Conservation of Angular Momentum -- Earth-Moon System -- Earth spin backwards? I am developing an exhibit for a museum.  We want to show how the Earth's spin rate changes as the Moon drifts farther and farther out from the Earth.  The visitor has a slider they can move to set the distance of the Moon from the Earth.
Assuming for a moment a closed system, we have conservation of total angular momentum in the Earth-Moon system:
$$ L_{tot}=L_{moon}+S_{earth}+S_{moon} $$
where $ L_{moon} $ is angular momentum of the Moon in orbit about the Earth, and $ S_{earth} $ and $ S_{moon} $ are the angular momentum of the Earth and Moon respectively due to their spin about their own axes.
$$ L_{moon}=m\sqrt{G(m+M)r} $$
where $G$ is the gravitational constant, $m$ mass of the Moon, $M$ mass of the Earth, and $r$ the orbital radius of the Moon about the Earth.
Angular momentum due to spin is calculated assuming uniform spheres spinning about their central axes:
$$ S_{earth}=\frac{2}{5}MR_{earth}^{2}\omega_{earth}$$
$$ S_{moon}=\frac{2}{5}mR_{moon}^{2}\omega_{moon}$$
To figure out the spin rate of the Earth at some new ${r}'$, I calculate ${L}'_{moon}$ and ${S}'_{moon}$ at the new radius as follows:
$$ {L}'_{moon}=m\sqrt{G(m+M){r}'} $$
$$ {S}'_{moon}=\frac{2}{5}mR_{moon}^{2}{\omega}'_{moon}$$
which, as long as we are within the tidal lock limit around 992,000 Km, ${S}'$ can be found from the orbital period of the Moon at that distance:
$${\tau}'_{moon}=2\pi\sqrt{\frac{r^{3}}{G(m+M)}}$$
and
$${\omega}'=\frac{2\pi}{{\tau}'}$$
Applying conservation of angular momentum, I then solve for ${S}'_{earth}$:
$$L_{tot}={L}'_{tot}$$
$$L_{tot}={L}'_{moon}+{S}'_{earth}+{S}'_{moon}$$
$${S}'_{earth}=L_{tot}-{L}'_{moon}-{S}'_{moon}$$
Using this setup, if I run the Moon in closer to the Earth to the distance at which astronomer's estimate the Moon originally formed, I get a spin rate for the Earth that matches what they estimate.  Setting the Moon at its normal orbit of ${r}'=60 R_{earth}$, I get the expected 24 hour period for the Earth and 27.3 hour orbital period for the Moon.  So, at this point everything looks fine and self consistent.
What happens though is that at a distance of roughly ${r}'=93.4 R_{earth}$ (595,700 Km) the Earth stops spinning.  Beyond ${r}'=93.4 R_{earth}$ the equations predict that to conserve angular momentum the Earth has to start spinning backwards!
This violates my gut intuition, and that of others, as to what would happen.
My question is, would the Earth actually start spinning backwards?
Is there a flaw in the analysis?  Specifically, am I violating the requirements of a "closed system" by allowing the visitor to play God and instantaneously move the Moon to a new orbital radius.  If so, how then should I account for the user intervening in the system, and correctly calculate the spin rate of the Earth for different Moon orbital radii?
 A: There's no (obvious) flaw in such a result as long as you take it for what it is, a constraint on the possible momenta, not a prediction of the future. 
If you had a device (located on the earth) that could move the moon to such a distance, it could only achieve its goal by spinning the earth backward.
This configuration won't evolve naturally.  Instead the limit is that the moon will regress until the earth's spin is tidally locked.  You would then need to add energy to the system to push the moon farther out (and slow/reverse the earth's spin).
Think of it this way:  The different angular velocities of the earth's spin and the moon's orbit creates an exploitable source of energy.  Any exploitation of that energy will serve to move the angular velocities closer.  This is the source of the current evolution of the system.  The moon's orbit is slowing down, but the earth's rotation is slowing down more.
Once the velocities are equal, there's no more energy to move the system away.  To make the earth spin backward, you'd have to add energy to the system.  Tidal forces would then act to bring it back to the locked rotation.
