Okay, I have a planet smaller than Earth that has a mean orbital radius of .9 AU, and it orbits twin suns. The larger sun has a mass of 1.15 solar masses, and the smaller sun has an mass of .45 solar masses. They orbit in the exact same direction, and the distance between the two suns is about .0134 AU. All orbits are stable, and we will assume that the gravity from other heavenly bodies is negligible. How eccentric will the orbit of the planet be?

  • $\begingroup$ To clarify what @HDE226868 said: you choose the eccentricity. You cannot compute it from the given data. $\endgroup$
    – AlexP
    Apr 22, 2021 at 19:56
  • $\begingroup$ 2 million km separation between those two stars is awesome close! If measured correctly, core-to-core, they are almost touching and will be swapping matter all the time. Not that this has any impact on your question, the planet can be in any eccentricity it wants, as long as the Perihelion is not too close to the stars. $\endgroup$
    – PcMan
    Apr 25, 2021 at 9:56

1 Answer 1


Given the information you've provided, there's no way to compute the eccentricity of the orbit; with this setup, it's a free parameter that you can set to whatever you want (although of course we need $e<1$ for a bound orbit). It would be possible to determine if we had some additional information, such as the energy of the planet $E$ and its angular momentum $\ell$, in which case the eccentricity would obey$^{\dagger}$ $$E=\frac{\gamma^2\mu}{2\ell^2}(e^2-1)$$ with $$\gamma\equiv GMm_p,\quad\mu\equiv\frac{Mm_p}{M+m_p}\approx m_p$$ and $M$ the combined mass of the stars and $m_p$ the mass of the planet. Alternatively, you could take a physical approach and specify the pericentric distance $r_p=a(1-e)$ or apocentric distance $r_a=a(1+e)$; either on its own would be enough to determine $e$.

That said, a circumbinary planet's eccentricity doesn't follow a uniform distribution. As you might expect, for a given semimajor axis, lower eccentricity orbits will survive for longer amounts of time, as lower eccentricities mean larger pericentric distances and therefore further closest approaches to the central stars. When the pericenter becomes comparable to the separation of the binary $s$, things get quite dicey, so ideally you want larger $r_p$ and therefore smaller $e$.

Some numerical studies have been done in this area; see for example Sutherland & Fabrycky 2016, which has an excellent title. They considered a binary star of masses $1M_{\odot}$ and $0.1M_{\odot}$ with a separation of 1 AU, and placed planets at varying distances from the binary's barycenter (semimajor axes between 1 AU and 4 AU, and eccentricities between 0 and 0.3). Not surprisingly, for a given $a$, the higher-eccentricity planets were ejected much quicker. Here's their Figure 2:

Image showing how long it took various orbits to be ejected

For example, consider planets with a semimajor axis of 2 AU. Almost all ejected planets with $e=0.3$ were ejected within 400 years, almost all ejected planets with $e=0.2$ were ejected within 2000 years, and planets with $e=0$ and $e=0.1$ largely survived significantly longer. Obviously this isn't the same as your setup, but the same principle applies. Granted, you have $s/a\approx0.014$, so this likely won't be an issue, but it could become problematic for extremely high eccentricities, i.e. $e\gtrsim0.9$.

The short answer, therefore, is that the vast majority of orbits will be just fine, but extremely high eccentricities will not be.

$^{\dagger}$This is taken from Classical Mechanics by John Taylor, which was the handiest reference I happened to have lying around with an example.

  • 1
    $\begingroup$ I wasn’t planning on having a very eccentric orbit, as I want my planet to be somewhat habitable all year round. $\endgroup$ Apr 23, 2021 at 2:13
  • $\begingroup$ @TysonDennis Well, that's certainly good to know. $\endgroup$
    – HDE 226868
    Apr 23, 2021 at 2:18

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