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I’m trying to create a system where a gas giant has two moons in binary orbit with each other. I’m trying to get that James Cameron look with the gas giant on the horizon, but also have a equally sized moon there two. I don’t like the idea of a horse shoe orbit as that would mean the other moon would be not in view for months to years at a time. Would a gas giant with a pair of earth sized moons in binary orbit be possible. If so what would be some complications with such system.

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  • $\begingroup$ they would rather quickly coalesce into one, i fear $\endgroup$ – Karl Dec 10 '18 at 21:01
  • $\begingroup$ @Karl Couldn't they be stable at Lagrange points relative to each other? $\endgroup$ – BrettFromLA Dec 11 '18 at 1:08
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    $\begingroup$ What do you mean by “binary orbit”? $\endgroup$ – cms Dec 11 '18 at 2:25
  • $\begingroup$ Much more likely to have them in normal orbits. Jupiter's Galilean moons aren't all that much smaller than Earth, and their closest approach distances to one another are comparable to the Earth-Moon distance. (And I think closer than L4/L5 distances...) $\endgroup$ – jamesqf Dec 11 '18 at 19:11
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    $\begingroup$ No, because the Lagrangian points are intrinsically unstable. If they don't impact each other, (or one is ejected, but there will probably not be enough energy present to do that) they will be separated into two separate orbits. $\endgroup$ – Karl Dec 11 '18 at 20:23
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It is sadly not possible to have them orbit one another, as you describe. The reason is because of tidal effects. If their orbit starts out as a perfect circle, then as soon as one gets closer to the gas giant than the other, the planet will pull a bit more on that moon than on the other one - making their orbit elliptical. This will continue with each round they make, until one of them smashes into the other one, into the gas giant, or is launched away from the system.

You can do better than horseshoe orbits though - you could have one of the moons in the L4 or L5 Lagrangian points with respect to the other, which are stable points in an orbit around the gas giant. Lagrangian points

Then, the moons would always be able to see one another.

EDIT: Whoops, it looks like everything below this line is false. I had misread a formula. It turns out the moons can be of similar size. I think I'll fix it when I have time.


One caveat is that their masses would have to differ significantly. The smaller moon must be 1/25th the mass of the larger one.

The difference between the sizes would be like the size of Earth compared with the size of the planet Mercury. Earth vs Mercury So it is clear that both are planets, but the size difference is quite apparent.

Now, how visible would they be from each other? For the sake of clarity, we'll call the larger moon Vericka and the smaller moon Valdera, whereas the gas giant would be Keizeria.

I will assume the most optimistic scenario. Vericka will be orbiting Keizeria just outside the Roche Limit (which is the closest a moon can theoretically orbit a planet before being ripped apart). Keizeria is as big as Jupiter, Vericka as big as Earth, and Valdera as big as Mercury. Using the simplest possible formula for the Roche Limit, I get an orbital distance of 54770 km.

Our Moon looks quite small in the night sky. It covers what one calls 1/2 degree, which means that you could put 720 Moons next to each other on the sky in a perfect line, without overlapping. Moon on the sky

You can do this 720 times before going back to the origin.

Now, how would Keizeria look from Vericka? The answer is that it would cover a whopping 103 degrees. If you look in a random direction in the night sky, the odds are 1/3rd that you would be staring in the face of the gas giant!

And finally, Vericka's long-lost partner moon - how big would it look on the night sky? The distance between Vericka and Valdera would be the same as the distance from Vericka to Keizeria, because they form an equilateral triangle. Putting Valdera's diameter in the same formula, you get a resulting apparent size of 5 degrees.

5 degrees is about ten times the apparent size of the Moon. You would be able to perceive very good details on the surface of the planet. Assuming Valdera has land and oceans like Earth (and Vericka), those with the keenest eyes would be able to see islands as small as 20 kilometers wide!

It would look a bit like this, in the sky: Valdera as seen from Vericka

Now, note that I took the most optimistic assumptions everywhere. In this model, the gas giant would be HUGE and the tidal forces would likely still cause regular earthquakes (Verickaquakes?) and make both planets uninhabitable. Perhaps it is better to cut the apparent size in half, to be on the safe side. But, all in all, this is the closest you are going to be able to get to a binary set of moons orbiting a gas giant, if you wish to obey the laws of physics.

If you wish to ignore them, you are free to make the planets actually self-correcting alien spaceships, or as dense as foam :-) But this is the science based answer that you requested.

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    $\begingroup$ I am no expert, but imho your assumption about necessary mass difference between moons is wrong. According to Wikipedia (lousy source, I know: en.wikipedia.org/wiki/Trojan_(celestial_body)#Stability), this 1:25 mass ratio is a minimum for primary trojan moon when compared to the planet, not another moon. Actually, given that planet is a gas giant orders of magnitude heavier than each of the moons, they can be equal in mass, sitting in each other's respective L4/L5 points. $\endgroup$ – Ijon Dec 12 '18 at 20:22
  • $\begingroup$ @Ijon Aaaaah... Okay, it looks like you are right. I misread my source (which is another Wikipedia page :P) It looks like the mass ratios are a bit complicated, but you are correct - Earth-sized moons around Jupiter can be of equal size in each others Lagrangian points... Boy, what to do now that I can throw the majority of my answer out of the window :P $\endgroup$ – KeizerHarm Dec 13 '18 at 13:29
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    $\begingroup$ @Ijon You guy's haven't fully read that section on wp. Firstly, the two moons can be of equal size if the central object is hypermassive (not sure what that is in numbers, but I'll guess a neutron star would do) AND if there are any other bodies present, you need much larger ratios than the minimum numbers given there. $\endgroup$ – Karl Dec 13 '18 at 18:45
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    $\begingroup$ Stating the equations you use to get results on the answer might be helpful, even if it is available on Wikipedia. Readers of your answer might not be as keen as you in equations, so walkthrough would be nice :) $\endgroup$ – Hendrik Lie Dec 14 '18 at 4:37
  • $\begingroup$ Right in the middle of exams at the moment, will fix all when I get the time ^^ Thanks for the feedback! $\endgroup$ – KeizerHarm Dec 14 '18 at 5:29
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You could have two Earth-sized planets circling the larger planet in the same path but not each other around the larger planet. They would eventually crash into each other or fly apart. To have them orbit around a center point and then a planet you would have to have some unstable physics.

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  • $\begingroup$ No you can't.Trojan orbits only work if the third object is very, very small. $\endgroup$ – Karl Dec 13 '18 at 6:56
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One way you could have near-equal size objects orbit their common barycenter, which orbits a much larger body, would be to have the smaller objects far enough from the parent body. Our own Earth-Moon system is a good example; we're so far from the Sun that its tidal effects don't make the Moon's orbit grossly unstable (though it is in fact technically chaotic, this has more do with the Earth's axial tilt relative to the plane of the ecliptic).

The closer the two worlds are to each other, the closer they can be to the primary without the problems with ejection or collision. If they're very close, they'll be visibly non-spherical (see Rocheworld or The Wooden Space Ships for some good illustrations), though the surfaces will be gravitationally level (i.e. sea level would follow the egg shape of each planet), and they could even be close enough to share an atmosphere (the "neck" between would be thin air, but passable with mid 20th century equivalent technology).

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