My moon planet is the moon to a gas giant in the habital zone of its red giant star, orbiting the same distance the earth orbits the sun. It is not tidally locked, being in opposite rotation of its planet, and far enough away that it still experiences seasons and tides without losing its atmosphere. It is abundant with life, including sentient life. This is a fantasy world birthed from the death of primordial beings, or so the sentient life assumes, so I have leeway with some of the physics in a sense. Within the home galaxy of my planetary system, is a far away blue dwarf that is also prominent in the night sky and provides some light to this planet though far away enough that it's heat and uv isn't so damaging, plus the gas giant provides protection.

Outside the elliptical galaxy in which this planetary system calls home, is another spiral galaxy closing in, a clash in the far away future, yet not so far away as the Milkyway from Andromeda are, the same scenario however. The alien galaxy closing in is a prominent purple galaxy in the night sky.

I cannot figure out day night cycles throughout a 365 day calendar in these settings. Like I said, the physics need to kind of work.

I figure there will be a long period of darkness, but the night sky will be illuminated by the outer galaxy and this neighboring blue dwarf.

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    $\begingroup$ "365 day calendar" doesn't quite gel with something orbiting in the habitable zone of a red giant, I think. You might want to rethink that requirement. $\endgroup$ Commented Mar 26 at 9:05

3 Answers 3


We can't give you hard numbers without orbital and rotation speeds, but we can give you advice about figuring out the solution. This is more valuable to the worldbuilding community at large, anyway. BTW, it's worth noting that we have a relevant question1 that should be peeked at when considering this issue.

  1. The orbit of the gas giant isn't particularly relevant to your question unless the orbit is elliptical enough to affect seasons, but as you're asking about the day/night cycle, seasons aren't important. On a side note, the habitable zone for a red giant is 50-70 AU. (More specifically, the habitable zone of our sun when/if it becomes a red giant is 50-70 AU. See [1]) Worse, the nature of the life cycle of a star is such that it's unlikely enough time exists for planets of a red giant in its habitable zone to evolve life (See [2]) Don't let that stop you, though. Personally, I think too much realism is boring.

  2. The orbit of your moon and its rotation are very much important. Let's ignore the 365-day issue. At best, that's an imposition of a "human" perspective. The odds of a moon orbiting a gas giant so slowly that it takes an actual Earth year are pretty bad (might not even be possible). To be honest, while the orbit of the moon could be considered a "year," it affects the day/night cycle much more strongly and would seem to me to be part of that and not part of the measurement of a "year." This will make more sense in a moment.

  3. The moon will only know light (kinda, see below, let's call if "full light") when it's on the day-side of the gas giant. During that time, the day side of the moon will be in full sunlight and the "night side" (the side facing away from the sun) will enjoy sunlight reflected off the gas giant. Thus, there are distinctly two kinds of "day."

  4. When the moon is on the night-side of the gas giant it will know night on the side of the moon facing away from the planet and will know something dim (but still illuminated, stronger than our full moon I expect) due to the refraction of light around the gas giant's perimeter on the planet-facing side of the moon. Thus, there are distinctly two kinds of "night."

  5. It is therefore reasonable to assume people on your moon would have four words describing the major divisions of light in the sky (I'm ignoring words like twilight and dusk, which describe transitions). Borrowing from "perigee" and "apogee" and for convenience in this answer we'll call them:

  • Periday: describing the side of the moon closest to the sun on the daylight side of the planet or "closest to the day."
  • Apoday: describing the side of the moon facing away from the sun on the daylight side of the planet or "furthest from the day."
  • Aponight: describing the side of the moon facing the planet on the night side of the planet or "furthest from the night."
  • Perinight: describing the side of the moon facing away from the planet on the night side of the planet or "closest to the night."
  1. Now, this assumes your moon is rotating with enough speed to clearly experience the entire four-stage day/night cycle. There is no "best" rotation speed anymore than there is a "best" orbital speed. But the ratio is important as it affects how much of each stage in the day/night cycle an inhabitant would experience. Assume for the moment that the moon is tidally locked. An inhabitant on one side of the moon would only ever experience apoday and aponight. On the other side of the planet someone would only experience periday and perinight. Tidal locking means the moon rotates once per orbit around the gas giant, so the ratio is 1:1. As the ratio increases (e.g., 2:1 or two rotations per orbit), inhabitants experience all four stages, but more of one set and less than the other until (simplistically) a balance between all four stages is experienced. Without sitting down and working out the math, I think that begins at 4:1. Note that the higher the ratio, the more important those other words we're ignoring (twilight... dusk...) become. I'm avoiding that 'cause I need to go to work pretty soon.

