# Homework Help: Moment of inertia of earth

1. Dec 6, 2009

### discordplus

1. The problem statement, all variables and given/known data

The density of Earth, at any distance r from it's center is approximately

p(r) = [14.2 - 11.6 (r/R)] * 10^3 kg*m^-3

where R is the radius of the Earth. Show that the density leads to a moment of inertia I = 0.330 MR^2 about an axis through the center, where M is the mass of the Earth.

2. Relevant equations

I = $$\int$$ r^2 dm
I of a hollow sphere = 2/3MR^2
I = $$\int$$ dI

3. The attempt at a solution

So what I believe is right is to sum up the contributions of the moments of inertia of infinitesimally thin hollow spheres, with limits of integration 0 to R.

So, I have a hollow sphere with thickness dr a distance r from the center of the Earth, whose moment is given by 2/3MR^2 so I have:

I = 2/3 $$\int$$ [R,0] r^2 dm

dm would be p(r) dV

dV should be 4/3$$\pi$$(r+dr)^3 - 4/3$$\pi$$r^3 because the all the mass is concentrated at the outside of the shell of thickness dr. This equals 3r^2dr.

So. 2 $$\int$$ p(r) * 3r^2 * r^2 dr = I

integrating with limits [R,0] and substituting in the actual values i got about 2 * 10^37 kg * m^2, but the answer is 8 * 10^37 kgm^2.

What did I do incorrectly?

2. Dec 6, 2009

### diazona

That's not correct. Check your math on that differential. (I mean: you're right that $\mathrm{d}V = \frac{4}{3}\pi(r + \mathrm{d}r)^3 - \frac{4}{3}\pi r^3$, you just computed the difference incorrectly.)

3. Dec 7, 2009

### discordplus

Oh, I forgot to move the 4pi/3 down. Wow. It came out to the right answer after that. Thanks.