If the total mechanical energy in a satellite's orbit (assuming circular) is greater when it is closer to the earth, and hence smaller when it is farther from the earth, then we can say that as the moon drifts from the earth, the moon loses energy in translational speed and gravitational potential energy. If only those two are taken into consideration, then there is a net energy loss from the moon.
I had first thought that the energy a satellite has increases as it goes on a larger orbit, but I ran some numbers and it didn't appear so. If I went wrong somewhere, please someone, correct me. Here are my numbers:
For a geostationary satellite (r = 42 164 km, v = 11 068 km/s, m = 1 kg), its total energy is PE + KE. PE = mgh, but g = 0.22416 m/s^2. The result is PE = 9 451.650 kJ, KE = 4 726.582 kJ
For a satellite at r = 45 000 km , m = 1kg, then v = sqrt(GM/r) = 2 976.06 km/s. g at that height is g = 0.19680 The result is PE = 8 856.094 kJ, KE = 4 428.047 kJ
At the larger orbit, both PE and KE are lower than if it was at a lower orbit. Is this right?
Now, the earth slows down its rotation, which allows the moon to go into a larger orbit by conservation of angular momentum. Since the moon goes into a larger orbit, it loses energy. But, since the spin of the earth has slowed down, it also loses energy. Moreover, the moon is still tidally locked with the earth, so its rotational speed isn't increasing.
All in all, there seems to be an energy loss that's going on. How is this being compensated? Is it in the translational speed of the moon (so that the moon is actually moving faster than it should be to maintain a stable orbit)? That seems reasonable - there could be an increase in translational and rotational speed to compensate for the energy loss, maintaining the moon to be tidally locked.
But that's just me. What really happens? How does the energy transfer occur, and are there mathematical equations describing this exchange?