassini used Venus for a gravity slingshot, the mass of Cassini had to increase (and the molecules seemingly coming from somewhere) and somehow decrease as it moved out farther from the earth. If they are changing, you still have the problem of mass either appearing or disappearing.

The problems with your idea are getting more and more convoluted. You now have G changing, M changing, and you now have to have some new phenomena to account for a chemical reactions that changes depending on G, when gravity is usually ignored when computing chemical reactions. All of this just to match observations that are consistant with a non-variable G. Is it really that hard to understand why most of us here aren't buying it?

nutant gene 71
2005-Jul-24, 02:24 PM
nutant gene 71 wrote:
Rocket firings would appear to be affected by variable mass in variable G, or so it would seem intuitively. But we don't know this, since there could be factors of which we are still ignorant: rocket fuel mass density in higher G may give different burn, so reaction force is greater than in Earth's 1G domain, for example.

So now you have to have another unobserved postulated effect, just to get your idea to match observations.
I'm looking into this right now and searching for clues of how would this work if a variable G were to be found real, which we don't have yet any reason to assume it is, so all hypothetical. So I'm checking out what I can find, rounding up the usual suspects, of what happens to electrons in a hypo variable G scenario. At first glance, it would appear that the proton mass, which really is the bulk of an atom's mass, would be affected, so in a higher G, for instance, the proton mass would increase. But I have no definitive carry over to proton's positive charge, which might affect electron velocity, hence energy, since the positive-negative charges have to remain equal, and only way I can figure it is that if proton's gravitational increases, the electron is put into higher velocity within its shell. We're talking here about a very minute change in variable G, still within domain of E-11 N.., but small within solar system, so effect of increased electron energy may not be big enough to matter? Then you have to consider the quantum effect of electron's energy, so a lot to look at. My primary suspect, at the moment in this police work, is: F = k(q1q2)/ R^2, which for the atom translates into: F = ke^2/R^2, where ke^2 = 2.307E-28 J.m, but have nothing else for now. I don't know if this trail is hot or not, but will keep looking into it. However, obviously, we know everything works as calculated regarding rocket thrust and electronics, and likely also spectrography, so these may be indicative of either of two possibilities: either there is no variable G, so the whole thing is an exercise in fantasy; or there is a variable G, but it does not translate into different kinetic and electronic mechanics, and we'd have to figure why not. Third possibility is that difference only show up in very high G, like at a neutron's star's level... or we have to look for these at the first nanoseconds of the Big Bang. :lol:

No, my idea is not more convoluted, though it remains hypothetical. But the way you see it is becoming something quite strange indeed. Thanks for even considering it, but these are mere possibilities for now, not real science or astronomy as we know it. But if we find evidence that G varies... ;)

Tassel
2005-Jul-24, 07:17 PM
If I take 1 kg of platinum-iridium of the size and volume of the French prototype, I know it is one kilogram of mass. But I took the same size and volume, still the same mass, to a hypo 10G domain, it would be 10 times heavier within the gravity of that domain than in Earth's 1G.
Therefore, 1kg in 1G is equivalent to 1kg in 10G. No need for "variable mass".

Conversely, if I took a one "kilogram" of platinum-iridium from Titan, in a hypo 10G, it would be 10 times lower volume and size than what we had in Earth's 1G.
What is your justification for the above statement? Mass is not a function of volume. How would you determine that you've collected 1kg of platinum-iridium on Titan, other than to compare to the platinum-iridium rod in France?

nutant gene 71
2005-Jul-25, 02:57 AM
If I take 1 kg of platinum-iridium of the size and volume of the French prototype, I know it is one kilogram of mass. But I took the same size and volume, still the same mass, to a hypo 10G domain, it would be 10 times heavier within the gravity of that domain than in Earth's 1G.
Therefore, 1kg in 1G is equivalent to 1kg in 10G. No need for "variable mass".

Conversely, if I took a one "kilogram" of platinum-iridium from Titan, in a hypo 10G, it would be 10 times lower volume and size than what we had in Earth's 1G.
What is your justification for the above statement? Mass is not a function of volume. How would you determine that you've collected 1kg of platinum-iridium on Titan, other than to compare to the platinum-iridium rod in France?
For starters, you might use the Equivalence Principle to measure off 1 "kg" on Titan. Then you might back that up with a spring scale set at 1 kg. Then count the molecules, or bring this new "kilogram" bar back to Earth and see what it is in our 1G, and compare its size and volume with the French bar.

I know, I know, the springs on Huygens.. etc. But don't put too much weight on that, since all those springs needed to do was RELEASE Huygens from Cassini, and then power Cassini out of there so it too wouldn't fall in to Titan. Once released, gravity did the rest, and Huygens made it into Titan's thick atmosphere, drag slowdown, parachute released, and it wobbled down to the surface... just as our friend Jerry described it. :)

Tassel
2005-Jul-25, 04:54 AM
For starters, you might use the Equivalence Principle to measure off 1 "kg" on Titan.
How would that work, exactly?

Then you might back that up with a spring scale set at 1 kg.
Spring scales measure weight (Newtons). How would I know when to stop putting matter on the scale? How many Newtons would the scale read when I had loaded 1kg on it and why?

