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Let us assume a sufficiently earth-like moon, like Europa. It orbits a gas giant. The whole setup is within the star's goldilocks zone, so the moon is theoretically habitable.

'Day' is one rotation of the moon around itself, 'month', for the sake of the argument, is one rotation of the moon around its planet, and 'year' is the planet going once around its sun.

Days would be completely dark for a certain portion of the month, I guess, since the planet would eclipse the sun, right? While completely light days would work similar to the way they do on Earth, provided there's no eclipse?
What I cannot figure out is how seasons would work, and whether they would exist at all. When would it be warmer/cooler?

Please, I would very much appreciate an answer in layman terms. I have found answers to similar questions filled with formulae I couldn't make heads or tails of.

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    $\begingroup$ The dark periods during the moon's day would be like a longer solar eclipse on Earth, or more accurately, like a lunar eclipse as seen from the moon. Just how often these eclipses occur would depend on the distance & diameter of the gas giant, and the orbital inclination of the moon. But in general, you might expect them to be about as common as lunar eclipses. It would take a very special alignment to have them happening every month. $\endgroup$ – jamesqf May 13 '18 at 18:30
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    $\begingroup$ This isn't what you're asking, but a terrestrial moon orbiting a gas giant is almost certainly tidally locked. The position of the gas giant is always the same in the sky, and for the side of the moon which faces away from it, it is not visible at all. This means that eclipses only happen to the near side. Also, one "day" is equal to one "month". $\endgroup$ – brendan May 13 '18 at 23:11
  • $\begingroup$ I tweaked the title of your question to try to better summarize the part you're interested in. Feel free to Edit further. $\endgroup$ – a CVn♦ May 14 '18 at 12:18
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Seasons are a product of the tilt of the body's axis relative to the star. Moons generally orbit near the equator of their parent body, so if the planet's axis is tilted relative to the star, the moon's axis have a similar tilt relative to the star. As the planet orbits the star, the moon will get the same sort of seasonal changes that the planet would get from its tilt.

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    $\begingroup$ The tilts are not related to each other. Rather, moons are embedded in a gravitational field where they perform a balance-like motion (libration). $\endgroup$ – rexkogitans May 13 '18 at 19:13
  • $\begingroup$ We can use Saturn and Titan as an example from our solar system. Saturn has a ~27 degree tilt. Titan orbits Saturn's equator and has a very small axial tilt relative to Saturn. As Saturn moves around the Sun, Titan has a ~27 degree tilt relative to the Sun and its hemispheres would get variable amounts of Sun based on where Saturn is in its orbit. Here we are not worried as much about Titan's motion relative to Saturn, we are looking at Titan being tipped relative to the Sun. $\endgroup$ – Futoque May 14 '18 at 16:48
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On a planet, seasons are dictated by the axial tilt. The colder seasons located in the hemisphere furthest from the sun, and the warmer seasons in the hemisphere closest to sun. Temperatures, and hence the seasons, in the different latitudes are determined by the angle of the sunrays hitting the land. If your moon had an axial tilt, this would be affected it in a similar manner.

However, your moon is also orbiting around the planet, moving further away from the sun and then closer to the sun and then back around the planet again. This orbit-change in distance from the sun would be much larger than the hemispheric-change in distance from the sun, and would affect both hemispheres of the moon at the same time. It would affect the overall climate and not just the seasons. Ie say it was what we would consider the "summer" hemisphere on the moon. It could be colder in that summer-hemisphere during the darker periods of that month when furtherest away from the sun, than compared to the winter-hemisphere during the bright days on the nearest sun-side of the planet.

This monthly moon orbit would be similar to both the more long-term Annual and even longer-term Milankovich Cycles which influences Earth's overall climate. The planet being closer and further away from the sun, aphelion and perihelion. The Milankovich climate cycle is thousands of years long and influences the development of ice-ages etc. Your moon would be experiencing the equivalent of a very much smaller-scale "Milankovich Junior" Cycle every lunar month, getting closer and further away from the sun as it orbited your planet. You would then also still have both the planet's Annual and Milankovich Cycles as well.

If you combine this moon orbit (climate) with a moon axial tilt (seasons) then you can have;

  • "winter" months with cooler weeks (sun-side) and very very cold dark weeks (far-side).
  • "summer" months with warmer weeks (sun-side) and still very cold dark weeks (far-side).
  • The moon's orbit around the planet would have much larger effect on your lunar climate than the seasonal axial tilt.

So now you just need to figure out if you want your moon to take 30 days to orbit your planet or 90 days!

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    $\begingroup$ This is not correct. Yes, seasons are dictated by the axial tilt. But no, it does not have to do with distance to the sun, so the orbit around the planet is irrelevant. The warm season is where the sun beams hit the surface in a steeper angle. $\endgroup$ – rexkogitans May 13 '18 at 19:10
  • $\begingroup$ @rexkogitans, we have aphelion and perihelion cycles where the planet is furtherest/closest positions to the sun. This cycle is known as the Milankovich cycle and does play a small part in temperature variation over the centuries. Yes, this actually does not affect the seasonal shift...on a planet. However it should play a part on the moon cycle (apogee/perigee) as it will be cycling closer to the sun more frequently than a planet and so is worth mentioning as a factor affecting temperature across the whole moon surface (not causing the hemispheric season shift), each month! $\endgroup$ – EveryBitHelps May 13 '18 at 19:40
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    $\begingroup$ The orbit of the Earth around the Sun is not perfectly circular. The distance between the Earth and the Sun varies in a year orders of magnitude more than what a moon would experience in one full lunar cycle. Yet the impact of such variation on our seasons is negligible. $\endgroup$ – Renan May 14 '18 at 2:27
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Axial tilt of the moon is the main factor when it comes to seasonal changes.

The orbit around the planet adds a small variation to the distance from the star, which is normally negligible.

Additionally this variation happens faster than the revolution around the star, therefore the thermal inertia of the planet can effectively mitigate it.

I.e. for the Moon-Earth system, the distance Moon-Sun varies of 2 light seconds around an average distance of 8 light minutes, which is $2/480 = 0.5\% $

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Enough people are discussing the tilt of the axis causing the seasons, but I'm wondering more about if the temperature disparity between lunar day and night might render such a body uninhabitable.

If, to pick a number, the moon orbits the planet in 28 days, it's going to be in complete darkness for about 8-10 days of that period, and semi-darkness a bit less than another 6-8 days. I'd be curious if such a body's atmosphere could hold enough heat for that long to remain liveable, seasons or no.

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  • $\begingroup$ For your 8-10 days per 28 days, the moon in question would have to be awfully close to the planet. Taking the Saturn-Titan example used in other answers: With a mean diameter of ~116464km and a minimum distance from Titan of ~1221830km, Saturn would have an angular size of 116464/1221830=0.095rad=5.46°. This is ten times wider than the moon's or the sun's seen from earth (with the apparent area growing quadratically), but it is nowhere near being large enough to put the moon in perpetual darkness for extended periods of time. In most orbits, Titan will miss Saturn's shadow completely. $\endgroup$ – 0range May 14 '18 at 19:18

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