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Staiduk
2004-Mar-24, 01:11 AM
Just popped into my head; now I can't get it out - got it while I was reading the thread on 'leap-seconds'. :)
Qustion: Why, after several billion years - is the Earth still rotating?
OK; I know there's no drag in space. Or at least; so little it's of no consequence. But: The Moon revolves in quite a different way - one revolution per day; so one face is always toward us. I seem to remember reading that this is because of Earth's gravities' signifigant effect - the pull of gravity slowly slowing the moon's rotation until it's fixed in reference to us. (This could be BA; if it is; that's likely the answer to the question. :) )
Anyhoo; the thing that troubles me is that yes; the Sun is 400 times farther away from us than the moon; but it's also 400 times larger - the reason solar eclipses look so flippin' cool; both bodies have the same angular size to us.
Gravity's effect drops much faster with distance; but I can't help but think that if the statement above - that Earth's gravity locked the Moon's spin into phase with us - is true; there's been plenty of time for the Sun's enormous gravity to do the same thing to Earth; and the other inner planets. But; we go on spinning happily away.
Why?
And of course; if the above premise is wrong; why is the moon rotating in phase with us?
Head hurts - tylenol time. ;)
Thanx!

Mars
2004-Mar-24, 01:22 AM
Could it have anything to do with Jupiter?

AGN Fuel
2004-Mar-24, 01:34 AM
Just popped into my head; now I can't get it out - got it while I was reading the thread on 'leap-seconds'. :)
Qustion: Why, after several billion years - is the Earth still rotating?
OK; I know there's no drag in space. Or at least; so little it's of no consequence. But: The Moon revolves in quite a different way - one revolution per day; so one face is always toward us. I seem to remember reading that this is because of Earth's gravities' signifigant effect - the pull of gravity slowly slowing the moon's rotation until it's fixed in reference to us. (This could be BA; if it is; that's likely the answer to the question. :) )
Anyhoo; the thing that troubles me is that yes; the Sun is 400 times farther away from us than the moon; but it's also 400 times larger - the reason solar eclipses look so flippin' cool; both bodies have the same angular size to us.
Gravity's effect drops much faster with distance; but I can't help but think that if the statement above - that Earth's gravity locked the Moon's spin into phase with us - is true; there's been plenty of time for the Sun's enormous gravity to do the same thing to Earth; and the other inner planets. But; we go on spinning happily away.
Why?
And of course; if the above premise is wrong; why is the moon rotating in phase with us?
Head hurts - tylenol time. ;)
Thanx!

The trick here is that it is the tidal influence that caused the locking you see with the moon, and the tidal influence drops in proportion to the cube of the distance.

So even though the sun vastly outweighs the moon, that advantage is lost by the fact that it is 400 times further away and so (proportionately, ignoring mass difference) 400^3, or 64,000,000, times weaker!

This is why the moon is the dominant force on tides on Earth, although the sun is certainly a contributor.

JohnOwens
2004-Mar-24, 01:51 AM
I seem to remember reading that this is because of Earth's gravities' signifigant effect - the pull of gravity slowly slowing the moon's rotation until it's fixed in reference to us. (This could be BA; if it is; that's likely the answer to the question. :) )
No, that's just right. GA there. :)


Anyhoo; the thing that troubles me is that yes; the Sun is 400 times farther away from us than the moon; but it's also 400 times larger - the reason solar eclipses look so flippin' cool; both bodies have the same angular size to us.
But the relevant feature is mass, rather than diameter, so for these purposes, we should consider the Sun to have about 27,000,000 times the mass of the Moon.


Gravity's effect drops much faster with distance; but I can't help but think that if the statement above - that Earth's gravity locked the Moon's spin into phase with us - is true; there's been plenty of time for the Sun's enormous gravity to do the same thing to Earth; and the other inner planets. But; we go on spinning happily away.
Why?
Yes, gravity drops according to the inverse square of distance, which is pretty steep. But the tidal effects drop with the inverse cube of distance, even steeper still. So that 400 times greater distance (I'm actually using the more precise figures for these numbers, like 389 here) causes gravity to be about 151,000 times weaker, but causes the tide to be about 59,000,000 weaker for the Sun than if it were at the Moon's distance (and we weren't inside it :wink: it'd have to be a neutron star or something). So the Sun's tides on the Earth are a bit less than half (1/2.18, or 0.459) those from the Moon.
Also, remember that what you really ought to be comparing is the Earth's tides on the Moon, which will be about 81 times those on the Earth from the Moon, and hence about 180 times those of the Sun on either the Earth or the Moon. This is the one which eventually did slow the Moon's rotation to 1 day/month, after all.
And to really throw a monkey wrench into the calculations, you have to consider the Earth's moment of inertia is greater than the Moon's not just by having more mass, but also by that mass having a greater average distance from the axis. For a sphere of homogenous density, I = (2/5)*m*r^2, where I is the moment of inertia. So, while the Earth might mass "only" 81 times the Moon, its moment of inertia is about 1100 times greater. And worst of all, I don't know just how the larger radius affects the tidal force exerted on the body. :-? It might not even have any affect at all, if the change in torque cancels with the change in moment of inertia.


