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I have read through previous questions here about seasons on a tidally-locked moon (I'm hoping this is far enough away from those to not be a duplicate question) and I think I understand the theory of it. What I'm struggling with is what those seasons would look like 'in practice', especially regarding my moon--more specifically, the supercontinent on which my worldbuilding takes place.

Planet Moon
Mass: 1.2 Jupiter 0.6 Earth
Radius: 1 0.926 Earth
Obliquity: 23° 6°
Rotational Period: 10.7 hour 6 days
Orbital Period: 1.76 years ^
Distance: 1.52 AU 0.00625 AU
Semi-major axis: 1.46 AU 0.00677 AU
/ Atmosphere 1.99 M/Earth, Pressure 1.7atm, Density 2.230 kg/m3

The supercontinent spans one of the poles, but that mass is smaller than that near/on the equator, and a part of it goes beyond the equator but not far enough away from it to become significantly cold again. All of it lies on the planet-facing side of the moon. I'm not at all concerned with the sun-facing side, so I don't need any information on that.

Now I know the seasons are caused by the tilt and that as a moon orbits its planet (and the planet the sun over its 1.76 year orbit), different 'ends' of the axis are closer or farther away from the sun, hence the seasonal change. I also know that regions near the equator have rainy and dry seasons but little temperature change over the year.

What I don't quite get is what all of this combined would look like. So, what the seasons are like on my supercontinent, on my moon. Are they the same as they would be on earth, only slightly longer? So my regions near the equator have a dry and a rainy season of 321.42 days each, while variations would be 'stronger' at the polar end, with a seasonal cycle akin to that of earth's near-polar regions?

enter image description here enter image description here

SaintDiabolus
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    What's the difference between the 'north' pole and the north pole without the scare quotes? How can a continent span the north pole and still have most of it on the equator? The obliquity of the axis of rotation of the satellite is 23°, hmm, all right, although I have a hard time believing that a tidally locked satellite has such a huge obliquity, but what is the inclination of its orbital plane? (For example, the axis of rotation of our own Moon is inclined about 6.7° with respect to its own orbital plane, but only about 1.5° with respect to the ecliptic.) P.S. A diagram would be helpful. – AlexP Jan 08 '24 at 18:27
  • I'll edit the difference, I wasn't consistent with the quotes. What I mean is that when I drew my continent and followed the tutorials by the youtuber artifexian on how to determine the climate of an area with the Hadley cells and similar, the upper part of the continent ended up being freezing cold, then gradually warmer at the centre, with a bigger continental mass being close to the equator than the cold part. I thought the obliquity of a tidally-locked satellite is for all intents and purposes equal to the planet, hence the high number. I'll edit with a (bad) diagram – SaintDiabolus Jan 08 '24 at 19:21
  • No, the obliquity of the axis of rotation of a tidally locked satellite is not linked to the obliquity of the axis of rotation of the planet. For example, the obliquity of the axis of rotation of our own Moon is about 6.7° with respect to its orbital plane, compared to the 23.4° of Earth. And then the orbit of the satellite is probably not exactly in the orbital plane of the planet, nor is it exacly on the equatorial plane of the planet. – AlexP Jan 08 '24 at 19:32
  • Aaah, I got it. Thanks for clarifying. I'll change it to a lower number – SaintDiabolus Jan 09 '24 at 07:43
  • What's the eccentricity of the orbit of the planet about the star? That would seem to give you the options for seasons I can't otherwise see happening here. – Escaped dental patient. Jan 13 '24 at 15:49
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    @Escapeddentalpatient Planet obliquity 23°, eccentricity (I didn’t change that number from its original in Universe Sandbox, so it's up for change) 0.254 – SaintDiabolus Jan 13 '24 at 22:58
  • This is in the VTC queue, but I'm not going to VTC. However, please note that it's unreasonable to ask about the seasons on a super-continent. Look at the massive variety of seasonal changes across just the U.S., Europe, or Russia. There are dozens if not hundreds of area-specific seasonal changes. There's nothing to stop all of those now-existing seasonal changes from existing on your super-continent other than differences in time due to the moon's orbital period and the planet's orbit causing its own seasons. But more to the point - humanity can't actually tell you the scientific truth ... – JBH Jan 17 '24 at 18:30
  • ... about this because we really only have one data point: Earth. Your best bet would be to use existing continental seasonal patterns and stretch them out or modify them based on the orbits of the two bodies. There comes a point when one is straining at a gnat. – JBH Jan 17 '24 at 18:31
  • The effect the tidal-locking would have was my main point of confusion, but for that other threads exist, I just need to figure out more closely (on my own) how a supercontinent would work with that. Thanks for the comment(s). – SaintDiabolus Jan 18 '24 at 07:39

