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DyerWolf
2007-Jan-31, 02:31 PM
Actually, I'm wondering whether the gravitational pull of the moon produces a measurable effect on a person's weight.

I remember my highschool science class had a scale in a glass box. The scale was so sensitive that when I placed my hand above it, the numbers changed - even though I wasn't touching the scale. (I thought that was really cool, but for whatever reason, doing so bugged the teacher). Do super accurate scales today have to compensate for the moon?

I know the moon affects the tides, but am I heavier when the moon is on the far side of the earth?

TIA

djellison
2007-Jan-31, 02:40 PM
F = G*M1*M2 / r^2

G is 6.67300 × 10-11 m^3 kg^-1 s^-2

M1 (moon) is 7.36 × 10^22
M2 (me) is 80 kg

r is 384,400,000 m (give or take the radius of the Earth)

SO.....

392906240000000 / 147763360000000000

is...

0.00266 Newtons of force applied to me by the Moon. Divide by 9.81 - that becomes 0.27 grams.

Weight Watchers probably the better option.

(I did this very very quickly - the maths may be wrong)

Doug

Cougar
2007-Jan-31, 04:38 PM
Actually, I'm wondering whether the gravitational pull of the moon produces a measurable effect on a person's weight.
Here's a vaguely related matter:

"Tree stem diameters fluctuate with tides." This was the title of a bit of scientific correspondence from Zurcher and Cantiani in the April 16, 1998 issue of Nature. They said "The diameter of tree stems... undergoes rhythmic fluctuations. We find a strong correlation between these fluctuations and the timing and strength of tides. This correlation suggests that the Moon is influencing the flow of water between different parts of the trees."

tony873004
2007-Jan-31, 06:38 PM
0.00266 Newtons of force applied to me by the Moon.
The math is good up to here.

a=F/m

a=0.002266 N / 80 kg = 0.000033 m/s^2

This acceleration is 1/300,000 of Earth's 9.8 m/s, so you'd weigh 299,999/300,000 of your regular weight.

Except....

The Moon also pulls the Earth you're standing on, thereby cancelling out most of this difference.

An easier way to do the problem is skip a step since the little m's cancel out, allowing you to directly use the formula for acceleration: GM/r^2.

6.673e-11 * 7.36e22 / 384400000^2 = 0.000033 m/s^2

The Moon will accelerate both you and the Earth by 0.000033 m/s^2 for a net difference of zero. There will still be a small difference, however, since the ground you're standing on is attached to the ground below it, which is further from the Moon, and subject to less acceleration from the Moon's gravity. Rigid forces resist the ground you're standing on from being pulled towards the Moon, but not by much. The difference is what causes tides, and pulls the Earth into an oval.

I haven't read the Nature article that Cougar talks about, but I find it very surprising that the Moon could influence the flow of water in a plant. The Moon has an easy time pulling tides on Earth because the size of the Earth is appreciable compared to the Earth / Moon distance. The side of the Earth nearest the Moon is 12,800 km closer to the Moon than the far side of the Earth. However, the top of the plant is only a few feet closer to the Moon than the bottom of the plant, rendering tidal forces on the plant insignificant.

I highly doubt there's a scale sensitive enough that it can measure gravity from your hand. If there were, there would be no need for Cavandish's elaborate setup to measure G. There must be another explanation as to why your hand caused the scale to move.

Amber Robot
2007-Jan-31, 06:51 PM
I haven't read the Nature article that Cougar talks about, but I find it very surprising that the Moon could influence the flow of water in a plant.

I took a quick look and the data look to be very high signal-to-noise. And it's not just the moon - the signature has that double-peaked, daily periodic signal that the tides have. It appears to be at a level of hundreds of microns or so, depending on the tree.

Kaptain K
2007-Jan-31, 07:08 PM
Does a person's weight fluctuate with the tides?
Depends on what kind of scale you use.

If you use a balance type scale, no, since both sides are affected. In fact, a balance type scale will give the same reading on the Moon or Mars or any other body.

If you use a spring type scale, yes, but, as has been pointed out, the difference is miniscule.

Amber Robot
2007-Jan-31, 08:11 PM
If you use a balance type scale, no, since both sides are affected. In fact, a balance type scale will give the same reading on the Moon or Mars or any other body.

That's because with a balance scale you are measuring mass, not weight.

publius
2007-Jan-31, 09:03 PM
That's because with a balance scale you are measuring mass, not weight.

