In the Circular 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?


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?

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    $\begingroup$ Are you asking about purely Newtonian gravity or also including relativistic effects? $\endgroup$ – mmeent Nov 15 '19 at 8:31
  • $\begingroup$ @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.. $\endgroup$ – uhoh Nov 15 '19 at 8:58

(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 2 3) 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) 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).

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    $\begingroup$ 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. $\endgroup$ – Mario Carneiro Nov 15 '19 at 17:58
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    $\begingroup$ @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. $\endgroup$ – Mario Carneiro Nov 16 '19 at 4:13
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    $\begingroup$ @uhoh See also 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. $\endgroup$ – Mario Carneiro Nov 16 '19 at 6:05
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    $\begingroup$ @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. $\endgroup$ – Mario Carneiro Nov 16 '19 at 11:03
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    $\begingroup$ @uhoh If you account for post-Newtonian effects, then when the masses are unequal, the sides are also not (quite) equal. $\endgroup$ – mmeent Nov 17 '19 at 21:25

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