Thread: Binary Stars - Planetary Orbit?

Let’s say there's a pair of binary stars – let’s call them Proxima and Alpha Centuri - and there’s a planet that orbits Proxima.


In theory: is it more correct to say that the planet orbits Proxima, or does it orbit a kind of gravitational average point between the stars?

(I’m aware that technically, everything orbits the net-gravitational force in the universe – but let’s discount that for now).



If so, what’s the term for this? ("The planet's orbits is...?")

Either is possible and have been shown to occur. It depends on how close to each of the stars are to each other, and how closely the planet orbits either one of the stars, or the barycenter of the whole system.

If the stars are far apart, a planet could orbit either one (though it's orbit would be affected by the other star). You would then say that the planet orbits either the A or B star.

If the planet orbits at a distance greater than the distance between the stars, you would say the planet orbits the binary system (or more correctly, the center of mass of the entire system, including the stars and any planets).

MW

Here's the thing, Proxima Centauri and A Centauri are separated by some 15,000 AU. It isn't even certain as to whether Proxima is bound to the A Centauri system. Even given that Proxima is a red dwarf, its Hill shpere would extend out to some 500 AU. More than enough for it have a planetary system of its own.

So let's consider A Centauri A and A Centauri B instead. Here we have two stars of roughly equal mass (1.1 and 0.9 solar masses respectively) orbiting at a distance of 17.57 AU from each other. This makes the Hill sphere of B ~ 11.39 AU and that of A ~ 13. The Hill spheres overlap so there is a "No Man's Land" where orbits would be unstable. The stable orbits would likely be within 2.28 AU of B and 3 AU within A, or as mentioned above, orbits far enough away that they orbit both stars.

For instance, here is a gravity sim representation of the A-B system showing two planets, with one 3 AU from A (magenta)and the Other starting 6 AU from A. (A is the reference point for this sim) The planet close in to A maintains a fairly stable orbit, while the further planet's path is wildly erratic until a close pass by B ends up ejecting it from the System. It is also fair to point out that this path has it crossing the closer planet's orbit several times, so it is conceivable that under a different starting configuration, it could make a close enough pass to the inner planet to disrupt its orbit enough to nudge it out its cozy orbit around A, sending it into its own erratic orbit, ending with it being ejected from the system.

Not an answer, really.

Actually, both replies are the only possible answers.

If you want a different one, you'll have to make it up yourself.

Originally Posted by Vexer

Not an answer, really.

Yes, you have been given the correct answers already.
The scenario you propose is non-trivial, because from a mechanical standpoint this is just a special case of a 3-body problem :

Three-body problem - Wikipedia, the free encyclopedia

The orbit of the planet would depend on the exact masses of the three bodies, as well as the distances and velocities of the bodies from each other. Depending on the initial conditions you may or may not get a stable orbit for your planet, but generally speaking, the closer the planet gets to the two central bodies, the more unorthodox and unstable its orbit will be. Your best bet will be to keep it as far away from the stars as possible ( perhaps just inside the Hill Sphere ), in which case it would orbit around the common gravitational centre of the binary stars to a good approximation - note that even then you are still not guaranteed a stable orbit over long periods of time.

Three-body problem - Wikipedia, the free encyclopedia... ah. Yes. I'd forgotten that. There is no answer.

Originally Posted by Vexer

Not an answer, really.

Not a polite response to serious efforts to address your question, but enormously useful in classifying you in the "Don't bother helping" category.

Originally Posted by Vexer

Three-body problem - Wikipedia, the free encyclopedia... ah. Yes. I'd forgotten that. There is no answer.

No that's no quite right. Depending on the initial conditions of the system there may be a stable orbit for your planet. The problem is that this is not guaranteed; the system could be stable, but it could also end up chaotic and unstable. It really depends on the exact configuration. Mathematically, this problem is always solvable, but not always in a closed analytic form.

Is it possible that this also works the other way around?
Does a binary star system offer better opportunities for capturing rouge planets than a single star system?

I don't think so. It would make it far less likely.

I did give this a quick go, using an old bit of code for a computer game particle system.
The results were very surprising.

I ran two versions of the same code one with a star and a large planet, in the second I upped the mass of the planet to 0.5 of a solar mass.

I did not bother modeling the chance that a rouge planet might get captured by airobreaking in the star's atmosphere or be slowed below escape velocity by collision with an asteroid.

I only did 40 runs of 100,000 iterations of each version, but in the two star version, the rouge planet ended at below escape velocity 7 times.

It looks like if the planet is not traveling that much above escape velocity it will always fall in to near the stars, and therefor has a high chance of getting a gravitational kick from them as they orbit the centre of mass.
There is about a 50/50 chance of it traveling in front of a moving star or behind it so the planet either leaves the system faster than it entered or gets captured.

The probability of this being an artifact of my coding is quite high, as my particle systeim uses some very dirty maths tricks for computer game performace levels.