Depends on what you mean by weight. Me, I'm just as liable as not to define weight as "gravitational mass", and declare a balance scale in a gravitational field measures "weight". The is actually the working, *legal* defintion of "weight", such as when you buy something with "net weight = X" on it. Yep, I liable to declare that's the meaning of weight until I need a word to actually mean the force required to hold a given mass stationary in field, and then I'll use weight for that too.

-Richard

Ken G
2007-Jan-31, 09:08 PM
I know the moon affects the tides, but am I heavier when the moon is on the far side of the earth?

A point that people have missed is that it matters if you are on a ship or on dry land. The ground responds to tidal forces much less than does water. I think tony873004 is more correct if you are on a ship, because then indeed most of the Moon's gravity is cancelled by the fact that you have a different centripetal acceleration and the scale won't notice a change in weight. But trees are on dry land, so the land does not respond as well to tidal force changes, and you stay (I believe) more or less on the same orbital path, with the same centripetal acceleration, regardless of the Moon's location. Thus the Moon's gravity really is closer to a pure change in your weight, so djellison'scalculation takes on new validity for trees. How much closer I really don't know, it depends on the contrast between how continents move and how oceans move, but we know there is a contrast, or we wouldn't have beach tides.

As for the tree rings, I'm with tony873004-- some ideas you just know are bogus. There is simply no way that direct tidal forces on the water in the tree can have anything at all to do with the diameter of the tree, the miniscule nature of the gravity effect makes that perfectly obvious. So the real questions are, is there actually a tidal correlation (or is the data itself bogus), and if it is for real, then what is the tidal cause? Saying it's the "flow of water" in the tree, in analogy to the way oceans react (which are... water, hmmm!), is laughable. Honestly. If it has anything to do with water, then it would have to be due to sloshing of actual tidal waters, like what happens in tidal pools and how that could affect a tree, but certainly not tidal forces on water in the tree! I'm sure water in a tree does not flow in an unfettered way, there are all kinds of capillary and structural pressures that connect the water to the tree, so it's a whole unified stress and strain problem. If it's actually tidal forces on the tree (and the gravity does vary by about 1 part in 10^5, as djellison showed, so there might conceivable by strain factors at the level of a hundred microns I can believe), then that's just what is-- changes in the response to strain, and how thick the tree needs to be to hold itself up. Then I'd ask-- what happens to the diameter if I hang my coat on the tree for a few hours? If nothing, well-- there you go.

Also, there are other factors the data should easily detect. Do the variations track spring and neap tides, which have quite noticeably different strengths? Do the variations depend on the tree locations, like some trees could be close to the coast where tidal waters come and go, and others might be farther away or at higher elevations? Is the effect once per day or twice? Amber Robot looked already and found that it's twice a day-- which is a big problem for the strain interpretation (which should be only once a day, the strength of the Moon's gravity varies once a day because it's different from a tidal variation, because it doesn't include centripetal force variation). So I conclude either that the data is a complete fraud (which I hesitate to conclude, but can't be ruled out), or their trees are near a coast and the sloshing of tidal waters is having an immediate effect on the trees.

Amber Robot
2007-Jan-31, 09:18 PM
As for the tree rings, I'm with tony873004-- some ideas you just know are bogus. There is simply no way that direct tidal forces on the water in the tree can have anything at all to do with the diameter of the tree, the miniscule nature of the gravity effect makes that perfectly obvious. If the effect is real, it is certainly due to sloshing of actual tidal waters, like what happens in tidal pools and how that could affect a tree, but certainly not tidal forces on water in the tree! But there are other factors the data should easily detect. Do the variations track spring and neap tides, which have quite noticeably different strengths? Do the variations depend on the tree locations, like some trees could be close to the coast where tidal waters come and go, and others might be farther away or at higher elevations? Good science would have no trouble determining if the effect is real or not, but it doesn't look like that has happened here.

Did you read the article? The variations do indeed track the two tides and have the same relative amplitude between them. Go read the article before you start stating that it is not good science.

publius
2007-Jan-31, 09:29 PM
Ken,

Superstitions about the moon persist to this day, and many times I get a chuckle about some local characters' beliefs about the moon. I just smile and nod, most of the time. :lol:

There is a strong belief that firewood, and even lumber, must be cut during a certain phase of the moon, otherwise it will not dry out properly. I can't remember what phase that supposed to be.

