# I Why the Moon always shows the same face

#### PeterDonis

Mentor
That's just an assertion of yours, I think
It's an obvious consequence of the fact that the semi-major axis and orbital period of the Earth-Moon system depend on the total energy.

Dissipation reduces the Energy situation
Obviously.

to a minimum
Only if there is one. What is the minimum total energy of the Earth-Moon system? There isn't one, except in the sense that if the Moon gets close enough to the Earth to be inside the Roche limit, it breaks up, and you no longer have a two-body system.

How can it increase the Energy in a system?
I never said it did. You appear to be reading things into my post that aren't there. Please read what I've actually said.

once the oscillations have died out there will be zero torque
Yes, until dissipation has reduced the total energy of the system and hence changed the zero torque equilibrium point.

there will be equilibrium of a sort because (stable) Equilibrium is the state where a disturbance will increase the Energy in a system and that will cause a restoring Force
This is not completely correct, because the "increase the energy in a system" part is not required. And it's not true for the Earth-Moon system, as we've already said multiple times in this thread.

What does that actually mean?
The quote you gave from me appears to be partially incorrect: I did not say "Please don't insist that a pendulum is the only alternative to tidal locking; I just chose it as a very simple situation - with very obvious differences from a planetary system." Perhaps you intended that to be part of your reply? (Also, the sentence before that is cut off.)

As for what it means, again, the fact that the equilibrium of zero torque for the Earth-Moon system depends on the total energy of the system is an obvious consequence of the fact that the semi-major axis and orbital period depend on the total energy, since the orbital period is what determines the rate of rotation required for the zero torque equilibrium.

It is not necessary to include Forces in an Energy argument.
If you're making an energy argument, yes. I'm not making an energy argument. The argument I'm making includes energy as a contributing factor, but it's not an "energy argument" in the sense of analyzing all of the system's dynamics using energy alone.

isn't this just the difference between Newtonian and Hamiltonian Mechanics? Your above argument is trying to mix the two together.
I don't know what you mean by this or how it relates to what I've said.

#### PeterDonis

Mentor
the semi-major axis and orbital period of the Earth-Moon system depend on the total energy
I should probably expand on this some. There are three main stores of energy (more precisely, mechanical energy) in the Earth-Moon system (if we view things from the standpoint of an inertial frame centered on the Earth): the Moon's orbital energy (kinetic + potential), the Moon's spin energy, and the Earth's spin energy. Heuristically, the processes involved in tidal locking and dissipation work like this:

If the Moon's rotation and revolution periods don't match, there will be torques that act to make them the same. This process conserves total mechanical energy, but transfers energy from the Moon's spin energy to the Moon's orbital energy (assuming that the system started with the Moon rotating much faster than it revolves around the Earth). There will also be dissipation within the Moon during this process, which transfers energy from the Moon's spin into heat, reducing the total mechanical energy of the system.

If the Earth's rotation period does not match the Moon's period of revolution around the Earth, there will be torques that act to make them the same. This process conserves total energy, but transfers energy from the Earth's spin energy to the Moon's orbital energy. There will also be dissipation within the Earth during this process, which transfers energy from the Earth's spin into heat, reducing the total mechanical energy of the system.

Once the Earth's rotation, the Moon's rotation, and the Moon's revolution periods all match, we have a zero torque equilibrium. The system might oscillate about this equilibrium, but that process conserves total mechanical energy. However, there will also still be dissipation in the system, both due to the oscillations, if any, and (a much smaller effect) due to the system's emission of gravitational waves, which happens even at the zero torque equilibrium. Dissipation acts to convert all three types of mechanical energy into either heat (if it's due to oscillation about the zero torque equilibrium) or gravitational wave energy. In either case, it reduces the total mechanical energy of the system.

During the first two phases described above, while dissipation is reducing the total mechanical energy of the system, there is also exchange of mechanical energy going on that is increasing the orbital energy of the Moon (at the expense of the Moon's and Earth's spin energy), which increases the semi-major axis and orbital period. So during these phases, the semi-major axis and orbital period are increasing while the total mechanical energy is decreasing.

During the third phase described above, there is no longer any net exchange of mechanical energy (I say "net" because there will still be some exchange during oscillations about the zero torque equilibrium, but they net out to zero over a complete cycle), so the reduction of total mechanical energy decreases the semi-major axis and orbital period. So during this phase, the semi-major axis, orbital period, and total mechanical energy are all decreasing. This is the main phase I was thinking of in the remark of mine that I quoted at the start of this post. And it should be evident that this implies that (1) the zero torque equilibrium point depends on the total mechanical energy, and (2) there is no minimum of the total mechanical energy (other than the Moon getting inside the Roche limit and breaking up).

#### alantheastronomer

PeterDonis said:
...if the Moon did not have an asymmetry which side of the Moon...faced the Earth as a result of tidal locking would have been a random result, since no side would have been preferred.
We know this is true because Venus is also tidally locked yet has no mass asymmetry...

#### PeterDonis

Mentor
Venus is also tidally locked
Not the way the Moon is. Venus rotates retrograde, not prograde, and its rotation period is longer than its period of revolution around the Sun. I believe the state it is in is a tidal resonance, but it's not the one usually referred to by the term "tidal locking".

#### alantheastronomer

Not the way the Moon is. Venus rotates retrograde, not prograde, and its rotation period is longer than its period of revolution around the Sun. I believe the state it is in is a tidal resonance, but it's not the one usually referred to by the term "tidal locking".
Yes that's certainly true, but assuming that it's rotation rate was much greater and prograde in the distant past, the same mechanism of tidal gravity is what's brought it into it's present resonant state.

