If V is the gravitational potential on the surface of the earth, then what is its value at the centre of the earth

The correct option is 3V/2.

Let us derive the relationship between the gravitational potential on the surface of the Earth and the gravitational potential at its center.

We model the Earth as a uniform solid sphere of mass $M$ and radius $R$.

The formula for the gravitational potential $V\left(r\right)$ inside a uniform solid sphere at a distance $r$ from its center (where $r\le R$) is given by:
$V\left(r\right)=-\frac{GM}{2R^3}(3R^2-r^2)$
where $G$ is the universal gravitational constant.

First, let's find the gravitational potential at the surface of the Earth. On the surface, the distance from the center is $r=R$. Substituting this into the formula:
$V_{\text{surface}}=-\frac{GM}{2R^3}(3R^2-R^2)$
$V_{\text{surface}}=-\frac{GM}{2R^3}\left(2R^2\right)$
$V_{\text{surface}}=-\frac{GM}{R}$
According to the question, the gravitational potential on the surface is $V$. Therefore, we can write:
$V=-\frac{GM}{R}$

Next, let's find the gravitational potential at the center of the Earth. At the center, the distance from the center is $r=0$. Substituting this into the general formula:
$V_{\text{center}}=-\frac{GM}{2R^3}(3R^2-0^2)$
$V_{\text{center}}=-\frac{GM}{2R^3}\left(3R^2\right)$
$V_{\text{center}}=-\frac{3}{2}\frac{GM}{R}$

Substituting $V=-\frac{GM}{R}$ into the equation for $V_{\text{center}}$:
$V_{\text{center}}=\frac{3}{2}V$

Thus, the value of the gravitational potential at the center of the Earth is $\frac{3V}{2}$.