Question

If the gravitational force has varied as ${{r}^{-5/2}}$instead of ${{r}^{-2}}$; the potential energy of a particle at a distance ‘r’ from the centre of the earth would be proportional to

& A.\,{{r}^{-1}} \\\ & B.\,{{r}^{-2}} \\\ & C.\,{{r}^{-3/2}} \\\ & D.\,{{r}^{-5/2}} \\\ \end{aligned}$$

Answer 1

We have to consider the formulae of the gravitational force and the gravitational potential energy. Represent the formula of the gravitational potential energy in terms of the gravitational force and equate the change in the value of the radius to obtain the required proportionality value.
Formula used:
${{F}_{g}}=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
$U=-\dfrac{G{{m}_{1}}{{m}_{2}}}{r}$

Complete answer:
From the given information, we have the data as follows.
The gravitational force has varied as ${{r}^{-5/2}}$instead of ${{r}^{-2}}$.
The formulae that we will be using to solve this problem are as follows.
The gravitational force, ${{F}_{g}}=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
The gravitational potential energy, $U=-\dfrac{G{{m}_{1}}{{m}_{2}}}{r}$
The value of the gravitational potential energy becomes less negative as the radius increases and the value of the gravitational potential energy becomes zero when the radius equals infinite.
Represent the gravitational potential energy in terms of the gravitational force.

& U=\int_{\infty }^{r}{F.\,dr} \\\ & \therefore U=\int_{\infty }^{r}{\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}} \\\ \end{aligned}$$ The above equation represents the gravitational potential energy in terms of the gravitational force when the radius is Replace $${{r}^{-2}}$$by $${{r}^{-5/2}}$$ $$\begin{aligned} & U=\int_{\infty }^{r}{\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{5/2}}}} \\\ & \Rightarrow U=-\dfrac{2}{3}\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{3/2}}} \\\ & \therefore U\propto \dfrac{1}{{{r}^{3/2}}} \\\ \end{aligned}$$ Thus, the gravitational potential energy is inversely proportional to the three-two times that of the radius. $$\therefore $$ If the gravitational force has varied as $${{r}^{-5/2}}$$instead of $${{r}^{-2}}$$; the potential energy of a particle at a distance ‘r’ from the centre of the earth would be proportional to . **Thus, option (C) is correct.** **Note:** The potential energy of the mass increases, when the mass is moved from the surface of the earth to infinite, as the work done will be against the gravitational force. Similarly, the potential energy of the mass decreases, when the mass is moved from infinite to the surface of the earth, as the work done will be along with the gravitational force.