4
$\begingroup$

Obviously, with so many stars and presumable planets, lots of things can happen for a short time, but I'm wondering if there is a stable configuration for either twin planets or a planet with a sizable moon where either moon or planet (or both) have a synchronous tidal lock to the sun and the moon and planet orbit each other more frequently - possibly in an integer spin-orbit resonance.

$\endgroup$

1 Answer 1

4
$\begingroup$

The negative (decelerating) tidal torque of the star is trying to synchronise the planet's rotation with its mean motion $n_p$ about the star. (Rare exceptions are close-in planets with a high eccentricity, which may end up in a higher spin-orbit resonances. The only confirmed case of the kind is Mercury in its 3:2 spin-orbit state.) The tidal torque from the moon is working to synchronise the planet with the mean motion $n_m$ of the moon.

These two torques are competing.

For a sufficiently distant planet (like our Earth), the lunar torque beats the solar torque, wherefore the Earth's rotation will eventually get synchronised with the mean motion of the Moon.

For a close-in planet, however, the situation may be different. We explored this topic in our recent work, and found that in some plausible scenarios, exomoons synchronise the rotation of their parent planets, overpowering the tidal action from the stars. Such exomoons can prevent inner exoplanets from spiraling into their host stars, thus extending these planets’ lifetimes.

This result may look counterintuitive because, for a close-in planet, the solar torque is expected to be stronger than the lunar torque. The outcome of their competition is, however, influenced by the fact that both torques bear a strong frequency-dependence. I can provide more detail if you are interested.

PS.
Contrary to a common stereotype, a moon can synchronise a planet not only while receding from it but also while descending. In the former case, synchronism is stable if attained before the moon crosses the reduced Hill sphere. In the latter case -- if attained above the Roche zone.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .