What are the day and night fluctuations for a moon orbiting a planet the size of Jupiter?

Question

In a nutshell, I'm trying to build a world with erratic day/night cycles.

To get the erratic days, I'm working with a habitable moon. It takes 3 full earth days (72 hours) for the habitable moon to rotate around the planet. 1.5 of those days (36 hours) are spent in the light, while another 1.5 (36 hours) is on the shadow side of the planet. However, the moon itself also rotates, though at this moment I am unsure at what speed. I'm thinking a full rotation would be between 20-36 hours, but I'm up for as much or little as needed if it helps with the warmth issue. Almost certainly the daylight and dark hours would be variable, aside from a periodic somewhat longer dark time behind the planet.

The planet would have a constant haze in the sky (it's a very basic life form that lives in the cloud layers of the atmosphere, kinda reddish in color--simple like a spore or plankton or other small thing).

A possible wild card is that another moon also orbits this planet, which is visible every 7 days to the habitable moon. Eclipses may be a thing, though I'm happy to place it in an orbit where that is rare, or never happens.

The planet itself is Jupiter-like, at least in size, though composition is up in the air. I have not figured out the orbiting time of this planet around the sun because I didn't think it was that important, but if it is I'm happy to let anyone play with those numbers freely.

I genuinely need help figuring out day/night fluctuations. I admit it, I'm not so hot with the maths. Can someone help with that, even a simple base pattern? I'm not married to day lengths (I was basing it loosely on Jupiter and a couple of it's moons) so whatever is easiest is fine, just let me know the new day lengths.

Answer 1

This is turning into a very long answer.

Any planet or habitable moon is likely to have a fairly constant average temperature.

At any one moment half of the Earth, for example, is in daylight and is being heated by the Sun, and the other half of the Earth is in night and is cooling off.

At any given moment - except during the equinoxes twice a year - one hemisphere of Earth will have longer days and shorter nights, and will be heating up from day to day, and the other hemisphere of Earth will have shorter days and longer nights and will be cooling off from night to night. Of course you say that your moon has no axial tilt and thus no seasons so there will be far fewer differences to average out.

Every difference averages out in the end, so the average temperature of Earth remains the same over time, except in so far as long term trends cause gradual warming or cooling.

An astronomical body will heat up with energy from its star until its thermal radiation out into space increases to equal the amount of radiation it receives from the star plus some more.

The extra heat will be internal heat left over from the formation of the planet plus internal heat from radioactive materials plus tidal heating which will be strong on a moon of a giant planet with possibly fellow moons. The astronomical body will heat up until its thermal radiation into space per second equals its total heat input from all sources per second, which includes radiation from its star, internal heat left over from its formation and internal heat from radioactivity plus plus tidal heating. Then the astronomical body will be in thermal equilibrium and its average total temperature averaged over its entire surface, hydrosphere, and atmosphere will remain constant except in so far as its heat sources increase or decrease over time.

So the length of the days and nights will not affect the average temperature of the moon at any one time or averaged over time. But it will affect how hot a particular spot on the planet will get during daytime and how cold a particular spot on the planet will get during nighttime.

The longer the day in a place, the hotter that place will get during the day. The longer the night in a place, the colder that place will get during the night. So the longer the days and nights are on your moon, the greater the temperature extremes between day and night will be.

On Earth, with axial tilt and seasons, there are major fluctuations in the lengths of day and night with time and also with latitude. By giving your moon no axial tilt, you have removed the major cause of fluctuations in the lengths of day and night on Earth. Day and night will always be the same length on every spot of the planet and at every time.

But because the moon in your story is a spheroid, light from the star will have different intensity at different latitudes. At the equator the star will appear straight up at 90 degrees, while way to the poles the star will appear at a 45 degree angle (thus spreading out he light rays less densely per cubic area of surface), and at the poles the star will appear at a zero degree angle at the horizon.

So the higher the latitude, the colder the surface, the water, and the air will be in day and in night. Thus there should be constant hot high winds blowing north and south from the equator toward the poles, and constant cold low winds blowing from the poles to the equator.