And of course; if the above premise is wrong; why is the moon rotating in phase with us?
Head hurts - tylenol time. ;)
One other point there: When the Moon's rotation was being slowed the most, it was closer to us, and so the Earth's tidal forces on the Moon were proportionately greater. The slowing of the Earth's rotation by the Moon has been (and still is) pushing the Moon further and further away. But when/if the Moon was at half the distance it is now, the tidal forces would have been eight times greater.

Here. Have some more Tylenol now. :wink:

Staiduk
2004-Mar-24, 03:26 AM
Yiiiii, I think I need the whole bottle.
Well; I was able to understand the gist anyway. Thanx. :)

George
2004-Mar-24, 04:55 AM
And worst of all, I don't know just how the larger radius affects the tidal force exerted on the body. :-? It might not even have any affect at all, if the change in torque cancels with the change in moment of inertia.

The moment of inertia varies as the square of the radius and torque only directly, so the waves might get a little bigger but the Earth's rotation would be less affected. The Moon would likely get more kick out of it. :wink:

George
2004-Mar-24, 05:02 AM
One other point there: When the Moon's rotation was being slowed the most, it was closer to us, and so the Earth's tidal forces on the Moon were proportionately greater. The slowing of the Earth's rotation by the Moon has been (and still is) pushing the Moon further and further away. But when/if the Moon was at half the distance it is now, the tidal forces would have been eight times greater.

Hmmm. This makes perfect sense, however, the Moon would not be all that old if this was the only story. My question is...what is the real story as YEC makes use of this assumption?

AGN Fuel
2004-Mar-24, 05:29 AM
One other point there: When the Moon's rotation was being slowed the most, it was closer to us, and so the Earth's tidal forces on the Moon were proportionately greater. The slowing of the Earth's rotation by the Moon has been (and still is) pushing the Moon further and further away. But when/if the Moon was at half the distance it is now, the tidal forces would have been eight times greater.

However, the Moon would not be all that old if this was the only story.

Why? :-s

tracer
2004-Mar-24, 07:42 AM
Why, after several billion years - is the Earth still rotating?
I'm guessing that your real question here is, "Why have tidal forces locked the moon into synchronous rotation with the Earth, but haven't locked the Earth into synchronous rotation with the moon?"

Well ... the fact that the Earth has 81 times the moon's mass has something to do with it, but it's not the whole story. The main reason the Earth's rotation hasn't braked to a halt yet is because it's covered in liquid. Namely, the oceans. When the Earth's tidal force grabbed on to the moon, it was grabbing on to the moon's solid surface, which is attached to the rest of the moon. When the moon's tidal force grabs onto the Earth, however, it's grabbing onto a bunch of liquid, which can slosh around without pulling on the solid bulk of the planet itself (much).

And actually, as your investigation of the Leap Second phenomenon may have revealed to you, the Earth's rotation is slowing down -- but very very slowly. I haven't seen the numbers, but I'm guessing the Earth'll slow down into synchronous rotation with the moon right about at the same time that the sun bloats up into a red giant and swallows the both of them.

Brady Yoon
2004-Mar-24, 07:46 AM
I haven't seen the numbers, but I'm guessing the Earth'll slow down into synchronous rotation with the moon right about at the same time that the sun bloats up into a red giant and swallows the both of them.
I heard it would take much longer-almost half a trillion years... I'm probably wrong though.

Brady Yoon
2004-Mar-24, 07:47 AM
Does anyone know how long the length on the day would be and how many days there would be when the Earth is in synchronized rotation with the moon? And how far away will the moon recede from us? Thanks.