Then count the molecules
If I count the number of molecules in the rod in France and find it is made up of X molecules, how many molecules would I count out on Titan to make a kilogram? X? If not X, why not?

or bring this new "kilogram" bar back to Earth and see what it is in our 1G, and compare its size and volume with the French bar.
Mass is not a function of volume. Many materials expand or contract (change volume) based on temperature. Obviously they are not gaining or losing mass based on temperature.

I know, I know, the springs on Huygens.. etc. But don't put too much weight on that, since all those springs needed to do was RELEASE Huygens from Cassini
Why not? If the release did not work as planned I'm certain you'd point it out and imply it's evidence Newtonian physics is all wrong. Why doesn't the fact that the release was flawless count against your "variable mass" idea?

pghnative
2005-Jul-25, 12:51 PM
(3) So why doesn't this show up in our orbitals or trajectories? Does it even matter as some suggested, that it is a null sum game? I think it does not show up, but it does matter.

G*M is what always rules, that the product of any (hypo) G adjusted for its native kilograms will always be the same. The Sun has an Earth derived mass of M = ~2E+30 kg (assuming Sun has 1G, which may not be true), so Saturn's mass (in hypo 10G) still answers to the inverse square law 1/R^2, because the product of G*M is still the same. And the same for the Sun responding to Saturn's gravitational attraction, where (10G*0.10m) is still the same product, so nothing changed with regards to their orbital relationship, the barycenter is unaffected, and neither would be their LaGrange points.
nutant
I'm sorry, but I don't understant your words. Let's talk again in numbers, if we could.

I would calculate the following:

1) Sun Mass: ~2 e 30 kg
2) G = 6.67e-11 m3 kg-1 s-2
3) Saturn gravitational mass: 5.7 e26 kg
4) Saturn inertial mass: 5.7e26 kg
5) Saturn dist from Sun (assume circle for now): 1.43 e12 m
6) Saturn "weight" (gravity force toward sun) = 3.7e22 N
7) Saturn acceleration toward sun: 6.52 m/s2

Could you calculate these seven numbers in your theory:
1) Sun Mass (I believe you said you agree that it's ~2 e 30 kg)
2) G : (I believe you said this is 66.7e-11, or 6.67e-10m3 kg-1 s-2)
3) Saturn gravitational mass
4) Saturn inertial mass
5) Saturn dist from Sun (assume circle for now)
6) Saturn "weight" (gravity force toward sun)
7) Saturn acceleration toward sun

BTW2, pgh, I think you have a typo in yours

"If Titan's mass is 1.35e23 in Earth's 1G measure, then it's mass is 1.35e23 in its 10G environment."

Did you not mean to say that it's mass in 10G environment its mass "m" = 0.135E+23 "kg"?No, that wasn't a typo --- you see, I still am not seeing why mass would change.

nutant gene 71
2005-Jul-25, 03:38 PM
No, that wasn't a typo --- you see, I still am not seeing why mass would change.
"Nough said. I've given you all the information you need to show how this works. I will reply only to specific errors in my math, or my reasoning. Work out your own math for any of the above. Please review my posts if you have further questions. Your repeating your questions, and I will not keep repeating my answers. Final word.

pghnative
2005-Jul-25, 03:52 PM
No, that wasn't a typo --- you see, I still am not seeing why mass would change.
"Nough said. I've given you all the information you need to show how this works. I will reply only to specific errors in my math, or my reasoning. Work out your own math for any of the above. Please review my posts if you have further questions. Your repeating your questions, and I will not keep repeating my answers. Final word.
The specific errors in your math have been pointed out quite ably by papageno. The key error is that you claim that if G is variable then the math shows that mass is also variable. Papageno correctly points out that they are two independant assumptions.

My questions were not repeats. They look similar, but the first set was designed to determine how, in your theory, mass, weight and acceleration are calculated for Objects around Titan in a (postulated) 10G environment. The second set of questions was designed to determine how, in your theory, mass and acceleration of Saturn around the Sun is calculated.

It should quickly become obvious that in order to correctly calculate Saturn's orbit around the Sun, you (in your theory) need to use the mass of the sun (~2e30 kg) and G = 1G (6.67 e-11 m3 kg-1s-2). Which means that an object at Saturn's orbit respond to the Sun's gravity as if G = 1G, but responds to Saturn's gravity as if G = 10G. This is patently absurd.

Final word.somehow, I suspect this isn't true either.

Tassel
2005-Jul-25, 04:21 PM
"Nough said. I've given you all the information you need to show how this works. I will reply only to specific errors in my math, or my reasoning.
What about actually providing some math that shows the necessity/correctness of your "variable mass" idea? In spite of your efforts, there hasn't been a single calculation (using correct equations...) in this thread that shows "constant mass" failing in a variable G scenario.

And I suppose I'm not going to get an answer to my questions as to how you would suggest we measure 1kg of mass in 10G on Titan in your "variable G/variable mass" universe.

We've gone over every conceivable "hypothetical" variation of location and value for G in the solar system, and yet we have never had to redefine the kilogram to model what you say we should observe in a variable G universe...even when you've changed your mind on what you think we'd observe! :)

nutant gene 71
2005-Jul-25, 08:17 PM
"Nough said. I've given you all the information you need to show how this works. I will reply only to specific errors in my math, or my reasoning.
What about actually providing some math that shows the necessity/correctness of your "variable mass" idea? In spite of your efforts, there hasn't been a single calculation (using correct equations...) in this thread that shows "constant mass" failing in a variable G scenario.

