Increasing distance between Earth and Moon I have a problem where a planet's rate of rotation is decreasing due to tidal effects of the moon. I know that the angular momentum of the system will be conserved. So, in order to conserve that the moon will recede away from the planet.
$$ L = mvr = m\omega r^2.$$
I'm not sure how to convert/translate this loss in angular momentum of the planet to the rate of recession of the moon.
 A: Whether the Earth slows down or speeds up isn't something you can just decide from basic logic. It is a mathematical question whose answer depends on the relative phase relationship of the driving force (the moon) and he driven object (the tide). If the frequency of the driving force is less than the natural frequency of the tides, the tidal bulge will lag and the effect will be to speed up the earth and pull the moon down into a lower orbit. (The lower orbit is actually faster, but the angular momentum goes down because the radius changes more than the velocity.) If the frequency of the driving force is greater than the natural frequency of the tides, the bulge will lead the moon and the effect will be to slow the earth down and drive the moon into a higher orbit. (The higher orbit will actually be slower, but the anuglar momentum goes up because of the radius.)
It is an experimental fact that the driving frequency of the moon (one rotation every 24 hours) is greater than the natural frequency of the tidal bulge (circe the earth in 2 or 3 days, or a wave velocity of several hundred mph). So the earth slows down and the moon gets farther away.
I analyze this in a series of blog posts starting here: How the Tides Slow Down The Moon
A: The key is, as you indicate, in the conservation of angular momentum.
First comment on the cause : the tides
the effect of tides (flows) causes a slowdown in the angular velocity of the Earth, that is to say, the Earth slows down by the tidal effects caused by the Moon ( the sun also causes but is much lower).
What happens? Let us consider the isolated system Earth-Moon


*

*The total angular momentum must be conserved.

*The angular velocity of the Earth decreases because of the effects of the tide, the period of rotation is increasing. That is to say, the angular momentum of the Earth decreases

*As a result, if the angular momentum of the Earth decreases, the angular momentum of the Moon has to increase to the total angular momentum is maintained, is the only option.

*Therefore, to increase the angular momentum of the moon, the orbit of the moon becomes higher, so, gradually, moon's orbit becomes larger= more eccentric = the distance increases with the Earth

Let me make you an approach.
Of the dynamics of the circular movement
$$  G \frac {M_e M_m}{d^2} = M_m w^2 d $$   where $w = \sqrt {G \frac{M_e}{d^3}}$
$M_e$ and $M_m$ are the mass of the Earth and Moon respectively.
Now, we consider the isolated system Earth-Moon, the total momentum $L$ must be conserved over the time, this is :
$$ L = L_e + L_m = L'_e + L'_m = L' $$
The moment of inertia of a sphere of mass m and radius R with respect to the axis of rotation that passes through its center is $2M_eR^2/5$ (obviously, for simplicity, we are considering that the earth is spherical), and the moment of inertia of a particle of mass m that is far d of the axis of rotation is $M_m d^2$
Another consideration, mass of the Earth is considerably greater than the mass of the Moon and that the distance between its centers is much greater than any of its rays, so we can consider the moon as a particle of mass $M_m = 8.99\times 10^{22} kg $ which describes a circular orbit of radius d around the Earth.
The last consideration, also for simplicity, we've considered the axis of rotation of the Earth perpendicular to the plane of the Moon's orbit.
Also, you know about the Earth : $M_e = 5.98 \times 10^{24} kg $ , $R_e = 6.37 \times 10^6 m$
After all these considerations and introduction, we have that :
$$ \frac 25 M_e R^2 \Omega + M_m d^2 w = \frac 25 M_e R^2 \Omega_1 + M_m d_1^2 w_1^2 $$
I have to leave now, anyway hope you understand my explanation. If you need, I can continue with the derivation but I think you'll be able to do that.
