Question

If the gravitational force had 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. $r^{-5/2}$

Answer 1

The correct option is C $r^{-3/2}$

The potential energy for a conservative force is defined as
$F=\frac{-dU}{dr}$ or $U={\int }_{\infty }^r\overline{F}.d\overline{r}$ $.......\left(i\right)$
or $U_r={\int }_{\infty }^r\frac{G{M}_1{M}_2}{r^2}dr=\frac{-G{M}_1{M}_2}{r}$ $.......\left(ii\right)$
$(\because U_{\infty }=0)$
If we bring the mass from the infinity to the centre of earth, then we obtain to work, 'so it has negative (gravitational force do work on the object) sign & potential energy decreases. But if we bring the mass from the surface of earth to infinite, then we must do work against gravitational force & potential energy of the mass increases.
Now in equation $\left(i\right)$ if $F=\frac{G{M}_1{M}_2}{r^{5/2}}$ instead of $F=\frac{G{M}_1{M}_2}{r^2}$ then
$U_r={\int }_{\infty }^r\frac{G{M}_1{M}_2}{r^{5/2}}dr=\frac{-2}{3}\frac{G{M}_1{M}_2}{r^{3/2}}$
$\Rightarrow U_r\propto \frac{1}{r^{3/2}}$