I got a really good laugh about that several years ago. My father and I built a new shed. We had a patch of pines that would make some good lumber, and these darn pine beetles are getting pretty bad around here, so we figured we'd save some money on lumber as well as getting some good out of some of those pines before the beetles go them.

Only problem is lumber jacking is some danged hard work, :) not to mention a wee bit on the dangerous side. You don't play when felling trees. Anyway, we got out there over the course of a week or so, and cut 'em down, "bucked 'em", and carried them to a small little saw mill.

A neighbor came up one day to shoot the bull while we were working, and remarked that "I'd always heard you had a pretty good head on your shoulders, and I see why". I wondered what he meant.

Well, by complete coincidence, I was cutting those trees exactly at the correct phase of the moon. :lol:

And it gets even stranger. There is a belief that there's a proper phase of the moon to dig holes in the ground. At the wrong time, you'll either have too much or not enough dirt to fill it back in. The phase of the moon controls how "fluffy" the loose dirt is somehow. Too early moon wise, and it's too fluffy, but too late and its not fluffy enough. Or vice versa.

That type of stuff goes way back in time and is passed on from generation to generation...........

-Richard

grav
2007-Jan-31, 09:43 PM
I think it would depend on the type of scale one used. The person would actually become lighter, and the scale would read that, but if the ground accelerated upward as well, then the scale would also be pushed from the bottom up, compressing it further, and adding it back. But even then, if the ground accelerated upward, then it would expand, bringing the person slightly further away from the center of the Earth, whereby subtracting the weight again when the ground has gone as far as it will go, since the ground can have no additional acceleration against gravity (toward us) unless it is actual moving. In fact, I'm thinking that when the ground has moved as far as it can, we would be even closer to the moon, and further from the center of the Earth, and weigh even less than we otherwise would before with just considering the reduction in weight due to the gravity of the moon alone.

publius
2007-Jan-31, 09:54 PM
Here's a little thought experiment about one gravitating body in a "background" field, due to other gravitating bodies.

Idealize the earth as a perfect, rigid sphere and consider test masses on the surface. Now, place that earth in a uniform gravitational field exactly equal to the surface gravity of the earth.

Think about the test masses on the surface. If we hold the earth stationary (align our coordinates along with the external field). On one side, you would weight twice as much, but on the other side you wouldn't weigh anything.

Now, let the earth free fall. Everbody now weighs the same. But look at things from the perspective of an observer stationary with the larger uniform field.

The geodesic of test mass on the front side of the earth is still "flat",
g is zero. He feels a force *because the earth, which is falling, is pushing him along* against his geodesic. This sort of puts an intereting spin "the earth pushing up" view of proper acceleration. It's quite literally happening, in a Newtonian sense. On the other side, g is twice the surface value, but the earth, again following its own geodesic pushes back at 1g, so the net force is still 1g.

Only the center of mass of the earth is following a geodesic. Now, when you transform to that accelerating frame (we're sticking with Galileo and Newton), the uniform field goes away essentially, and you're back to the normal "picture" of the earth's gravity alone.

Now, give the external field some tides, let it vary a little. To keep it simple, just let it vary linearly in one direction. Now, what forces do test masses on either side feel as the fall with the earth? You'll see its all about the tides.

-Richard

grav
2007-Jan-31, 09:54 PM
That type of stuff goes way back in time and is passed on from generation to generation...........
Well, if it goes back to a time before we even knew that the moon creates gravity or causes tidal expansion, then I wonder if there might be something to it. :think: :)

Ken G
2007-Jan-31, 10:03 PM
Amber Robot:

Sorry, but I still know they are wrong, without reading the article. All you've done is rule out some of the ways they could be wrong, which I was indeed just guessing about above and would have said differently after more thought (forget what I said about djellison's calculation, the weight variation on a scale, whether on a ship or dry land, is about 1 part in 10^12, along the lines of what tony873004 said). You say they do track spring and neap tides, and are twice a day, so we have narrowed it down to either fraudulent data or something that depends on nearby beach tides. The magnitude of actual tidal forces on the tree we have already seen calculated, but let's do it again. It is so small that we may as well be discussing astrology here.

