# Why the torque exerted by the moon upon the Earth makes the Moon to increase his orbit? [duplicate]

For what I know, this torque exists because of the misalignment of the tidal bulge with the apsidal line Earth-Moon a certain angle $$\alpha$$, which makes the earth rotation to slow down a bit. But, how that pulls away the moon? I thought this "loss" of rotational momentum transferred into tidal heating. I fail to see how this energy is transferred into more orbital momentum, because I thought the conservation of angular momentum didn't have to do with the rotation of the objects. I mean, isn't true that $$c$$ remains constant in the two-body problem (being $$c = \vec{r}\times \dot{\vec{r}}$$)? How the rotation plays into that?

• The Earth's tidal bulge remains ahead of the apsidal line, therefore it pulls the Moon forward in its orbit (while, it drags down the Earth's spin). Jun 20 '19 at 22:07

because I thought the conservation of angular momentum didn't have to do with the rotation of the objects.

The angular momentum of the system is conserved. As rotation can contribute to angular momentum, it does affect the equation. You can ignore it for point-like objects. In the case of the earth-moon system, the earth's rotation is a significant portion of the total momentum.

A not-too-wrong version would be to assume an axis centered on the earth. Then the total angular momentum of the system could be found by summing the individual momenta for:

• the earth's rotation
• the moon's rotation
• the moon's orbit

Assuming the moon is tidally locked to the earth and has a nearly circular orbit, this gives you only two variables. You can then calculate that, if you slow the earth's rotation, the only way to keep angular momentum of the system constant is to increase the angular momentum of the moon by an equal amount. This would correspond to a larger orbit.

Further if you calculate the KE from both configurations, you find the second one has less energy. So there must be an energy loss to transition between them.