<p>(Ignoring that all orbits are technically unstable due to the emission of gravitational waves.)</p>

<p>There are known solutions to the gravitational three body problem that can be shown to be stable. Lagrange found a three body solution for general masses where all three orbit the common center of mass in an equilateral triangular formation. Gascheau proved in 1843 that this solution is stable if the component masses satisfy</p>

<p><span class="math-container">$$ \frac{m_1 m_2+ m_1 m_3 + m_2 m_3}{(m_1+m_2+m_3)^2} &lt; 1/27$$</span></p>

<p>More recently, it was shown by Kei Yamada and collaborators (<a href="http://arxiv.org/abs/arXiv:1212.0754" rel="noreferrer">1</a> <a href="http://arxiv.org/abs/arXiv:1505.04534" rel="noreferrer">2</a> <a href="https://arxiv.org/abs/arXiv:1612.08361" rel="noreferrer">3</a>) that if one includes the first order Post-Newtonian correction, this solution is modified by a (small) correction to the  triangle legs based on the component masses, i.e. it is only an equilateral triangle if all masses are equal. The general effect of the 1PN interactions is to decrease the region of parameter space where this solution is stable, but for suitably far separated systems there will still be masses for which the system is stable. Moreover, they also proved that (<a href="https://arxiv.org/abs/arXiv:1512.01087" rel="noreferrer">4</a>) these solutions are stable under the emission of gravitational radiation (i.e. a triangular system will adiabatically evolve into another triangular solution).</p>

<p>The effects of 2PN interactions (and beyond) on the stability of these triangular solutions is currently unknown (as far as I am aware).</p>

CC BY-SA 4.0
2019-11-15T09:41:50.347Z
This is excellent, thank you very much! I'll have a look at these this weekend.
uhoh
CC BY-SA 4.0
2019-11-15T17:58:16.450Z
Just a remark, regarding the equation, for the less math savvy: This constraint basically means that one of the masses is much heavier than both of the other two. If they are all the same the fraction comes out to $1/3$, and it goes to $0$ as one mass dominates; if the masses are $m,am,bm$ then $a+b\le 0.041$, that is, the other two masses together contribute only 4% of the largest mass.
Mario Carneiro
CC BY-SA 4.0
2019-11-16T02:14:40.630Z
If the masses are not equal, are they still in an "equilateral triangular formation"?
uhoh
CC BY-SA 4.0
2019-11-16T04:13:26.393Z
@uhoh They can still be in an equilateral triangle, but they spin around a point which is not the center of the triangle, but is weighted toward the heavier object.
Mario Carneiro
CC BY-SA 4.0
2019-11-16T04:49:12.690Z
@MarioCarneiro okay that must have been a pleasant surprise for Gascheau to have derived! I'll go off and read more and then run a numerical simulation to quell any lingering (and apparently unfounded) doubts. Thanks!
uhoh
CC BY-SA 4.0
2019-11-16T06:05:20.703Z
@uhoh See also https://en.wikipedia.org/wiki/Trojan_(celestial_body) - when one mass is much smaller than the other two, it is basically hanging around the L4/L5 lagrangian points of the orbit generated by the other two.
Mario Carneiro
CC BY-SA 4.0
2019-11-16T08:17:53.210Z
@MarioCarneiro I understand that those are *roughly 60 degrees* but I am not sure that they are exactly for all mass triplets that satisfy this inequality. I'll read further over the next day or two, thanks!
uhoh
CC BY-SA 4.0
2019-11-16T11:03:11.997Z
@uhoh I don't know about Gascheau's analysis, but for Lagrange's, where one mass is negligible and the other two are large, they are at exactly 60 degrees, assuming the two major bodies are in a circular orbit. The stability of the arrangement says that if you aren't at exactly 60 degrees then you will be pushed back toward 60 degrees, resulting in the so called tadpole orbits around the Lagrangian points. So an exactly equilateral arrangement is the equilibrium, and real orbits are slightly deformed triangles that wobble around equilateral.
Mario Carneiro
CC BY-SA 4.0
2019-11-16T11:22:03.543Z
@MarioCarneiro I'm really familliar with the CR3BP and I thought that **I'd gotten that approximation out of the way** with the introduction to my question. I'm only asking about three massive bodies, let's not discuss CR3BP further.
uhoh
CC BY-SA 4.0
2019-11-17T21:25:42.700Z
@uhoh If you account for post-Newtonian effects, then when the masses are unequal, the sides are also not (quite) equal.
TimRias
CC BY-SA 4.0
2020-05-10T02:22:30.040Z
I've just asked separately: [What kind of triangle is formed by three unequal masses in a circular restricted three body orbit?](https://astronomy.stackexchange.com/q/36121/7982)
uhoh
CC BY-SA 4.0
2020-11-30T22:53:21.850Z
Do units matter?
WarpPrime
CC BY-SA 4.0
2020-11-30T23:22:36.677Z
@fasterthanlight No, the relevant equations are phrased in dimensionless ratios.
TimRias
CC BY-SA 4.0
2019-11-15T09:15:31.190Z
2019-11-15T09:15:31.190Z
TimRias
<p><strong>Background:</strong></p>

