Question

The gravitational field due to a mass distribution is given by $E=-K/x^3$ in $x-$direction. Taking the gravitational potential to be zero at infinity, find its value at a distance $x$.

• A. $\frac{K}{2x^2}$
• B. $\frac{-K}{3x^2}$
• C. $\frac{-2K}{3x^2}$
• D. $\frac{-3K}{2x^2}$

Answer 1

The correct option is A $\frac{K}{2x^2}$

The change in potential in going from the initial to find position in a gravitational field $E$ is given by

${\int }_i^{f}dV=-{\int }_i^f\overrightarrow{E}\cdot \overrightarrow{dr}$

Since the change in field is only in the $x-$ direction

$\therefore {\int }_{\infty }^{x}dV=-{\int }_{\infty }^{x}Edx=-{\int }_{\infty }^x\frac{K}{x^3}dx$

$\Rightarrow V_x-V_{\infty }=-K{\int }_{\infty }^x\frac{1}{x^3}dx=K{\left[\frac{1}{2x^2}\right]}_{\infty }^x$

As we know that $V_{\infty }=0$

$\Rightarrow V_x=\frac{K}{2}[\frac{1}{x^2}-\frac{1}{{\infty }^2}]=\frac{K}{2x^2}$

Hence, option $\left(a\right)$ is correct.