JohnOwens
2004-Mar-24, 08:21 AM
Well ... the fact that the Earth has 81 times the moon's mass has something to do with it, but it's not the whole story. The main reason the Earth's rotation hasn't braked to a halt yet is because it's covered in liquid. Namely, the oceans. When the Earth's tidal force grabbed on to the moon, it was grabbing on to the moon's solid surface, which is attached to the rest of the moon. When the moon's tidal force grabs onto the Earth, however, it's grabbing onto a bunch of liquid, which can slosh around without pulling on the solid bulk of the planet itself (much).
Good point there, and damn, I can't believe I didn't bring that up myself. I remembered it in another post I made, about the tidal pushing being a conceivable (but unlikely) explanation for Jovian planets being pushed away from their suns, whereupon I raised the objection (to myself) that the Sun (or any other sun) was too fluid for tides to have the same effect. And yes, it should be true also that with Earth being covered with fluid, it would have much less effect than on a largely solid body such as the Moon.
In one of those counter-intuitive sorts of things, though, even though a more solid body like the Moon is slowed more by tidal forces than a fluid one, a hypothetical totally rigid body wouldn't be slowed down at all by tidal forces, since there won't be any bulge for the other body to pull on, right? :-k

JohnOwens
2004-Mar-24, 09:07 AM
Does anyone know how long the length on the day would be and how many days there would be when the Earth is in synchronized rotation with the moon? And how far away will the moon recede from us? Thanks.
OK, I'll have to work on this a moment, but it will amount to working out the equilibrium with the Earth's current angular momentum and the Moon's angular momentum, and finding at which point they'll have the same period/angular velocity....

I = (2/5)*m*r^2
Id = I + m*d^2

where I is the moment of inertia around a sphere's natural axis, Id is the moment of inertia around another axis, and d is the distance between the natural axis and another axis it might spin around (such as a barycenter).
Hmm...
The barycenter of the Earth-Moon system is about 4,671 km from the Earth's axis, so about 379,730 km from the Moon's. Earth's mass about 5.98*10^24 kg, radius about 6,371 km, gives us I = 9.70*10^37 kg*m^2 for Earth on its own axis. Work out Earth's angular velocity, let's use, say, radians for the moment. 2*pi radians/(23h 56m 4s) (we're using sidereal days here) = 7.29*10^-5 rad/s. Multiply, and you get about 7.08*10^33 kg*m^2/s (or rad*kg*m^2/s, if you prefer, but I'd like to keep things "simple", relatively speaking).
For the Moon, you need to add its angular momentum for rotation and revolution. For rotation, the mass is 7.35*10^22 kg, radius is 1,738 km, I = 8.88 kg*m^2, angular velocity is (pardon while I look up the sidereal Lunar day...) 2*pi rad/(27d 7h 43.7m) = 2.66*10^-6 rad/s, multiply and you get angular momentum of 2.36*10^29 kg*m^2/s. Revolution, I'm not so sure I'm doing this entirely correctly, but it shouldn't be off by more than I, which will become rather insignificant in this case: Id = 8.88 kg*m^2 + m*d^2, Id = 8.88*10^34 kg*m^2 + 1.06*10^40 kg*m^2 (see why I becomes insignificant?), Id = 1.06*10^40 kg*m^2. Angular velocity is exactly the same as for rotation, 2.66*10^-6 rad/s, multiply, and get angular momentum of 2.82*10^34 kg*m^2/s.
Now add those all up (Earth's rotational angular momentum, Moon's rotational angular momentum, Moon's revolutional momentum), and you get 3.528*10^34 kg*m^2/s for the system. If you want to be really anal retentive about it, we should include the angular momentum for the Earth's revolution around the barycenter, too. Let's see if it even shows up with four decimal places. Id = I+m*d^2, Id = 9.70*10^37 kg*m^2 + 5.98*10^24 kg * (4,671 km)^2, Id = 9.70*10^37 kg*m^2 + 1.30*10^38 kg*m^2, Id = 2.27*10^38 kg*m^2. The period for this is the sidereal month again, so angular momentum is 2.27*10^38 kg*m^2 * 2.66*10^-6 rad/s = 6.05*10^32 kg*m^2/s. Add that in, and you're now looking at 3.588*10^34 kg*m^2/s. I guess it even changed the third digit significantly!

OK, now comes the tricky part: figuring out what angular velocity will give these bodies the right distance and the right barycenter while still having the same total angular momentum. (We'll neglect the fact that the Sun is slowing down the Earth's rotation at the same time, if that's OK with you. :wink: ) I'll have to think about this long enough that I might as well include it in a Part II, I think, perhaps better written after I sober up a little bit more. (Yes, I'm like this even when I'm a bit tipsy. :-? )

kjavds
2004-Mar-24, 10:42 AM
Interesting thread. The reason the Earth hasn't gone into lock step with the Sun is because the Sun is too fluid. As other posters hinted, the fluidity inherent in our oceans may also contribute to this. Remember also: the Sun only makes one revolution every 11 years (or is it 22?), which may too be a factor -- ie. Sun and Earth RPMs are too disparate. The rigidity thing is big, I know. Well, there's my 2 cents.