And I suppose I'm not going to get an answer to my questions as to how you would suggest we measure 1kg of mass in 10G on Titan in your "variable G/variable mass" universe.

We've gone over every conceivable "hypothetical" variation of location and value for G in the solar system, and yet we have never had to redefine the kilogram to model what you say we should observe in a variable G universe...even when you've changed your mind on what you think we'd observe! :)

See this post, July 9, 2005 (http://www.badastronomy.com/phpBB/viewtopic.php?p=500623#500623), pg. 2 of this thread.

pghnative
2005-Jul-25, 08:42 PM
"Nough said. I've given you all the information you need to show how this works. I will reply only to specific errors in my math, or my reasoning.
What about actually providing some math that shows the necessity/correctness of your "variable mass" idea? In spite of your efforts, there hasn't been a single calculation (using correct equations...) in this thread that shows "constant mass" failing in a variable G scenario.

And I suppose I'm not going to get an answer to my questions as to how you would suggest we measure 1kg of mass in 10G on Titan in your "variable G/variable mass" universe.

We've gone over every conceivable "hypothetical" variation of location and value for G in the solar system, and yet we have never had to redefine the kilogram to model what you say we should observe in a variable G universe...even when you've changed your mind on what you think we'd observe! :)

See this post, July 9, 2005 (http://www.badastronomy.com/phpBB/viewtopic.php?p=500623#500623), pg. 2 of this thread.
Yes, I can see that post. That post is effectively debunked by

a) my analogous post regarding Saturn above, which demonstrates that you need G to be = 1G at Saturn's orbit to correctly calculate its orbital velocity.
b) Tassel's subsequent post on pg. 2 showing that you used the wrong mass in the equation "GM = rv2". You used M = mass of jupiter; in reality, M=mass of sun in that equation.

Note that we've pointed out an error in your mathematics, as per:

I will reply only to specific errors in my math, or my reasoning. So would you now be able to answer my questions about Saturn's orbit? If you do, it will be patently obvious where your mathematical errors are.

nutant gene 71
2005-Jul-26, 02:37 AM
"Nough said. I've given you all the information you need to show how this works. I will reply only to specific errors in my math, or my reasoning.
What about actually providing some math that shows the necessity/correctness of your "variable mass" idea? In spite of your efforts, there hasn't been a single calculation (using correct equations...) in this thread that shows "constant mass" failing in a variable G scenario.

And I suppose I'm not going to get an answer to my questions as to how you would suggest we measure 1kg of mass in 10G on Titan in your "variable G/variable mass" universe.

We've gone over every conceivable "hypothetical" variation of location and value for G in the solar system, and yet we have never had to redefine the kilogram to model what you say we should observe in a variable G universe...even when you've changed your mind on what you think we'd observe! :)

See this post, July 9, 2005 (http://www.badastronomy.com/phpBB/viewtopic.php?p=500623#500623), pg. 2 of this thread.
Yes, I can see that post. That post is effectively debunked by

a) my analogous post regarding Saturn above, which demonstrates that you need G to be = 1G at Saturn's orbit to correctly calculate its orbital velocity.
b) Tassel's subsequent post on pg. 2 showing that you used the wrong mass in the equation "GM = rv2". You used M = mass of jupiter; in reality, M=mass of sun in that equation.

Note that we've pointed out an error in your mathematics, as per:

I will reply only to specific errors in my math, or my reasoning. So would you now be able to answer my questions about Saturn's orbit? If you do, it will be patently obvious where your mathematical errors are.

RE: GM = Rv^2 for Saturn, it would be the same methodology as worked out for Jupiter here (http://www.badastronomy.com/phpBB/viewtopic.php?p=501708#501708), July 11, 2005, pg. 3 of this thread.

We know GM = rv^2 is a short form, a reduced simplified equation. We also know that what's on the left has to be equivalent to what's on the right, which is what an equation equals. One way to see this is to reconstruct from whence this simplified equation came from. We know the following:

F = GMm/ r^2, and we know F = ma, which is also F = M*(Gm/r^2), where a = (Gm)/ r^2, but we can also see Newton's orbital equation one more way, as:

F = M*(Gm)/r^2 = mv^2/ r.

I showed the M separate, since it represents the Sun, a fixed value of mass, unaffected by G. The (Gm) is together because these are the variables, hypothetically, that need to be addressed. And on the right (mv^2/r) is a version of the same equation. When M*(Gm)/r^2 = mv^2/r is netted out, it gives us:

GM = rv^2

(Please note that in the "short form", big M is now the planet, since the Sun's mass drops out from the "long form".) --this is wrong, M remains Sun's mass, will redo below to correct error. GM=rv^2 cannot be used for variable G, since Sun's G and M are invariable.--n.g.71

Taking Saturn's mass and distance from the Sun, adjusted for hypo 10G at Saturn:

Remember (critical point) neither Saturn's orbit nor its mass are changed, only how we measure the mass in terms of an increased G. So let's plug in some numbers for Saturn as we know them in (Earth's) 'universal constant' G, where G = 6.67E-11 m^3 kg^-1 s^-2 and Saturn's m = 5.685E+26 kg:

M * (Gm)/r^2 = mv^2/r, so that M*(6.67E-11 m^3 kg^-1 s^-2)(5.685E+26 kg)/ r^2 = m v^2/r, which is M*(37.92E+15 m^3 s^-2)/r^2 = m v^2/r

(Remember the right side is in Earth units, so does not change.)

