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Fo my question, I'd like the moon to orbit around a saturn-like planet without rings.

This moon would be located at a million kilometers away the host planet. the moon's size would be about 12 000 kilometers wide.

The moon would also be habitable and shelter a wide biodiversity and a stable climate.

Knowing that it wouldn't be tidally locked, it would have a 24 hour day and would take a few months to orbit around it's planet. How could it be able to do so? I hope you can answer me with a lot of details and thank you.

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    $\begingroup$ The moon would be tidal locked depending on distance and the size of the planet. If you are flexible with that "million Kilometer" distance, you can make it not tidal locked. The further away from the planet the less tidal forces the moon will feel and the less chances it will be tidal locked. 1 million km might not be far enough from a saturn sized planet. $\endgroup$
    – A. C. A. C.
    Commented Jul 20, 2017 at 18:36
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    – Secespitus
    Commented Jul 20, 2017 at 19:02
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    $\begingroup$ Phoebe is not tidally locked, but it is about 13 Gm (over 8 million miles) away from Saturn $\endgroup$
    – Gary Walker
    Commented Jul 20, 2017 at 19:34
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    $\begingroup$ As I don't have the astrophysics to help with the actual question, I'll just offer that "habitable" might be a generous word for a moon which spends one or more months out of every few months freezing in the shadow of its host planet. $\endgroup$
    – Henry Taylor
    Commented Jul 20, 2017 at 20:08
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    $\begingroup$ If the moon is relatively new, it will still be rotating. Tidal locking happens over time. $\endgroup$
    – BrettFromLA
    Commented Jul 20, 2017 at 20:08

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Yes. Tidal locking is the end of a process (of tidal acceleration) that might take millions of years. A moon (or any other body orbiting a larger body) will avoid tidal locking…

  • the more massive it is,
  • the faster it rotates around its axis to begin with, and
  • the farther it orbits from the larger body to begin with.

Tides diminish in strength at rate proportional to the cube of the distance between the bodies, so the "distance" component of the equation is by far the most important.

If the moon was captured recently (in geological terms) it will take some time to become tidally locked. It will take a lot longer if it was captured in a faraway orbit. And even longer if, when captured, it was spinning rapidly. Note, however, that as tidal acceleration begins to act, the slowing down of the moon's spin will induce friction. Expect to have a geologically active moon, churned by the planetary tides.

For reference, an Earth-sized moon orbiting a Saturn-sized planet at one million km would take only about 12 days to complete an orbit. For a smaller gas giant (with the mass of Neptune), you could have an Earth analogue in orbit 2 million km away with a period of 76 days. Just play around with the numbers in the equations (WolframAlpha is your friend!).

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    $\begingroup$ This doesn't seem very friendly to life. $\endgroup$
    – user25818
    Commented Jul 20, 2017 at 20:46
  • $\begingroup$ It depends on how forceful the churning is. :) $\endgroup$
    – pablodf76
    Commented Jul 20, 2017 at 21:05
  • $\begingroup$ Yes that could be bad, but I mean if the moon is only a few million years old that seems like a pretty short time for biodiversity. Taking "biodiversity" to mean multi-celled life. $\endgroup$
    – user25818
    Commented Jul 20, 2017 at 21:10
  • $\begingroup$ I just laid out some ideas to slow down tidal locking so that those few million years could turn into a few hundred, or more. OTOH the question didn't specify the moon's life forms had to have evolved there. $\endgroup$
    – pablodf76
    Commented Jul 20, 2017 at 21:15
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Yes, but things can't be exactly as you want them.

With the given values for orbital radius and orbital period (assuming 60 days for "a few months"), you can use Kepler's Third Law to derive that the parent body's mass must be about 2.2*10^25 kg, which is about a quarter of the mass of Uranus. This number would be smaller if you want a larger orbital period. Gas giants of that size should be possible, but it is unknown if any exist, and it won't be close to Saturn, which is about 25 times larger than this gas giant.

Because of this, it is impossible for you to have all three values where you want them to be. You have to pick whether to change the orbital radius, the orbital period, or the parent body mass.

If you change the orbital radius, to orbit much farther than a million kilometers, but have the specified orbital period and parent body, then it will be in a similar situation as Iapetus, but a little closer. Iapetus is tidally locked to Saturn, so it is safe to assume that another moon out there would also be.

If you change the orbital period, to orbit more quickly, but have specified orbital radius and parent body, then that is similar to Titan. Titan is also closer to what you want in terms of size, as it is larger by volume than Mercury, but it still is tidally locked, so any other moon out there is probably still tidally locked.

If you decrease the mass of the parent body, then there are no similar examples because there are no objects of that size to look at in our solar system. However, the "moon" would, assuming the same density of Earth, have a mass of 4.0 * 10^25 kg. In this case, it wouldn't be a moon, it would be the planet of this system, and the gas giant "planet" would be its moon, and this system would resemble the Pluto-Charon system.

Unfortunately, none of those situations give you a moon that isn't tidally locked, which is what you requested. However, there is one alternative which might work for your story:

Mercury is close enough to the sun that tidal forces should dominate its rotation, but it has a high eccentricity of 0.2 that prevents true tidal locking. Instead, Mercury has a 3:2 spin-orbit resonance. Your moon could be moved to a closer orbit, so its orbital period is a few days, and then given some orbital resonance due to a high eccentricity. If the orbital period is 60 hours, then a 5:2 spin-orbit resonance would be appropriate for a 24 hour day. Now, this will disrupt the wanted orbital period, and the distance will fluctuate quite a bit, but it will have a moon near to the gas giant with a 24 hour day and a longer orbital period.

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If the planet or moon has an eccentric orbit, like Mercury, it may continue to rotate. Mercury has a 3:2 resonant lock - it rotates three times for every two orbits. Other resonant locks are possible as long as the same, or the opposite, hemisphere faces the primary at each periapsis.

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  • $\begingroup$ There might be a way to get rather long spin-orbit ratios by having the moon tidally locked to another moon that's in an orbit-orbit resonance and where the tidal effects of them passing dominates over that of the primary. Having one of the moons be a captured body in a retrograde orbit might be handy. $\endgroup$
    – BCS
    Commented Dec 6 at 20:20
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It is not really plausible for a habitable moon to be orbiting a gas giant and not be tidally locked. As pablodf76, who is more optimistic on this, says, the timescale for becoming locked is on the millions of years. This has to be compared to the billions of years it took intelligent life to evolve on the earth.

It is true that the timescale becomes much longer if you are farther away, but, as Jarred Allen says, Iapetus is far from Saturn (more than 3.5 million km) and still tidally locked. The fact is, every large moon in the solar system is tidally locked.

You can imagine a capture scenario, but remember that you want life to have time to evolve on the moon. So the moon and the gas giant must both be in the habitable zone for billions of years, before the moon finally truly becomes a moon by being captured. Not impossible but freakishly unlikely.

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  • $\begingroup$ You might be able to have the planet spending a long time in a marginally stable Trojan type orbit and only recently have been captured. That would allow an eccentric orbit for a spin-orbit resonance or explain why things are still spinning down. $\endgroup$
    – BCS
    Commented Dec 6 at 20:28

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