It appears that there are gas giants in habitable zones of stars, and likely some of them should have moons that are big enough to have an Earth-like world, so that part should be plausible enough.

But tidal locking or too strong tidal forces seem to be a problem for having a stable habitable moon. Tidal locking basically means that same side of the moon will be facing the gas giant, in other words day will be as long as one round around the gas giant, and the gas giant will always be visible in roughly same direction in the sky. The latter doesn't really matter, but the former would probably be a problem if the day-night cycle was more than a few Earth days.

All large moons in the solar system are AFAIK tidally locked, but it will take some time before it happens. When you take formula for the timescale and insert formula for a into it, you get

$$t_{lock} = 3 G^2 T^4 r \mu / (8 \pi^4 m_s) \cdot 10^{10} years$$


$$T = \sqrt[4]{8\pi^4 m_s t_{lock} / (3G^2R\mu \cdot10^{10} years)} $$

Where G is the gravitational constant, t_lock the required time in years, m_s mass of the moons in kilograms, R radius of the moon in metres and µ the rigidity of the moon.

ie. mass of the planet does not affect the time it will take, unless I brainfarted something.

https://en.wikipedia.org/wiki/Tidal_locking#Timescale https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body

Let's say we want at least two billion years before the moon is tidally locked. Also, assume the mass of the moon to be half of that of Earth and radius about 4800km. When you insert the values, you get required orbital period of 22.2 million seconds or 257 days, which is clearly far too much for the moon's orbit to stay stable in the habitable zone. Though this is curiously more than Venus' orbital period around the Sun, so either the formula isn't all that accurate or I bungled it up...

Anyways, Io's orbital period around Jupiter is only 1.77 days, so the moon could just orbit it tidally locked and still have a relatively short day. But wouldn't a close orbit like this cause uncomfortably strong tidal effects on the moon?

Alternatively, could spin-orbit resonance allow moving the moon further away from the gas giant, without increasing length of day-night cycle, and with the added bonus of not having the gas giant in the same direction? For example, Mercury is locked in 3:2 spin-orbit resonance, meaning it spins three times for every trip around the sun. Apparently higher order resonances like 5:2 are also possible.


EDIT: According to this, having an atmosphere could help a planet avoid being tidally locked. It doesn't consider moons, but I'd imagine that the same would apply to moons if they orbit further away from the gas giant, like at a 50 day orbital period? Even if it requires a bit thicker atmosphere than on Earth, like a few bars, that shouldn't be an issue.


  • $\begingroup$ A tidally locked circular orbit of Jupiter causes 0 tidal effects. It's nearby orbit of other moons disturbing this orbit into the elliptical that cause the tidal heating of Io. Remove those and you don't have an issue. $\endgroup$ Commented Dec 31, 2019 at 16:03
  • $\begingroup$ For example, Enceladus has an orbital period of 1.3 days and does not have an excessive tidal heating problem with a surface temperature of only -198 degrees. $\endgroup$ Commented Dec 31, 2019 at 16:11
  • $\begingroup$ I guess that would work. I didn't realize that Io's tidal heating and volcanic activity was due to disturbance from other moons... $\endgroup$
    – Joku_242
    Commented Dec 31, 2019 at 16:13
  • $\begingroup$ What do you mean with the other approximation for locking time? I calculated the required orbital period, which doesn't seem to depend on m_p. Anyways, formula for a was given directly in one of the Wiki articles: $$a = \sqrt[3]{Gm_p T^2 / {4 \pi^2}}$$ When you put this into the formula, $m_p ^2$ ends up being both on top and bottom of the equation, so they cancel out each other. $\endgroup$
    – Joku_242
    Commented Dec 31, 2019 at 16:26
  • $\begingroup$ @Joku_242 ahh, my bad I'm getting the two meanings of $\mu$ mixed up. $\endgroup$ Commented Dec 31, 2019 at 16:33

3 Answers 3


There are other questions on this site that discuss various aspects of habitable moons.

After being here for a few years I learned how to search for topics.

My answer here The major effects on the habitability of exomoons orbiting this Brown Dwarf1 says in part:

There are many previous questions with answers about possible habitable moons of gas giant planets (or sometimes brown dwarfs) in the habitable zones of stars.

Here is a link to apparently 667 posts on that topic:


And I think that someone interested in the habitability of exomoons, moons of planets in other star systems, should check out:

"Exomoon Habitability Constrained by Illumination and Tidal heating" by Rene Heller and Roy Barnes, Astrobiology, January 2013.


I believe that the article by Heller and Barnes states that a moon of a gas giant planet would have its orbit rapidly circularized and shifted to the planet's equatorial plane and also become tidally locked to the planet in just a few million years, so that is something that you should check out.