If the moon in your story has things that make it interesting for most types of stories, such as advanced multi celled lifeforms like plants and animals, intelligent natives, or habitability for humans (requiring oxygen in the atmosphere among other things), it should have existed with fairly steady temperatures for billions of years, since it took Earth billions of years to achieve any one of those conditions, and any large temperature changes during that period would have killed off all life on the planet.

So the star in your system has to have spectral type that enables it to shine fairly steadily for billions of years before turning into a red giant and then a white dwarf, destroying all life on its planets.

So if your world was a planet orbiting a star, it should have had a stable elliptical orbit with low eccentricity - closely approaching a circle - around its star for billions of years.

But if your world is a moon orbiting a planet, then it is the planet that has to have a stable elliptical orbit with low eccentricity - closely approaching a circle - around its star for billions of years.

If a gas giant planet forms a moon or captures a wandering body and makes it a moon, tidal interactions between the gas giant planet and the moon will quickly - in mere millions of years - turn the moon's orbit into an almost circular one around the equator of the planet, and should also adjust the axial tilt of the moon so that it matches that of the planet.

So for the axial tilt of the moon to be zero, the axial tilt of the gas giant planet should also be zero.

The axial tilts of solar system planets are 82.23 degrees (Uranus), 28.32 degrees (Neptune), 26.73 degrees (Saturn), 25.19 degrees (Mars) - all higher than earth's - 23.44 degrees (Earth), 3.13 degrees (Jupiter), 2.64 degrees (Venus), and 0.03 degrees (Mercury) - all lower than Earth's.

So it seems quite possible for a giant planet your moon orbits to have a very low axial tilt, and thus for your moon to have a very low axial tilt.

But you wrote:

To get the erratic days, I'm working with a habitable moon. It takes 3 full earth days (72 hours) for the habitable moon to rotate around the planet. 1.5 of those days (36 hours) are spent in the light, while another 1.5 (36 hours) is on the shadow side of the planet. However, the moon itself also rotates, though at this moment I am unsure at what speed. I'm thinking a full rotation would be between 20-36 hours, but I'm up for as much or little as needed if it helps with the warmth issue. Almost certainly the daylight and dark hours would be variable, aside from a periodic somewhat longer dark time behind the planet.

However, the same process of tidal actions which would circularize the orbit of a moon in just a few million years should also tidally lock the rotation of the moon to the planet in just a few million years. After just a few million years one side of the moon would always face the planet and one side of the moon would always face away from the planet. And that would be billions of years before natural process made the moon interesting enough for your story.

So your idea of having an orbital period of about 72 Earth hours and also a rotation period of about 23 to 36 hours is not very plausible.

With a tidally locked moon with an orbital period and rotation period both of 72 hours, the outer side of the moon facing away from the planet would have sunlight for 36 hours when the moon was in the half of the orbit closer to the star, and would be in darkness for 36 hours when the moon was in the half of the orbit farther from the star.

With a tidally locked moon with an orbital period and rotation period both of 72 hours, the inner side of the moon facing toward the planet would have sunlight for 36 hours when the moon was in the half of the orbit farther from the star, and would be in darkness for 36 hours when the moon was in the half of the orbit Closer to the star.

But the 36 hours of daylight when the inner side of the moon was farther from the star would be probably be interrupted by an eclipse when the moon passed into the shadow of the giant planet. That eclipse should be comparatively short. I think that when answering some other question I figured because of the relatives sizes of the planet and the moon's orbit that the eclipse could be no more than about a quarter or a third of the period that the moon was on the far side of the planet away from the sun, and easily much less than that.

The inner side of your moon could sometimes be eclipsed by one or more large inner moons that may get between your moon and the star when your moon is father from the star than the planet.

Either side of your moon could sometimes be eclipsed by one or more large outer moons that may get between your moon and the star regardless of where your moon is relative to the star and the planet.

