# Lunar Orbital Angular Momentum

1. Aug 23, 2009

### Blabacus

1. The problem statement, all variables and given/known data
In lectures and textbooks the lunar orbital angular momentum is given as:

A = m/(1 + m/M)xR2x$$\omega$$L

But if the distance r of the moon from the centre of mass of the Earth-moon system is given by:
mxr = Mx(R-r); so r = MR/(M +m) = R/(1 + m/M)

and so the moon's angular momentum should be

mxr2x$$\omega$$L = m/(1 + m/M)2xR2x$$\omega$$L

Explain the discrepancy.

2. Relevant equations

m = Mass of Moon.
M = Mass of Earth.
$$\omega$$L = Angular orbital velocity of Moon
r = distance of Moon from the CoM.

3. The attempt at a solution

I figured that the discrepancy is due to the fact that Earth's contribution to the orbital angular momentum of the system has not been accounted for. However we are expected to show mathematically how the first equation A = m/(1 + m/M)xR2x$$\omega$$L holds true when the Earth's contribution is added. I can't for the life of me get it to work.

I have tried adding Earth's contribution to both sides of the equation, but cannot arrive back at the first equation.

mxr2x$$\omega$$L + MxR2x$$\omega$$L = mxR2x$$\omega$$L/(1 + m/M)2 + Mxr2x(1 +m/M)2

Any help would be much appreciated.

Last edited: Aug 23, 2009
2. Aug 23, 2009

### queenofbabes

You have added the earth's contribution wrongly. It should be M * (R-r)^2 * w, keeping in mind that you have used R to define the total distance between earth and moon, not between earth and CoM. Try it, it should work out.

P.S. If your teacher approves of the reduced mass method, it becomes a whole lot easier. I know some profs don't :P