JohnOwens
2004-Mar-24, 10:51 AM
Interesting thread. The reason the Earth hasn't gone into lock step with the Sun is because the Sun is too fluid. As other posters hinted, the fluidity inherent in our oceans may also contribute to this. Remember also: the Sun only makes one revolution every 11 years (or is it 22?), which may too be a factor -- ie. Sun and Earth RPMs are too disparate. The rigidity thing is big, I know. Well, there's my 2 cents.
There is no revolution of the Sun in anything like 11 or 22 years, that's the sunspot & solar activity cycle you're thinking of, and I suppose could be considered a period of its magnetic field, but nothing relevant to what we're considering here.
Properly, the Sun's revolution is 226 million years or so, the time it takes to go around the galactic core. But the relevant time for these calculations is the Sun's rotation period of 25 days, which would indeed be fast enough to slow the Earth's orbit down and send us further out, however slowly, if it weren't for the aforementioned fluidity of the Sun.

In other news, still working on the final solution to the big equations for the Earth-Moon system above....

milli360
2004-Mar-24, 10:52 AM
Also, remember that what you really ought to be comparing is the Earth's tides on the Moon, which will be about 81 times those on the Earth from the Moon, and hence about 180 times those of the Sun on either the Earth or the Moon.
Not 81, but more like 20 times. The tidal force is also proportional to radius of the body. Since the moon has a quarter of the radius of that of the earth, the tide raised by the earth has another factor of one fourth involved.

Well ... the fact that the Earth has 81 times the moon's mass has something to do with it, but it's not the whole story. The main reason the Earth's rotation hasn't braked to a halt yet is because it's covered in liquid. Namely, the oceans. When the Earth's tidal force grabbed on to the moon, it was grabbing on to the moon's solid surface, which is attached to the rest of the moon. When the moon's tidal force grabs onto the Earth, however, it's grabbing onto a bunch of liquid, which can slosh around without pulling on the solid bulk of the planet itself (much).
The oceans are a tiny portion of the earth mass, so there is no way that that can be true. The "earth tide", the distortion of the solid rock of the earth is just a bit less than half of what if would be if the rock weren't solid. If it distorted freely, it would bob up and down like the ocean tides, and we would not even notice that there were tides.

The oceans actually contribute to the tidal friction, by "running into" the continents, and lagging behind the curve. The tremendous tides at the Bay of Fundy are an example. The point directly beneath the moon passes around the Earth at a rate of many hundreds of miles per hour, but there is no tidal surge that can go that fast.

JohnOwens
2004-Mar-24, 11:21 AM
Also, remember that what you really ought to be comparing is the Earth's tides on the Moon, which will be about 81 times those on the Earth from the Moon, and hence about 180 times those of the Sun on either the Earth or the Moon.
Not 81, but more like 20 times. The tidal force is also proportional to radius of the body. Since the moon has a quarter of the radius of that of the earth, the tide raised by the earth has another factor of one fourth involved.
Apples and oranges. When I was saying "tidal force", "tidal effects", or just plain "tides", I was usually (or maybe always - I can hope!) referring to the force difference per distance, what you might express as dF/dx if x is distance (or 2*G*M/r^3, if you prefer). Unfortunately, our language doesn't seem well equipped with different terms for these different tidal concepts. :-? I'll agree with you that the force at the closest & furthest points on the Moon would experience about a quarter of that 81 times. But I took that into account instead by describing it as having a lesser moment arm, therefore lesser torque. I just did the allowance for it at a different step than you would have.

Added: OK, my brain needs a rest. Maybe someone else can find the equilibrium of the angular momentum meanwhile. If not, I should have something definite by afternoonish (CST).

George
2004-Mar-24, 03:29 PM
However, the Moon would not be all that old if this was the only story.

Why? :-s


FOOTNOTE
* The moon was probably created close to its present distance from the earth. Over 10,000 years, lunar recession amounts to less than one kilometre.

For the technical reader: since tidal forces are inversely proportional to the cube of the distance, the recession rate (dR/dt) is inversely proportional to the sixth power of the distance. So dR/dt = k/R^6, where k is a constant = (present speed: 0.04 m/year) x (present distance: 384,400,000 m)^6 = 1.29x10^50 m^7/year. Integrating this differential equation gives the time to move from Ri to Rf as t = 1/7k(Rf^7 - Ri^7). For Rf = the present distance and Ri = 0, i.e. the earth and moon touching, t = 1.37 x 10^9 years

This is from AIG...>>> here [ (http://www.answersingenesis.org/docs/1294.asp) <<<

[the bold is mine]

You know a lot more about the Roche limit and related issues than me.

