# Relationship between angular momentum of Earth and recession rate of the Moon

So the problem goes like this:

Two masses $m_1$ and $m_2$ orbit each other with semimajor axis $a$. The orbit is circular, and $m_1 \gg m_2$. The body $m_1$ has a rotational moment of intertia $I_1$ (about an axis that passes through its center of mass) and a spin angular frequency of $\Omega_1$. Treat $m_2$ as a point mass.

So first, they want me to write the total angular momentum $L$ of the system, which I've determined to be:

$L \simeq m_2\sqrt{Gm_1a} + I_1\Omega_1$

So this seems fine. But then they want me to take the time derivative of $L$ and set it to 0 (since of course, net torque is 0) and relate $\dot a$ to $\dot{\Omega}_1$.

So....

$\dot L = [m_2\sqrt{Gm_1}]\frac{1}{2\sqrt{a}}\dot a + I_1\dot\Omega_1 = 0$

$\dot a = - (\frac{2I_1}{m_2\sqrt{Gm_1}})\dot \Omega_1 \sqrt a$.

This looks ok, except for that $\sqrt a$ in there. Am I doing something wrong here?

-
I'm confused. Your title talks about Earth and Mars, but you start by discussing physics when "Two masses $m_1$ and $m_2$ orbit each other" which does not apply to the named bodies. –  dmckee Jan 31 '13 at 19:32
Thanks, I meant the moon. –  Ashwin Iyengar Apr 8 '13 at 9:46

So it turns out that this is fine; the rate at which the semimajor axis decreases definitely depends on the current value of the semimajor axis.

-