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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).

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    $\begingroup$ What you've drawn is reminiscent of Ringworld, which was eminently unstable even with a large stellar mass at the system CM . $\endgroup$ Feb 7, 2022 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.

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    $\begingroup$ Ah makes sense, thanks for the answer! $\endgroup$
    – ROODAY
    Feb 5, 2022 at 20:14
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    $\begingroup$ @ROODAY Related, Klemperer rosette, which involves unequal masses but similar concepts. en.wikipedia.org/wiki/Klemperer_rosette $\endgroup$ Feb 6, 2022 at 5:43
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    $\begingroup$ @James K a followup, does your answer mean that the proper definition of a stable orbit is one that is resistant to perturbations? $\endgroup$
    – ROODAY
    Feb 6, 2022 at 8:01
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    $\begingroup$ 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?' $\endgroup$
    – Sacha
    Feb 6, 2022 at 11:18
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    $\begingroup$ @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. $\endgroup$
    – 8bittree
    Feb 7, 2022 at 19:25

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