Now let's plug in the new Saturn's G', tenfold, approxmately G' = ~66.7E-11 Nm^2 kg^-2:

M*[(~66.7E-11 m^3 kg^-1 s^-2) * m'] /r^2 = m v^2/r, and to conserve the product (G'm') = 37.92E+15 m^3 s^-2, divided by 66.7E-11 kg, m' becomes = 0.5685E+26 kg (which is one tenth the mass of what Saturn was in a G = 6.67E-11 N.. equation).

M, r^2, and mv^2/r remain the same as before. So we don't see any difference in Saturn's orbit with a different G. You cannot tell from the orbital "short form" GM=rv^2 because adjusted G' and m' are not shown, as they were canceled out in a 1G universal constant scenario. If you want to continue and plug in the M, r, r^2, in the "long form" you'll still get the same results as before.

The same applies to all the other values, since they are all derived from Earth's observations, so M and mv^2/r remain the same, figured in Earth units. Only (G*m) are adjusted for each other, while the Sun's "m" on the right side remains the same as it had been figured in the original. Why? Because all our math is derived from Earth based units of kilogram mass (as well as meters and seconds). If you were using Saturn based "kilograms", then you would not need to adjust (G*m) since it would all be figured "as if" that was the "universal constant" in its own kilograms. The orbital equation would still yield same, given the new units for G' and m'. It's just we seem to like Earth's G units for mass better, for better or worse. Consequently, our inherent "terracentrism" has given us a bias to believe G is a universal constant. ;)

So the new orbital would be, for Saturn at hypo 10G:

GM = rv^2, plugging in the adjusted G and M (which is now Saturn):

(66.7E-11 m^3 kg^-1 s^-2)(0.5685E+26 kg) = rv^2

37.92E+15 m^3 s^-2 = rv^2.

Like I said before, you have all you need to figure this out. Whether or not you do, or whether you believe it, is up to you.

[Edited, for explanation of M vs. m, as Sun's mass, cannot use short form of G*M, also typo errors.]

Tassel
2005-Jul-26, 04:22 AM
M*[(~66.7E-11 m^3 kg^-1 s^-2) * m'] /r^2 = m v^2/r, and to conserve the product (G'm') = 37.92E+15 m^3 s^-2, divided by 66.7E-11 kg, m' becomes = 0.685E+26 kg (which is one tenth the mass of what Saturn was in a G = 6.67E-11 N.. equation).

Well, m' is 5.685E+25kg, but other than that, this is pretty much right.

The problem is, they aren't "Saturn Kilograms". They're just kilograms. If we were wrong about G at Saturn, and it is 10x higher than we think, then we have overestimated Saturn's mass by 10x. So, we recalculate Saturn's mass and it's 5.685E+25kg and we're done. We can now use that number for Saturn's mass in any equation we like.

pghnative
2005-Jul-26, 12:40 PM
RE: GM = Rv^2 for Saturn, it would be the same methodology as worked out for Jupiter here (http://www.badastronomy.com/phpBB/viewtopic.php?p=501708#501708), July 11, 2005, pg. 3 of this thread.

We know GM = rv^2 is a short form, a reduced simplified equation. We also know that what's on the left has to be equivalent to what's on the right, which is what an equation equals. One way to see this is to reconstruct from whence this simplified equation came from. We know the following:

F = GMm/ r^2, and we know F = ma, which is also F = M*(Gm/r^2), where a = (Gm)/ r^2, but we can also see Newton's orbital equation one more way, as:

F = M*(Gm)/r^2 = mv^2/ r.

I showed the M separate, since it represents the Sun, a fixed value of mass, unaffected by G. The (Gm) is together because these are the variables, hypothetically, that need to be addressed. And on the right (mv^2/r) is a version of the same equation. When M*(Gm)/r^2 = mv^2/r is netted out, it gives us:

GM = rv^2

(Please note that in the "short form", big M is now the planet, since the Sun's mass drops out from the "long form".)

I'm sorry, nutant, but this is where you go wrong. The Sun's mass does not drop out of the equation. It is Saturn's mass which drops out. See below.

So the new orbital would be, for Saturn at hypo 10G:

GM = rv^2, plugging in the adjusted G and M (which is now Saturn):

(66.7E-11 m^3 kg^-1 s^-2)(0.5685E+26 kg) = rv^2

37.92E+15 m^3 s^-2 = rv^2.

OK, so let's solve for Saturn's orbital velocity. On average r = 1.4e12 m for saturn (see here (http://www.nineplanets.org/saturn.html).

Using your calculated value for GM (37.9e15 m3s-2) this gives v = 165 m / s, equivalent to an orbital period of about 1700 Earth years. This is wrong. Average orbital velocity for Saturn is ~9800 m/s for an orbital period of about 29 years.

nutant gene 71
2005-Jul-26, 05:15 PM
Right, GM = rv^2 does not work for a variable G, since M is the Sun's mass, which is invariable. So the "short form" as worked out above cannot be used for variable G.