I believe the BA explained that their (AIG) view was too simplistic and did not take into account other variables which affect the rate of orbit change for the Moon. However, he did not elaborate (probably to keep his book such an easy read).

So I figured maybe now is a good time to address it. 8) However, if this is redundant please direct me to the proper thread or just tell me to search.

[edited to clarify exponents]

tracer
2004-Mar-24, 04:30 PM
Careful there, George -- copying-and-pasting the text directly from the AIG webpage destroys the exponent information. E.g., the last equation on the actual page reads "t = 1.37 x 10^9 years", not "t = 1.37 x 109 years".

milli360
2004-Mar-24, 05:29 PM
Apples and oranges. When I was saying "tidal force", "tidal effects", or just plain "tides", I was usually (or maybe always - I can hope!) referring to the force difference per distance, what you might express as dF/dx if x is distance (or 2*G*M/r^3, if you prefer). Unfortunately, our language doesn't seem well equipped with different terms for these different tidal concepts.

I won't let you get away with it, though. Force (or even force difference) per distance is not force. It has units of mass per time squared.

For you standing there, that force per distance value that you are talking about, is less from the moon than it is from the person standing next to you. That was a point made in the famous Culver and Ianna criticism of astrology. But the reason, in both the case of the moon and in the case of the person standing next to you, that you don't feel a tide within your body is that you are not large enough. Makes all the difference.

Plus, when you multiply by the radius of the affected body, that quantity has units of force.

SeanF
2004-Mar-24, 05:58 PM
But the reason, in both the case of the moon and in the case of the person standing next to you, that you don't feel a tide within your body is that you are not large enough.

You sure you don't want to restate that? In regards to why I don't feel tidal forces from the person standing next to me?

Kevinito
2004-Mar-24, 06:02 PM
The Moon revolves in quite a different way - one revolution per day; so one face is always toward us.

Correct me if I am wrong, but I believe the Moon orbits around the Earth once per 28 days and has a rotation period of 28 days, so that the same hemisphere is always facing toward us.

-Kevin

George
2004-Mar-24, 06:08 PM
Careful there, George -- copying-and-pasting the text directly from the AIG webpage destroys the exponent information. E.g., the last equation on the actual page reads "t = 1.37 x 10^9 years", not "t = 1.37 x 109 years".

Thanks. I cleaned it up, hopefully. I also did not check their math, but, I do hope to get input on variations to their idea of such a young moon, or, are they in line with conventional wisdom? [Edit:I am refering to the billion years from a hypothetical "touch" point and not the 10,000 years.]

aurora
2004-Mar-24, 06:24 PM
I believe the BA explained that their (AIG) view was too simplistic and did not take into account other variables which affect the rate of orbit change for the Moon. However, he did not elaborate (probably to keep his book such an easy read).


Long time poster Tim Thompson has written a lot on this subject. Try a web search, or search here, or search the talk.origins web site to find a link to his article(s).

milli360
2004-Mar-24, 06:28 PM
But the reason, in both the case of the moon and in the case of the person standing next to you, that you don't feel a tide within your body is that you are not large enough.

You sure you don't want to restate that? In regards to why I don't feel tidal forces from the person standing next to me?
The point I was making was that JohnOwens was calling GM/R^3 the tidal force, whereas it's not even a force. But GMr/R^3 would be. His quantity GM/R^3 is smaller for the moon than for the person standing next to you. The difference between the moon's effect on the Earth and the moon's effect on you is that factor of r.

SeanF
2004-Mar-24, 07:18 PM
But the reason, in both the case of the moon and in the case of the person standing next to you, that you don't feel a tide within your body is that you are not large enough.

You sure you don't want to restate that? In regards to why I don't feel tidal forces from the person standing next to me?
The point I was making was that JohnOwens was calling GM/R^3 the tidal force, whereas it's not even a force. But GMr/R^3 would be. His quantity GM/R^3 is smaller for the moon than for the person standing next to you. The difference between the moon's effect on the Earth and the moon's effect on you is that factor of r.

