Question

If $V_s$ is the gravitational potential on the surface of the earth, then what is its value at distance ${\left(\frac{1}{4}\right)}^{th}$ of the radius of the earth from the centre of the earth?
(Assume that mass distribution is uniform)

• A. $\frac{49}{32}{V}_s$
• B. $\frac{47}{32}{V}_s$
• C. $\frac{23}{32}{V}_s$
• D. $\frac{25}{32}{V}_s$

Answer 1

The correct option is B $\frac{47}{32}{V}_s$

Let the mass of the earth is $M$ and radius is $R$.

The gravitational potential due to a uniform solid sphere, at a distance $r$ from centre is given by

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

Gravitational potential at the surface of the earth is
Substitute $r=R$ in equation $\left(i\right)$

$\Rightarrow V_s=-\frac{GM}{R}\dots \left(ii\right)$

Gravitational potential at distance $r=\frac{R}{4}$ from the centre of the earth is

$V=-\frac{GM}{2R^3}(3R^2-{\left(\frac{R}{4}\right)}^2)$

$\Rightarrow V=-\frac{47GM}{32R}$

$\Rightarrow V=\frac{47}{32}{V}_s$ (By equation $\left(ii\right)$

Hence, option (b) is correct.