I will go back into my above to work out the "long form" where a variable G shows up. As Sun is a fixed point of reference M, variable G does not apply. I'll get back when have some time.

nutant gene 71
2005-Jul-26, 06:07 PM
CORRECTED, long form for Saturn:

F = M(G*m)/ R^2 = mv^2/R

M(66.7E-11 m^3kg^-1s^-2)(0.5686E+26kg) = R(5.685E+26kg)v^2

M(37.92E+15 m^3 s^-2)/ (5.685E+26kg) = Rv^2

M(6.67E-11 m^3 kg^-1 s^-2) = Rv^2

this is the correct representation of variable G for Saturn using the long form. Notice the right side mass for Saturn is in Earth kilograms in 1G (must equal the left side of equation), while left side is Saturn kilograms in 10G. Sir Isaac, that clever fellow, simply netted this out using a constant G, so he could make orbital equation in the short form.

If G proves to be variable, we are stuck using the long form. Same results, different conditions, which did not exist in Newtons days. Since Sir Isaac knew nothing of gas giant atmospheres, or Pluto and Enceladuss atmospheres, or netron stars super gravity, nor of the Pioneers anomaly, though I am sure he would have relished 21st century astronomical data, he had little reason to tamper with his universal G. Given what we know already, and yet to discover, there is no need to keep ourselves stuck in the 17th century forever, especially if G is found to vary in the universe.

pghnative
2005-Jul-26, 06:22 PM
CORRECTED, long form for Saturn:

F = M(G*m)/ R^2 = mv^2/R

M(66.7E-11 m^3kg^-1s^-2)(0.5686E+26kg) = R(5.685E+26kg)v^2

M(37.92E+15 m^3 s^-2)/ (5.685E+26kg) = Rv^2

M(6.67E-11 m^3 kg^-1 s^-2) = Rv^2
.
edited whole message to reword:
Gee -- funny how that works. By twisting and turning your theory (and violating the equivalence principle in equation 2, by the way) you ultimately provide to us OUR equation, which is that
M-sun * G = r(saturn) * v(saturn)^2

Where G = a nicely constant 6.67e-11 m3kg-1s-2.

edited a second time to remove editorial comment

nutant gene 71
2005-Jul-26, 09:08 PM
(and violating the equivalence principle in equation 2, by the way)

Not so.

1 Saturn kg (@10G) = 10 Earth kg @1G), where each molecule in Saturns 10G has ten times the gravitational draw and inertial mass of an Earths 1G molecule. This is a fundamental tenet of variable G, equivalence is never broken. Saturns 0.5685E+26 kg is still equivalent to Earths 5.686E+26 kg; it would take the same force to move 0.10 kg on Saturn as it would take to move 1.0 kg on Earth. But if you want to move 1.0 kg on Saturn, youd better be ready with 10 times the force.

And yes, as I explained before, the orbital parameters are not affected, nor is the inverse square law violated. The only difference is that it would take fewer molecules on Saturn to make the same mass on Earth. If you take 10% of platinum-iridium from Saturn and test it against the Equivalence Principle, you would discover that in 10G, it behaves as if it were 100% of Earths equivalent platinum-iridium matter. The molecules were never lost, but were never needed in the first place.

A natural question arises: What is the best way to test for variable G, using the Equivalence Principle, away from our 1 AU?

Can we go to Mars, at 1.5 AU, and see a measurable difference in G? What kind of experiments should be designed to test for this. Assuming we cant bring back samples (at this time) and can only use remote sensing instruments to register anomalous G readings, what instruments should be sent over there?

Thats the next level.

pghnative
2005-Jul-26, 10:05 PM
(and violating the equivalence principle in equation 2, by the way)
Saturns 0.5685E+26 kg is still equivalent to Earths 5.686E+26 kg;
0.568 does not equal 5.68.

For your equation to be valid, the units must be different.

For instance, 1 does not equal 2.2. But 1 kg =~ 2.2 pounds.

So what you are saying is that all 8 pages of this thread is about unit conversion.

Hey everyone! I can prove G is variable. On Earth it is 6.67e-11. On Jupiter it is 1.57e+03.

(that would be in units of feet cubed per pound per fortnight squared)

frogesque
2005-Jul-26, 10:09 PM
nutant gene 71 wrote:

A natural question arises: What is the best way to test for variable G, using the Equivalence Principle, away from our 1 AU?

'Twas done when Hugens and Cassini separated as expected using springs, masses and moments of inertia calibrated here on Earth. I know you refuse to accept this as one item of evidence (amongst a host of others such as reaction wheel behavior) but that does not negate the evidence that for all practical purposes of interplanetary work within the solar system G can be regarded as constant.

nutant gene 71
2005-Jul-27, 08:14 PM
Whats wrong with this picture?
Here are some of our solar systems moons, with their radii, and whether or not they have atmospheres:

Big moons:
Earth Moons radius: 1,737 km  has virtually no atmosphere (http://www.solarviews.com/eng/moon.htm)

Saturn's Titan: 2,575 km  has thick atmosphere (http://www.solarviews.com/eng/titan.htm), at 1.6 bar
Jupiter's Io: 1,821 km  outgassing volcanoes (http://www.solarviews.com/eng/io.htm), but no atmosphere
Jupiter's Europa: 1,560 km  possible volcanis (http://www.solarviews.com/eng/europa.htm), but no atmosphere?
Neptunes Triton: 1,350 km - has thin atmosphere (http://www.solarviews.com/eng/triton.htm), with clouds

Small moons with atmospheres detected:
Pluto's Charon: ~505 km  may have atmosphere (http://www.space.com/scienceastronomy/050726_charon_pluto.html), still unconfirmed

Small planets:
Mercury: 2,440 km  has virtually no atmosphere (http://www.solarviews.com/eng/mercury.htm)
Pluto: 1,151 km  has surprising atmosphere, global warming (http://www.space.com/scienceastronomy/pluto_warming_021009.html) as planet increases distance from the Sun -??