A satellite will have a stable orbit only if it is outside of the planet's Roche Limit and inside of the planet's Hill Sphere. The size of the planet's Hill Sphere is calculated from the relative masses of the planet and the star and the distance between them.

I believe that a moon's orbit will not be stable over billions of years unless it is within about the inner third of the radius of the Hill Sphere.

So for a planet with a specified mass, specified pass of the star, and distance between the planet and the star, the distances where a large moon could orbit will be relatively simple to calculate.

And if the mass of the planet is known, and the distances range where moons can orbit with long range stability, the minimum and maximum orbital periods can be calculated.

Heller and Barnes may also discuss the distances at which a moon might have too much tidal heating and may have formulas to calculate the distance.

Heller and Barnes suggest that large, potentially habitable exomoons might have orbital periods between that of Io and that of Titan.

I also note that it is possible that if a tidally locked world has enough hydrosphere and atmosphere it will have enough circulation to spread heat fairly evenly and so long days might not be a problem for habitability. That is rather speculative.

  • $\begingroup$ A quick look at the eccentricities and inclinations of some of the moons of the solar systems show that there's some wiggle room in Heller and Barnes' assumptions. $\endgroup$ Commented Dec 31, 2019 at 19:22
  • $\begingroup$ Something like 2-3 day day length along the lines of Io or Europa should be fine, I guess. So that should work. Just curious, what do you think about the paper I found about atmosphere possibly allowing planet to avoid tidal locking, could it also apply to a moon on a more distant orbit? $\endgroup$
    – Joku_242
    Commented Jan 1, 2020 at 15:29


mass of the planet does not affect the time it will take

it does constrain the range of values that $T$ can take, of course (via the size of the planet's Hill sphere and the Roche limit of the system). I'm not sure if the approximation holds true when the relationship is more like a binary planet rather than a little moon-massive planet one. Too hard to work that out, though, and it is probably within a couple of orders of magnitude which will do.


wouldn't a close orbit like this cause uncomfortably strong tidal effects on the moon?

For want of better terminology, tidal effects are caused by the motion of a body within a gravitational field beyond simple circular orbital motion and a day length equal to its orbital period. For an obvious example: if you magically tidally locked the Earth to the Sun, you'd no longer get any tides.

The tidal force experienced by a body is a slightly fiddly thing to compute, but (loosely speaking) the tidal acceleration experienced by a point on the surface of a satellite facing its parent planet in the direction of the planet is

$$a_t \approx {2rGM \over R^3}$$

where $r$ the distance of the point from the centre of mass of the satellite along the vector pointing towards the parent, $G$ is the gravitational constant, $M$ is the mass of the parent planet and $R$ is the current orbital distance. Changes in $a_t$ will be associated with tidal effects. An eccentric orbit (changing $R$) and a rotating satellite (effectively changing $r$ as the point moves away from the planet as the satellite rotates) both experience a changing $a_t$ for given points on their surface.

For a tidally locked body that is in an eccentric orbit, like Io, $R$ and hence the strength of this force will vary over the orbital period, causing tidal bulges to rise and fall and so work is still done. This might normally act to circularise the orbit of the body (though I'm not sure of the magnitude of the effect), but Io is in turn tweaked by the other large Galilean moons and so has kept some measure of its eccentricity.

The comments on the original question mention Enceladus as a short-period body which doesn't experience serious tidal heating, but consider that:

  1. Saturn is about a third the size of Jupiter.
  2. Enceladus is about a seventh of the size of Io.
  3. Enceladus is made of ice (which, as you can see from the tidal locking link in your question is "squishier" than rock).
  4. Enceladus is quite a bit further from the Sun than Io is.
  5. Enceladus has a very high albedo, making it much easier to shed heat.

All this means that despite the fact that it has an orbital radius ~5/9ths of Io, it experiences less than a quarter of the tidal forces that Io does and any heating will be balanced by greatly increased cooling. Enceladus is suspected to have some degree of tidal heating, contributing to its cryovulcanism. If you put Io in Enceladus' orbit, it would experience about 50% higher tidal forces than at its current position.

But what's to be done?

As Slarty pointed out, you could just handwave away all other large moons that might tweak your moon out of a circular orbit. Note that this isn't the same as vanishing every other moon; so long as they're neglible in mass compared to the moon of interest, they can stay. This isn't entirely implausible... Triton is substantially more massive than any of the other moons of Neptune, for example, and it has a highly circular orbit.

