Angular momentum of the earth

1. Apr 22, 2005

vpea

I'm stuck on the second part of a problem and can't seem to get the right answer:

Calculate the magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles. Treat the earth as a uniform sphere of radius 6.38*10^6 that makes one revolution in 24.0 hours. (m=5.97*10^24)

I got the answer 1.77*10^34 but it's wrong :(

2. Apr 22, 2005

Staff: Mentor

3. Apr 22, 2005

vpea

I found the moment of Inertia (mr^2) to be = 2.43*10^38. Then found the angular velocity to be (2pi)/(24*60*60). Using the equation I=m*r^2, I got 1.77*10^34

4. Apr 22, 2005

Staff: Mentor

That's the moment of inertial of a particle, not a solid sphere about its center. The one you want is $I = 2/5 m r^2$.

5. Apr 22, 2005

vpea

Thanks for that! I tried looking on the internet for the moment of Inertia of the earth and was getting all sorts of weird numbers no wonder it wasn't working.

6. Apr 22, 2005

Staff: Mentor

Every "General Physics" textbook I've ever seen has a table of formulas for moment of inertia of various shapes of objects. What kind of a course are you taking?