# Reconciling geosynchronous orbits and why the moon is moving away

First time on PSE, and forgive me if the question doesn't make sense.

Everything I've read, and the explanations make perfect intuitive sense, for why the moon is moving away from the earth is because there is a transference of energy from the phase shift in the oceans caused by the water's viscosity to the orbit; that the interplay between the faster spinning earth, the viscosity of the oceans, and the tidal pull from the moon is basically pulling the moon faster, hence sending it to a higher orbit.

My question is thus: assuming the earth has plenty of energy, shouldn't the moon be approaching a geosynchronous orbit, which is almost 1/10 of the current distance to the moon? Thus why isn't the moon being pulled toward the earth instead? I'm sure I'm overlooking some elementary fact.

Maybe my question is boiled down to "how do you move a satellite into a lower orbit?"?

• You've got that backwards: the Earth is approaching a selenosynchronous rotational period (though the red giant sun will consume the pair before they achieve mutual lock). Dec 15, 2015 at 2:38
• @dmckee If the Moon was in geosynchronous orbit, would that be stable (under perturbations of the orbit) ? Dec 15, 2015 at 4:29

In a sense the Moon is approaching a geosynchronous orbit, however the radius of a geosynchronous orbit (call this $r_g$) depends on the angular velocity of the Earth's rotation, so $r_g$ changes as the Earth's rotation changes. Specifically it increases as the Earth's rotation slows.
Currently the angular velocity of the Earth's rotation is faster than the angular velocity of the Moon's orbit, and this means the Earth's rotation is being slowed by the tidal losses induced by the Moon. As the rotation slows this means $r_g$ increases, so in effect the geosynchronous orbit is growing outwards towards the Moon.
• thank you, this makes sense and was kind of what I was thinking, but if you visualize $r_g$ and $r_{moon}$ as concentric circles approaching each other, why are they approaching each other in a increasing/increasing fashion (obviously $\frac{d}{dt}r_g>\frac{d}{dt}r_{moon}$ to make this work), rather than $r_{moon}$ decreasing toward an increasing $r_g$? Dec 16, 2015 at 2:20
• @charlestoncrabb: as the Earth's rotation slows, its rotational kinetic energy is transferred to the Moon i.e. the Earth loses energy and the Moon gains energy. As you add energy to an orbiting body it moves outwards and (unexpectedly) slows down. That's because the total energy is the sum of the potential and kinetic energies, and the two are related by the virial theorem $2K = -U$. To go into this in detail would really need a new question, though I'm sure something like this must have been asked here already. Dec 16, 2015 at 6:07