# Angular momentum of a moon

1. Nov 14, 2005

There is a moon orbiting an Earth-like planet. The mass of the moon is 4.77e22 kg, the center-to-center separation of the planet and the moon is 616000 km, the orbital period of the moon is 25.6 days, and the radius of the moon is 1590 km. What is the angular momentum of the moon about the planet?

I found the period to be 2211840 seconds.

I then used the foruma v = 2πr/T :
2π(616000000)/2211840 = 1749.874 m/s

I changed the translational velocity to angular velocity with v=rω:
1749.874=616000000ω = 2.8407e-6 (Did I use the correct radius? Maybe that's my error)

I found inertia with I=(2/5)mr^2:
I=(2/5)(4.77x10^22)(1590000^2) = 4.8236x10^34

Finally I found momentum with L=Iω:
L=(4.8236x10^34)(2.8407x10^-6) = 1.370 x 10^29

My answer is wrong. I'm assuming that the moon is spherical so I used that particular equation for inertia.

2. Nov 15, 2005

### Tide

I didn't check your numbers but I think you could have done it all in one step with $L = mr^2\omega$.

Last edited: Nov 15, 2005
3. Nov 15, 2005

How is that true? Why would I=mr?

4. Nov 15, 2005

### SpaceTiger

Staff Emeritus
This is the moment of inertia for the rotation of the moon. It asked you to find the angular momentum of the orbit, not the rotation. I think Tide's post had a typo, the angular momentum of the orbit is given by

$$L=m\omega r^2=mvr$$

5. Nov 15, 2005

### Tide

Thanks for catching that, ST! I'll correct it.