Oh, I know what the point of your post was, but that statement in particular . . . I just don't think a 100 kilogram body is going to produce noticeable tidal effects on me no matter how big I am.

daver
2004-Mar-24, 07:34 PM
Oh, I know what the point of your post was, but that statement in particular . . . I just don't think a 100 kilogram body is going to produce noticeable tidal effects on me no matter how big I am.
Possibly not, but if i didn't slip a decimal, a 1 kg object 1 meter from you will exert about 770 times the tidal force that the moon does.

milli360
2004-Mar-24, 07:48 PM
Oh, I know what the point of your post was, but that statement in particular . . . I just don't think a 100 kilogram body is going to produce noticeable tidal effects on me no matter how big I am.
If your radius doubles, the effect doubles. It's just that at a certain point you get too big to stand next to him. With the moon, that's no problem. :)


Possibly not, but if i didn't slip a decimal, a 1 kg object 1 meter from you will exert about 770 times the tidal force that the moon does.
That's odd, I got 52.

SeanF
2004-Mar-24, 08:04 PM
Oh, I know what the point of your post was, but that statement in particular . . . I just don't think a 100 kilogram body is going to produce noticeable tidal effects on me no matter how big I am.
If your radius doubles, the effect doubles. It's just that at a certain point you get too big to stand next to him.

Semantics. I have no problem with describing the Earth and I as "standing right next to each other," and I have no illusions about how much of a tidal effect I'm generating on the Earth. The Earth ain't feeling it.

However, making the other person denser could make me feel tidal forces from 'em . . . say, anybody ever stood next to [name deleted to protect the dense]?

:)

milli360
2004-Mar-24, 08:15 PM
Semantics. I have no problem with describing the Earth and I as "standing right next to each other," and I have no illusions about how much of a tidal effect I'm generating on the Earth. The Earth ain't feeling it.
You may have no problem with it, but I do. In the context of this discussion, the distance would be represented by R. That R is the distance between your centers-of-mass, so you and the Earth are not that close--but she does speak well of you.

SeanF
2004-Mar-24, 08:23 PM
Semantics. I have no problem with describing the Earth and I as "standing right next to each other," and I have no illusions about how much of a tidal effect I'm generating on the Earth. The Earth ain't feeling it.
You may have no problem with it, but I do. In the context of this discussion, the distance would be represented by R. That R is the distance between your centers-of-mass, so you and the Earth are not that close--but she does speak well of you.

Alrighty. So, I ask again if you want to rephrase your statement:


But the reason, . . . in the case of the person standing next to you, that you don't feel a tide within your body is that you are not large enough.

Given that "standing next to you" and "large enough" are apparently mutually exclusive? :)

milli360
2004-Mar-24, 08:36 PM
So, I ask again if you want to rephrase your statement:


But the reason, . . . in the case of the person standing next to you, that you don't feel a tide within your body is that you are not large enough.
OK, "your gf wants you to answer the spam" :)

SeanF
2004-Mar-24, 08:45 PM
So, I ask again if you want to rephrase your statement:


But the reason, . . . in the case of the person standing next to you, that you don't feel a tide within your body is that you are not large enough.
OK, "your gf wants you to answer the spam" :)

I . . . you . . . that . . . you know what? That one actually makes more sense! ;)

daver
2004-Mar-24, 09:17 PM
That's odd, I got 52.
The moon is 384e6 meters away, right? And it masses 7.35e22 kg. So the 1 kg object will have 384e6^3/7.35e22 times the influence as the moon.

milli360
2004-Mar-24, 09:49 PM
That's odd, I got 52.
The moon is 384e6 meters away, right? And it masses 7.35e22 kg. So the 1 kg object will have 384e6^3/7.35e22 times the influence as the moon.
I found my mistake. I didn't cube the 3.86, so I was off by 3.86^2.

JohnOwens
2004-Mar-24, 10:34 PM
Apples and oranges. When I was saying "tidal force", "tidal effects", or just plain "tides", I was usually (or maybe always - I can hope!) referring to the force difference per distance, what you might express as dF/dx if x is distance (or 2*G*M/r^3, if you prefer). Unfortunately, our language doesn't seem well equipped with different terms for these different tidal concepts.

I won't let you get away with it, though. Force (or even force difference) per distance is not force. It has units of mass per time squared.
I understand that, and I concede it, in case you didn't notice. That's why I'm so wishy-washy flipping between calling it "tidal force", "tidal effects", and "tides". If there's a proper term for dF/dx or (2*M*G/r^3), I'd be more than happy to use it exclusively if you'll share it with the rest of the group, especially if it isn't just plain "tide" by itself. I suppose if you wanted to get relativistic about it, you could refer to it as "locally positive curvature of space along a vector", or something like that. But that's quite a mouthful. :-? I've always wanted a word for dF/dx, though, ever since I first derived it and realized its significance.

