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Old Nov8-09, 03:03 PM                  #1
SidT

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Effect of Volume on Revolution

Let us consider the revolution of the moon around the Earth. If the moon was expanded like a balloon so that it had the same mass but different volume (and therefore lower density), would it effect the moon's revolution in any way?
I probably sound like a retard right now...
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Old Nov8-09, 05:15 PM       Last edited by Gerenuk; Nov8-09 at 06:18 PM..            #2
Gerenuk

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Re: Effect of Volume on Revolution

For perfect spheres there wouldn't be a difference in gravitational attraction so the orbit would be the same I suppose.

It's easy to show that the potential of a spherical object is the same as the one of a point. I just did a calculation and apparently even the force on a homogeneous sphere due to a potential is the same as the force on a point particle with the same mass at the center of the sphere. So the gravitational attraction between spheres is the same as between points.

http://en.wikipedia.org/wiki/Shell_theorem

But somehow there are also rotational effects which cause the moon to sync with the earth. Not exactly sure how that works, but a different radius would make a difference.
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Old Nov8-09, 08:11 PM                  #3
SidT

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Re: Effect of Volume on Revolution

Thanks a lot. :)
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Old Nov8-09, 09:43 PM                  #4
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Re: Effect of Volume on Revolution

Originally Posted by Gerenuk View Post
But somehow there are also rotational effects which cause the moon to sync with the earth. Not exactly sure how that works, but a different radius would make a difference.
The effect is called tide-lock. I'm certainly no expert, but it seems to me that it depends upon the distribution of mass within one or both of the component bodies. That would, for instance, imply that the near side of the moon is slightly more massive than the far side.
Best that one of the Astronomy or Astrophysics experts elaborate. You can also Google tidal lockage.
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