Something is wrong here, none of this makes any consistent sense, where some bodies of greater density and size have virtually no atmospheres, while others (mostly in outer solar system) with smaller size and lower density have atmospheres. Is astronomy sweeping something under the atmosphere rug here?

Why dont our Moon, Mercury, and Io have atmospheres, but Pluto and Enceladus do? (Pluto is smaller with lower density than our Moon. Enceladus is off the charts!) Or why Titan, slightly larger than Mercury, has an atmosphere thicker than Earths?

Or is the outgassing explanation for why small bodies have an atmosphere merely ad hoc, and perhaps are we witnessing circumstantial evidence of a greater G out there?.. and just not talking about it?

What are the official scientific explanations for these anomalous atmospheres?

pghnative
2005-Jul-27, 09:09 PM
Temperature.

Celestial Mechanic
2005-Jul-27, 09:16 PM
Whats wrong with this picture?
[Snip!]
Something is wrong here, none of this makes any consistent sense, where some bodies of greater density and size have virtually no atmospheres, while others (mostly in outer solar system) with smaller size and lower density have atmospheres. Is astronomy sweeping something under the atmosphere rug here?

Why dont our Moon, Mercury, and Io have atmospheres, but Pluto and Enceladus do? (Pluto is smaller with lower density than our Moon. Enceladus is off the charts!) Or why Titan, slightly larger than Mercury, has an atmosphere thicker than Earths?
[Snip!]
When all your handwaving has failed you, why not try the obvious? All the bodies that you cited with low densities and atmospheres are at or beyond Saturn's orbit--AND BABY, IT'S COLD OUT THERE! Heat means higher kinetic energy of gas molecules which means greater chance of escape. For example, the Earth's atmosphere is hot enough that hydrogen and helium "leak" out and we do not see significant amounts of these gases.

nutant gene 71
2005-Jul-28, 01:44 AM
I realize there's BIG CHILL out there. That's why I included Io and Europa in the list. How "warm" is Europa? Io has active volcanoes, so maybe "gassing out" should give us some atmosphere?

pghnative
2005-Jul-28, 02:29 AM
I realize there's BIG CHILL out there. Did you? That is far from obvious. Your post made great mention of masses and atmospheres. You also made sure to wonder if "astronomy is sweeping something under the atmosphere rug here". Yet you made no mention of temperature being a mitigating factor.

nutant gene 71
2005-Jul-28, 03:28 AM
I realize there's BIG CHILL out there. Did you? That is far from obvious. Your post made great mention of masses and atmospheres. You also made sure to wonder if "astronomy is sweeping something under the atmosphere rug here". Yet you made no mention of temperature being a mitigating factor.
Which temps would you like, pgh?

For Jupiter, black body temp is 110.0K, vs. Earth's 254.3K; solar irradiance is 50.50 W/m^2, vs. Earth's 1367.6 W/m^2. Quite a difference, no? For Saturn, it's 81.1K and 14.90 W/m^2. Can we assume that their moons are in similar temp conditions? .. very cold compared to Earth's.

Shouldn't Jupiter's temps be enough to have large moons like Io and Europa have some atmosphere? Especially for Io which is actively "gassing out"? Or is a hypo G domain of only 5G not enough, while Saturn's hypo 10G is sufficient?

:-?

pghnative
2005-Jul-28, 03:43 AM
Golly, you've convinced me. Yup. Saturn must be 10G.

nutant gene 71
2005-Jul-28, 05:15 PM
Golly, you've convinced me. Yup. Saturn must be 10G.
No, not really. :)

I was thinking more like six of one, half dozen of the other. The combo of colder temps, with stickier G, might just be the recipe needed to cook up those small body atmospheres. There's a baker's dozen of them, so maybe the combo would explain why molecules can't reach escape velocities on some moons, while blown away on others... ;)

Here's a cross-reference resource for comparrisons: Galilean Fact Sheet (http://nssdc.gsfc.nasa.gov/planetary/factsheet/galileanfact_table.html)

nutant gene 71
2005-Jul-29, 02:11 AM
...the Earth's atmosphere is hot enough that hydrogen and helium "leak" out and we do not see significant amounts of these gases.

Are moons "gassing out", a small rodent's behind? How long before they're all gassed out? Shouldn't they be all shriveled up like prunes by now?

Over billions of years, Earth's hydrogen is locked up in with oxygen, H2O, in water.

I find our theories on "gassing out" small bodies with atmospheres, cold or not, seriously lacking substance. It's a lot of hot air. :lol:

nutant gene 71
2005-Jul-31, 02:36 PM

The JPL News release says:

"The ion and neutral mass spectrometer and the ultraviolet imaging spectrograph found the atmosphere contains water vapor. The mass spectrometer found the water vapor comprises about 65 percent of the atmosphere, with molecular hydrogen at about 20 percent. The rest is mostly carbon dioxide and some combination of molecular nitrogen and carbon monoxide. The variation of water vapor density with altitude suggests the water vapor may come from a localized source comparable to a geothermal hot spot. The ultraviolet results strongly suggest a local vapor cloud."

