Question

The gravitational field due to a mass distribution is $E = \dfrac{K}{{{x^3}}}$ 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. $\dfrac{k}{x}$

B. $\dfrac{k}{{2x}}$

C. $\dfrac{k}{{{x^2}}}$

D. $\dfrac{k}{{2{x^2}}}$

Answer 1

Hint Use the definition of gravitational field as, the Force Exerted by a mass distribution on a unit mass, given by $E = \dfrac{K}{{{x^3}}}$
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 $ $W = \dfrac{{mK}}{{{x^3}}}.dx$ [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 = \int\limits_\infty ^x {\dfrac{{mK}}{{{x^3}}}} dx$

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

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

$ \Rightarrow P = \dfrac{1}{m}\int\limits_\infty ^x {\dfrac{{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 = \dfrac{1}{m}mK\int\limits_\infty ^x {\dfrac{1}{{{x^3}}}} dx$

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

Hence we get the gravitational potential as:

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

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

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

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

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

This is of the form,

$ \Rightarrow P = \dfrac{k}{{2{x^2}}}$

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