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I know of three sets of stable orbits in a binary star system: orbiting closely around star A, orbiting closely around star B, or orbiting distantly around both stars (and their mutual center of gravity) at once.

Is there a fourth set of stable orbits, around the mutual center of gravity, but inside both stars' orbits?

Two stars orbiting their mutual center of gravity, with a planet orbiting that same point

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  • $\begingroup$ Can you draw a picture of what this would look like? $\endgroup$
    – HDE 226868 ♦
    Jul 26, 2015 at 13:25
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    $\begingroup$ I've added an image. $\endgroup$
    – Mark
    Jul 26, 2015 at 19:45
  • $\begingroup$ I strongly suspect that's not stable long term, even though the planet does orbit the 2 stars center of gravity. I find it hard to believe that wouldn't rather quickly destabilize. There's a handful of calculated 3 body scenarios that work, my favorite is the figure 8 (this one is 3 equal bodies) ams.org/featurecolumn/images/simo03.gif and more here: news.sciencemag.org/physics/2013/03/…. Your question is slightly different, but I still feel pretty confident in saying that wouldn't be stable for long. $\endgroup$
    – userLTK
    Jul 27, 2015 at 8:15
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    $\begingroup$ You can't have an orbit like that. There is no force toward the barycenter. This planet would be shot out of the binary system at high speed in no time flat. $\endgroup$
    – David Hammen
    Jul 28, 2015 at 18:26

1 Answer 1

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The point you appear to refer to, is called the Lagrangian point $L_1$. This point is a saddle in the field of gravity, hence not to be considered to be stable in the strict sense. Two other Lagrangian points, called $L_4$ and $L_5$, can be stable, provided the considered orbiting objects are of small mass in comparison to the two main bodies of the system, and if the masses of the binary components are sufficiently different.

According to theorem 4.1 of this paper, $L_4$ and $L_5$ are stable in all directions, if and only if the mass ratio of the two main binary components $\frac{m_1}{m_2}\geq\frac{25+3\sqrt{69}}{2}\approx 24.9599$. According to theorem 3.1 of the same paper all Lagrangian points are stable in z-direction, which is the direction perpendicular to the orbital plane of the binary system. (Credits for this corrected version go to user DylanSp.)

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  • $\begingroup$ For a binary star system, L4 and L5 are generally not stable: the mass ratio between the two stars isn't high enough. $\endgroup$
    – Mark
    Dec 4, 2015 at 9:57
  • $\begingroup$ It's the mass ratio between the third body and the two other bodies, which needs to close to zero for L4 and L5 to be stable, not the ratio between the two stars. $\endgroup$
    – Gerald
    Dec 4, 2015 at 16:00
  • $\begingroup$ added equal mass binary case $\endgroup$
    – Gerald
    Dec 4, 2015 at 16:12
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    $\begingroup$ @Gerald According to these notes from NASA, the ratio between $M_1$ and $M_2$ also needs to be sufficiently high for $L_4$ and $L_5$ need to be stable. Your source doesn't go into sufficient detail on the conditions for stability of $L_4$ and $L_5$. $\endgroup$
    – DylanSp
    Mar 3, 2016 at 18:56
  • $\begingroup$ @DylanSp That's a good point. However, if the paper you referenced is read with a strict mathematical interpretation, it says "This will be true if", not "This will be true iff", with "iff" meaning "if and only if". So I'm not sure, whether the conclusion in the other direction holds. The figure on page 2 of the paper I referenced, shows a minimum for L4 and L5 in the equal-mass binary case. Out of the hip, I can't decide which interpretation is correct. $\endgroup$
    – Gerald
    Mar 3, 2016 at 22:24

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