JohnOwens
2004-Mar-25, 02:42 AM
Does anyone know how long the length on the day would be and how many days there would be when the Earth is in synchronized rotation with the moon? And how far away will the moon recede from us? Thanks.
OK, finishing up my big, equation-laden figuring (http://www.badastronomy.com/phpBB/viewtopic.php?p=228996#228996), now....

We left off with an angular momentum of 3.588*10^34 kg*m^2/s. At this point, I need to use a symbol for angular velocity, which is conventionally the Greek lowercase omega, but a "w" looks close enough and we aren't using it for anything else here, so we'll use it. L is angular momentum.
L_E = w_E*I_E : I_E = (2/5)*m_E*r_E^2
L_Ed = w_M*I_Ed : I_Ed = I_E + m_E*d_E^2
L_M = w_M*I_M : I_M = (2/5*m_M*r_M^2
L_Md = w_M*I_Md : I_Md = I_M + m_M*d_M^2
(the colons are just separating equations) where _E terms describe Earth's rotation on its own axis, _Ed terms are Earth's revolution around the barycenter, _M are for the Moon's rotation around its own axis, and _Md are for the Moon's revolution around the barycenter. w_E is the speed of the Earth's rotation, w_M is the speed of the Moon's rotation, the Moon's revolution, and the Earth's revolution around the barycenter, all at once. d_* is the distance from the barycenter.
For the solution for what we'll finally end up with for when we're synchronized, we set w_E = w_M. Since L_total = L_E+L_Ed+L_M+L_Md, when w_E=w_M we can use the above equations to get L_total = w_M*(I_E+I_Ed+I_M_I_Md), or L_total/w_M = I_total = I_E+I_Ed+I_M+I_Md. Substituting for I_Ed & I_Md gives us I_total = 2*I_E + 2*I_M + m_E*d_E^2 + m_M*d_M^2. Using one of the formulas relating to barycenters, d_E = d_total*(m_M/(m_E+m_M)), and substituting and canceling and manipulating, we get I_total = 2*I_E + 2*I_M + d_M^2*m_M*(1+m_M/m_E). Since I_E and I_M are both known, let's call (2*I_E+2*I_M) I_sum, to shorten things a bit, so we have I_total = I_sum + d_M^2*m_M*(1+m_M/m_E).
Now we go back to the old classic orbital equation, T^2*M*G=4*pi^2*r^3, manipulate a bit, and we get w=sqrt(M*G/r^3). The "r" there is equal to d_E+d_M, so we have to manipulate a bit to get it in terms of d_M. We end up with a nasty thing I'll set apart here,
w = sqrt( ((m_M+m_E)*G*m_E^3) / (d_M^3*(m_M+m_E)^3) )
w = sqrt( (G*m_E^3) / (d_M^3*(m_M+m_E)^2) )
w = sqrt( (G*m_E^3/(m_M+m_E)^2) / d_M^3 )
Plugging in known values, we get
w = sqrt( 3.891*10^14 m^3/s^2 / d_M^3)

Plugging that into L_total = w_M*(I_E+I_Ed+I_M_I_Md) = w_M*(I_sum+d_M^2*m_M*(1+m_M/m_E)), along with some known values, we end up with this:
3.588*10^34 kg*m^2/s = sqrt(3.891*10^14 m^3/s^2 / d_M^3) * (1.942*10^38 kg*m^2 + d_M^2*7.439*10^22 kg)
3.588*10^34 kg*m^2/s=(sqrt(3.891*10^14 m^3/s^2/d_M^3)*1.942*10^38 kg*m^2)+(sqrt(3.891*10^14 m^3/s^2/d_M^3)*d_M^2*7.439*10^22 kg)

At this point, there may be some more elegant way to use something quadratic-equation-like to solve for d_M, but I just plugged it into my calculator to have it solve by reiteration (I think it uses Newton's method, but I don't know the details). I got d_M = 589,100 km, and using that and some of the above gibberish, I got the distance between the Earth and Moon to be 596,300 km. The Earth will be 7,240 km from the barycenter, which will finally be above the surface. Using the orbital equations, the day/month or whatever you call it will be 52d 17h 1.2min long (where a "day" equals 86400 seconds, of course).

In the really long run, things will get complicated as the Sun transfers angular momentum to the Earth in the same way, the Earth's year gets longer, the Sun's day gets longer, etc., and let's not bring the Sun's future expansion into it, please!, and the Sun slows down the Earth's rotation, which is accelerated in turn by the Moon, etc. etc. If it's all the same with you, I'd rather not go into that right now. :-? :wink:

milli360
2004-Mar-25, 08:48 AM
I understand that, and I concede it, in case you didn't notice.
That made me think that I'd missed something, so I went back and read the thread. What concession?