The change in the gravity at the top of the tree compared to the bottom, all fictitious effects included, is still always going to be of order the height of the tree compared to the distance to the Moon, times the gravity from the Moon (not the Earth gravity, mind you!). No observations can deny that simple fact. The distance to the Moon is 4x10^8 m, and the size of the tree is, let's say, 40 m. That's a difference in scale of 10^-7, times the Moon's gravity at Earth! We already saw that the Moon's gravity at Earth is about 10^-5 compared to Earth gravity. So put it together-- the change in gravity at the top of the tree, compared to the bottom, and that is what is varying over the day, measured in units of the local g, is about 10^-12. That's the acceleration difference you have to work with. Now use that to get a 100 micron difference in tree diameter, which presumably is 1 part in 10^-5 in diameter. Knock yourself out if you want to waste your time, but I'm telling you, a 1 part in 10^12 variation in the stress on the tree is not going to give you a 1 part in 10^5 change in diameter.

Now, you tell me, if I had wasted my time reading the article, would it have shed any new light on this calculation? If the data is real, it ain't tidal stress on the tree, but maybe it could be due to nearby tidal waters.

grav
2007-Jan-31, 10:14 PM
Here's a little thought experiment about one gravitating body in a "background" field, due to other gravitating bodies.

Idealize the earth as a perfect, rigid sphere and consider test masses on the surface. Now, place that earth in a uniform gravitational field exactly equal to the surface gravity of the earth.

Think about the test masses on the surface. If we hold the earth stationary (align our coordinates along with the external field). On one side, you would weight twice as much, but on the other side you wouldn't weigh anything.

Now, let the earth free fall. Everbody now weighs the same. But look at things from the perspective of an observer stationary with the larger uniform field.

The geodesic of test mass on the front side of the earth is still "flat",
g is zero. He feels a force *because the earth, which is falling, is pushing him along* against his geodesic. This sort of puts an intereting spin "the earth pushing up" view of proper acceleration. It's quite literally happening, in a Newtonian sense. On the other side, g is twice the surface value, but the earth, again following its own geodesic pushes back at 1g, so the net force is still 1g.

Only the center of mass of the earth is following a geodesic. Now, when you transform to that accelerating frame (we're sticking with Galileo and Newton), the uniform field goes away essentially, and you're back to the normal "picture" of the earth's gravity alone.

Now, give the external field some tides, let it vary a little. To keep it simple, just let it vary linearly in one direction. Now, what forces do test masses on either side feel as the fall with the earth? You'll see its all about the tides.

-RichardThat's an interesting way of looking at it. So if we were in freefall with the earth toward the moon, whereas all points on the earth would fall at the same rate, the only force we would feel is that of the earth itself, and there would be no difference in our weight. That is also because the center of the earth is falling with us, and we would always remain at the same distance from it. And then for the earth-moon revolution, the distance from the barycenter to the center of the earth remains the same, but only we would accelerate as that side of the earth expands, but not along with its center, whereby moving us further away from Earth's center and closer to the moon, and feeling a lesser force toward Earth. If the earth remained a rigid body, then we would only feel the difference in the acceleration from the center of the moon at that distance.

Amber Robot
2007-Jan-31, 10:46 PM
Now, you tell me, if I had wasted my time reading the article, would it have shed any new light on this calculation? If the data is real, it ain't tidal stress on the tree, but maybe it could be due to nearby tidal waters.

All I was saying was that you should read the article before bashing it. They don't explicitly state that it is tidal forces that are causing the variations, only that it strongly correlates with the tide. I was under the impression that Nature was a reputable journal, so it's probably unlikely, though not impossible, that their data is fraudulent.

As far as being from nearby tidal waters, they do state that they measured the changes in trees grown in "containers" as well as in "stem sections that are disconnected from the root system and crown."

This is quite far from my area of expertise, so other than reading the Nature article, and relaying its contents, I can't do much more. I would agree with you that the tidal forces would seem to be exceedingly small, and I would hope that they would, in some future work, attempt to come up with some kind of physical mechanism.

publius
2007-Jan-31, 11:16 PM
Grav,

No, when the earth('s center of mass) free falls in the moon's (and the rest of the solar system's), tidal forces do make for a variation in weight. That was what the last part of the exercise was to see.

In a uniform field, you weigh the same as normal, but in a non-uniform tidal field, there are variations.