  2. Finally, I mentioned seasons in #1 due to the shape of the gas giant's orbit. Two other ways to great seasons are the axial tilt of the moon and the eccentricity of the moon's orbit around the gas giant. This really complicates things because those "seasons" would be experienced on a per-lunar-orbit basis, which is really fast. In fact, as I think about it, it might be too fast to actually experience a "season," but it would seriously change the way day and night are experienced, and so I'm going to advocate ignoring that completely. Unless you're a glutton for punishment! Because what the tilt and eccentricity would do is change where on the moon the four stages are experienced per-rotation and to what degree. Seriously... if you intend to do this, have pizza and cream soda available. Lots of pizza and cream soda. I'm just sayn'

1If you're tempted to vote to close this question as a duplicate of that question, keep in mind that while the other question was asked first, this question's ambiguity (not normally an advantage!) means the answers to this question are more generally applicable, which logically would mean that the earlier question is a duplicate of this one. However, though I'm not going to worry about it until after I get home from work — assuming I remember to come back to it — I believe one of the two questions should be closed as a duplicate of the other.

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    $\begingroup$ This has helped make sense of my dilemma! Thank you for taking the time to help me understand and arrive to a conclusion! $\endgroup$
    – SRL
    Commented Mar 26 at 20:59

The moon orbits the planet, that's the most logical choice for the moon's year. The planet orbiting around the central star is not the primary choice, for someone living on the moon. More or less as we don't have a strong interest in computing the "year" defined by the Sun closing a whole loop around the core of the Milky Way (sure, some scientists do care about it and bla bla, but it's nowhere on our timekeeping artifacts.)

The moon day will be defined by its rotation around its own axis, which you have not specified in the question.

By the way, if with

being in opposite rotation of its planet

you mean that the planet rotates E -> W and the moon orbits the planet W -> E, be aware that this is an unstable configuration.

  • $\begingroup$ This makes more sense of things, thank you for taking the time to answer my question! $\endgroup$
    – SRL
    Commented Mar 26 at 20:58
  • $\begingroup$ The reason a year is interesting to us is not inherently because it's related to the orbit of our ball of rocks around its parent, it's that its parent is specifically a star that provides the majority of light and heat. While I'm not an expert, and I expect that the gas giant would provide a significant amount of light via reflection, I would still assume that the sun is the primary source and thus the seasonal year of the gas giant's rotation around the sun is more relevant than the orbit of the moon around its parent (though that may be a perturbing factor). $\endgroup$
    – Miral
    Commented Mar 27 at 5:25

You've got a bunch of unknowns here, some of which are more important than others.

  1. The size and mass of the primary star, which determines how hot it is, and how big it looks from your moon.
  2. The size and mass of the planet about which your moon orbits.
  3. The inclination of your moon's orbit, relative to the plane of its parent planet's orbit.

(1) determines where the habitable zone of your star system lies. This is a complex thing to compute, so I'm not going to try. Wikipedia has a brief introduction which serves to highlight how brief the habitable period might be, but for a star of sunlike mass, the habitable zone lasts longest at a jupiter-like orbital distance. Jupiter's orbital period is nearly 12 years, which means your comment about a "365 day calendar" can't work. An object close enough to a red giant to orbit that quickly is going to get cooked or even entirely engulfed by its parent.

The combination of (1) and (2) determines the size of the Hill sphere of your planet, which in turn restricts how far away from your planet your moon can be. You've got some wiggle-room here given your comment about leeway with physics, but the key takeaway is this: orbits too far away from a planet are unstable, and the moon will fly away into an astrocentric orbit, or in to interstellar space. Orbits too close to a planet result in tidal locking. The tidal-locking timescale is proportional to the 6th power of the orbital radius of your moon, so you want to maximize that distance. In order to do that without falling out of the Hill sphere, your planet needs to be very large indeed... probably a brown dwarf, tens of times heavier than Jupiter.

