2
$\begingroup$

Tidal heating of a tidally locked moon is relatively straight forward to calculate, even though details of its internal structure is hard to work out in the first place.

By contrast, tidal heating due to tides of a rotating body is much more involved.


These two processes aren't completely equivalent. For instance, for a tidally locked body, the heating depends on the eccentricity, with no heating at zero eccentricity. By contrast, a rotating body can very much experience tidal heating even when in a completely circular orbit.

As a naive assumption, a rotating body will still receive heating from the deformation caused by an eccentric orbit, with some additional heat added through the periodic deformation caused by the rotation. As such, a rotating body will always receive more heat than a non-rotating one.

Which has the obvious use case that one can obtain a lower bound for the tidal heating through a simple calculation.

But is this assumption anywhere close to the truth?


Or, to put it in another way, two scenarios of moons in identical orbits. One tidally locked and the other not, which experience more tidal heating?

$\endgroup$
4
  • 1
    $\begingroup$ Could you clarify what your specific question is? Do you have a particular "simple calculation" in mind, and you are asking if it is accurate? (If so, maybe post what that calculation is - I don't think tidal heating is ever particularly simple to calculate.) Also, when you say "non-rotating" above, I think you mean to say "tidally locked" - an orbiting body that is not rotating relative to a fixed background frame is rotating relative to the object it orbits. $\endgroup$ Jul 31, 2020 at 1:56
  • $\begingroup$ I assume you mean two scenarios of moons in identical orbits. One tidally locked and the other not. Maybe you could clarify that. $\endgroup$ Jul 31, 2020 at 2:24
  • $\begingroup$ Just noting, A tidally locked moon, even in a perfectly circular orbit can still be heated if its spin axis is different than its orbital axis. Although maybe at a certain difference between spin axis and orbit axis, we might stop calling the moon tidally locked. $\endgroup$ Jul 31, 2020 at 2:30
  • $\begingroup$ @KeithReynolds tidal stress due to axis misalignment just needlessly complicates things, and may be ignored. $\endgroup$ Jul 31, 2020 at 10:04

1 Answer 1

2
$\begingroup$

The answer is: yes.

In a synchronised moon, the tidal bulge is aimed, approximately, at the planet. Approximately -- because for a finite eccentricity the position of the planet as seen from the moon is slightly librating. Consequently, the position of the bulge on the moon is slightly librating. As a result of this, the moon is experiencing tidal heating of order $e^2$.

In a nonsynchronous moon, the tidal bulge is perennially running over its circumference -- which is causing tidal heating of order $e^0$.

For details, see this paper and also this one.

PS.
Additional tidal heat emerges due to physical libration. Usually, this does not add much to the story. However, in manifestly triaxial bodies (Phobos, Mimas, Enceladus) physical libration can boost the tidal effects considerably. This pertains especially to Epimetheus whose physical libration boosts tidal heating by a factor of 26, see the penultimate line in Table 2 in this paper.

$\endgroup$

You must log in to answer this question.

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