# Which mass distributions guarantee two bodies have non-Keplerian orbits? Which non-spherical distributions still allow noncircular Keplerian orbits?

The trajectories of two point masses or spherically symmetric masses with respect to their center of mass are conic sections or Kepler orbits.

Consider that the bodies have finite size with respect to their separation and not necessarily uniform, or even spherically symmetric mass distributions.

In that case what are the constraints on their mass distributions and orientations such that their orbits are still Keplerian? Or does any deviation from spherical symmetry of one or both body immediately result in a non-Keplerian orbit?

This answer to the Astronomy SE question Which point in an orbiting body most closely follows its Keplerian trajectory? explains:

If one or both of the bodies have non-spherically symmetric density distribution, the orbits will no longer be keplerian.

but without citing sources nor using math. I can certainly imagine "... are no longer necessarily Keplerian" but is this always true in absolute?

Just for example couldn't two tidally locked ellipsoids have circular orbits about their center of mass?

If so, perhaps that's the only exception, but perhaps not. If they orbited in non-circular orbits would their apparent libration also mean that their centers of mass no longer followed strict Keplerian orbits? Would they differ from ellipses, or still be elliptical but no longer exactly follow equal area per unit time trajectories? Would there be another point within the bodies besides their centers of mass that still did follow a Keplerian orbit?

Question: Which mass distributions guarantee that two bodies will always have non-Keplerian orbits? Which non-spherical distributions still allow for noncircular Keplerian orbits?

• @Qmechanic thanks for the edit; can you help me understand why the classical-mechanics tag does not apply? I'm thinking that those following the tag might be able to answer this fairly simple mechanics problem, and they might not also be following celestial-mechanics. Thanks! – uhoh Jan 20 '19 at 6:15
• If there are a tag X and a subset tag Y that both apply, the general guideline is to use just the subset tag Y. – Qmechanic Jan 20 '19 at 7:05
• @Qmechanic It's not a very "celestial" question, just two masses with a 1/r^2 force (could as well be +/- charges), but I've got Kepler in there, so okay. – uhoh Jan 20 '19 at 7:07

This answer is trivial, but Keplerian orbits are assured as long as the gravitational fields are spherically symmetric and thus decay as $$1/r^2$$. There are an infinite number of mass distributions that can generate a given field. For example a cubic shell of side $$l$$ with a mass density $$(l/2)/((l/2)^2+x^2+y^2)^{(3/2)}$$ on each square surface will produce a spherically symmetric field. In this example, the mass density at the corners is about 1/5 of that at the centers of the faces.
Some insight can be gained by analogy with electric field/flux. 'Tubes' of gravitational flux must terminate on elements of mass. If the incoming flux is spherically symmetric, then the mass density of the shell must go as $$1/r^2$$ where $$r$$ is its distance from some interior point. In addition, since the surface will generally be at an angle to the incoming radial tube, the mass density gets reduced by the cosine of that angle. This is exactly how the result for the cubic shell is derived.
So, in theory, any shell that completely encloses a central point can produce a spherical field. This result can obviously be extended to solid bodies.
It is interesting to speculate whether a field can go as $$1/r^2$$ only in one (say equatorial) plane but not in other planes. I think the answer is 'no' because the far-field from any object is spherically symmetric and this boundary condition seems to preclude such solutions except by approximation over a limited range of radii (e.g. perhaps by sticking a dumbell through the hole in a donut).

• This sounds very convincing! Since matter is "transparent" to gravity do tubes of flux really terminate on the gravitational "charges" in the same way that they might on electrical charges on an electrical conductor's surface? After coffee I"ll go off an check this numerically. You may also be interested in What would a Helmholz coil-like mass configuration look like? (produces locally uniform gravity field) – uhoh Oct 17 '20 at 22:01
• @uhoh I've just found this wonderful article on a cubic planet complete with lakes and mountains and non-Keplerian satellites: arxiv.org/pdf/1206.3857.pdf – Roger Wood Oct 17 '20 at 22:05
• @uhoh I added a nice Helmholz-ish answer to your other question – Roger Wood Oct 20 '20 at 5:40
• I see that, thank you for that! As you already knew it would and must this answer checks out numerically. Thank you for your answer here with a concrete example. I'll do the ice cream cone and the donut + donut hole next. – uhoh Oct 20 '20 at 7:53
• @uhuh this answer would seems to extend to solid 3D objects ok. For the cube example, the shells can be stacked one inside the other to form a solid cube. Insisting on constant density is a bit more tricky, but the thin shell of varying density could be replaced with a shell of varying thickness. All that is required is that the incoming converging gravitational flux tubes terminate on the appropriate amount of mass. For constant density this means the appropriate volume of material. But it seems inevitable that a constant density object will be hollow, otherwise it has to be a solid sphere. – Roger Wood Oct 25 '20 at 6:29

There are infinitely many such shapes that work, even ones without anything like spherical symmetry at all. To give a simple example, recall the method of images for spheres in electromagnetism. When you move a point charge near a conducting sphere, it induces a non-symmetric charge distribution on the sphere. But the electrostatic field of that charge distribution is exactly mimicked by a fictitious point charge inside the sphere, called the image charge.

That means, conversely, that the gravitational field of a point mass can be exactly mimicked by a non-symmetric mass distribution encasing it. By superposing multiple such distributions, you can get a mass distribution that looks as weird as you want, but whose field has exactly $$1/r^2$$ behavior outside it.

• @uhoh Why wouldn't the analogy apply? All the equations that matter are the same. The fact that a certain charge distribution on a sphere has the same external field as an image point charge follows from nothing besides Coulomb's law. So the fact that the analogous mass distribution on a sphere has the same external field as a point mass follows from nothing besides Newton's law of gravity. They're both inverse square laws, so what's the problem? – knzhou Oct 18 '20 at 23:46
• @uhoh Consider a grounded conducting sphere in the presence of a negative point charge. Positive charge is induced on the sphere to shield the field lines from the negative charge. All I am saying is that you can turn that positive charge distribution to a positive mass distribution, and it'll also look like it's made by an image point mass. I'm not saying that you can shield gravitational charge, because you can't even set up the same situation in gravity; there are no negative masses. – knzhou Oct 19 '20 at 0:33
• @uhoh But that's not what I'm doing. I'm saying that the technique of image charges in electromagnetism, which does not at all apply to gravity, still yields a useful particular example of a charge distribution. That particular example of a charge distribution has an analogous mass distribution, which has an analogous gravitational field because Coulomb's law and Newton's law of gravity are both inverse square. – knzhou Oct 19 '20 at 0:51
• Got it now, yes, thanks! – uhoh Oct 19 '20 at 0:54
• "the electrostatic field of that charge distribution is exactly mimicked by a fictitious point charge inside the sphere, called the image charge". This statement is true for a flat plane, but not for a sphere. Spherical aberration causes the (virtual) image of a point to be spread into a caustic curve. – Roger Wood Oct 19 '20 at 6:12