Question

A small body is at a distance $r$ from the centre of Mercury planet, where $r$ is greater than the radius of Mercury. The energy required to shift the body from $r$ to $2r$ measured from the centre is $E$. The energy reqiured to shift it from $2r$ to $3r$ willl be

• A. $E$
• B. $\frac{E}{2}$
• C. $\frac{E}{3}$
• D. $\frac{E}{4}$

Answer 1

The correct option is C $\frac{E}{3}$

Let $M$ be the mass of mercury and $m$ be the mass of body

The energy required to shift a body from one point to another will be equal to the difference in mechanical energy between the two points.

When the body is moved from $r$ to $2r:$

Change in potential energy of body will be

$\Delta U_1=U_{2r}-U_r=-\frac{GMm}{2r}+\frac{GMm}{r}$

$\Rightarrow \Delta U_1=\frac{GMm}{2r}$

Since, body is in rest at initial and final point.

$\therefore \Delta K{E}_1=0$

So, change in mechanical energy is given by

$\Delta E_1=\Delta U_1+\Delta K{E}_1=\frac{GMm}{2r}+0=\frac{GMm}{2r}$

$\Rightarrow \Delta E_1=\frac{GMm}{2r}=E\left(given\right)$

Similarly, when the body is moved from $2r$ to $3r:$

Change in potential energy of body will be

$\Delta U_2=U_{3r}-U_{2r}=-\frac{GMm}{3r}+\frac{GMm}{2r}$

$\Rightarrow \Delta U_2=\frac{GMm}{6r}$

Since, body is in rest at initial and final point.

$\therefore \Delta K{E}_2=0$

So, change in mechanical energy is given by

$\Delta E_2=\Delta U_2+\Delta K{E}_2=\frac{GMm}{6r}+0=\frac{GMm}{6r}$

$\Rightarrow \Delta E_2=\frac{GMm}{6r}=\frac{E}{3}$

Hence, option $\left(c\right)$ is correct.