Lets try an assemble some actual numbers. Lets assume a solar-mass red giant, with a 60-jupiter-mass brown dwarf orbitting at a jupiter-like distance of 5.2 AU. The red giant's radius is now ~1 AU. From that distance, the star will have an angular diameter of ~21.7°. That's big... they don't call them giant stars for nothing. Compare with the regular moon and sun, which are more like 30 arc minutes across... more than 40 times smaller. Wikipedia suggests that 20 degrees of arc is about the apparent width of your spread hand at arm's length. That's a lot of star.

The Hill radius of the brown dwarf is also big... nearly 1.4 AU. Orbits are only stable within about a third of that, so your moon might orbit at ~70 million kilometers. Brown dwarfs are much more dense than gas giants, so if it were a bit like COROT-15b it might only have a jupiter-like radius. At that distance, it will only have an apparent angular diameter of about 7 minutes of arc... that's a little larger than the apparent size of the Mare Serenetatis.

A visualization of the 7 minutes of arc over a photo of the moon

(The full moon over Santorini, with a crude visualization of how big a disc that was 7 arc minutes across might look. The full size of the red giant simply wouldn't fit, as it would be 4 times wider than the whole image. Original image credit Klearchos Kapoutsis)

A visualisation of 20 degrees of arc over a photo of the moon above a city skyline

(The full moon over Tokyo, with a crude visualisation of how big the red giant star would look in the sky by comparison. Original image credit Markus Winkler)

This would mean that the planet could never eclipse the star, so there's never a dark period at all if your moon is rotating about its own axis. Assuming your moon is at its maximum possible distance from the planet to minimize the chance of it being tidally locked, it'll have an orbital period of ~481 days. If you wanted it to have an orbital period of an Earthlike 365.25 days, that gives a distance of ~57.7 million kilometers.

And that leads to a very interesting calendar.

Firstly, you have the "little year" it takes for the moon to orbit the brown dwarf. During that time, its distance from the star varies by nearly 0.78 AU. That means that it has seasons where things are hotter and brighter in the summer, and colder and darker in the winter, but the day length, and the time of sunrise and sunset, don't change very much.

Secondly you have the "grand year" it takes for the brown dwarf to orbit the star, which is more like 12 years long. This will have seasons if the moon's orbit about the brown dwarf is inclined relative to the orbital plane of the brown dwarf (see this other recent answer of mine: Would a moon that is tidally locked to a gas giant rotate at an axis?). These will "stack" with the summers and winters of the little year, but will also have changes in day length and sunrise/sunset times and so seem more like the winters and summers that Earth gets, only much longer.

In the end, the day/night cycles are the simplest thing... nights will get a little extra illumination from the other objects in your setting, including the brown dwarf, but will otherwise be as long and more-or-less as dark as a moonless night on Earth could be. Sunrises and sunsets will be particularly dramatic gven the size of the star which will be a dominating presence in the daytime sky.

There's a second option, which involves a tidally locked world orbiting a Jupiter-like gas giant very closely, in order to get the eclipsed periods you're interested in.

For the planet to eclipse the star, your moon would need to be more like 371758 km away from it (that's 5.2 jupiter radii). That's closer than Io orbits Jupiter in the real world. There's little chance for your moon to anyting but tidally locked, and it would have an orbital period of about 35 hours. That's short enough that the tidal locking isn't really a problem, because nights aren't too long. You get a spectacular view from the gas-giant-facing side, when once a day the massive planet above you eclipses the equally massive star, which will be a pretty awe-inspiring event (at least for visitors, maybe the locals would get jaded). On the non-planet-facing side, you just get a fairly regular day-night cycle, albeit with a giant sun during the day.

At that distance, there's no "little year", only the "grand year", as the orbital distance is too low for there to be a significant change in solar radiation received. That means you can't wrangle a 365-day calendar. You can still have seasons if the moon's orbital plane is inclined relative to the planet's.

So, sorry if that wasn't quite what you were hoping, but I also hope you can see there are other potentially interesting features that such a setting brings.

  • $\begingroup$ Thank you so much for the detailed input, this has helped me a great deal and I'll be taking your answer into consideration! $\endgroup$
    – SRL
    Commented Mar 26 at 20:57

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