Question

if the gravitational force had varied as r-7/2 instead of r-2 the potential energy of a particle at a distance r from centre of earth would be proportional to​

Answer 1

Given that,

If the gravitational force had varied as $r^{-\dfrac{7}{2}}$ instead of $r^{-2}$ the potential energy of a particle at a distance r from centre of earth

The potential energy for a conservative force is defined as

$F=-\dfrac{dU}{dr}$

$U=-\int_{\infty}^{r}{F\cdot dr}$....(I)

Put the value into the formula

$U=-\int_{\infty}^{r}{\dfrac{GM_{1}M_{2}}{r^2}}dr$

$U_{r}=-\dfrac{GM_{1}M_{2}}{r}$.....(II)

We know that.

$U_{\infty}=0$

If we bring the mass from the infinity to the center of earth, then the work will be negative because the gravitational force do work on the object so the potential energy decreases,

If we bring the mass from the surface of earth to infinite, then the work will be against gravitational force and potential energy of the mass increases.

We need to calculate the potential energy

Using formula of force

If $F = \dfrac{GM_{1}M_{2}}{r^{\frac{7}{2}}}$ instead of $F=\dfrac{GM_{1}M_{2}}{r^2}$

Then, $U_{r}=\int_{\infty}^{r}{\dfrac{GM_{1}M_{2}}{r^{\frac{7}{2}}}}$

$U_{r}=\dfrac{-2}{5}\dfrac{GM_{1}M_{2}}{r^{\frac{5}{2}}}$

$U_{r}\propto\dfrac{1}{r^{\frac{5}{2}}}$

$U_{r}\propto r^{\frac{-5}{2}}$

Hence, The potential energy of a particle is proportional to​ $r^{\frac{-5}{2}}$

Answer 2

Explanation:

hope this will help you...................