2 Answers2

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I would say that your continent would be rather cold, but given that you can modify the atmosphere and temp all you want, I suppose it could really be any temperature. Whatever you end up doing with that, the weather will be relatively similar to earth, with a similar seasonal cycle to boot. Although it should be noted that the eclipse period would have a significant effect on weather (especially since it happens in the middle of the local 'day'). This may lead to significant formation of tropical storms, with the oceans retaining warmth from the midday sun while the atmosphere dramatically cools due to the cold snap. Depending on how large your continent is, the side closest to the tropics might experience tropical storms. This would be GREATLY MAGNIFIED if your atmosphere has lots of greenhouse gases or is very humid. As for the majority of the continent which would likely be further from the tropics, tropical storms would still be basically unheard of (as on earth in northern europe/canada). However, these places would still have unique weather patterns from the eclipse, as the sudden cooling would likely destabilise weather systems and cause somewhat unpredictable weather during and following the eclipse.

Also, the longer day length might lead to greatly decreased ice formation even in areas which have cold enough temperatures that, on earth, they would likely be very snowy and icy. I imagine ice and snow would only become very common much closer to the poles than on earth. Snowfall and frost would still be present in your continent, but would be more like a periodic event or an early-morning phenomenon. The extended day length might make many warmer areas very uncomfortable, with temperature compounding across the very long days, possibly going from freezing to comfortable to scalding hot. As such, life might find a balance between areas which are too cold to survive the long night and areas which are too hot to bear the afternoons. This might be mitigated on the more tropical areas of the continent, however, as the aforementioned tropical storms (while violent and dangerous), could provide some much needed cooling, with the storms following the eclipse softening the afternoon heat.

The climate would be complex and there will probably be things I haven't considered, but that's a general gist of what I think.

Hope that helps!

Valitenci
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  • From my prior research (i.e. google, articles and questions here regarding the topic), especially regarding Titan and Saturn, moons tidally locked to a planet that has seasons would also have seasons. – SaintDiabolus Jan 10 '24 at 06:59
  • You're absolutely right, it seems that moons get a sort of 'effective axial tilt' from their planet, so the moon would have the same underlying seasons as the planet. I misread some materials when writing my answer. I will edit my answer and swap out the last paragraph with a proper response regarding what your seasons will be like. My apologies. – Valitenci Jan 11 '24 at 02:47
  • No problem! I might be misunderstanding something (or I'm too much influenced by the visuals of James Cameron's Avatar), but when I drew my diagram, I had three days of night but illumination from the planet, then three days of daylight because the moon 'faces' the sun in its rotation around the planet. There's a period where the planet blocks the sunlight, but besides that, there still would be some. This could 100% be wrong or too simplistic, but from what I've seen with other questions on here, earth-life should still be possible on a tidally locked moon, with climates and all like earth – SaintDiabolus Jan 11 '24 at 13:35
  • Ah, I think there might be some confusion. You seem to have stated that your continent lies 'entirely on the planet facing side of the moon'. This side is the one without any sunlight at all. Life IS possible on a tidally locked moon, but only on the side facing away from the planet, where sunlight can reach, leading to an elongated but habitable day-night cycle. – Valitenci Jan 13 '24 at 02:21
  • That is true, yeah, but sunlight also reaches the planet-locked side, doesn't it? I thought that when it is night the sun and eclipse the planet blocks the sun from the moon, but besides then, when the planet-siding face 'faces' in the direction of the sun, some light would come through? – SaintDiabolus Jan 13 '24 at 10:55
  • Sorry for late reply. You do have a point, and your graphic is very useful. Your continent would indeed get significant light, almost as much as the opposite side, although with a notable period of eclipse. Your continent would likely be rather cold, but very habitable, maybe like northern europe or canada. Will modify my answer, my apologies. – Valitenci Jan 15 '24 at 00:33
  • No worries! Thanks for answering and clarifying – SaintDiabolus Jan 15 '24 at 06:53
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Not sure if it's possible to ignore the moon's sun-facing side, because the moon's planet-always-facing side still rotates relative to the sun, becoming the new sun-facing side, once every day.

One approximation using these online tools, assuming the planet is 380 earth masses and the moon is one earth mass, also assuming circular orbits, tubular shadows and irrelevance of direction of planet's axis of rotation, shows that:

Orbital velocity of the moon around planet should be 12.5km/s

Angular diameter of planet seen from moon's surface should be ~8.5° (16 times lunar disk diameter in earth's sky)

Planet's diameter being Jupiter-like 140000km, longest eclipse duration should be ~3h rather than 22h.

