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I.e. a system that has 3 objects of equal mass, rotating around the system's center of gravity like so: enter image description here

Please excuse the crude drawing, but I've just been reading The Three-Body Problem book by Liu Cixin and it made me wonder if such a system like this could be possible (I don't know much about astronomy).

ROODAY
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    What you've drawn is reminiscent of Ringworld, which was eminently unstable even with a large stellar mass at the system CM . – Carl Witthoft Feb 07 '22 at 13:03

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Yes and No. — It depends on what you mean by "stable". To be precise, "stable" means immune to small perturbations. "Equilibrium" can be either "stable" or "unstable," as shown in any first-year Calculus text.

Can you balance a pencil on its point?

If you have three ideal spherical bodies (and an otherwise empty universe), moving according to Newtonian gravity (or perhaps even Einstein's relativity), and each body is moving exactly at the right velocity, then this system could exist.

However if you perturbed it by even the smallest amount, then it would gradually deviate from this orbit and probably end up with either a collision, or an ejection of one of the planets. In this sense it is not stable.

It is like balancing a pencil on the point. It is possible in theory, but in practice the pencil will always fall down. Similarly, this is possible in theory (or in a computer model) but could not exist in practice.

The best known stable solutions to the three-body problem are hierarchical. Either a "sun" is orbited by a "planet" which is orbited by a "moon", or two "suns" are in a tight orbit, which is orbited by a "planet". In these configurations there is a clear structure, and the orbits of each level can be approximated by Keplerian ellipses.

This solution was found by Lagrange, and it is a special case of the L4 and L5 orbits, in which the three bodies move in an equilateral triangle. Other solutions of the three-body problem are known However non-hierarchical solutions that are not only periodic, but resistant to small perturbations, don't exist when then the three bodies have equal mass.

Carl Witthoft
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James K
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    Ah makes sense, thanks for the answer! – ROODAY Feb 05 '22 at 20:14
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    @ROODAY Related, Klemperer rosette, which involves unequal masses but similar concepts. https://en.wikipedia.org/wiki/Klemperer_rosette – user3067860 Feb 06 '22 at 05:43
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    @James K a followup, does your answer mean that the proper definition of a stable orbit is one that is resistant to perturbations? – ROODAY Feb 06 '22 at 08:01
  • I think it is useful to distinguish between, for example, Sun+planet+L4 trojan (which is a stable three body configuration) and things like Sun+planet+L2 (JWST orbit) This latter orbit is not "stable" and will require the JWST to use rockets to remain in its halo orbit. – James K Feb 06 '22 at 09:14
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    The explaination and example describe an unstable equilibrium, as opposed to a stable equilibrium. For clarity and to avoid mixing up terms, the short answer at the top should just answer 'no' to the question: 'is it stable?' – Sacha Feb 06 '22 at 11:18
  • James, could you please clarify the very last statement: is it provable that a 3-body system of equal masses cannot be stable? I'm thinking of a tight binary with the 3rd body on a near-circular orbit with a very large radius w.r.t. the distance between the first two bodies. – kkm -still wary of SE promises Feb 06 '22 at 16:30
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    I probably need to say something like "Non-heiracrical solutions". – James K Feb 06 '22 at 17:03
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    @Sacha I tend to agree, but I've seen plenty orbits described as "stable" which are at unstable equilibrium. The "figure-8" orbits are often described as "stable", whereas I think "periodic" would be a better description. – James K Feb 06 '22 at 20:53
  • @JamesK, thank you, non-hierarchical does it! BTW, I don't even know if there is a formal criterion for a problem to be considered [non]hierarchical! :-) Do you? Re stable and periodic, I would rather separate the concepts of equilibrium (essentially, the "static stability") and periodicity (as the "dynamic stability"). The Poincare-Bendixon is the prime example of an attractive basin absent equilibria, albeit for a planar field case (and a potentially infinite time to reach it, but that's a technicality :-) ). – kkm -still wary of SE promises Feb 06 '22 at 23:01
  • https://en.wikipedia.org/wiki/Three-body_problem#Special-case_solutions would you describe this orbit as non-heirarchical? Its three bodies, equal masses, and stable (https://arxiv.org/pdf/math/0011268.pdf) – QCD_IS_GOOD Feb 07 '22 at 07:34
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    and in your computer model, numerical instability might perhaps be a problem (depending on everything of course) – ilkkachu Feb 07 '22 at 11:57
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    @J... There certainly is a "depends." It depends on whether the person using the term is using its scientifically precise definition, or if they are a layman, who may not be aware of such a precise definition, using what limited knowledge and vocabulary they have on the subject to convey a general idea. James K's answer does a good job of pointing out that there is a precise definition, and that it might not have been the term ROODAY meant, while still answering what was probably meant. – 8bittree Feb 07 '22 at 19:25
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    @J... Ah, fair enough. – 8bittree Feb 07 '22 at 20:04
  • The "resistant to perturbations" thing is pretty much the definition of "stable", though, so while I appreciate having the nuance in the answer, I think it should be a bit clearer that the answer is "no". (Also, L4 is only stable as long as the mass there is so small it's essentially a test particle in a 2-body system... Lagrange points are only really valid in that case, actually). – Thriveth Feb 09 '22 at 22:11