There are already a lot of questions on WB:SE about habitable gas giant moons, so I'll try and be specific:

  • I have a gas giant of mass approximately $2\cdot10^{27}\text{ kg}$ and radius approximately $80,000 \text{ km}$, so it's a super-Jupiter
  • The gas giant has some moons in very tight orbits that are tidally locked as well as a large Titan-like moon that is too far away to be tidally locked (see Stix, Hydra, etc. orbiting Pluto) that I'm wondering about specifically in a roughly-circular orbit ($0<e<0.001$) and semi-major axis $739,170\text{ km}$ so that its orbital period is almost exactly 96 hours
  • The Titan-like moon (just "the moon" throughout the rest of the question) isn't tidally locked to the gas giant since the gas giant's core, which makes up most of its mass, rotates once every six hours, and the moon rotates on its axis every twelve hours, and has a radius of roughly $3,000\text{ km}$
  • The moon has some axial tilt that causes it to have seasons like on Earth, but since the gas giant's orbital period is roughly 4,000 days they take a while to cycle
  • Most important: the moon's exactly lined up with the ecliptic plane, so once every eight days the gas giant eclipses it.

For reference, this is what the orbital situation looks like:

orbital diagram

Here, the blue circle is the orbit of the moon, the red circle is the "surface" of the gas giant, the black area is the shadow cast by the gas giant, and the red points indicate when a "new day" starts on the moon. Every "week" (eight days here), there's a roughly-3-to-4 hour period in which the moon is in the shadow of the gas giant and therefore completely devoid of sunlight.

Here's the question: does the "long night" at the end of every week significantly change the temperature on the moon? The question is important because the moon is supposed to be habitable and have species living there, and if the moon just freezes over every eight days it's going to be difficult to justify life if the temperature just decides to go to -200 C every eight days.

  • 2
    $\begingroup$ why would a fast spinning core prevent tidal locking? $\endgroup$
    – L.Dutch
    Mar 29 at 14:02
  • $\begingroup$ The idea is that it isn't impossible for moons to be tidally locked to the gas giant, but because it's rotating so fast the tidal forces on neighboring bodies are so great that the equilibrium is only stable for objects very close to it; it's possible I'm wrong about this but I think that would mean that more distant objects have smaller stability domains for tidal locking than ones very, very close to the gas giant ... $\endgroup$ Mar 29 at 14:06
  • $\begingroup$ @controlgroup do you have any references for this theory? I don't see any mention of the spin rate of the parent planet having any significant influence on the tidal spindown of moons, but I've not read that many papers on the subject... $\endgroup$ Mar 29 at 14:11
  • $\begingroup$ It's entirely possible I have dramatically misunderstood the concept of tidal forces; let me update the question... $\endgroup$ Mar 29 at 14:14
  • 2
    $\begingroup$ The question @L.Dutch is asking is what is the relationship between Earth's rotation and the Moon's tidal lock. The Moon is tidally locked to Earth in general, not to any particular point on Earth. (Oh, and your satellite is only about one a half times as heavy as the Moon, while the primary is some 300 times as heavy as Earth. The satellite is extremely likely to be tidally locked to the primary.) $\endgroup$
    – AlexP
    Mar 29 at 14:17

1 Answer 1


Radiative cooling times for moon-sized things are long. The formula looks something like this:

$$t_{cooling} = \frac{Nk_B}{8\sigma\epsilon\pi r^2}\left[ \frac{1}{T_{final}^3} - \frac{1}{T_{initial}^3} \right]$$

$t_{cooling}$ is the time it takes to fall from initial temperature $t_{initial}$ to final temperature $t_{final}$. $r$ is the radius of the moon.

(technically your moon isn't a homogenous isothermal sphere, but lets just imagine it is one for the moment, and it only needs to pretend to be one for a few hours during the eclipse)

You can get $N$ (the number of particles) from $\frac{mN_A}{M}$ where $m$ is the mass of your moon, $N_A$ is Avogadro's number, $\sigma$ is the Stefan-Boltzmann constant, $\epsilon$ is the emissivity of the moon (lets assume 1 for simplicity) and $M$ is the molar mass of whatever your moon is made of. If we handwave in a molar mass of 30 g/mol and a Titan-like mass of 1.4 x 1023 kg, N ends up being ~2.81 x 1048.

With a bit of rearranging, you get

$$\frac{8\sigma\epsilon\pi r^2 t_{cooling}}{Nk_B} + \frac{1}{T_{initial}^3} = \frac{1}{T_{final}^3}$$

Assuming a Titan-like radius of 2575 km, and an initial temperature of 280 K, over a cooling period of 12000 seconds you end up with a $T_{final}$ of.... (drumroll)

About 280K.

(if you're suspicious about my working, you can use the calcular on this page of hyperphysics to get a similar answer)

The temperature drop is obviously not going to be zero, but in all honesty it is so low that if you chuck a bit more greehouse gas into your atmosphere it'll basically be unnoticable. If you want your world to maintain its temperature, that'll be fine.

Your tidal locking assumptions are much more suspicious, so I'd turn your attention towards those instead!

  • $\begingroup$ OK, so what you're saying (the actual answer to the OP's question) is that the moon won't cool appreciably at all over the four hour span. Right? And if we assume it's habitable (OP's expectation), then it's even less, right? $\endgroup$
    – JBH
    Mar 29 at 17:46
  • $\begingroup$ That seems to be what I gather - a transition from 280 K to 280 K represents a virtually-negligible temperature change $\endgroup$ Mar 29 at 18:34
  • 1
    $\begingroup$ @JBH a few hours extra of darkness a week just ain't that much, really. I suspect if it were modelled correctly you'd find that the average temperature across the whole year was lower than a non-eclipsing world, and that'll have climatic effects, but not enough to tip it into uninhabitability. $\endgroup$ Mar 29 at 19:39

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