Question

If the gravitational field intensity at a point is given by $g=\dfrac{GM}{{{r}^{2.5}}}$, then the potential at a distance $r$ is
A. $\dfrac{-2GM}{3{{r}^{1.5}}}$
B. $\dfrac{-GM}{{{r}^{3.5}}}$
C. $\dfrac{GM}{3{{r}^{1.5}}}$
D. $\dfrac{GM}{{{r}^{3.5}}}$

Answer 1

Hint: This problem can be solved by using the definition of gravitational potential and its relation with gravitational field intensity. Gravitational potential of a point is the amount of work done per unit mass to bring the object from infinity to that point without any acceleration.
The gravitational field intensity is the gravitational force exerted per unit mass on an object.

Formula used:
$V=\int\limits_{\infty }^{r}{I.dr}$
where $V$ is the gravitational potential of a point, $I$ is the gravitational field intensity and $dr$ is the infinitesimal displacement of the body in the direction of the gravitational field intensity.

Complete step by step answer:
The gravitational potential of a point is the total amount of work done to bring a unit mass from infinity to that point in the presence of a gravitational field.

Hence,
$V=\int\limits_{0}^{W}{d{{W}_{\text{unit mass}}}}$ --(1)
Where $d{{W}_{\text{unit mass}}}$ is the infinitesimal amount of work done upon the unit mass.

Gravitational field intensity $\left( I \right)$ is the gravitational force per unit mass. Hence, the work done by the gravitational field to displace a unit mass by an infinitesimal displacement $dr$ will be
$d{{W}_{\text{unit mass}}}=I.dr$ --(2)
Hence, using (1) and (2), we get,
$V=\int\limits_{\infty }^{r}{I.dr}$ --(3)
where $V$ is the gravitational potential of a point, $I$ is the gravitational field intensity and $dr$ is the infinitesimal displacement of the body in the direction of the gravitational field intensity.
Now, let us analyze the question.
Given, $I=\dfrac{GM}{{{r}^{2.5}}}$ --(4)
Putting (4) in (3), we get,
$V=\int\limits_{\infty }^{r}{\dfrac{GM}{{{r}^{2.5}}}dr=GM\int\limits_{\infty }^{r}{{{r}^{-2.5}}dr}}=GM\left[ \dfrac{{{r}^{-1.5}}}{-1.5} \right]_{\infty }^{r}$ $\left( \because \int{{{x}^{n}}dx}=\left[ \dfrac{{{x}^{n+1}}}{n+1} \right] \right)$
$\therefore V=\left[ -\dfrac{GM}{\dfrac{3{{r}^{1.5}}}{2}} \right]_{\infty }^{r}=\left[ -\dfrac{2GM}{3{{r}^{1.5}}} \right]_{\infty }^{r}=\left[ -\dfrac{2GM}{3{{r}^{1.5}}}-\left( -\dfrac{2GM}{3{{\left( \infty \right)}^{1.5}}} \right) \right]=\left[ -\dfrac{2GM}{3{{r}^{1.5}}}-0 \right]=-\dfrac{2GM}{3{{r}^{1.5}}}$ $\left( \because \dfrac{1}{\infty }=0 \right)$
Hence, the required gravitational potential is $-\dfrac{2GM}{3{{r}^{1.5}}}$.
Hence, the correct option is A) $-\dfrac{2GM}{3{{r}^{1.5}}}$.

Note: It is always better to solve these types of problems by strictly following their mathematical definitions. By following the mathematical definitions and proceeding with the calculation, there is very little chance of making a mistake. Students must also take care of the signs in the mathematical definitions. Many mathematical definitions in the topic of gravitation include a negative sign which is very important especially in cases like gravitational potential.
For example, in the above problem, the negative sign in the answer is indicative that the force acting on the body is attractive and external work has to be done to move the body in a direction opposite to the force. A positive sign would have suggested a repulsive force.