If the giant planet your moon orbits orbits just slightly beyond the orbit of an inner giant planet, the two orbits being much closer together than the orbits of any planets in our solar system, then whenever the slightly faster moving inner planet catches up with the planet the moon orbits, it might cast a shadow large enough and long enough to reach the outer planet and the habitable moon and eclipse them. Depending on the orbital factors, this might happen between once every Earth week or so and once every few Earth years.

If you want your habitable moon to not yet be tidally locked to the giant planet and not have a synchchronous rotation, it will have to be very young and not yet have a naturally developed biosphere habitable for humans.

So maybe it could have an artificially and unnaturally developed biosphere habitable for humans, created by super advanced aliens who terraformed the moon and seeded it with advanced lifeforms like animals and plants.

Or maybe the moon was originally an independent planet orbiting its star in a nearly circular orbit for billions of years, developing a biosphere habitable for humans, until it approached too close to the giant planet and was captured by the giant planet.

The moon could have had a rotation period of 23 to 36 Earth hours before being captured into a 72 hour orbit around the giant planet. The giant planet would be gradually slowing down the rotation of its new moon to match it's 72 hour orbital period, but it might have slowed it down only slightly so far.

Since you want your world to be a habitable moon of a larger planet you should look up all the questions about such habitable moons.

My answer here has links to other questions about habitable moons and even a scientific paper.

What is the maximum orbital time for my moon around my planet?1

Note that a large habitable moon of a gas giant planet would probably have an orbital period of a few Earth days long. In our solar system the large moons of giant planets have orbital periods ranging from 1.77 Earth days (Io of Jupiter) to 16.69 Earth days (Callisto of Jupiter) and 15.945 Earth days (Titan of Saturn).

Also note that is is calculated that the year of a planet has to be at least 9 times as long as the month of that planet's moon for the moon to have stable orbit.

So if the year of your moon's planet is exactly 9 times as long as the month of your moon it would be between 15.23 Earth days and 150.21 Earth days long, and of course the year of your moon's planet could be much more than 9 times as long as the month of your moon. The year of your moon's planet could probably get to be up to a few Earth years long if that planet and moon orbit in the habitable zone of their star.

By comparison, known exoplanets that are probably orbiting in the habitable zones of their stars have years that range from 4.05 Earth days to 384.8 Earth days long.

https://en.wikipedia.org/wiki/List_of_potentially_habitable_exoplanets2

PS See my answer to this question:

Longest possible eclipse in double star system3

Answer 2

Another option for erratic day / night cycles might be a non-spheroidal body, like Mars' moon Phobos, that is rotating around a strange axis (rather than one in line with the plane of the ecliptic). I haven't done the calculations, but your day / night cycles may not be of consistent length in that scenario, though I'm pretty sure they would repeat regularly.

(For what it's worth, I pictured Darth Vader's TIE fighter tumbling off into space after it's ejected from the trench of the Death Star in 1977.)

Answer 3

Start with your moon. As others mentioned, it is highly likely that its orbital and rotation periods will be locked together. So with the 72 hour revolution, it also has a 72 hour rotation.

However, this does not translate to a boring 72 hour day/night cycle for the entire moon. Where you are on the moon will have a profound effect on the light/dark cycle.

The area of the planet that comes closest to a normal day/night would be the back, the side that always faces away from the planet. During the half of the moon's orbit when it is on the "day-side" of the planet, the sun will move through the sky for the 36 hour "day". On the "night-side" of the planet, this side of the moon will likely have the only true night sky. Both day and night skies will sometimes have the other-moon visible, the exact schedule of which would depend on what kind of resonance the two moons (or more!) have with each other. The magnitude of the solar eclipse will depend on the size of and distance to the other-moon. If you assume they are on the same plane, then a daily eclipse can be assured. Depending on how the orbital planes align, you may never actually have a "Full Moon" in the sky.

The "front" of your moon has a much more varying light cycle. For one, the planet will always be in the sky. It will be 25 (or more) times larger than the Moon in Earth's sky. While on the "day-side" of the planet, it will technically be local "night" as the sun be shining on the other side of the moon. However, the massive "Full" planet in the sky will light up the landscape quite effectively. During dawn and twilight, you will have the sun in the sky, but it will be eclipsed for most of the day as the moon moves to the "night-side" of the planet.

