Let us consider a solid sphere of radius \( a \) and mass \( M \).

So the gravitational potential at a point on the surface of the sphere is \( V_a=-\frac{GM}{a}\tag{1} \)

Gravitational potential at a point inside the sphere at a distance \( r \) from the centre of the sphere is \( \displaystyle{-GM\frac{3a^2-r^2}{2a^3}} \).

Gravitational Potential Due To A Solid Homogeneous Sphere At A Point (i) Outside, (ii) On The Surface, And (iii) Inside A Point Of The Sphere. Read In Detail.

At the centre of the sphere is \( r=a \). So the gravitational potential at the centre of the sphere is \( V_0=-GM\frac{3a^2}{2a^3}=-\frac{3}{2}\frac{GM}{a} \tag{2} \).

Dividing equation 2 by equation 1, we get

\( \frac{V_0}{V_a}=\frac{3}{2} \\or,\ V_0=\frac{3}{2}V_a \)

Hence potential at the centre is \( \frac{3}{2} \) times that on the surface.