Distortion of volume of bodies by gravity

[grav]

We just had a very interesting discussion in the Spinning Moon (http://www.bautforum.com/showthread.php?t=43086) thread. It appears that the gravitational attraction of Earth on the moon will create bulges in the front and back of 2x, as well as a squeezing effect of -x around its circumference as viewed from the Earth. x is the ratio of the radius of the moon to the distance between the centers of the Earth and moon, or rm/dE,m. The centrifugal force of its orbit will also create an equatorial bulge of y around the equator, where y is the ratio of the radius of the moon to the distance from the center of the moon to the barycenter, or rm/dbary=(rm/dE,m)*[(ME+Mm)/ME]. To find the acceleration for either of these, we simply multiply them times the acceleration at the center of the moon at its distance from the Earth, which is considered 1. This acceleration is GME/dE,m2.

Someone mentioned in another thread that one could not find the actual distance for the bulges without actually knowing the make-up of the moon. I now believe that we can. With no outside influences, and no rotation, the moon would be a perfect sphere with radius rm. The gravitational acceleration on its surface would match the internal outward acceleration that keeps it from collapsing, and that would be gm=GMm/rm2. An extra gravitational influence on its surface of x*gE(d), then, would give it a total inward acceleration of x*gE(d)+gm=GMErm/dE,m3+GMm/rm2. The surface would then collapse to about gm/(x*gE(d)+gm) of its former size. Let's define x*gE(d)/gm as x', which is [(GME/dE,m2)*(rm/dE,m)]/(GMm/rm2)=(ME/dE,m3)/(Mm/rm3)=(ME/Mm)*(rm/dE,m)3=(ME/Mm)*x3. And y'=y*gE(d)/gm=(ME/Mm)*x2y. The formula is now 1/(1+x'), which is approximately 1-x' times smaller. The total contraction (for x is small), then, is x'*rm=rm2/dE,m for the lateral squeezing, twice that for the tidal bulges on the front and back, but expanding, and a equatorial bulge of y'*rm=rm2/dbary=(rm2/dE,m)*[(ME+Mm)/ME].

As far as the volume goes, it would, of course, be (4pi/3)rm3 with no additional forces acting on it. With just the gravitational influences, it would become a prolate ellipsoid with a volume of (4pi/3)ab2, where a is rm(1+2x') and b is rm(1-x'). The entire volume would then become V=(4pi/3)r3(1-3x'2+2x'3), so that it is smaller by approximately 3x'2 than it would be for a perfect sphere. With only the centrifugal forces being considered, it becomes an oblate ellipsoid, with a volume of (4pi/3)a2b, where a is rm(1+y') and b is rm. The volume, then, for a distorted rotating sphere only, then, is V=(4pi/3)rm3(1+2y'+y'2), for an increase in volume of approximately 2y'. For the moon in orbit, both of these will take place at the same time and create an ellipsoid of volume (4pi/3)abc, where a is rm(1+2x'+y'), b is rm(1-x'+y'), and c is rm(1-x'). This gets a little long, however, but if we determine the distance to the barycenter to be about the same as the distance beteen the centers of the Earth and moon (ME>>Mm), then x' and y' are about equal. So now we have a volume of V=(4pi/3)rm3(1+3x')(1)(1-x')=(4pi/3)rm3(1+2x'+3x'2), which is greater than a perfect sphere by about 2x'. Since the gravitational component alone would subtract 3x'2 from the volume and centrifugal alone would add 2y', and since x' is close to y' and the total volume is greater by 2x' (or 2y'), then the main contributor to the extra volume is the centrifugal force, since the difference in volume is about the same with or without the gravitational force, even though the tidal bulges extend twice as far, because the equatorial bulge extends all the way around the moon and up its sides.

Well, that's it for a minute. I just thought all of this was interesting. If anybody else has anything else to add, please do. I think I'll give my fingers a rest for now.

[Tensor]

We just had a very interesting discussion in the Spinning Moon (http://www.bautforum.com/showthread.php?t=43086) thread.

Snip...

Well, that's it for a minute. I just thought all of this was interesting. If anybody else has anything else to add, please do. I think I'll give my fingers a rest for now.

The point about having to know the composition, before knowing how large the bulges are, is very valid. The amount of the bulge is dependent on the equations of state of the material making up the bulge. This effect is rather obvious on the Earth. If you consider the atmosphere, the water, and the solid land, the greatest bulge is in the atmosphere. The next would be the water and the smallest would be the solid land.