<p>In the <a href="https://en.wikipedia.org/wiki/Three-body_problem#Restricted_three-body_problem" rel="noreferrer">Circular Restricted Three Body Problem</a> (CR3BP, CRTBP) some halo orbits are mathematically stable. That means that <em>the orbit of the third body</em> is closed, periodic, and stable against small perturbations as long as the two primary bodies are in circular orbits around their common center of mass. See answers to <a href="https://space.stackexchange.com/q/15268/12102">Are some Halo Orbits actually Stable?</a></p>

<p><strong>Question:</strong></p>

<p>But for <em>three massive bodies</em> such as stars, <strong><em>if they are</em></strong> interacting only with each other and only gravitationally, are all orbits still at least technically unstable? </p>

<p>Yes some systems may last longer than the lifetime of the individual stars, or even the age of the universe, but strictly mathematically are there some stable configurations or are they all mathematically unstable, i.e. one member can eventually be ejected <a href="https://space.stackexchange.com/q/15364/12102">like this</a>?</p>

CC BY-SA 4.0
2019-11-15T08:31:34.070Z
Are you asking about purely Newtonian gravity or also including relativistic effects?
TimRias
CC BY-SA 4.0
2019-11-15T08:58:31.210Z
@mmeent that's a very good point! While I was writing the question I think I actually wrote "Newtonian gravity" in there somewhere but it doesn't seem to be there now. I'm *primarily interested in* answers that address Newtonian physics, but I wouldn't want to rule out someone posting an interesting answers that either includes relativistic effects or talks about them separately. I can always ask a separate question about that if someone wants to answer elsewhere..
uhoh
<orbital-mechanics><multiple-star-systems>
Are the orbits of all triple star systems at least technically unstable?
uhoh
# Are the orbits of all triple star systems at least technically unstable?

(Asked by: username_0 on 2019-11-15)

**Background:**

In the [Circular Restricted Three Body Problem](https://en.wikipedia.org/wiki/Three-body_problem#Restricted_three-body_problem) (CR3BP, CRTBP) some halo orbits are mathematically stable. That means that *the orbit of the third body* is closed, periodic, and stable against small perturbations as long as the two primary bodies are in circular orbits around their common center of mass. See answers to [Are some Halo Orbits actually Stable?](https://space.stackexchange.com/q/15268/12102)

**Question:**

But for *three massive bodies* such as stars, ***if they are*** interacting only with each other and only gravitationally, are all orbits still at least technically unstable?

Yes some systems may last longer than the lifetime of the individual stars, or even the age of the universe, but strictly mathematically are there some stable configurations or are they all mathematically unstable, i.e. one member can eventually be ejected [like this](https://space.stackexchange.com/q/15364/12102)?

**Tags:** orbital-mechanics, multiple-star-systems
> **Comments:**
> **username_1**: Are you asking about purely Newtonian gravity or also including relativistic effects?  
> **username_0**: @mmeent that's a very good point! While I was writing the question I think I actually wrote "Newtonian gravity" in there somewhere but it doesn't seem to be there now. I'm *primarily interested in* answers that address Newtonian physics, but I wouldn't want to rule out someone posting an interesting answers that either includes relativistic effects or talks about them separately. I can always ask a separate question about that if someone wants to answer elsewhere..  

