Might a planet perform figure-8 orbits around two stars?

I'm thinking that if the two stars were equal mass (and not orbiting each other) then a planet that were to go right between them would continue in a straight line, with no preference for either star. But since the two stars would in fact be orbiting each other, the system would be rotating and thus there would be a Coriolis preference for one of the stars. Might that preference be made to alternate stars?

Another possibility would be if each star were in turn orbited closely by another planet, which would perform three orbits for each orbit of our planet of interest. Then things could be timed where on one pass star A's inner planet were aligned right to pull the planet of interest into an orbit, and on another pass star B's inner planet would be aligned right to pull the planet of interest into an orbit. So we have a system of five bodies, two massive (stars) each orbited by a minor, and one minor performing figure-8s.

Is this at least plausible if contrived?


3 Answers 3


It would be possible, but very unlikely, since the orbits wouldn't be stable.

Try to take a look at this visualization of the gravitational potential of a binary star system (from the Wikipedia Roche Lobe entry):

Roche potential

If the planet orbits just one of the stars, its orbit will be inside one of the lobes of the thick-lined figure eight at the bottom part, analogous to a ball rolling around inside one of the "bowls" on the 3D-figure. Such an orbit will be stable, just like the Earth's around the sun (bar perturbations from other planets, but let's leave them out for now), and there will be many different orbital energies for which this is true.

The same goes for an orbit around both stars: the planet will have many different energy levels at which it would simply experience the two stars' gravity combined as the gravity of one single body (and in which case the figure wouldn't apply, since it would be practically unaffected by the two stars orbiting each other).

In order to orbit in a figure eight, you have to imagine that the ball has to roll across the ridge between the two indentations in the 3D part of the figure. It is clear that this is possible, but also intuitively clear that this would only be possible for a narrow range of orbital energies (a little less and it would go into one of the holes, a little more and it would simply just orbit them both), and that it would not be a stable orbit. The ball would have to roll in an orbit where it exactly passes the central saddle point at the ridge (L1) in order to stay stable, the tiniest little imperfection will get it perturbed even further away from its ideal trajectory.

Your 5-body system could possibly be timed in such a way that it would work, but it would suffer the same fundamental flaw, and as far as I can see, it would also introduce even more sources of instability into the system.

This is, by the way, the gravitational potential in the rotated coordinate system, and you can see from the symmetry of the system that the coreolis preference you mention is not present. A simple symmetry argument should convince you of the same, though: Assume the system is rotating clockwise. This should allegedly give you a preference for one of the stars. But if you now let the system continue, while you rotate yourself 180 degrees up/down, it will now be rotating counter clockwise, which should give a coreolis preference for the other star, which of course cannot be the case, since there is no preferred up/down direction in a system like this.

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    $\begingroup$ Thank you, the illustration makes it clear exactly why this system is unstable. $\endgroup$
    – dotancohen
    Commented Jun 6, 2013 at 18:00
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    $\begingroup$ Is it just me or does the planar projection look like guy wearing goggles. $\endgroup$ Commented Oct 9, 2015 at 20:34
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    $\begingroup$ It is definitely a Ninja Turtle. $\endgroup$
    – Thriveth
    Commented Oct 10, 2015 at 9:57
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    $\begingroup$ I am wondering what such an unstable orbit would behave like. Assuming a planet with insufficient orbital energy to reach a stable orbit around both stars yet too much energy to stay in a stable orbit around just one of them, how would it behave? I can see two possible outcomes. Either it will alternate between the figure-8 orbit and orbits around each star, or it will get a gravitational slingshot out of the system. But which is it? Are there other possible outcomes? $\endgroup$
    – kasperd
    Commented Mar 27, 2016 at 22:58
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    $\begingroup$ Somewhere, there's intelligent life on a planet that has a figure 8 orbit saying "with billions of solar systems, clearly there are many many that make a figure 8 orbit, which we know is an orbit that is likely to have intelligent life". $\endgroup$
    – Anthony
    Commented Jan 8, 2019 at 15:57

Considering that there are no other answers, I will answer.

From user9886's links I have found several types of figure-eight orbits. Here is one nice one: http://www.ams.org/samplings/feature-column/fcarc-orbits1

However, I don't know if the two orbits postulated in the OP are possible.



Generally speaking, binaries cannot have shared planets orbiting at close distances. Neither dynamics nor kinematics holds.

If a binary has a common planet, the planet must be far enough, that is, the radius of the circle is large enough to be far greater than the distance between the two stars. This approximates two stars as one star at the center of mass.

The acceleration of the winding provides centripetal force, so the system can be stable. It's impossible to circle two stars at close range.

Each star of the binary can have its own planet, in this case, the planet need to close to its own star and far away to other star

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    $\begingroup$ Dear Cang Ye. It is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) $\endgroup$
    – Qmechanic
    Commented Jul 10, 2019 at 3:51

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