Ratio of gravitational potential between surface and centre of a solid sphere is

Question

A.-2/3 B.-3/2 C.2/3 D.3/2

 

Answer 1

Using the convention that the potential has limit 0 infinitely far away, we have the following:

 

The potential on the outside of the sphere is V(r) = -GM/r, where r is the distance from the center.

 

Thus if the sphere has radius R, the surface potential is V(R) = -GM/R

 

Next, the gravity field inside the sphere is a direct variation with the discance from the center, that is:

 

Φ(r) = Φ_surface × r/R = GM/R² × r/R = GMr/R³

 

And the potential inside is the integral of this:

 

V(r) = GMr²/(2R³) + C.

 

We can solve for C by insisting that V(r) is continuous at r = R.

 

GMR²/(2R³) + C = -GM/R

 

GM/(2R) + C = -GM/R

 

C = -GM/R - GM/(2R) = -3GM/(2R)

 

So for r < R:

 

V(r) = GMr²/(2R³) - 3GM/(2R).

 

This means V(0) = -3GM/(2R).

 

So the ratio is:

 

V(R) / V(0) = (-GM/R) / (-3GM/(2R)) = 2/3