The moon's orbit around the planet being tilted 23° relative to the planet's orbit around the sun should prevent eclipses from occuring during winter and summer, and contribute to earth-like looking seasons. (by being roughly the same inclination as earth)

Conversely there would be more eclipses and total eclipses during spring and fall, probably messing a bit with those two seasons, making them colder than without eclipses.

Having the moon's rotation plane tilted 6° relative to moon's orbital plane around the planet, should mean 12° amplitude oscillation of planet's elevation in the moon's sky, that could cause daily wind patterns globally.

Edit: here's an image of moon and planet to scale

enter image description here

Edit2: One year for the planet-moon system is ~118 moon's days. each day is ~136h There aren't many days in a year where the 23° orbital tilt of the planet-moon system allows for eclipses at all.

using CAD to geometrically approximate this, there are two windows for eclipses to happen, each lasting ~8 days, starting 4 days before equinoxes and ending 4 days after.

So within spring and fall is "eclipses time", consisting of 8 star eclipses alternating with 8 planet eclipses in each window, which makes a total of roughly 32 occultation events each year.

  • Regarding the eclipse, the calculations I did based on another question on here got me the 22h I mentioned (AlexP's answer link. I can 100% believe I did the math wrong and wouldn't be upset if it was. Star: Diameter 702098, Radius 1404196. Planet: 139822, 69911. Moon: 11799.092, 5899.546. I also didn't consider the tilt of the planet and moon when I did those calculations. 2 π × 92,373.5km in 6 days, 96733.3km per day. 92,373.5/96733.3 = 0.955 days, 22.92 hour traversal of umbra. – SaintDiabolus Jan 11 '24 at 13:29
  • "because the moon's planet-always-facing side still rotates relative to the sun, becoming the new sun-facing side, once every day." I'm not sure that's right? My understanding of tidal locking is that the same side always faces the planet, so the planet-facing side would always face the planet, but get light from the sun and see it when it faces that direction as it faces the planet? – SaintDiabolus Jan 11 '24 at 13:30
  • How often would the eclipses be, on average, in summer/winter and spring/autumn? I think I read that for spring/autumn it would be an almost daily thing? – SaintDiabolus Jan 11 '24 at 13:32
  • I think your planet and moon look very much like Ganymede and Jupiter, regarding sizes, distances, and revolution period, you can get an idea of occultation time and why it is not 22h-ish, using "Nasa's eyes on the solar system" –  Jan 11 '24 at 15:14
  • Will check that out, thank you! – SaintDiabolus Jan 12 '24 at 15:43
  • Thank you also for the edit and the diagram. I must have done the math completely wrong at some point to get 22 hours. If I wanted more frequent eclipses, I would have to change the tilt of both the planet and the moon, correct? I would assume to a lesser tilt, which would mean no or very limited seasons? – SaintDiabolus Jan 12 '24 at 15:45
  • i think since angular diameter of planet seen from moon is ~8.5°, then if moon's orbit around planet is tilted half that angle or less, relative to orbital plane of the planet around its star, then there should be ecplises all year. having the tilt above that mid value (say 5°) would make eclipses happen all year except one or two days at summer and winter solstices with no eclipses. no clue if 5° tilt is enough to get a clear distinction between seasons, and all of this is for simple circular orbits. –  Jan 12 '24 at 16:14
  • I know that's not what this question was actually about, but I followed the steps in the link again and again got 22 hours for the eclipse, so I'm very confused and wondering where I've gone wrong. When I follow the steps with the numbers from that answer, the result is the same as AlexP's, but when I enter my numbers, I get 86470km for the umbra, it travels 2π × 86470, so 172940 π in six days, so 90551 per day. Dividing the umbra with the distance it travels in a day, I get 0.954931 = 22.918344 hours. – SaintDiabolus Jan 12 '24 at 17:20
  • Regarding the seasons and eclipse (in a mathematically correct world), thank you again for the explanation. I'll probably stick to the possible eclipse times near the two equinoxes from your original second edit – SaintDiabolus Jan 12 '24 at 17:22
  • well i may be fully wrong too and I d'love to know why; to me 22h is roughly 1/6 of a moon's day, and i don't think the planet is large enough or close enough to the moon to mask 1/6 of moon's orbital circonference, given your values. –  Jan 12 '24 at 17:44
  • I gave the numbers to a friend and he might have solved it. I've been using the planet's, sun's, and moon's radius and diameter. He says I should have used the respective orbits and put in the radius and diameter of those – SaintDiabolus Jan 12 '24 at 18:31