I'm not entirely sure that a spin-orbit resonance will help you, because the only way you'll get it is to have much more massive moons in more distant orbits, and they'll either give you the sort of orbital eccentricities that lead to tidal heating or require such a massive and complex system of moons that you'll run out of space in your planet's sphere of influence, or your planet will end up needing to be so massive it'll turn into a dwarf star.

A suggestion in an older question on this site (I can't recall which right now; might hunt it down another time) was that if your moon had its own submoons, the angular momentum of the system could be large enough to prevent tidal locking. In that specific example, the planet was a brown dwarf which would have been capable of supporting a large terrestrial moon with its own satellites in relatively distant orbits whilst still remaining in the habital zone of the primary star. I don't know how you'd compute the locking time of a moon-submoon system, so good luck with that.

Finally, a large watery world will suffer fewer ill effects from tidal forces than an icy or rocky world, and you might expect to find water worlds around close-in gas giants. Life (and the evolution thereof) on a water world is slightly less convenient than on something with a harder surface, but the problems are not necessarily insurmountable.

  • $\begingroup$ Btw, why is Mercury's spin-orbit resonance stable, then? The closest planets to it are very far away and relatively small in mass. $\endgroup$
    – Joku_242
    Commented Jan 1, 2020 at 0:17
  • $\begingroup$ @Joku_242 they're big enough and close enough to have Quite Interesting effects on Mercury's orbit, that's for sure. Mercury is also a pretty heavy and dense object (remember tidal forces scale with distance from centre of mass; smaller objects are less affected) which is comparatively far from the Sun compared to the relative distance of gas giant moons to their planet. It experiences something like three orders of magnitude lower tidal accelerations as a result. Its orbit is also quite eccentric in a way that gas giant moons are not; they are (mostly) circularised quite quickly. $\endgroup$ Commented Jan 1, 2020 at 10:26
  • $\begingroup$ Starfish Prime, when you expand R as function of T in formula for tidal acceleration, you get $$a_{t} = {8\pi^2 r} / {T^2}$$ Ie. again the mass of the planet doesn't matter. If $r/{T^2}$ is kept constant, the tidal acceleration stays the same. So the orbital period would have to be $$T_m = sqrt{T_{e}^2r_{m} / {r_e}}$$ With radius of 4800km for the hypothetical moon, you get orbital period of about 6 days, which is a bit long. Though Europa has orbital period of 3.55 days, and it doesn't have excessive tidal issues like Io despite its somewhat eccentric orbit. $\endgroup$
    – Joku_242
    Commented Jan 1, 2020 at 12:16
  • $\begingroup$ @Joku_242 you can hide the planetary mass term behind the orbital period, but it doesn't go away. The range of values of $T$ which give long-lived and stable orbits are still quite constrained. Europa is smaller and further away from Jupiter than Io, reducing the strength of tidal forces (to about 21%), and is less dense, reducing the severity of tidal effects. Tidal heating does still occur, of course, but the above factors mitigate it. $\endgroup$ Commented Jan 1, 2020 at 20:50

You need to remove all of the disturbing influences from your moons environment. Firstly make the moon have a circular orbit around the planet and the planet have a circular orbit around the star. Then remove all other moons or at least make them small and far out. That should do the trick. If necessary the moon could also start life with a large spin for some reason that way any residual effects would take even longer to stop the rotation.

  • 2
    $\begingroup$ Downvotes without comment will never get answers improved, folks. $\endgroup$ Commented Dec 31, 2019 at 16:56
  • $\begingroup$ Yes thankyou Starfish Prime whats the issue and I will try to improve it. Inaccurate? Too easy? Not enough technical detail? wrong tags? $\endgroup$
    – Slarty
    Commented Dec 31, 2019 at 17:08
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    $\begingroup$ @StarfishPrime Downvoting with a comment is how I got a serial down voter following me across the site, wrecking all of my old posts. SE really needs to allow an anonymous comment function for downvoters. $\endgroup$ Commented Jan 2, 2020 at 4:15
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    $\begingroup$ @SurprisedDog yeah, there's a lot to dislike about the current system. I've had similar things happen to me in the past, but they were mostly undone by mods. $\endgroup$ Commented Jan 2, 2020 at 8:03
  • $\begingroup$ @SurprisedDog Yes, that can happen. It does not happen frequently, though. For me it's still once in a lifetime. Most people around here know that a downvote shouldn't be something personal, and should not be taken as such. And some actively try to learn from the downvotes they get. I, for one, am deeply indepted to quite a number of people who did downvote my posts, and left an explanation. There's nothing better for learning than constructive criticism! $\endgroup$ Commented Jan 11, 2020 at 23:20

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