JEE Main 2023 — Gravitation Question with Solution

The gravitational potential (V) due to a sphere of uniform density at a distance r from its center is given by:

$$V(r)=\frac{GM}{2R^3}\left(3R^2-r^2\right)$$

At the surface of the sphere (r = R), the gravitational potential is:

$$V=\frac{GM}{R}$$

Now, let's find the gravitational potential at the center of the sphere (r = 0):

$$V(0)=\frac{GM}{2R^3}\left(3R^2-0^2\right)=\frac{3GM}{2R}$$

The gravitational potential at the center of the sphere is $$\frac{3}{2}$$ times the potential at the surface of the sphere. Therefore:

$$V(0)=\frac{3V}{2}$$