# Gravitation of non-uniform density

1. Apr 8, 2007

### Mindscrape

1. The problem statement, all variables and given/known data
A gaseous planet, with radius R, has a radially dependent density function
$$\rho (r') = \rho_0 [\frac{r'}{R}]^2$$
where r' is the distance from the center planet. Find the magnitude of force for a mass m inside and outside of the planet.

2. Relevant equations
$$F = -\frac{GmM}{r^2}$$
$$F = -G m \int_V \frac{\rho(r') \hat{e_r}}{r^2}dv'$$

3. The attempt at a solution
I'm pretty sure I did it right, but would like some confirmation. The mass inside the planet, taking a sample shell, would be

$$M = \int_0^r 4\pi r'^2 dr' \rho(r')$$

which would give

$$M = \int_0^r 4\pi r'^2 dr' \rho_0 (\frac{r'}{R})^2$$

and evaluates to

$$M = \frac{4}{5} \pi r^5 \frac{\rho_0}{R^2}$$

such that the force inside will be

$$F = \frac{-Gm r^3}{r^2}\rho_0 \hat{e_r}$$

Outside

the same idea applies

$$M = \int_0^R 4 \pi r'^2 dr' \rho(r')$$

and gives

$$M = \frac{4}{5} \pi R^5 \frac{\rho_0}{R^2}$$

so that the force is

$$F = \frac{-GmR^2}{r^2}\rho_0 \hat{e_r}$$

Both answers, more or less, make sense. Inside, as you move further away that the force will increase. Outside, the force will decrease as you move further away.

Last edited: Apr 8, 2007
2. Apr 8, 2007

### StatMechGuy

It's the same problem as a Gauss' law problem for a ball of non-uniform charge, and that's how the solution goes, so that all looks right.