By the way, the bulges and squeezes are an example of tidal gravity. It's those effects that are used in gravitational wave detectors.

[grav]

The point about having to know the composition, before knowing how large the bulges are, is very valid. The amount of the bulge is dependent on the equations of state of the material making up the bulge. This effect is rather obvious on the Earth. If you consider the atmosphere, the water, and the solid land, the greatest bulge is in the atmosphere. The next would be the water and the smallest would be the solid land.
That's what I thought at first, too. But it turns out (as far as I can tell) that it makes no difference at all. Whatever the make-up, it is initially balanced by its own gravity, and that gives us the acceleration to distance ratio, similar to a spring. The ratio for the moon is (GMm/rm2)/rm=GMm/rm3, which is proportional to Dm, so if other moons or planets have about the same density, then they should have about the same "spring" ratio. This is because no matter what the make-up is, it is still counterbalanced by its own gravity at rm (when not rotating or in another gravity field). I did not consider how different layers would act, however, for solid, liquid, and gas, as you are talking about with Earth. I suppose that might be quite different. I am actually only considering that the body's make-up is more or less consistent, as the moon should be.

[hhEb09'1]

What blows me away grav is the total lack of units across your calculations. :)

When you say "The gravitational acceleration on its surface would match the internal outward acceleration that keeps it from collapsing" it makes it sound as if you were matching a force there instead of accelerations. :)

[grav]

What blows me away grav is the total lack of units across your calculations. :)I like to work in unitless quantities. I find I can determine so much more this way. I can spot relationships so much easier and pick out equalities at a glance.

When you say "The gravitational acceleration on its surface would match the internal outward acceleration that keeps it from collapsing" it makes it sound as if you were matching a force there instead of accelerations. :)
No. Just accelerations. The force would depend on the mass of the atoms being accelerated. I do sometimes refer to the force, since it still exists, of course, but it would really just be a hinderance to calculate for an unknown mass, and is unnecessary and therefore not used in the calculations.

[grav]

I will now calculate for the equatorial bulge. The extra distance past the normal radius for a perfect sphere would be simply y'*rm, and y'=(ME/Mm)*x2y, where x is rm/dE,m and y is (rm/dE,m)*[(ME+Mm)/ME]. The entire distance, then, is dy=rm*(ME/Mm)*(rm/dE,m)3*[(ME+Mm)/ME)=rm*[(ME+Mm)/Mm]*(rm/dE,m)3, where ME=5.9736*1024 kg, Mm=7.34*1022 kg, rm=1.737*106 m, and dE,m=3.844*108 m. We get dy=rm*(6.047*1024 kg/ 7.34*1022 kg)*(1.737*106 m/ 3.844*108 m)3=7.6014*10-6*rm=13.2 m.

Grant Hutchinson,
If you read this, I believe you mentioned in another thread that the theoretical difference between the equator and the poles is about 500 meters, which is much different than what I have here. Are you sure that's correct? Well, it is just theoretical, so I'm sure it's difficult to say. Even if I made the error somewhere of not multiplying by (ME/Mm) somehow, then I would get twice that difference.

Does anyone know of another large solid body with a known equatorial bulge that I can try this with?

[hhEb09'1]

I like to work in unitless quantities. I find I can determine so much more this way. I can spot relationships so much easier and pick out equalities at a glance. In your calculation of the precession of Mercury's orbit, you calculated a value, and it equaled the precession, but how do you know they have anything to do with each other?

[grav]

In your calculation of the precession of Mercury's orbit, you calculated a value, and it equaled the precession, but how do you know they have anything to do with each other?
I don't yet. So far it was just a good guess that happened to work out, although I did think through it logically and it came out right the first time, so I think that's a good sign. But I recently asked a question in Q&A to see if anyone knew the precession of the moon after the sun, Earth, and all of the other planets are accounted for, so I can try it with that as well. I haven't had much luck searching, and I haven't received an answer yet.

[grav]

As you well know, hhEb09'1, ;) the centrifugal acceleration caused by revolution is exactly the same as that for rotation. So I've decided to try that as well. Since we are only considering quantities for the moon only in the case, values for the Earth need not be considered. The extra equatorial distance, then, would be dy=r*[g/(g+a)]-r, which becomes dy=r*(1-a/g)-r=r*(a/g) for small a/g. a is v2/r=(2pir/P)2/r=4pi2r/P2, and g=GM/r2. We now have dy=r*[(4pi2r/P2)/GM/r2)]=4pi2r4/GMP2. r=1.737*106 m, M=7.34*1022 kg, and P=2.36*106 s. We get as a result, dy=13.2 meters, just as before.

