Jupiter has so many moons, and those moons have a gravitational pull on it. Does that mean, that over time, the length of Jupiter's days will get shorter?


1 Answer 1



A moon that has an orbit below the synchronous orbit $a_{synch}=\sqrt[3]{GM_{Jupiter}T_{Jupiter}^2/4\pi^2}$ (for Jupiter, 160,000 km) will move around the planet faster than it rotates and hence cause a tidal bulge that trails the moon, pulling at it (making the moon slowly spiral in) while speeding up the rotation of the planet. Only Adrastea and Metis are in this position, and they are tiny.

Moons outside the orbit will instead cause tidal bulges that are ahead of them (since the planet turns faster) and will be pulled forward, slowly moving to higher orbits while depleting the rotation rate. So the moons are very slowly increasing the length of Jupiter days.

In practice there is a lot of interaction here: the tidal bulges affect all moons, and the moons also affect each other and have orbital resonances. Io, that has the largest tidal effect (since the tidal force $\propto m/a^3$ is largest), has an elliptic orbit that means it is alternately speeding up and slowing down.

So the above answer is just a first approximation: ideally one needs to do a very careful simulation of the interactions in the system over long time, but I would be very surprised if the answer was anything else. The reason is the virial theorem: in interacting systems of this kind the average kinetic energy (rotation and moon velocity) tends to approach half of the average potential energy. Over time this tends to move more and more angular momentum into ever more remote orbiting objects. The main thing that can defeat this is if there is a lot of energy dissipation slowing the entire system. But the mass of the moons compared to the planet is very small, so there is likely less energy dissipation per kilogram of matter here than between e.g. Earth and the Moon.


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