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For purposes of realistic, but otherworldly visualisations of foreign worlds I'm wondering what the fastest speed is that a moon will track through the visible sky, as observed from the body of a planet (or other moon).

As in, the moon rises, the observer watches it move across the sky, and the moon sets on the opposite horizon.

My intuition says that there exist a bunch of limits here... Orbital (angular) velocities depend on orbital height, and there are limits to that depending on the masses of two bodies. Then the rotation of the observers planet/moon plays a role too, but I assume there are limits here as well in terms of what is realistic or physically stable for celestial bodies.

How quickly could a moon appear and disappear after tracking once through the sky overhead? Hours? Minutes? Less?

Assume:

  • The planet/moon the observer is standing on can support... well a human observer standing on it. At least in a spacesuit - so this gives certain upper/lower bounds in terms of g-force/size/mass.
  • The orbital system is stable (more or less) - so no rogue moon flashing by.
  • Assume the moons track in the sky traverses through the zenith (so not just shortly peeking over the horizon for a second, as the sun does in arctic fall/spring).
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  • $\begingroup$ Phobos completes an orbit in about seven and a half hours, and traverses the Martian sky in "4 hours and 15 minutes or less" (west to east, of course). The International Space Station, which is sufficiently large to be quite often visible with the naked eye (and could easily be much larger), completes an orbit in about an hour and a half; but because of its low altitude needs to be boosted periodically. What answer does the question expect which is not already provided in the usual places? $\endgroup$
    – AlexP
    Commented Dec 9, 2022 at 13:13
  • $\begingroup$ @Igysin Habitable Planets for Man, 1964, Stephen H. Dole, discusses the limits of how fast or slow a planet habitable for humans could rotate. rand.org/content/dam/rand/pubs/commercial_books/2007/… For a completely airless planet, the orbital period of a satellite at the surface is the shortest possible orbital period, but for any planet & moon combination, the Roche Limit of how close that moon can get before breaking up can be calculated. I note that even thousand of miles above the surface of Mars, there is enough air to lower the orbit of Phobos. $\endgroup$
    – M. A. Golding
    Commented Dec 9, 2022 at 18:07

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Somewhere in the area of an hour, for large moons.

If you want a nice cinematic view with a large moon, it has to be orbiting relatively far away otherwise it'll be destroyed by tidal forces. Orbiting far away means it orbits slower, and so it moves slower.

The Moon moves across the sky primarily because the Earth rotates, and not really because of the Moon's orbit as it's orbiting so slowly. For a large satellite, the planet can rotate a lot faster before centrifugal force pulls it apart than the satellite can orbit closely before it gets ripped apart.

Using Universe Sandbox (which is a fantastic tool for mostly accurately simulating systems like this), it shows that the Earth would need to spin approximately once every 70 minutes or so to rip itself apart. The Moon at just above the Roche limit would only orbit every seven hours; much slower. This is similar for most bodies I tried out.

Whilst a planet spinning at close to its own escape velocity would be... an experience, to say the least, it's definitely survivable especially at the upper latitudes.

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I am misusing the prefix "geo-" to refer to this other planet rather than earth.

assuming this moon is natural and not the result of a geolocically recent event it is at or above geosynvchronous altitude, this puts an asymptotic lower limit on orbit time at 1 local day as observed.

Below geosynchronous orbit tidal drag would decellerate* the moon causing it to spiral down and crash over geological time.

  • by decellerate I mean apply a force in opposing its relative motion, not make it go slower, orbital mechanics is perverse.
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  • $\begingroup$ What you say is correct and true and you have my upvote. However, we can redefine the length of a day for some really fun results (see Lura Skye's answer). $\endgroup$
    – The Square-Cube Law
    Commented Dec 11, 2022 at 9:03
  • $\begingroup$ Yes, I neglected to mention that, density and radius of the planet have an effect too. $\endgroup$
    – Jasen
    Commented Dec 11, 2022 at 9:06
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Consider Jasen's answer for a limit on systems that are stable for billions of years. The stable limit will have you see the moon go over the sky in about half a day.

Also consider Lura Skye's answer for a lesson in cruel astrophysics. If your day lasts a few minutes, well...

Let me add a different scenario, one that we have in the real world. Phobos orbits Mars really fast:

Phobos orbits 6,000 km (3,700 mi) from the Martian surface, closer to its primary body than any other known planetary moon. It is so close that it orbits Mars much faster than Mars rotates, and completes an orbit in just 7 hours and 39 minutes. As a result, from the surface of Mars it appears to rise in the west, move across the sky in 4 hours and 15 minutes or less, and set in the east, twice each Martian day.

For reference, a Martian day lasts just 37 minutes more than an Earth day,

Since Phobos orbits Mars below the synchronous limit, its orbit is decaying. According to some estimates it will reach the Roche limit and be destroyed in 43 million years. The wikipedia article I linked above elaborates further on this.

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Short answer: A few hours seems to be the lower limit regarding air resistance

I feel like your observer is probably fine, assuming the masses of the earth and its moon. The gravity of the earth is always higher than the moons gravity, even if they would touch each other. The observer might not want to live there and experience a few strange moments and spectacular tides, but should be fine otherwise.

I would guess that air resistance is actually the limiting factor. Low-earth-orbit Satellites and space stations orbit the earth roughly 400 km above the surface and take ~90 min for it. The main problem here is air resistance. Even at this height, they have to accelerate occasionally, to compensate for it. The orbit of the moon at this height would not be stable since the velocity would decrease over time and it would fall. Honestly, I don't really know at which height the air resistance can be neglected so the orbit would be stable for the foreseeable future. A very close moon will also cause "atmospheric tides", so the atmosphere will be thicker at the side facing the moon. A few thousand kilometers (wild guess) distance might be enough to have a stable orbit for a while, this would correspond to a few hours per cycle. I'm intentionally vague here, since this is so speculative.

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    $\begingroup$ @Matthais The orbital periods of the innermost moons of solar system planets are not speculative but are measured. Especialy the period of Phobos, the innermost moon of Mars, snce Mars is a planet with a solid surface someone could stand on. Air resistance is lowering the orbit of Phobos, which is expected to reach its Roche Limit and break up in about 30 to 50 million years. $\endgroup$
    – M. A. Golding
    Commented Dec 9, 2022 at 18:05
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Frameshift: What sort of "world" is permitted?

I'm thinking of a "moon" that's actually a planet orbiting just inside a ringworld. Flyby velocity will be in the ballpark of .001c.

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