My planet, is roughly as big as Earth, but has two moons. The first moon, Luna, is essentially the same as Earth's moon, but the second, much, much smaller moon, Selene, actually orbits Luna. Selene is about the size of Ceres (it's a captured asteroid) and orbits at about (10,000-60,000) km from Luna.
So what would the lunar cycles look like in terms of length, and such? Taking into account the fact that Selene passes behind Luna. Secondly is this setup feasible at all?

Edit: Because of the Roche limit, the distance has now been changed to 10,000-60,000 km.

  • $\begingroup$ Just a note, normally you should avoid changing details about your question in such a way that it makes answers no longer true. You can comment on other people's answers to explore different possibilities, or if necessary, create a new question that has those changes from the start. $\endgroup$
    – Cadence
    Jul 7, 2018 at 2:08

2 Answers 2


Is it feasible? Not as stated (KerrAvon2055's answer explains both why, and what will happen). However, if the orbital distance is at least 10,000 km, it will be safely outside the Selene/Luna Roche limit. The maximum orbital distance is governed by Luna's Hill sphere - the radius that divides "places where you will orbit Luna" from "places where you will orbit Earth" - and is about 60,000 km. So there's plenty of non-destructive orbits to be had.

In this scenario, Selene will eventually be perturbed into an Earth-centric orbit (or something else; it might get ejected and end up orbiting the Sun, for instance), but "eventually" in this context means over thousands or millions of years, so you could have it hang around for a good long while.

What would it look like? Luna would look pretty much how you would expect: it's too massive for its orbit to be meaningfully changed by Selene. (Ceres is actually about as massive compared to the moon as the moon is to Earth, about 1.2% the mass.) Luna's orbital cycle would control the position and appearance of both moons, and it would have the same length as our own moon.

Selene's radius would be about 1/4 Luna's, so it would occupy 1/16th as much sky. Luna's visible diameter is about 32 arcminutes or half a degree; Selene's would be about an eighth of a degree. Its albedo is somewhat lower, so it wouldn't be as bright, but it should be visible and clearly not a star or planet.

Selene would stay near Luna in the sky - its maximum orbital radius around Luna is about ~1/6 of Luna's own orbital radius, which means (if I remember my trigonometry correctly) it should remain within about 10 degrees of Luna, less if its orbit is closer. Because of that, it should have more or less the same phases at the same time. While in the right configuration, with Selene closer to the Sun, you might actually see Lunashine on its dark side.

I'm not clear on whether Selene would experience tidal locking to Luna, and if so, how long that would take. I assume it can't be tidally locked to Earth while still in its orbit. So it would probably not have a "near side" and "far side"; the portion of its surface visible from Earth would rotate over time, at least for awhile.

Depending on the specifics of their orbit, Selene and Luna might eclipse one another. Even at its closest approach on its furthest (from Luna) orbit, Selene isn't nearly big enough to completely eclipse Luna. The reverse is not true; Luna could completely hide Selene if their orbits happen to coincide. Like solar eclipses, Selenian eclipses might only be visible from parts of Earth, albeit much larger ones. Selene isn't big enough to eclipse the Sun, but it might have interesting transits.

  • $\begingroup$ Because I apparently have a hard time with mass vs. density, the Luna/Selene Roche limit would actually be smaller, about 2,500 km. However, that's a lower bound and you probably don't want to get too close, so a 10,000+ km orbit still works. $\endgroup$
    – Cadence
    Jul 7, 2018 at 2:20
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    $\begingroup$ Tidal locking would happen quickly with a small orbit near the Roche limit, because tidal forces are greater there, so they will do more work and the dissipation will be faster. Additionally, a small body has greater deviations from being a perfect sphere, which will shorten the time scale also. (Man-made satellites can be tidally locked to Earth just by making them long and thin.) The time scale calculation is not a simple one, since you can't just do it with orbital mechanics, you have to model the moons' interiors' response to stresses as well. But my guess is tidal locking is plausible. $\endgroup$
    – N. Virgo
    Jul 7, 2018 at 7:41
  • $\begingroup$ (That's tidal locking to Luna - you're correct that it can't be tidally locked to Earth.) $\endgroup$
    – N. Virgo
    Jul 7, 2018 at 7:42
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    $\begingroup$ Also worth noting: from Selene's point of view, Luna and Earth appear about the same size in the sky, because the Moon-Selene system is basically a scaled down version of the Earth-Moon system. But the Earth has a higher albedo, meaning it's more reflective. (The Moon is about the same colour as a tarmac road, while the Earth has a lot of white clouds.) So while Lunashine might sometimes be visible on Selene's dark side, it will be much darker than Earthshine, so it'll be harder to see. $\endgroup$
    – N. Virgo
    Jul 7, 2018 at 7:48
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    $\begingroup$ Another note: there are two more types of eclipse that could happen. Selene can pass into the Moon's shadow, and Luna can pass into Selene's shadow. During the first kind, Selene will disappear from the sky unless Earthshine makes it visible. During the second kind, Selene's shadow will be visible as a circle moving across Luna's surface. $\endgroup$
    – N. Virgo
    Jul 7, 2018 at 7:50

Very little changes - on Earth

Ceres has a mass of slightly over 1% of the moon. As far as Earth is concerned, it is now being orbited by a mass with a barycentre slightly offset from the Moon's centre of mass. But that will change soon because...

Selene is well within the Roche Limit for the moon. So shortly after it arrives in orbit it will disintegrate as a result of the tidal effects from the moon and large parts will probably form a ring system. Eventually orbital perturbations may result in all of the mass in the ring hitting the moon - I look forward to hearing from anyone who can calculate this. So the only effect on earth will be a spectacular ring surrounding the moon in the night sky. The moon will get done over big time by falling rocks.

Do not buy lunar real estate if Selene is on its way!

Edit: time to disintegrate. Wild guesses here, but once Selene's near approach is within the Roche limit it will go pretty quickly, probably in a matter of hours. It will not break down to tiny rocks all at once, it will break into smaller and smaller chunks progressively. As the chunks get closer and closer to those permitted by the Roche limit the rate of disintegration will slow.

  • $\begingroup$ Oh, dear! How long will it take to disintegrate? That could actually be interesting if it was a long enough period. $\endgroup$ Jul 7, 2018 at 2:00
  • $\begingroup$ @TheImperial calculating the exact disintegration time is above my pay grade. Editing to add best guesses. I would also like to observe - from a safe distance. $\endgroup$ Jul 7, 2018 at 2:21

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