...Question I: I don't question your math or physics but I do wonder what the mechanical link might be such that the transfer of energy from earth to moon directly affects the moon's rotational velocity or results in a new orbit trajectory. Energy being a scalar it isn't obvious to me how a transfer of energy can alter the orbit or rotation unless the energy transferred is "directional" in some way. ...
I think I know the answer to your question. The Earth is not a rigid body. The moon's gravity gradient cause tides, which are bulges of mass that tend to keep their same relationship to the moon. Think of them as fixed relative to the moon and Earth rotating under these bulges. There are different land masses spinning underneath these bulges so they cannot keep exactly the same relationship, but this is not important if we think of their long term average.
Now to make it more clear, lets conceptually divide the Earth into two parts: One is a perfect rigid sphere and the other, to be extreme in our model, is just a dumbbell, (rotating with the Earth's 24hour day but fixed wrt the moon, representing the tides).
I will type draw the dumbbell as: #----------#
And the moon as: O
This crude arrow:
^
|
|
indicates the moon's velocity.
Now because I think you will agree that the rigid sphere part of the Earth cannot accelerate the moon, to change it orbit, even if it is spins, let’s ignore it. I.e. all of the moon's orbit change must come from the dumbbell, but it is not spinning when viewed from the moon. Wobbling a little bit, due to the effect of the land masses the tides encounter but I will make discussion simpler by having a model of Earth where there is no land mass - just a uniformly deep single ocean covering the Earth. Most of this ocean water is also spherical, so it too can be ignored as not able to change the moon's orbit.
SUMMARY (thus far): Only the tidal dumbbell is available to change the orbit of the moon and your question is how does it do it? What is the mechanism?
If the spatial fixed relationship of the moon and dumbbell were like this:
^
|
|
O...........................................................................................................................................a#----------#b
where I have separately labled the two masses of the dumbell for later reference. (Ignore the ........... as they are needed to keep Sciforum's computer from compressing many spaces down to only one.)
Then the dumbbell also cannot change the moon's orbit as 100% of the forces from both of the # masses acts only along the line from Earth to Moon; Helping make it orbit the Earth. (Moon's velocity turning clockwise in my typed drawing but not changing in speed.) However, that is not their spatial relationship to the moon because it takes some time to lift up a tidal bulge. I.e. the max tide is NOT on the Earth / moon line but delayed. I.e. the near moon bulge peak is a little west of the Moon Earth line as seen from the moon. (The moon is traveling Eastward. I had to give the view as if seen from far below the south pole as I needed a point for my arrow and only the ^ is available.* Thus, in this "southern view," drawing the Earths spinning surface and the moon are going "clockewise.") Thus the typed drawing should be more like:
...........................................................................................................................................................................#b
^
|
|
O
.................................................................................................................................................................a#
Recall the crude arrow pointing up above the moon indicates the moon's velocity. Also note the delay required for the bulge to reach its peak and Earth's rotation have rotated the dumbbell counter closkwise wrt the first drawing's in line position ( a#----------------#b )
Now I need not show the dumbbell’s bar -------------- to keep them apart and it would be hard to get correct even if I used: / to make a sloped line between the two #s because what I type on my computer displays very differently when posted.
Also imagine that the two #s were about 100 yards more to the right and thus both making a very small angle triangle with the moon at the apex.
The scale here is very wrong as the distance between the two #s is the Earth's diameter and the ....... needs to be very much larger as it is the moon Earth separation. I.e. both #s are only a small angle off the moon Earth, center to center, line with the near one's peak displaced a little west of the Earth moon line.
Now both of the #s do exert a force on the moon in the direction of its travel. The more distant # is trying to speed moon's velocity and the closer # is trying to slow it down. Because the closer one has the stronger force, it wins and the moon's velocity slows down.
With this slower velocity the "centrafugal force" is less so to still be in a circular orbit that is stable the moon must be in weaker Earth's gravity (or further away). BTW please note that there is no such thing as "centrafugal force." - It just feels that way on a merry-go-round, etc.
