Can a moon orbit its planet faster than its planet rotates?

Can a moon orbit its planet faster than its planet rotates?

Can a moon orbit its planet more than once per the planet's day? It seems possible but I'm not sure.

Are there any known examples of this situation?

• Good question, welcome to Stack Exchange! By "faster" you mean in angular velocity, not linear velocity. Earth's equator rotates at about 500 m/s and Earth's Moon's orbital velocity is about 1000 m/s, but of course it takes almost 30 times longer for the Moon to go around once than it does a point on the equator, because the moon's angular velocity is slower.
– uhoh
Nov 19 '20 at 1:32
• There's no reason why Venus could not have moons, right? If it had one (or several), it would orbit with a period much shorter than the planet's day. Nov 19 '20 at 22:30
• Venus takes more than 200 earth-days to rotate once. So if a hypothetical moon existed around Venus, it would definitely have an orbital period less than 200 days. Nov 21 '20 at 0:53

What immediately springs to mind is the Martian moon Phobos, orbiting the planet in 7 hours 39 minutes. That's a fair bit quicker than the 24 hour 37 minute sidereal period of Mars.

From the surface of the planet, Phobos and Deimos will therefore appear to cross the sky in opposite directions.

Other solar system examples include the small inner Jupiter moons Metis and Adrastea.

Orbital period is determined by the semi-major axis, which for a circular orbit is just the radius as:

$$T = 2\pi \sqrt{\frac{a^3}{GM}}$$

The fastest a moon could orbit is if this radius is above the ground (and atmosphere) of the parent planet. We can use this to determine a condition for when your described scenario is possible.

First, we turn the orbital period equation around to get the semi-major axis from the period:

$$a =\sqrt[3]{GM\left(\frac{T}{2\pi}\right)^2}$$

Which for the boundary case $$T = P_{sideral}$$ has to be greater than the radius of the planet:

$$\sqrt[3]{GM\left(\frac{P_{sideral}}{2\pi}\right)^2} > r_{planet}$$

More intuitively, this is likely universally true since if orbital velocity was lower than the rotational velocity of the surface, the "surface" would be slingshot into space.

With one caveat:

For larger moons, you have to substitute $$r_{planet}$$ with the Roche limit of your moon, since it can not keep structural integrity below that. This would for any substantial moon put the limit a fair bit higher, giving the inequality from earlier some purpose.

• Orbits lower than "stationary" (faster than the planet's rotation) tend to decay lower and lower. It is projected that Phobos will fall in ~20 milion years. On the other hand, orbits higher and slower than the stationary, "decay" higher. Our Moon goes few cm higher every year. Nov 19 '20 at 9:40
• Fun fact, you can rearrange the last equation into $P>2\pi \sqrt{\frac{1}{G}\frac{r^3}{M}}$; the $r^3/M$ term is inversely proportional to the average density of the planet, so the minimum period of an orbit around a body depends only on that body's average density! Nov 19 '20 at 16:41
• "Phobos and Deimos will therefore appear to cross the sky in opposite directions". Space is weird. Nov 19 '20 at 22:39
• @Schwern the moons are so close to the surface that for large parts of the planet they are never even visible. Nov 20 '20 at 5:12
• @OrganicMarble Is there an easy way to make out regions on mars, where one would actually be able to see both moons cross the sky regularly? Because if there is a place it should get consideration for the firs martian settlement Nov 20 '20 at 9:49

Are there any known examples of this situation?

Yes!

In addition to Phobos mentioned in this answer and from Astronomy SE:

Though it's a stretch, GNSS satellites in MEO like GPS, Galileo, GLONASS, etc... are examples of such "moons" orbiting planet Earth, since they're positioned at an altitude lower than geostationary orbit but are still high enough for atmospheric drag to be so low that they can maintain that orbit without needing additional sources of propulsion/correction.

• Do operating GPS satellites ever make orbital maneuvers for station-keeping? They do for purposes of maintaining orbital phasing, but mostly not due to atmospheric drag. There are other perturbing forces, like gravity from the Moon and Sun and photon pressure as well.
– uhoh
Nov 19 '20 at 1:57
• @uhuh Fair enough, that might indeed contradict my statement. It's not clear to me though if those maneuvers are needed to maintain altitude, or are primarily required for more lateral position fine-tuning.
– Will
Nov 19 '20 at 2:05
• Pretty much any satellite in LEO, no? The ISS? Wait for it - that's not a moon, it's a space station Nov 19 '20 at 2:16
• @uhoh Ah, thanks. Also, your edits are much appreciated, especially since I don't really have time to improve this answer myself right now anyway.
– Will
Nov 19 '20 at 2:38
• @OrganicMarble Many LEO satellites are in retrograde orbits, so compared to the rotation of the planet, they orbit in negative time. ;-)
– gerrit
Nov 19 '20 at 9:27

This is equivalent to asking whether it's possible for a moon to orbit lower than a geosynchronous orbit, and there are of course satellites in orbit around the Earth that are lower. It's also equivalent to whether it's possible for a planet to rotate slower than its moon orbits, and clearly there's no lower limit on how slowly a planet can rotate (at least not as far as a direct result of physical laws; as a practical matter if the matter that the planet coalesces from has angular momentum, that will be preserved in the planet's rotation).

However, if the planet has an atmosphere, that puts a lower limit on moon orbital radius (and keep in mind that even if an atmosphere seems negligible, drag over millions of years can add up). Another lower limit is the Roche limit. Also, tidal forces will cause the moon to spiral in, and as it spirals closer, the effect gets even larger.

If a planet has a large moon that causes significant tides on the planet, then the Moon will spiral inward if it rotates faster than the planet and will then eventually collide with the planet. This is because the tidal bulges caused by the moon are going to be pulled back relative to the moon by the planet's slower angular rotation. Th force of gravity exerted by these bulges on the moon cause the moon to be a tiny bit slower in its orbit than it should be, causing it to spiral inward. There is then a transfer of angular momentum from the moon to the planet, but the moon will actually orbit the planet faster as its orbit decays. The moon will end up being destroyed when it's orbital radius becomes less than the Roche limit.