Notice how very light hydrogen still lingers in the atmosphere. The polar "hot" spots is really a mystery though. Volcanoes might be replenishing Enceladus atmosphere, though same volcanism for Jupiter's (larger) moon Io does not seem to have same effect on atmospheric retention. Very strange... Polar hotspot at 110K is about temp of Jupiter's surface, which is 4 AU closer to the Sun.

If this small moon is close to Saturn's rings, could tidal forces from the rings, combined with Saturn's gravitational, cause internal heating? But at temps of 100K, not much room for anything like normal "water vapor cloud".

PatKelley
2005-Jul-31, 07:33 PM
The rings are not likely a source of gravitational interaction or internal heating for Enceladus, but rather an example of the forces that are heating it internally. The rings are within 2.44 planetary diameters of Saturn, and roughly a similar density (less than or near that of water) - which means that within that range objects tend to get pulled differentially inner region to outer region. If they are not pulled completely apart within this range, it is reasonable to surmise that slightly more distant objects would suffer the less destructive fate of tidal stressing, and that depending on density and size the objects fate might be different than being torn apart.

nutant gene 71
2005-Aug-04, 06:29 PM
SHORT SUMMATION, since this hypothetical mass thread seems to have fallen asleep, so'd like to make a few notes here.

What I found most useful in these discussions is how my focus was narrowed from a broader range of what a hypothetical variable G would do to orbital and gravitational behavior to how it is more meaningful to think narrowly of what such a hypo variable G does to mass: mass is mass, but in a higher G, it takes fewer molecules to have the same mass as in a lower G; in effect, my tentative conclusion is that each molecule is affected by how it will display G in its own gravitational domain.

When I first stumbled on this possibility, of G not being a universal constant, my first thought was that any spacecraft flying into a higher G domain would be accelerated faster, but now I can see this is not the case. Mass is mass, even in a higher G region, so though the gravitational mass may increase, its response to the (formerly computed as universal) 1G mass relationships do not change (in relation to formerly computed masses per 1G), same as a billiard ball will accelerate just as fast as a bowling ball in a (same G) gravitational field. We had computed all orbitals and masses at a distance per a constant G, so that's the canvas on which all our drawings of astronomy is laid out. However, if we should find in the future that G is a variable, then the canvas will have to be stretched out differently, where a very low G makes mass very gravitationally light-mass close to a hot source of e.m. energy, and conversely a higher G making mass gravitationally heavier-mass with distance from this hot radiant source. By 50 AU, gravity should be great indeed, though my calculations that it is 50 times as great as in Earth's location at 1 AU is merely computational conjecture (where G grows at linear rate of 1G per 1AU), as derived using what I call the "Axiomatic equation" (http://www.humancafe.com/cgi-bin/discus/show.cgi?70/108.html), a hybrid of Planck-deBroglie, Einstein equations. (The inverse square law remains the same, whether in 1G or variable G, since mass is mass, and mass computed in 1G is merely adjusted (as shown through this thread) for greater G; and orbits remain same, as do gravity assist trajectories.) I cannot get to 50 AU to test for this G (my new on-order-hybrid-vehicle can't get there, my rocket ship's still in the shop :) ), so I am left to wait and see what our space probes sent out to measure for inertial mass, and per equivalence G, out in those great regions away from the Sun's hot energy, and solar wind, to get a reading in situ. What I suspect we will find, if G is variable as here hypothesized, is that flywheels spin faster due to increased centripetal force, increased gravitational (molecular) mass will make the spacecraft act as if they were more massive (not merely heavier, but gravitationally more powerful) and pulled back towards the Sun by this proportion, and the gravitational proportional between objects out there would be greater, so mass attraction is increased. This would mean that far ranging comets, those with highly elliptical orbits, would be seen to gather material of dust, water molecules, gases, very far out in the solar system's hypo higher G, and shed that same accumulation on their return trip back into a much lower G domain of the inner solar system. Hence, closer to the Sun they would gas out, or at least drop a few dust balls.

But the most interesting thing will be to see if we get the same deep space gravity effect on onboard clocks in a greater G as we do for atomic clocks within Earth's gravitational domain. We know atomic clocks slow while traveling through gravity (which is now interpreted relativistically as "time dilation"), so it should be important to measure for this same slowing effect in a greater G region. And if we find that atomic oscillations slow in like proportion to G, then we could take it one step further and expect light to redshift in like fashion in greater G. This is what the above Cosmic Lightshift post (http://www.badastronomy.com/phpBB/viewtopic.php?p=504217&amp;sid=4b03d68a03a222aa1c63159 c20c0d9ae#504217) was about, showing how much more gravity was needed to effect a cosmic light redshift at 1z; the subsequent post (http://www.badastronomy.com/phpBB/viewtopic.php?p=504364#504364) showed how a similar reading for G was arrived separately using the "cut-off" wavelength of the photoelectric effect. Though this is purely a hypo computational result, it could mean that space is isotropic and homogeneous at some five orders of magnitude higher G than experienced in our inner solar system; if this were so, then cosmic light as it travels through the gravitational mass of that rarified space, which is 99.99% of the cosmos, would redshift naturally. Net result could be that both oscillating atoms and light "slow" in higher G, if this is so. Of course, if this is so, there are serious consequences to present day cosmology theories, especially space expansion and Big Bang. The good side of this is that if G is a variable inversely-proportional-to e.m. energy received (net of Sun's overwhelming radiant effect), it may give us a way to harness it as a new power source, one that is continuously accelerative and virtually cost free.