#### PeterDonis

Mentor
assuming that it's rotation rate was much greater and prograde in the distant past, the same mechanism of tidal gravity is what's brought it into it's present resonant state.
That's not possible. If it were rotating prograde faster than it revolved around the Sun, tidal torques would drive its prograde rotation rate to be the same as its rate of revolution around the Sun. (Or it could possibly get stuck in another resonance before that, as Mercury is.) Tidal torques at that point would stop changing the rotation rate; they certainly woudn't reverse it.

#### alantheastronomer

That's not possible. If it were rotating prograde faster than it revolved around the Sun, tidal torques would drive its prograde rotation rate to be the same as its rate of revolution around the Sun. (Or it could possibly get stuck in another resonance before that, as Mercury is.) Tidal torques at that point would stop changing the rotation rate; they certainly woudn't reverse it.
Yes one would think so... but if -dw/dt (where w is angular rotational velocity) were nonzero, then when it reached corotation it is possible for it to overcompensate and become retrograde. A similar effect occurs in accretion - if the angular momentum of the accreting matter is such that v^2/r is less than GM/r^2, then the deceleration -dv(r)/dt is nonzero and when it reaches the equilibrium radius, it overshoots and continues on to smaller radii until the centripetal acceleration overpowers gravity and it begins to move outward again. Of course then it will collide with other infalling matter and it's outward motion will be damped. I'm not saying this is definitely the case for Venus - it's also possible that it started out with a retrograde rotation with it ending in it's current resonance, or that it started out in that resonance when it first formed, or it's also possible that it formed at some other radii and migrated to it's present location. All I'm saying is that Venus' situation (and, as you point out, Mercury's) illustrates that, as you say, tidal gravity doesn't need to depend on an asymmetric mass distribution.

#### PeterDonis

Mentor
one would think so
Not just "one would think so". It is so based on the torques vanishing when the rotation and revolution rates are equal. That has already been discussed at length in this thread.

if -dw/dt (where w is angular rotational velocity) were nonzero, then when it reached corotation it is possible for it to overcompensate and become retrograde
First, rotating prograde a bit more slowly than the revolution rate is not at all the same as retrograde rotation.

Second, if prograde rotation overshoots and gets slower than the revolution rate, then the torques are now opposite and will act to speed up the rotation rate until it is equal to the revolution rate. If it overshoots again, you simply get oscillations about the zero torque equilibrium. That has already been discussed at length in this thread as well.

All I'm saying is that Venus' situation (and, as you point out, Mercury's) illustrates that, as you say, tidal gravity doesn't need to depend on an asymmetric mass distribution.
That's certainly not all you are saying. This part of what you are saying is correct. But you are also saying other things that are not correct. See above.

#### alantheastronomer

PeterDonis said:
That's certainly not all you are saying.
It is in post #78. Also, "all I'm saying is.." is just an expression; it means "my main point is..." or "the gist of my statement is...", not "the sum total of my account is..."
First, rotating prograde a bit more slowly than the revolution rate is not at all the same as retrograde rotation.
That's very true but - let me give you one more example in physics, along with the accretion situation, where counterintuitively, motion seemingly miraculously appears to reverse; are you familiar with the concept of the magnetic mirror? A charge particle moves in a circular motion in a magnetic field...as it moves towards a region where the magnetic field gets stronger, it's perpendicular velocity increases and it's kinetic energy gets larger. Since energy must be conserved, where does this increase in kinetic energy come from? It "steals" it from the particle's forward motion and it decelerates until the field strength becomes so great that it actually reverses direction!
if prograde rotation overshoots and gets slower than the revolution rate, then the torques are now opposite and will act to speed up the rotation rate until it is equal to the revolution rate. If it overshoots again, you simply get oscillations about the zero torque equilibrium.
Yes that's true, unless the planet has time to "relax" and lose it's tidal bulge so that there are no more restoring torques to bring it back to equilibrium. Venus now is perfectly spherical, so it is pretty safe to say that at some time in it's past, it's shape was restored to it's present condition.... If I can go a little off topic for a moment, something else you said caught my interest....From post #65
Dissipation...will mainly be friction inside the Earth...
and post #70
...only a small fraction of the total rotational kinetic energy lost by the Earth as it's rotation slows is transferred to the Moon along with angular momentum; the rest is converted to heat.
I remember reading somewhere that given the best estimates of the amount of uranium in the Earth's core, it's still not enough to keep the outer core fluid and it should have solidified long ago and so there needs to be another source of heat otherwise there would be no magnetic field. Is it possible that tidal friction could be the source of this heat, and that tidal torques on the liquid outer core drives the convection that is necessary for the dynamo responsible for the earth's magnetic field?

#### PeterDonis

Mentor
It is in post #78.
No, it isn't. Either own everything you said, or don't say it in the first place.

are you familiar with the concept of the magnetic mirror?
This concept is irrelevant to the discussion here, since there is no gravitational analogue to it.
that's true, unless the planet has time to "relax" and lose it's tidal bulge so that there are no more restoring torques to bring it back to equilibrium.
The planet won't "relax" and lose its tidal bulge unless it is no longer subject to tidal forces. See further comments below.

Venus now is perfectly spherical
To the accuracy we can measure. But since it is subject to the Sun's tidal gravity, it must have a tidal bulge of some extent. It's just much smaller than Earth's, because Venus has no ocean and no Moon. But this will have been true throughout its history if it's true now; it's not the result of any "relaxation" process.

Since the thread has run its course anyway, the thread is now closed. If you want to discuss further your question about the heat source of the Earth's core, please PM me and I will spin that part of your post off into a new thread.

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