---

## 1 answers

### Answer by username_1 on 2019-11-15. Score: 11 | Accepted answer

(Ignoring that all orbits are technically unstable due to the emission of gravitational waves.)

There are known solutions to the gravitational three body problem that can be shown to be stable. Lagrange found a three body solution for general masses where all three orbit the common center of mass in an equilateral triangular formation. Gascheau proved in 1843 that this solution is stable if the component masses satisfy

$$ \frac{m\_1 m\_2+ m\_1 m\_3 + m\_2 m\_3}{(m\_1+m\_2+m\_3)^2} < 1/27$$

More recently, it was shown by Kei Yamada and collaborators ([1](http://arxiv.org/abs/arXiv:1212.0754) [2](http://arxiv.org/abs/arXiv:1505.04534) [3](https://arxiv.org/abs/arXiv:1612.08361)) that if one includes the first order Post-Newtonian correction, this solution is modified by a (small) correction to the triangle legs based on the component masses, i.e. it is only an equilateral triangle if all masses are equal. The general effect of the 1PN interactions is to decrease the region of parameter space where this solution is stable, but for suitably far separated systems there will still be masses for which the system is stable. Moreover, they also proved that ([4](https://arxiv.org/abs/arXiv:1512.01087)) these solutions are stable under the emission of gravitational radiation (i.e. a triangular system will adiabatically evolve into another triangular solution).

The effects of 2PN interactions (and beyond) on the stability of these triangular solutions is currently unknown (as far as I am aware).

> **Comments:**
> **username_0**: This is excellent, thank you very much! I'll have a look at these this weekend.  
> **username_2**: Just a remark, regarding the equation, for the less math savvy: This constraint basically means that one of the masses is much heavier than both of the other two. If they are all the same the fraction comes out to $1/3$, and it goes to $0$ as one mass dominates; if the masses are $m,am,bm$ then $a+b\le 0.041$, that is, the other two masses together contribute only 4% of the largest mass.  
> **username_0**: If the masses are not equal, are they still in an "equilateral triangular formation"?  
> **username_2**: @uhoh They can still be in an equilateral triangle, but they spin around a point which is not the center of the triangle, but is weighted toward the heavier object.  
> **username_0**: @MarioCarneiro okay that must have been a pleasant surprise for Gascheau to have derived! I'll go off and read more and then run a numerical simulation to quell any lingering (and apparently unfounded) doubts. Thanks!  
> **username_2**: @uhoh See also https://en.wikipedia.org/wiki/Trojan_(celestial_body) - when one mass is much smaller than the other two, it is basically hanging around the L4/L5 lagrangian points of the orbit generated by the other two.  
> **username_0**: @MarioCarneiro I understand that those are *roughly 60 degrees* but I am not sure that they are exactly for all mass triplets that satisfy this inequality. I'll read further over the next day or two, thanks!  
> **username_2**: @uhoh I don't know about Gascheau's analysis, but for Lagrange's, where one mass is negligible and the other two are large, they are at exactly 60 degrees, assuming the two major bodies are in a circular orbit. The stability of the arrangement says that if you aren't at exactly 60 degrees then you will be pushed back toward 60 degrees, resulting in the so called tadpole orbits around the Lagrangian points. So an exactly equilateral arrangement is the equilibrium, and real orbits are slightly deformed triangles that wobble around equilateral.  
> **username_0**: @MarioCarneiro I'm really familliar with the CR3BP and I thought that **I'd gotten that approximation out of the way** with the introduction to my question. I'm only asking about three massive bodies, let's not discuss CR3BP further.  
> **username_1**: @uhoh If you account for post-Newtonian effects, then when the masses are unequal, the sides are also not (quite) equal.  
> **username_0**: I've just asked separately: [What kind of triangle is formed by three unequal masses in a circular restricted three body orbit?](https://astronomy.stackexchange.com/q/36121/7982)  
> **username_3**: Do units matter?  
> **username_1**: @fasterthanlight No, the relevant equations are phrased in dimensionless ratios.  

---

