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Keplerian orbits can describe the motion of a negligible mass around a massive fixed central body, but they can also describe the motion of two bodies around their center of motion... with some constraints on their mass distributions.

But which point within each body is the point that traces out it's orbit? And which two points define the distances used to determine the center of motion? Are they the center of mass, or the center of gravity, or in fact a different point that needs its own definition?

Each body has a significant, non- negligible size and could have a non-uniform, or potentially a non-spherically symmetric density distribution.

uhoh
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If one or both of the bodies have non-spherically symmetric density distribution, the orbits will no longer be keplerian. The gravitational force will not only be dependent on the distance to the central body (or barycentre in the case of two bodies with non-negligible mass) but also on the relative orientation of both bodies.

For two spherically symmetric bodies, the c.o.m. (or c.o.g.) will trace out the orbit around the barycentre. The barycentre is determined by the same points.

Alexander Vandenberghe
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  • Thank you for your answer! Can you find some sources to support your conclusions? Also, how is the center of gravity (c.o.g.) of an object defined when it sits in the gravity field of another body? Is it really identical to the center of mass? – uhoh Jan 10 '19 at 13:29
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    It’s defined to be that point where if, hypothetically, all the mass were located, then the net gravitational force would be the same as it really is given the true mass distribution. In general it’s not identical to the center of mass, but it is for spherically symmetric bodies (see Gauss’s Law). – David Schneider-Joseph Jan 10 '19 at 15:27
  • @DavidSchneider-Joseph is that true when the spherically symmetric body is in a gravity field with a nonzero gradient as well, such as 1/r (or more complicated) potential from another nearby body? – uhoh Jan 11 '19 at 02:40
  • @DavidSchneider-Joseph fyi I"ve just asked Constraints on the mass distribution within each body such that their mutual orbits are Keplerian? – uhoh Jan 11 '19 at 02:55
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    It’s true for a spherically symmetric body in any gravitational field produced by mass located on the exterior of the body. – David Schneider-Joseph Jan 11 '19 at 08:00
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    It’s also true for any body (spherically symmetric or not) in a uniform gravitational field. – David Schneider-Joseph Jan 11 '19 at 08:00
  • @DavidSchneider-Joseph would you be inclined to expand on that in the form of an answer? It wouldn't need to be very long, just definitive. Thanks! – uhoh Jan 24 '19 at 00:41
  • @AlexanderVandenberghe * If one or both of the bodies have non-spherically symmetric density distribution, the orbits will no longer be keplerian.* I guess the only possible exception to that could be a mathematical oddity of a perfectly circular orbit with perfect tidal locking, and the probability of a perfect anything is zero in the real world. Thanks! – uhoh Feb 04 '19 at 08:33
  • One lingering question I have is if c.o.g. is actually the same as c.o.m. for non-spherically-symmetric bodies, or if c.o.g. is even defined in orbital mechanics or ends up being a rabbit-hole. – uhoh Feb 04 '19 at 08:35