Answer 4

Solution

To get fluctuations in the day night pattern you need a moon which is not tidally locked. For this you could use a setup similar to Pluto where Charon (the closest moon of Pluto) has a bit more than 10% of the mass Pluto possess. Charon disturbs the orbits of the other Pluto moons. Which causes them to not be tidally locked with Pluto.

Calculations

You said you needed help with the math so here is an example which shows you a possible setup. For a different setup I will tell you what to change.

(Needed) Formulas

• Gravitational Force: $F = G \cdot \frac {m}{r^2};$ F = Force, G = $6.673 \cdot 10^{-11}$(gravitational constant), m1 = planet mass, r = distance between planet and moon (normally there is another $m$ for the mass of the moon but the second equation uses the same, so we can cut it)
• Centripetal Force: $F = \frac {v^2}{r} = \frac {4 \pi^2}{T^2} \cdot r;$ v = speed of the moon, T = time to complete one orbit (normally there would be the $m$ for the moon, but we cut it from the other equation, so we have to do the same here)
  • The Lenght in meters of the orbit: $d = 2\pi r$ d = orbit length

This is every thing we need to determine fluctuations.

Example

• Main planet mass: $1.899 \cdot 10^{27}$ (Jupiter)

• Distance between Planet and moon 1 000 000 km/ 1 000 000 000 m (if our solar system is anything to go by most moons with the size to be inhabited by similar life forms are around that distance)

For the Gravitational Forec we get this : $F = 6.673 \cdot 10^{-11} \frac{1.899 \cdot 10^{27}kg}{1000000000^2m} = 0.1267N$

This value for F we will now use to determine our speed:

First we take the equation from the Formula section: $F = \frac {v^2}{r}$ and switch it to get our speed: $v^2 = F \cdot r$

And now we plug in our numbers: $v^2 = 0.1267N \cdot 1000000000m = 126 700 000$ and because we want the actual speed not the speed squared we take the root of it and arrive at 11 256.11 m/s.

Now we need the length of the orbit $2\pi 1000000000m = 2\pi \cdot 10^{9}$ when we divide the distance by the speed we get the time it takes for one orbit: $\frac {2\pi \cdot 10^{9}m}{11256m/s}= 558202.22s$ which is 6.46 days or pretty exact 155 hours. To increase or decrease the time simply change the distance r from the planet to the moon (smaller shorter than 155 hours/higher longer than 155 hours).

Conclusion

If we assume that the moon rotates every 30 hours around itself this would cause the time of the day/nighttime to be shifted by 5 hours every 5 days (30*5 = 150 5 hours short of 155). And assuming that every time the moon is behind the planet it would block out the sun. This means the pattern would reoccur every 30 days (5*6 = 30 => shifted the light/dark a full day of light)

Because the moon would enter the behind the planet spot 5 hours later compared to the day time of the moon. And after 6 rotations around the planet it would be the same as 30 days before.

I hope this last part is understandable but I cannot come up with a better explanation. Also the time it takes for the moon to rotate around itself can be freely chosen.

Answer 5

If your habitable moon is tidally locked to the giant planet, then day/night duration will be governed by moon's rotation period. For example, for Earth's Moon it lasts 29.5 Earth days.

If moon orbits the giant very closely, then (for its planetary side only) star eclipses would be a regular things, whereas at night the skies will be beautifully illuminated by the giant's reflected light. For the "dark side" of the moon, day/night cycle would be unaffected.

Other moons may contribute to the amount of light at night, but that would be a relatively minor thing, except for the night sky pictures.

If the moon is NOT tidally locked to the giant, its own rotation will make the days shorter or longer, depending on rotation speed. However, having a non-locked moon under the stated conditions is scientifically doubtful.

Also it is important to say that day/night hours would NOT be variable, but rather very regular. Only on moon's planetary side star eclipses may be quite irregular.