Question

The gravitational field due to a mass distribution is $E={\displaystyle \frac{K}{x^3}}$ in the x-direction. (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is:

A. ${\displaystyle \frac{k}{x}}$

B. ${\displaystyle \frac{k}{2x}}$

C. ${\displaystyle \frac{k}{x^2}}$

D. ${\displaystyle \frac{k}{2x^2}}$

Answer 1

Hint Use the definition of gravitational field as, the Force Exerted by a mass distribution on a unit mass, given by $E={\displaystyle \frac{K}{x^3}}$
We will start with a test mass $m$experiencing the given field at $dx$ distance and integrate it from infinity to x to compute Gravitational Potential.

Complete step by step answer
Given a test mass $m$ on the x-axis, the work done $W$ to move it a distance $dx$ in the given gravitational field will be:


$\Rightarrow$ $W={\displaystyle \frac{mK}{x^3}}.dx$ [Since, $W=F.S$, where F is the force and S is the displacement]

Hence, the work done in bringing the test mass from infinity to x will be

$\Rightarrow W=\underset{\mathrm{\infty }}{\overset{x}{\int }}{\displaystyle \frac{mK}{x^3}}dx$

Since gravitational potential is required work done to bring a unit mass from infinity to x,

$\Rightarrow P={\displaystyle \frac{W}{m}}$
Hence, for our case, gravitational potential $P$ will be,

$\Rightarrow P={\displaystyle \frac{1}{m}}\underset{\mathrm{\infty }}{\overset{x}{\int }}{\displaystyle \frac{mK}{x^3}}dx$

Since, test mass m and constant K are independent of the variable x, we can bring it outside the integral,

$\Rightarrow P={\displaystyle \frac{1}{m}}mK\underset{\mathrm{\infty }}{\overset{x}{\int }}{\displaystyle \frac{1}{x^3}}dx$

Cancelling $m$ from the numerator and denominator and solving the integral using$\int x^{-3}dx={\displaystyle \frac{x^{-3+1}}{-3+1}}$ , we get
$\Rightarrow P=K{\left[{\displaystyle \frac{x^{-2}}{-2}}\right]}_{\mathrm{\infty }}^x$

Hence we get the gravitational potential as:

$\Rightarrow P=-K{\left[{\displaystyle \frac{1}{2x^2}}\right]}_{\mathrm{\infty }}^x$

Putting the limits in the differential and using gravitational potential at $\mathrm{\infty }$ = 0, we get

$\Rightarrow P=-K\left[{\displaystyle \frac{1}{2x^2}}-0\right]$

This gives us the final value for potential at x, for the given electrical field as,

$\Rightarrow P=-{\displaystyle \frac{K}{2x^2}}$

This is of the form,

$\Rightarrow P={\displaystyle \frac{k}{2x^2}}$

Note Alternative method – Use the formula for the Gravitational Potential as the Gravitational field from infinity to x as $P=\underset{\mathrm{\infty }}{\overset{x}{\int }}{\displaystyle \frac{K}{x^3}}dx$. This method does not include the term for test mass m but is less intuitive.