Gravitation acceleration question

[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...)

 

[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.

 

[quasar987]

One of the position is going to be outside the sphere and the other is going to 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 &gt;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.