Question

The gravitational field in $x$ direction due to some mass distribution is $E=\frac{k}{x^3}$, where $k$ is a constant. Assuming the gravitational potential to be zero at infinity, then its value at a distance $x$ will be

• A. $\frac{k}{x}$
• B. $\frac{k}{2x}$
• C. $\frac{k}{x^2}$
• D. $\frac{k}{2x^2}$

Answer 1

The correct option is D $\frac{k}{2x^2}$

Gravitational potential energy is given by the relation,
$V=-\int E.dx$
$V=-\int \left(\frac{k}{x^3}\right).dx$
$V=\frac{k}{2x^2}$