Do the example. Let the external field be some g(x) = (g0 + kx). Hold the earth stationary at x = 0 in that field. What is the weight on opposite sides. Now let the earth free-fall in this linear tidal field. Now do the weight calculations. You'll see differences in the external field do make a difference in the weight of something on the surface.

And this is a Newtonian example of the one main points about gravity that I ramble on and on about. When you're free-falling, the only thing you can locally determine is the tidal field, the *differences* in g.

Forget about the earth's gravity at first. Imagine a "ruler" in that tidal field that is free falling. The center of mass is what follows the geodesic -- every other point feels a real acceleration proportional to the tidal differences.

With a self-gravitating body, you add the field of the body itself to the tides. And that is a vector sum of course.

-Richard

Ken G
2007-Feb-01, 12:29 AM
All I was saying was that you should read the article before bashing it. That is certainly true in principle, but whenever someone says that, I ask: was I wrong? You read the article, does it present a remotely plausible case that direct tidal stresses from the Moon could have caused their observations? Are these people closet astrologers that you don't realize? It is even possible to measure a tree diameter to a few microns accurately using their technique? Are any of those questions answered in the article?

They don't explicitly state that it is tidal forces that are causing the variations, only that it strongly correlates with the tide. Yes they do, it's right in the abstract in this thread. Their suggestion in the abstract is nothing short of laughable, I'm afraid that's just the truth.
I was under the impression that Nature was a reputable journal, so it's probably unlikely, though not impossible, that their data is fraudulent.
You did not say the article was refereed, it was reported as a "communication". But I agree, fraud is the last resort here, and I'd only tend to go there if the authors had an astrological axe to grind. More likely, there is a simple uncontrolled effect going on, happens all the time. It could even be "cherry picked" data, out of a much larger set, though that also leans toward fraud.

This is quite far from my area of expertise, so other than reading the Nature article, and relaying its contents, I can't do much more. I would agree with you that the tidal forces would seem to be exceedingly small, and I would hope that they would, in some future work, attempt to come up with some kind of physical mechanism.
I'm not shooting the messenger here, I appreciate your passing on what the article is saying. Are you saying that this is a refereed article by established researchers presenting data that is known to be from a reliable method, and it is getting this result? I'll bet you 100 bucks it fails at least one of those. Because any reasonable referee with the least knowledge of how gravity and tides work would know that the measured effect is not due to Moon-induced flows of water inside their tree.

grav
2007-Feb-01, 01:36 AM
Grav,

No, when the earth('s center of mass) free falls in the moon's (and the rest of the solar system's), tidal forces do make for a variation in weight. That was what the last part of the exercise was to see.

In a uniform field, you weigh the same as normal, but in a non-uniform tidal field, there are variations.

Do the example. Let the external field be some g(x) = (g0 + kx). Hold the earth stationary at x = 0 in that field. What is the weight on opposite sides. Now let the earth free-fall in this linear tidal field. Now do the weight calculations. You'll see differences in the external field do make a difference in the weight of something on the surface.

And this is a Newtonian example of the one main points about gravity that I ramble on and on about. When you're free-falling, the only thing you can locally determine is the tidal field, the *differences* in g.

Forget about the earth's gravity at first. Imagine a "ruler" in that tidal field that is free falling. The center of mass is what follows the geodesic -- every other point feels a real acceleration proportional to the tidal differences.

With a self-gravitating body, you add the field of the body itself to the tides. And that is a vector sum of course.

-RichardOh. You wanted the tidal forces in a non-linear field during freefall. That would just be the same thing as Ken G was talking about in the Spinning Moon thread, where it is the difference in gravitational potential along the radial direction (or the gravity gradient) which creates a difference in forces between the front and back which stretches it along the line of motion (radially).

publius
2007-Feb-01, 02:18 AM
Oh. You wanted the tidal forces in a non-linear field during freefall. That would just be the same thing as Ken G was talking about in the Spinning Moon thread, where it is the difference in gravitational potential along the radial direction (or the gravity gradient) which creates a difference in forces between the front and back which stretches it along the line of motion (radially).

Grav,

Yes, however you mean "non-uniform", not "non-linear". It is differences in gravitational acceleration that are the only thing truly locally measurable, absent some "stationary reference point" which is arbitrary, although in some cases very meaningful.

-Richard

Amber Robot
2007-Feb-01, 04:38 AM
Are you saying that this is a refereed article by established researchers presenting data that is known to be from a reliable method, and it is getting this result?