The only thing I noticed is the post (http://www.badastronomy.com/phpBB/viewtopic.php?p=229028#229028) where you claim that you took the additional factor of 1/4 into account because of the reduced moment arm, but that's not true--not only is the moment arm reduced, but the force is too, in addition. The factor is multiplied.


That's why I'm so wishy-washy flipping between calling it "tidal force", "tidal effects", and "tides". If there's a proper term for dF/dx or (2*M*G/r^3), I'd be more than happy to use it exclusively if you'll share it with the rest of the group, especially if it isn't just plain "tide" by itself.
Usually, it's expressed as a differential, 2MG/r^3 times dr, where dr is assumed to be much smaller than r. Leaving off the dr means you are talking about oranges when you should be talking about apples.

Sawicki, in his paper (http://www.jal.cc.il.us/~mikolajsawicki/Tides_new2.pdf), calculates the 50 days or so period of "earth lock", and adds that the solar tide on the system would cause the moon to spiral back in to the earth. Much further in the future.

Kaptain K
2004-Mar-25, 05:43 PM
The Earth will be 7,240 km from the barycenter, which will finally be above the surface. Using the orbital equations, the day/month or whatever you call it will be 52d 17h 1.2min long (where a "day" equals 86400 seconds, of course).
I read (many years ago) that when the Earth becomes tidally locked to the Moon, that the "day/month" would be 55 (current) days long. Definitely within a reasonable margin of error! =D>

JohnOwens
2004-Mar-25, 09:53 PM
I understand that, and I concede it, in case you didn't notice.
That made me think that I'd missed something, so I went back and read the thread. What concession?

The only thing I noticed is the post (http://www.badastronomy.com/phpBB/viewtopic.php?p=229028#229028) where you claim that you took the additional factor of 1/4 into account because of the reduced moment arm, but that's not true--not only is the moment arm reduced, but the force is too, in addition. The factor is multiplied.

So, while the Earth might mass "only" 81 times the Moon, its moment of inertia is about 1100 times greater. And worst of all, I don't know just how the larger radius affects the tidal force exerted on the body. :-? It might not even have any affect at all, if the change in torque cancels with the change in moment of inertia.(emphasis added)
Now I'm going to have to figure out what the original question was all over again, it's been through so much since then. :wink:


Usually, it's expressed as a differential, 2MG/r^3 times dr, where dr is assumed to be much smaller than r. Leaving off the dr means you are talking about oranges when you should be talking about apples.
OK, but short version: what's the English name for it? Or is it something only expressed mathematically, without a unique name?


I read (many years ago) that when the Earth becomes tidally locked to the Moon, that the "day/month" would be 55 (current) days long. Definitely within a reasonable margin of error! =D>
Well, I was considering bringing up the non-uniform density distribution, which should decrease the Earth's rotational moment of inertia, hence increasing the eventual day/month length. But I thought things were fairly complicated enough as it was without doing nonlinear integration on it. :wink:

Andreas
2004-Mar-26, 01:55 AM
The Moon revolves in quite a different way - one revolution per day; so one face is always toward us.

Correct me if I am wrong, but I believe the Moon orbits around the Earth once per 28 days and has a rotation period of 28 days, so that the same hemisphere is always facing toward us.
In other words, a moon day is 28 earth days and that is equal to the time for one revolution around the earth. Maybe you got tripped over the confusion of revolution and rotation. :)

milli360
2004-Mar-26, 10:45 AM
The Moon revolves in quite a different way - one revolution per day; so one face is always toward us.

Correct me if I am wrong, but I believe the Moon orbits around the Earth once per 28 days and has a rotation period of 28 days, so that the same hemisphere is always facing toward us.
In other words, a moon day is 28 earth days and that is equal to the time for one revolution around the earth. Maybe you got tripped over the confusion of revolution and rotation.
It looks to me more likely was confused by the use of "day" to mean "moon day". In the quoted passage, "one revolution per day" without the qualifier can (should?) be interpreted as one revolution per 24 hours.

hedin
2004-Mar-28, 06:58 PM
Im not gonna get into the math since Im not smart enough. Just wanted to say that there are no dumb questions-only dumb answers. :D

Brady Yoon
2004-Mar-28, 08:18 PM
Will the Earth become tidally locked to the moon before the sun becomes a red giant or after?

milli360
2004-Mar-28, 09:40 PM
Will the Earth become tidally locked to the moon before the sun becomes a red giant or after?
Probably not, or are you looking for some sort of guarantee? :)