Question

A body of mass $m$ is raised to a height $10R$ from the surface of the earth, $R$ where is the radius of the earth. The increase in potential energy is ($G$=universal gravitational constant, $M$=mass of the earth and $g$=acceleration due to gravity)
A. $\dfrac{{GMm}}{{11R}}$
B. $\dfrac{{GMm}}{{10R}}$
C. $\dfrac{{mgR}}{{11G}}$
D. $\dfrac{{10GMm}}{{11R}}$

Answer 1

Use the formulae for the potential energy of the body on the surface of the earth and at a particular height from the surface of the earth. Then calculate the change in potential energy of the body.
Formula used:
The potential energy $U$ of an object on the surface of the Earth is
$U = - \dfrac{{GMm}}{R}$ …… (1)
Here, $G$ is the universal gravitational constant, $M$ is the mass of the earth, $m$ is the mass of the object and $R$ is the radius of the earth.
The potential energy ${U_h}$ of an object at a height $h$ from the surface of the Earth is
${U_h} = - \dfrac{{GMm}}{{R + h}}$ …… (2)
Here, $G$ is the universal gravitational constant, $M$ is the mass of the earth, $m$ is the mass of the object and $R$ is the radius of the earth.

Complete step by step answer:
A body of mass $m$ is raised to a height $10R$ from the surface of the earth.
The potential energy $U$ of the body on the surface of the earth is
$U = - \dfrac{{GMm}}{R}$
Calculate the potential energy ${U_h}$ of the body at a height $10R$ from the surface of the earth.
Substitute $10R$ for $h$ in equation (2).
${U_h} = - \dfrac{{GMm}}{{R + 10R}}$
$\Rightarrow {U_h} = - \dfrac{{GMm}}{{11R}}$
The change in potential energy $\Delta U$ of the body is the difference of the potential energy ${U_h}$ of the body at a height $10R$ from the surface of the earth and the potential energy $U$ of the body on the surface of the earth.
$\Delta U = {U_h} - U$
Substitute $- \dfrac{{GMm}}{{11R}}$ for ${U_h}$ and $- \dfrac{{GMm}}{R}$ for $U$ in the above equation.
$\Delta U = \left[ { - \dfrac{{GMm}}{{11R}}} \right] - \left[ { - \dfrac{{GMm}}{R}} \right]$
$\Rightarrow \Delta U = - \dfrac{{GMm}}{{11R}} + \dfrac{{GMm}}{R}$
$\Rightarrow \Delta U = \dfrac{{ - GMmR + 11GMmR}}{{11{R^2}}}$
$\Rightarrow \Delta U = \dfrac{{10GMmR}}{{11{R^2}}}$
$\Rightarrow \Delta U = \dfrac{{10GMm}}{{11R}}$

Therefore, the change in potential energy of the body is $\dfrac{{10GMm}}{{11R}}$.

So, the correct answer is “Option D”.

Note:
The gravitational potential energy of the body increases with increase in height from the surface of the Earth as for the body of a fixed mass the gravitational potential energy is inversely proportional to the negative of the height of the body from the earth’s surface.