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Hungry4info
2010-Jul-31, 10:17 PM
A tidally locked body on an eccentric orbit will orbit about its primary in a pseudosynchronous rotation, where the rotation rate is synchronised at periapsis, but rotates faster than the mean motion everywhere else. Over time the slight difference betwee the rotation and orbital rate (this is, after all, pseudosynchronous rotation) will result in other parts of the tidally locked body being presented to the primary.

According to this (http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1981A%26A....99..126H&data_type=PDF_HIGH&type=PRINTER&filetype=.pdf) (PDF) paper, the rotation rate of a tidally locked body is

http://www.oklo.org/wp-content/images/pseudocirc.jpg


Here's the age old "Why does the moon always present the same face to Earth" question. The moon's orbit is not circular, so I would expect pseudosynchronous rotation to slowly bring to us the far side of the moon over the course of some years. Could it be that the difference in tidal forces between Earth and the moon at its periastron and apastron are insufficient to make a difference? Perhaps some sort of sun-Earth-moon interaction drown out the effect?

whimsyfree
2010-Aug-01, 12:53 AM
A tidally locked body on an eccentric orbit will orbit about its primary in a pseudosynchronous rotation,

It depends on how you define tidally locked. There does not appear to be a generally accepted definition, so statements such as yours should be avoided. Many people consider Mercury to be tidally locked, and its rotation is not pseudosynchronous. You seem to regard pseudosynchronous rotation as inevitable. It is not.

neilzero
2010-Aug-01, 01:06 AM
Also the density of our moon is not symmetrical, so the creeping of the observed side may not occur, unless the moon is sufficiently plastic to adjust the locations of it's density anomalies. The moon does rock like a pendulum. Does anyone know the period of the rock = liberation? Neil

Hornblower
2010-Aug-01, 01:13 AM
The findings in this paper were primarily concerned with stars, which are fluid and do not have any frozen elongations in their shapes. The authors mentioned Mercury as being locked in a 3:2 resonance, and they stated that a fluid body in the same position would evolve away from that state.

The Moon's orbital eccentricity is much lower than that of Mercury, enabling it to be in a stable 1:1 pattern with just a few degrees of libration around its mean orientation with respect to the Earth.

Hungry4info
2010-Aug-01, 04:27 PM
You seem to regard pseudosynchronous rotation as inevitable. It is not. I don't see pseudosynchronous rotation as inevitable. Of course it isn't, the moon is an example. In cased you failed to read my post, I am asking why. I'm trying to understand the mechanics of it. Your recent habit of picking apart the wording of my posts while completely missing their point isn't doing either of us any good. I humbly ask you to cease.


Also the density of our moon is not symmetrical
The authors mentioned Mercury as being locked in a 3:2 resonance, and they stated that a fluid body in the same position would evolve away from that state.

That makes sense. I appreciate it.

Ken G
2010-Aug-01, 05:13 PM
Hornblower's point is the key. Apparently, for a gaseous body, tidal effects will favor a pseudosynchronous orbit at periastron, whereas rigid bodies favor integer commensurate rotation rates like 1:1 and 3:2. The explanation I've heard for Mercury is that its football shape is maintained as it orbits (it doesn't reconform to pointing at the Sun all the time), so as long as its "point" is aimed at the Sun at every periastron, it's "happy." Any ratio like n:2 allows that (since there are two points), so my guess is, a rigid body simply picks out the integer n that allows it to come as close as possible to pseudosynchronous, whereas a fluid body can achieve exact pseudosynchronicity because its shape is constantly deforming in its own frame. Whether the body counts as "rigid" or "fluid" must depend on the comparison of its tidal deformation timescale compared to its orbital period. This makes sense to me, though it's only my reaction to your interesting question-- I've never heard this answer put quite this way, so caveat emptor (translation: I'm too lazy to research into whether or not this is really right.)

Hungry4info
2010-Aug-01, 05:44 PM
That makes sense. Trying to envision Prometheus orbiting Saturn in an eccentric orbit such that its long axis points toward the planet at periapsis, I can see how it would get stuck in an n:2 spin-orbit resonance. In a more circular orbit (but e still > 0), it's still going to want to point its long axis toward the planet, only way to reasonably do this is a 1:1 spin-orbit resonance.

whimsyfree
2010-Aug-01, 09:58 PM
I don't see pseudosynchronous rotation as inevitable. Of course it isn't, the moon is an example. In cased you failed to read my post, I am asking why.


I read your post but you apparently didn't read the paper you linked to. It mentions that non-gaseous bodies may deviate from pseudo-synchronization due to asymmetries on page 130, using the example of Mercury.