[hhEb09'1]

I don't yet. So far it was just a good guess that happened to work out, although I did think through it logically and it came out right the first time, so I think that's a good sign. But I recently asked a question in Q&A to see if anyone knew the precession of the moon after the sun, Earth, and all of the other planets are accounted for, so I can try it with that as well. I haven't had much luck searching, and I haven't received an answer yet.There's an idea. Make your calculation first, and post it, before you know the answer. :)

[hhEb09'1]

As you well know, hhEb09'1, ;) the centrifugal acceleration caused by revolution is exactly the same as that for rotation. I disagree. What you've done is confuse the centrifugal force of rotation, and think it is caused by revolution.

[grav]

I disagree. What you've done is confuse the centrifugal force of rotation, and think it is caused by revolution.
Well, I mean it is the same calculated either way, so that's what I did. :)

[grav]

Well, I mean it is the same calculated either way, so that's what I did. :)
And since that's the case, I will calculate it for the Earth as well. It shouldn't matter that it is revolving with the moon. So the formula is dy=4pi2r4/GMP2, where r=6.371*106 m, M=5.9736*1024 kg, and P=86164 s. From this I get dy=21979 meters, or about 22 kilometers (44 km across the diameter) difference between the equator and the poles. Is that correct?

[grav]

The point about having to know the composition, before knowing how large the bulges are, is very valid. The amount of the bulge is dependent on the equations of state of the material making up the bulge. This effect is rather obvious on the Earth. If you consider the atmosphere, the water, and the solid land, the greatest bulge is in the atmosphere. The next would be the water and the smallest would be the solid land.
Well, I just looked up the equatorial bulge of the Earth, and it comes very close to what I came up with. Apparently it does not matter too much what the composition is or how it is layered after all. :)

[Tensor]

That's what I thought at first, too. But it turns out (as far as I can tell) that it makes no difference at all.

But it does make a difference. If it didn't matter, we wouldn't have tides, as the water and the land would move the same amount. They don't, the water moves more, which we see as tides.

[hhEb09'1]

Well, I just looked up the equatorial bulge of the Earth, and it comes very close to what I came up with. Apparently it does not matter too much what the composition is or how it is layered after all. :)This wiki article (http://en.wikipedia.org/wiki/Figure_of_the_Earth) says 21.20 km, vs. your 21.98 km. I don't think you're going to be able to get close enough to pull out the effect of GR. :)

[grav]

But it does make a difference. If it didn't matter, we wouldn't have tides, as the water and the land would move the same amount. They don't, the water moves more, which we see as tides.Well, yes. That's true. I wonder if the average density of the oceans is different than the average density of the Earth as a whole.

[grav]

This wiki article (http://en.wikipedia.org/wiki/Figure_of_the_Earth) says 21.20 km, vs. your 21.98 km. I don't think you're going to be able to get close enough to pull out the effect of GR. :)
I wasn't doing any GR calculations here. :confused: It is purely Newtonian. :)

[grant hutchison]

That's what I thought at first, too. But it turns out (as far as I can tell) that it makes no difference at all.Radial mass distribution is important. A self-gravitating mass with much of its mass concentrated at the core will behave differently under tidal or rotational stress from one that has a homogeneous mass distribution, since the radial distribution of resultant forces will be different.

If you read this, I believe you mentioned in another thread that the theoretical difference between the equator and the poles is about 500 meters, which is much different than what I have here. Are you sure that's correct?That was for the moon? Pretty much. It comes from a standard formula that takes the mass distribution into account using the measured J2 term of the gravitational field of the moon.
You can find details in any of the standard references, such as Murray & Dermott's Solar System Dynamics, which is where I got it.

Grant Hutchison

[grav]

Well, apparently I've been beat to the punch once again, of course. Oh well, it's still a learning experience for me. In the website, Equatorial Bulge (Wikipedia) (http://en.wikipedia.org/wiki/Equatorial_bulge), at the very bottom, where it says Mathematical Expression, they have apparently come up with the same formula I have. It is 3pi/2GDP2, where D is the density and P is the period as described this way. The density is M/[(4pi/3)r3], so the formula becomes 3pi(4pi/3)r3/2GMP2, or 2pi2r3/GMP2. When multiplied by the diameter, one gets 4pi2r4/GMP2, exactly as I have. What confuses me, then, is how a theoretical equatorial bulge of the moon could be 500 meters, then, when the formula comes out to 13.2 meters. Is there something else that's not being accounted for? I will have to look into it.