Initially it seems this is wrong as Janus said the Earth was transferring energy to the moon (and that is true) yet I am claiming the moon is slowing down! I.e. that the moon's kinetic energy has Decreased!
Resolution of this paradox requires a little more understanding of orbits. I.e. the moon is in a gravitational well made (by the Earth). It takes energy to climb partially out of that well and be further away. I.e. the Moon gained Potential Energy, PE, and lost Kinetic Energy, KE is the answer to the paradox. Here it is in detail:
Well it turns out that the total Energy, E, is negative for any bound orbit (that is what "bound" means.) and for circular orbits the magnitude of the PE is twice that of the KE. So to make this very clear I will switch to a numerical example with the moon assumed to be in a circular orbit (to keep it simple):
A long time ago when the moon was closer and more tightly bound to the Earth, here is the "energy story":
E = -51 = PE + KE = -102 + 51
Now the Energy story is:
E =-50 = PE + KE = -100 + 50
Note that the moon, now farther away, is only 50 energy units bound. I.e. it is escaping from the Earth's "gravity grasp" and has gained one unit of energy (from the Earth's spin - The days are getting longer.). The moon has move up out of the potential well by 2 units of energy and lost one unit of kinetic energy.
I hope and trust you now understand both the mechanism and that energy are being transferred to the moon, despite it slowing down.
You may not know that Pluto is going slower than the Earth. But this follows from Keppler's law that the square of the period is proportional to the cube of the orbit radius (actual the semi major axis of the ellipse, but I am sticking with circular orbits to keep it simple.) I will write that as:
T^2 ~ R^3 although instead of R one traditionally uses "a" for the semi major radius)
Thus if R were doubled, then T is 2(square root of 2) or more than doubled.
Yet the circumference around the twice larger orbit is only doubled. I.e. it takes more than twice as long to make the full trip around but the path around has only doubled. Thus, the planet (or moon) must be going slower in the larger orbit.
I.e. the whole "story" is consistent and here is
a quick summary:
Because the two tides are not on the Moon / Earth line and the near to moon one is making a stronger force on the moon than the further one and a component of that force is slowing the moon down the moon is moving to a more distant orbit and the day is getting longer. There is energy being transferred to the moon, two units of which are decreasing the negative potential energy and one unit is decreasing the positive kinetic energy for a net transfer to the moon of one positive unit of energy, which previously was in the rotational energy of the Earth's spin.
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*I could have given a view from far above the north pole if I had put the moon on the right edge of your screen, and then that same "up arrow" would give counter clockwise rotations one often sees in such drawings, but that is very hard to get correct when typing and using ............ to defeat the sciforum's computer compressing spaces down to only one. The V could have been used as the arrow point down instead of up and keep the moon O at the left edge still but I could not get it to center on the | line. Besides I live in the southern hemisphere now so it seems OK to me.
PS I am sure Janus 58 could explain it perfectly with 1/10 the words but some people may follow a long winded and detailed version better.
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On your first question: What is measured is the slowing of Earth's spin and the increasing in the Earth moon separation. (Two independent measures that agree but either could be used to calculate the other - so this is a very good test.) I imagine by assuming perfect consevation of total angular momentum this data allows a good measure of the Earth's moment of inertia to fall out.
The laser retro-reflectors left on the moon and very brief (or modulated for precise time ref) laser reflections lets one measure the earth moon separation.
The cumulative change in the spin rate times the total time lapsed (an angle) is best measured, I think from old total moon eclipses. For example, assume there is a constant spin, rate and calculate in which Chinese city an ecllipse of ~3,000 years ago was visible. Well it was not seen there, but in a citry X miles to the west according to the temple astronomers records and local stories of dragon that ate the moon etc.
From X and the radius of the Earth, you get the "correction angle" the Earth failed to rotate thru in these 3,500 years because it has not been spinning at a constant rate, but slowing.