And finally, there is the issue raised by papageno on chemical bonding and bandstructure (http://www.badastronomy.com/phpBB/viewtopic.php?p=505884#505884) of semiconductors, as well as danger of extending the deBroglie formula to relativistic particles. I think a parallel issue is being discussed in Rolf's thread on Electromagnetic paradox (http://www.badastronomy.com/phpBB/viewtopic.php?t=23314), where the relativistic considerations are taken at 50 AU, for example. This is an interesting area of study, because there appears to be virtually no relativistic effects on electrodynamics (other than perhaps some numerically chaotic computational drift -?) to directly affect electron behavior at great distances. However, this is perhaps still too early in our approach to solving this dilemma, though Rolf Guthmann seems to have tackled it head on, to his credit. Whether or not we can define this a priori, before we actually measure for these effects at great AU distances, should be of great interest to us, or at least to me.

So this is a summation of sorts, where we are now on this hypothetical mass discussion, though agreement on any of these issues is conspicuous by its absence... I kind of hate sitting around and waiting for results to come in from the outer field our of solar system, but until my rocket ship's out of the shop outfitted with its new gravity-assist-drive, I guess I'll just have to cool my heels here on Earth. Thanks for all of our fine thoughts and contributions, even if, especially if, you disagreed with me. :)

Tassel
2005-Aug-04, 07:11 PM
mass is mass, but in a higher G, it takes fewer molecules to have the same mass as in a lower G; in effect, my tentative conclusion is that each molecule is affected by how it will display G in its own gravitational domain.
Your conclusion is wrong. Mass is mass. In a higher G, all other things being equal, the same mass will weigh more.

Mass is mass, even in a higher G region, so though the gravitational mass may increase
It would not. All other things being equal, the weight of a given mass would increase.

By 50 AU, gravity should be great indeed, though my calculations that it is 50 times as great as in Earth's location at 1 AU is merely computational conjecture (where G grows at linear rate of 1G per 1AU), as derived using what I call the "Axiomatic equation" (http://www.humancafe.com/cgi-bin/discus/show.cgi?70/108.html)
However we know through observation that this is not true. We have observed objects whose orbits take them from the inner solar system, out towards the edge of the solar system and do not see deviations from the orbits calculated using Newtonian physics and a constant G. You choose to ignore these observations.

so I am left to wait and see what our space probes sent out to measure for inertial mass
Which probes would those be?

What I suspect we will find, if G is variable as here hypothesized, is that flywheels spin faster due to increased centripetal force, increased gravitational mass will make the spacecraft act as if they were more massive
We have flywheels currently working in the outer solar system and the effects you suspect we will find are not seen. You choose to ignore these observations.

I kind of hate sitting around and waiting for results to come in from the outer field our of solar system
It appears you also hate the results that have already come in. Presumably because they disagree with your ideas.

Scientists use all the data available to them in their attempts to better understand nature. Pseudoscientists pick and choose which data to present based on what they think supports their ideas.

nutant gene 71
2005-Aug-04, 07:29 PM
Scientists use all the data available to them in their attempts to better understand nature. Pseudoscientists pick and choose which data to present based on what they think supports their ideas.
Ah, so quick to judge and cast labels. So your quick accusation seems to be "pseudoscientist". Yet your lack of keen interest here is glaring, and perhaps a dose of greater tolerance for those who have a greater curiosity than yours should be a lauded goal? The rest of it... snipped.. is just a rehash of all that's been said before. I don't know why you think you have the final verdict on all this, but I am patient for further study. Hence, this is why this thread was titled "hypothetical". 8)

pghnative
2005-Aug-04, 07:41 PM
Tassel
I'm sure you are going to want to reply to NG, but I'd urge you not too. It is obvious from all of the recent thread bumps that NG is just looking for attention. I'd suggest just letting it die --- the lack of mathematical reasoning in his arguments is already obvious to any lurkers.

Tassel
2005-Aug-04, 07:59 PM
Tassel
I'm sure you are going to want to reply to NG, but I'd urge you not too. It is obvious from all of the recent thread bumps that NG is just looking for attention. I'd suggest just letting it die --- the lack of mathematical reasoning in his arguments is already obvious to any lurkers.
You're right. If I had thought a little more before responding, I would have realized I'm just feeding a troll. My apologies for bumping the thread.

nutant gene 71
2005-Aug-04, 08:21 PM
Tassel
I'm sure you are going to want to reply to NG, but I'd urge you not too. It is obvious from all of the recent thread bumps that NG is just looking for attention. I'd suggest just letting it die --- the lack of mathematical reasoning in his arguments is already obvious to any lurkers.
You're right. If I had thought a little more before responding, I would have realized I'm just feeding a troll. My apologies for bumping the thread.
Let it rest. Goodnight. 8)

nutant gene 71
2005-Aug-05, 04:57 PM
Postscript: incase you missed the "no math" allegedly missing in mine per:

...the lack of mathematical reasoning in his arguments ...

rule of thumb for var mass (http://www.badastronomy.com/phpBB/viewtopic.php?p=507348#507348)
deep space gravity, pt. 2 (http://www.badastronomy.com/phpBB/viewtopic.php?p=504364#504364)
space redshift gravity, pt. 1 (http://www.badastronomy.com/phpBB/viewtopic.php?p=504217#504217)