No, I'm not saying that at all. I thought that Nature was a reputable journal, but I could be wrong. It's not my field.

Ken G
2007-Feb-01, 08:38 AM
Nature is certainly a very reputable Journal, but that extends more to the refereed articles than the communications, which are basically letters to the editor as far as I know. They must have thought the data was credible to publish the communication, but they would have also known that the suggestion of measurable effects due to tidal forces on water in the tree is absurd.

astromark
2007-Feb-01, 09:15 AM
Here am I sitting on the fence quite undecided as to this question; The Moon does have a tidal effect on planet Earths oceans, So much less than the sun does. Tides. Thats acceptable to me. The fact that Earths tectonic plates are distorted by the force of gravity is measurable. Does it altar how much we might weigh,? Only with the strictest of control methods could you detect that difference and, only with the most sensitive of scales could this be detected. Tacking off your tie might make more difference. You have been entertaining however, I thank you.

DyerWolf
2007-Feb-02, 12:57 PM
Let me see if I've got this right:

The moon pulls earth's water toward it, causing the water to "mound up" toward the moon.

If the earth and moon were tidally locked, we would experience no tides, and the water would have a "permanent mound" in the direction of the moon.

Tides occur because while the water is moving toward the moon, the earth's rotation causes the continental land masses to "stir" the water. (Undisturbed, the water would mound up; because the land passes "under" and "through" the mound zone, but contains the water, tides occur.

Ken G
2007-Feb-02, 03:52 PM
Close, but there are a few corrections to make. The "pull" of the Moon is what makes the Earth orbit, that by itself is not what causes tides. What causes tides is a football shaped deformation that appears because of the contrast between what the Moon is doing to the Earth along the line to the Moon, and what it is doing perpendicular to that line. Many explanations you'll see will only contrast what the Moon is doing to the near and far sides of Earth, but they are just wrong (an explanation that is woefully incomplete is a wrong explanation). Along the line to the Moon, the fact that the Moon's gravity gets weaker with distance causes a stretching effect, whereas along perpendicular directions, the fact that the force points at the center of the Moon causes a squeezing effect. The contrast between stretching and squeezing generates a football shape (no matter how the Earth is rotating).

OK, so now we have a football shape, which means there is a mound both toward and away from the Moon (two high tides per day!). Then the Earth's rotation comes in, which can be thought of as moving a particular beach through the mounds and troughs of ocean over the course of the day (it is mostly the oceans that respond to the tidal stress, disconnecting somewhat from the land). You are right that a particular patch of ocean cannot move up and down without some "stirring", and the way that stirring interacts with coastline features can cause a lot of different tidal effects. Some, like the Bay of Fundy, experience a huge tidal surge.

NEOWatcher
2007-Feb-02, 04:07 PM
Close, but there are a few corrections to make.
Forgive me for questioning, because you have some excellent points explained. These are things that a layperson such as me tend to forget the details of and boil it down to the "pull" but I'm stuck on this phrase...

The "pull" of the Moon is what makes the Earth orbit, that by itself is not what causes tides.
The "moon causing the Earth's orbit". Am I reading that wrong? Or is there a wrong word or a word missing for me.

BioSci
2007-Feb-02, 09:40 PM
That is certainly true in principle, but whenever someone says that, I ask: was I wrong? You read the article, does it present a remotely plausible case that direct tidal stresses from the Moon could have caused their observations?..

Actually from reading the article they do not claim that the direct tidal stresses cause their observations - but are only correlated with the tides:

The diameter of tree stems growing under open and controlled conditions undergoes rhythmic fluctuations independent of daily periodic factors such as light, temperature and humidity. We find a strong correlation between these fluctuations and the timing and strength of tides. This correlation suggests that the Moon is influencing the flow of water between different parts of the trees.

The bolded conclusion talks about "flow of water". From a plant physiology perspective what this means is that the osmotic potential or pressue with the stem cells is changing and they claim some sort of direct relation to the Moon. This is not a direct "tidal movement" of water but a physiological, cell-based, response that they claim is correlated with the Moon.

Are these people closet astrologers that you don't realize? It is even possible to measure a tree diameter to a few microns accurately using their technique?

The article does not give any indication how such small changes in stem diameter were measured.

