# Gravitation acceleration question

1. Nov 23, 2004

### jenavira

A uniform solid sphere with radius R produces a gravitational acceleration a(g) on its surface. At what two distances from the center of the sphere is the gravitational acceleration a(g)/3?

I know that gravitational acceleration = GM/r^2, and that on the surface of the sphere, a(g) = (4 pi G rho /3)R. Beyond that...I'm kinda stumped. (I managed to find an explanation of this somewhere, but it didn't really help. I get that you can replace (4 pi G rho/3) with a constant, but...)

2. Nov 23, 2004

### Tide

You should realize or be able to figure out that

$$g(r) = g_0 \left(\frac {R}{r} \right)^2$$

when r > R and

$$g(r) = g_0 \frac {r}{R}$$

when r < R.

3. Nov 23, 2004

### quasar987

One of the position is gonna be outside the sphere and the other is gonna be inside.

For outside: One of the distance is going to be outside the sphere because the gravitationnal force, and therefor the gravitationnal field (=F/m) decreases continuously with the distance. So there must be a distance $r_1 >R$ somewhere where the field is g/3.

We know $MG/R^2 = g$, and we want to find $r_1$ such that $MG/r_{1}^2 = g/3 = MG/3R^2$. Solve for $r_1$.

For inside: You have to know that a shell of uniform matter density produces no net gravitationnal field inside of it. With that in mind, you can regard a point a distance $r_2$ inside a uniform sphere as being inside a shell of thickness $R-r_2$ and at the surface of a sphere of radius $r_2$. Therefor, only the matter of the sphere exerts a net gravitationnal force at $r_2$. You must also know that a sphere of uniform density produces the exact same gravitationnal field at every distance at its surface (and beyond) as a point particle located at its center would. Work out a formula for the mass of the sphere. How does it relate to M? Solved for $r_2$ just like for outside.