4
$\begingroup$

The Moon's rotation is firmly tidally locked to the Earth and the Earth's rotation is firmly tidally unlocked with respect to the Moon. I gather that Mercury's rotation is tidally locked in a 3:2 resonance - which sounds a little precarious. Are there any bodies which are on the verge of being tidally locked (very nonuniform rotation) or are barely tidally locked (very large libration)? Are there any bodies whose rotation is actually chaotic for such reasons?

[Edit] Note, this question is about the physical libration rather than the apparent or optical libration caused by the body being observing from different perspectives.

$\endgroup$
10
  • 2
    $\begingroup$ Libration is not an expression of the "quality" of tidal locking. It is an expression of the excentricity and obliquity of the orbit of the tidally-locked body. $\endgroup$ – planetmaker Dec 11 '20 at 8:18
  • 1
    $\begingroup$ Great question! This is not the same thing but it's quite an interesting related situation: Is Venus in some way tidally locked to… Earth? $\endgroup$ – uhoh Dec 11 '20 at 14:52
  • 1
    $\begingroup$ @planetmaker that's not correct. You've only described optical libration. Now read about true libration which is what the OP is describing. Our Moon does both, the latter is small now, but it would have been quite wild at some point the past. $\endgroup$ – uhoh Dec 11 '20 at 14:59
  • 1
    $\begingroup$ @uhoh That paper on the Moon's libration is surely a classic. There's something special about papers like this that were written well before the age of computers. $\endgroup$ – Roger Wood Dec 12 '20 at 4:14
  • 2
    $\begingroup$ The magnitude of tidal forces decreases with a power of 7..8 of the distance of two bodies (as far as I remember). This means that tidal locking either occurs very quickly (on astronomical time scales) or never. The chances to find an object in the middle of the transition from unlocked to locked is thus very small. $\endgroup$ – Hartmut Braun Dec 12 '20 at 8:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.