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I have two earth clones, in essence, separated by 16550 miles (26350 kilometers). They are, of course, tidally locked, and orbit each other once every 24 hours. These planets orbit a sun identical to ours in the same time as earth. Now, I want to add a little bit more to the system. A moon with half the mass and size of ours.

But I definitely don't want this to happen.

enter image description here

Is it possible to add a moon to this system and have it be stable for at least 10 billion years? Also, at what distance would this moon have to be? How fast would it orbit these two planets? Bonus points if you can make a diagram of its orbit.

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

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    $\begingroup$ phet.colorado.edu/sims/my-solar-system/my-solar-system_en.html $\endgroup$ – King-Ink Feb 21 '16 at 4:30
  • $\begingroup$ Looks interesting, but I won't be able to see that, or pretty much any pictures for three days. Vacation means iPad which means no flash player and no pictures :( $\endgroup$ – Xandar The Zenon Feb 21 '16 at 4:36
  • $\begingroup$ Be aware that while your planets aren't quite close enough to each other to break up, the effective surface gravity will vary a good deal due to tidal forces, and they won't be spherical. Since the rock is more rigid than the water or atmosphere, I'd expect big oceans and dense atmosphere at the points directly under the other planet, and their antipodes, and no water and thin/no atmosphere at the "equatorial" rim. $\endgroup$ – Mike Scott Feb 21 '16 at 8:04
  • $\begingroup$ @MikeScott Please go here and tell me more. $\endgroup$ – Xandar The Zenon Feb 21 '16 at 15:18
  • $\begingroup$ Two tidelocked earth clones whose orbital period is 24 hours? They need to be 106,400 km from one planet center to the other. Semi major axis would be 53,200 km. $\endgroup$ – HopDavid Feb 24 '16 at 1:26
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Is it possible? Absolutely. Pluto and Charon are tidally locked, and yet Pluto has four other moons1. Charon is relatively massive in comparison to Pluto - about one twelfth its mass. Indeed, the center of mass of the system lies outside Pluto. I see no reason why the stability should differ for two binary planets.

This does not, however, give us an answer of the inner range of the orbits. I am not aware of in-depth analyses of the dynamics of natural satellites of binary planets. However, analysis of circumbinary planets orbiting binary stars does exist, and we can use the same orbital mechanics here.

Welsh et al. (2013) state2

The stability criterion requires the planet to orbit outside roughly ∼2-4 times the binary semi-major axis, or periods ∼3-8 times the binary period.

There you go. So the moon must have

  • A semi-major axis of at least 26,350 kilometers (two times the semi-major axis, which is half of the separation distance) beyond the orbit of the planets.
  • A period of at least 72 hours (three times the orbital period of the planets).

Here is a diagram, as you wished, for bonus points (although a little bit off - I did this in Paint):

Here, $a$ is the semi-major axis of the planets.


1 Three of them are in resonance with one another, and perturbations by other bodies do ensure that the system is chaotic, but it does not seem that their orbits are unstable.
2 I will admit that the criterion may be different because of the mass difference - the moon will be much more massive relative to the planets than a circumbinary planet would be to a binary star - but I don't think it will make a huge difference.

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Yes, it is.

You want your moon to be stable for 10 billions years. Let's take a look to our Moon: it orbits our Earth at a distance of 384,400 km, with a period of 29 d, 12 h and 44 m and a median velocity equal to 1022 m/s. As the system transfers angular momentum from the Earth's spin into the Moon's orbit, the Moon is slowly moving away (4 cm in a year). After the Earth tidally locks to the Moon, the system will begin transferring angular momentum from the Earth-Moon system into the Earth-Sun system and the Moon will begin moving closer again. However, the Sun will engulf them both before the Moon collides with the Earth.

Now, let's put our Moon in your system and consider stability. I don't know all the Maths, but, knowing our Moon will probably be stable for some another billions years, it's safe in this case to assume it will be stable also here: basing my assumptions on the fact the planets orbit each other with a distance of 26,350 km, then their orbit is too close to give the Moon's orbit a problem. I would just reduce the size of the Moon to make it less affected by an impact between the planets (if you want it) and to make the system more stable (though I think it should be stable).

I wrote this answer assuming that the mass of the planets is Earth-sized or a bit more and that your planets don't orbit a M or K star. If it's not so, then the stability is not guaranteed. I know I've only written about as specific situation, but I hope it's still useful.

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You want your moon to orbit outside the Roche limit and inside sphere of influence.

A number of issues, (1) you would have to calculate the orbital parameters of the Earth clones and (2) you would have to calculate the orbital parameter of a moon orbiting a mass the size of two Earths given a certain orbital period.

It is impossible to calculate (2) without knowledge of the time taken for this Moon to orbit. The first, however, can be done.

One day is 86400 seconds. The mass of the Earth is 5.972 × 1024 kg.

$$T = 2 \pi \sqrt{\frac{2 \cdot R^3}{G \cdot 2 M}}$$

Since we know T, the gravitational constant G, and M, solve for R. The answer is 42109.77324 kilometres. So you'll have to put it further than that. Now, the Roche limit of each Earth can probably be approximated from the Roche limit of the actual Earth by itself. That is 18,470 km.

So, put it further than 26165.8 + 18470 km, or 44 000 kilometres. Whatever it is, you can put it where the Earth is now with stability. Naturally, this is a complicated three-body problem now, so that issue applies.

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  • $\begingroup$ There's a two in your numerator that shouldn't be there. $\endgroup$ – HopDavid Feb 24 '16 at 1:30

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