Are any of those questions answered in the article?
Yes they do, it's right in the abstract in this thread. Their suggestion in the abstract is nothing short of laughable, I'm afraid that's just the truth.You did not say the article was refereed, it was reported as a "communication". But I agree, fraud is the last resort here, and I'd only tend to go there if the authors had an astrological axe to grind. More likely, there is a simple uncontrolled effect going on, happens all the time. It could even be "cherry picked" data, out of a much larger set, though that also leans toward fraud.

The article is poorly written and although Nature claims to "review" such types of communications, I think that they sometimes publish material for its "strangeness" and not because the data is sound.
It is conceivable that osmotic pressure of stem cells could vary enough to cause a measurable change in stem diameter on a daily basis. There are well-established mechanisms in plants that could cause such variations. However, such mechanisms are circadian (or metabolic) in nature and certainly not tidal. I expect that this data is simply artifactual or as suggested "cherry-picked".

The article is a good example of "junk science" and "against the mainstream" that made its way into a premier science journal. :sad:

Ken G
2007-Feb-02, 09:41 PM
The "moon causing the Earth's orbit". Am I reading that wrong? Or is there a wrong word or a word missing for me.

That's what I meant to say, but I see that I need to clarify that I don't mean the Earth's orbit about the Sun, but rather about the Moon (an orbit on top of an orbit, if you like). The Earth does in fact orbit in the Earth/Moon system-- imagine a rod between the Earth and the Moon, and the "balance point" (or center of mass) on that rod. You could set just that one point on your finger, and rotate the system, and that motion would describe the orbit of both the Moon and the Earth (but the Earth has a different rotation than this "tidally locked" picture, which allows us to have ocean tides, but that doesn't change the key elements of the football shape). So the Earth goes in a little circle, just like the Moon, only it is smaller in proportion to the mass ratio, and all this is happening on top of the big circle around the Sun (that last bit I'm just ignoring here).

tony873004
2007-Feb-04, 01:38 AM
Animation of the Earth and Moon orbiting the barycenter of the Earth/Moon system:
http://www.orbitsimulator.com/gravity/articles/barycenter.html

mugaliens
2007-Feb-05, 10:01 PM
Mass? No.

Weight (Mass under the influence of gravity)? Of course, same as the water pulled about by the tides.

briandunning
2007-Feb-07, 06:13 PM
The scale was so sensitive that when I placed my hand above it, the numbers changed - even though I wasn't touching the scale.

Why would this happen? I can't believe your hand's gravity could have affected it.

01101001
2007-Feb-07, 06:35 PM
Why would this happen? I can't believe your hand's gravity could have affected it.

Assuming the hand was inside the glass box, slight shaking of the hand could induce pressure waves in the air that would strike the measuring plate. The warm hand could make air above the plate less dense and cause convection. Things like that would be a reason a sensitive scale has a glass box: to keep air movement from affecting the readings.

Here's is probably the sort of analytical balance, a sensitive scale, that DyerWolf was talking about: Wikipedia: Weighing scale :: Laboratory balances (http://en.wikipedia.org/wiki/Weighing_scale#Laboratory_balances).

The weighing pan(s) of a high accuracy (0.1 mg or better) analytical balance are inside a see-through enclosure with doors so dust does not collect and so any air currents in the room do not affect the delicate balance. Also, the sample must be at room temperature to prevent natural convection from forming air currents inside the enclosure, affecting the weighing.

Living hands are not at room temperature -- for typical rooms.

Professor Illwill
2007-Feb-09, 01:52 PM
I've read through this thread, and if this has been mentioned before please forgive me. I don't think it has.

The moon doesn't just blink into and out of existence overhead. Multiple calculations need to be performed with the moon in different positions. I'd say matching to the tides should suffice.

The weight change will not only be to make a body lighter when it is overhead, but also to make one heavier when it is in other positions.

I have never understood how the tide opposite the moon can be high. It's been explained, but it's just so counter-intuitive. I don't dispute it. I won't dispute the math. But it's bothersome. I wonder if the same would apply to objects on land, or if they would get heavier.

Wait! It just came to me! The bulk of the earth is closer to the moon than the far side, so it's accelerating faster than objects on the far side. Woo-hoo! Learned something new.
Good topic.

Also need to point out that a balance-type scale must be used to obtain mass for equasions, which will predict spring-type scale readings. Both would have to be very accurate.

Otherwise just experiment with spring type and save the math.