Non-deadly gas giant moon "weeks"?

Question

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:

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.

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!