[grant hutchison]

What confuses me, then, is how a theoretical equatorial bulge of the moon could be 500 meters, then, when the formula comes out to 13.2 meters. Is there something else that's not being accounted for? I will have to look into it.It's to do with the mass distribution, as I've said. You just need to get yourself a decent textbook ... it would save you a lot of work and blind alleys.

Grant Hutchison

[grav]

In your calculation of the precession of Mercury's orbit, you calculated a value, and it equaled the precession, but how do you know they have anything to do with each other?
Why did you post this on here? Did you somehow know they had something in common? :eh: It is the spin that causes centrifugal bulges and the spin that causes precession as well, according to my theory, anyway. But I could not figure out exactly what it was about the spin itself that would cause the precession to begin with. I thought it might have something to do with a Doppler effect for gravity, but it is only determined across the gradient of the equator itself. If some of the mass of the sun is displaced through its spin, causing an equatorial bulge, then we can probably no longer calculate by considering it as a point mass. This only works for a perfect sphere, as far as I know. Since there is a definite relation between the spin of a body and its equatorial bulge, regardless of its composition (as long as the composition is about uniform), then there might definitely be a relation between the displacement of mass due to rotation and the precession of the orbitting bodies.

[grav]

It's to do with the mass distribution, as I've said. You just need to get yourself a decent textbook ... it would save you a lot of work and blind alleys.

Grant Hutchison
So the moon's equatorial bulge is larger because its center of mass is offset, perhaps? That would probably cause the moon to oscillate back and forth, right? So maybe it throws more of its mass in one direction instead of the other, creating a larger equatorial bulge, but really just greater on one side? Or maybe the moon was much closer to the Earth at one time and revolved around it faster, and its equatorial bulge "froze" that way? :)

[grav]

Well, the equatorial bulge does indeed seem to be related to the precession. But in a strange way, since it is apparently related to the Schwarzchild radius as well. Let's take the equatorial bulge of the sun, which is found with 4pi2r4/GMP2=v2r2/GM=14515.579 meters. Okay, now let's take the amount of the precession due to its spin as found by my theory. This is vr/c=4629.90718 meters. This is the distance a planet will precess during each orbit. The ratio of these two distances is 14515.579 m/ 4629.90718 m=3.135177108. Now let's take the Schwarzchild radius. This is GM/c2=1476.760968 meters. Einstein predicted three times this value would equal the precession, which is 4430.282904 meters per orbit. Okay, now let's divide the precession according to my theory by the Schwarzchild radius. It is 4629.90718 m/ 1476.760968 m=3.135177107, exactly the same as that for the ratio of the equatorial bulge to the precession according to my theory. So the precession from spin, then, would equal the square root of the the multiple of the Schwarzchild radius and the extra radius of the equatorial bulge, or dspin=(dSchw*dequa)1/2.

[hhEb09'1]

Well, yes. That's true. I wonder if the average density of the oceans is different than the average density of the Earth as a whole.The average density of water is around 1, the average density of the Earth, including the water, is 5.5! Is there something else that's not being accounted for? I will have to look into it.After the formula that you reference, at the bottom of this wiki page (http://en.wikipedia.org/wiki/Equatorial_bulge), it says "The approximation is valid in the case of a fluid planet of uniform density"

The average density of granite is around 2.5, and that is about the average for the crust of the earth. The core is much more dense.

[grant hutchison]

It's to do with the mass distribution, as I've said. You just need to get yourself a decent textbook ... it would save you a lot of work and blind alleys.So the moon's equatorial bulge is larger because its center of mass is offset, perhaps?
As I've said before, you need to visualize the problem before you start flinging algebra at it.
Your calculations so far are based on the potential of a point mass. As you've found, the distortion is roughly proportional to the ratio of centrifugal to gravitational force at the equator.
But real objects have extended mass, and when they distort in response to the potential, they induce distortion in the potential. There's a feedback loop. If the object's mass is concentrated near its centre, it'll behave more like a point mass (the bulk of its mass will undergo slight distortion, and so feed back less on the potential). If an object's mass is extended, it'll feed back more.
So your calculation is "naive", in that it fails to address a significant aspect of reality.

Grant Hutchison