<p>Yes and No.  — It depends on what you mean by &quot;stable&quot;. To be precise, &quot;stable&quot; means immune to small perturbations.  &quot;Equilibrium&quot; can be either &quot;stable&quot; or &quot;unstable,&quot; as shown in any first-year Calculus text.</p>
<p>Can you balance a pencil on its point?</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p>The best known stable solutions to the three-body problem are hierarchical. Either a &quot;sun&quot; is orbited by a &quot;planet&quot; which is orbited by a &quot;moon&quot;,  or two &quot;suns&quot; are in a tight orbit, which is orbited by a &quot;planet&quot;.  In these configurations there is a clear structure, and the orbits of each level can be approximated by Keplerian ellipses.</p>
<p>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 <a href="https://en.wikipedia.org/wiki/Three-body_problem#Special-case_solutions" rel="noreferrer">solutions of the three-body problem are known</a> 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.</p>

CC BY-SA 4.0
2022-02-05T20:14:42.600Z
Ah makes sense, thanks for the answer!
ROODAY
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2022-02-06T05:43:44.377Z
@ROODAY Related, Klemperer rosette, which involves unequal masses but similar concepts. https://en.wikipedia.org/wiki/Klemperer_rosette
user3067860
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2022-02-06T08:01:10.823Z
@James K a followup, does your answer mean that the proper definition of a stable orbit is one that is resistant to perturbations?
ROODAY
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2022-02-06T09:14:26.067Z
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
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2022-02-06T11:18:33.473Z
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
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2022-02-06T16:30:16.023Z
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 mistrusts SE
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2022-02-06T17:03:50.723Z
I probably need to say something like "Non-heiracrical solutions".
James K
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2022-02-06T20:53:17.800Z
@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
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2022-02-06T23:01:58.990Z
@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 mistrusts SE
CC BY-SA 4.0
2022-02-07T07:34:23.737Z
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
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2022-02-07T11:57:56.503Z
and in your computer model, numerical instability might perhaps be a problem  (depending on everything of course)
ilkkachu
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2022-02-07T19:25:06.900Z
@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
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2022-02-07T20:04:31.077Z
@J... Ah, fair enough.
8bittree
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2022-02-09T22:11:45.253Z
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
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2022-02-05T19:48:46.327Z
2022-02-07T13:05:46.777Z
2022-02-07T13:05:46.777Z
James K
<p>I.e. a system that has 3 objects of equal mass, rotating around the system's center of gravity like so:
<a href="https://i.sstatic.net/WZYL5.png" rel="noreferrer"><img src="https://i.sstatic.net/WZYL5.png" alt="enter image description here" /></a></p>
<p>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).</p>

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2022-02-07T13:03:39.380Z
What you've drawn is reminiscent of Ringworld, which was eminently unstable even with a large stellar mass at the system CM .
Carl Witthoft
<orbit><gravity><n-body-simulations>
Is it possible to have a stable 3 body system that orbits in a perfect circle?
ROODAY
# Is it possible to have a stable 3 body system that orbits in a perfect circle?

(Asked by: username_0 on 2022-02-05)

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](https://i.sstatic.net/WZYL5.png)](https://i.sstatic.net/WZYL5.png)

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).

**Tags:** orbit, gravity, n-body-simulations

> **Comments:**
> **username_1**: What you've drawn is reminiscent of Ringworld, which was eminently unstable even with a large stellar mass at the system CM .  

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## 1 answers

### Answer by username_2 on 2022-02-05. Score: 34 | Accepted answer

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](https://en.wikipedia.org/wiki/Three-body_problem#Special-case_solutions) 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.

> **Comments:**
> **username_0**: Ah makes sense, thanks for the answer!  
> **username_3**: @username_0 Related, Klemperer rosette, which involves unequal masses but similar concepts. https://en.wikipedia.org/wiki/Klemperer_rosette  
> **username_0**: @username_2 a followup, does your answer mean that the proper definition of a stable orbit is one that is resistant to perturbations?  
> **username_2**: 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.  
> **username_4**: 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?'  
> **username_5**: username_2, 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.  
> **username_2**: I probably need to say something like "Non-heiracrical solutions".  
> **username_2**: @username_4  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.  
> **username_5**: @username_2, 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 :-) ).  
> **username_6**: 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)  
> **username_7**: and in your computer model, numerical instability might perhaps be a problem  (depending on everything of course)  
> **username_8**: @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. username_2's answer does a good job of pointing out that there is a precise definition, and that it *might* not have been the term username_0 meant, while still answering what was probably meant.  
> **username_8**: @J... Ah, fair enough.  
> **username_9**: 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).  

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