Question

A particle of mass $m$ is taken from the earth’s surface to a height $h$. Find out the change in its potential energy for $h < < R$ and prove.

Answer 1

Hint
From the formula for the potential energy of a body on the earth’s surface, we can find the potential at a height $h$ from the surface of the earth. Then by finding the difference between the potential energies of the 2 bodies, we can calculate the change in the potential energy.
In the solution, we will be using the following formula,
$\Rightarrow U = - \dfrac{{GMm}}{R}$
where $U$ is the potential energy of the body
$G$ is the universal gravitational constant
$M$ is the mass of the earth
$m$ is the mass of the body and
$R$ is the distance from the center of the earth.

Complete step by step answer
For this problem let us consider the radius of the earth is given by $R$. So a body on the surface of the earth has a potential energy due to the gravitational field of the earth. From the formula for the potential energy of a body at a distance of $R$ from the center of the earth, that is on the surface of the earth, is given by,
$\Rightarrow U = - \dfrac{{GMm}}{R}$
Similarly for a body that is at a distance of $h$ from the surface, that is at a distance $\left( {R + h} \right)$ from the center of the earth is,
$\Rightarrow U' = - \dfrac{{GMm}}{{\left( {R + h} \right)}}$
Therefore the change in the potential energy of the body when taken from the surface of the earth to a height $h$ is the difference between the final and initial energy.
Therefore,
$\Rightarrow \Delta U = U' - U$
Substituting the value we get
$\Rightarrow \Delta U = - \dfrac{{GMm}}{{\left( {R + h} \right)}} - \left( { - \dfrac{{GMm}}{R}} \right)$
On taking the numerator common in both the cases,
$\Rightarrow \Delta U = - GMm\left[ {\dfrac{1}{{\left( {R + h} \right)}} - \dfrac{1}{R}} \right]$
On taking the LCM as, $R\left( {R + h} \right)$, we get
$\Rightarrow \Delta U = - GMm\left[ {\dfrac{{R - \left( {R + h} \right)}}{{R\left( {R + h} \right)}}} \right]$
So we simplify it as,
$\Rightarrow \Delta U = - GMm\left[ {\dfrac{{R - R - h}}{{R\left( {R + h} \right)}}} \right]$
The $R$ gets cancelled and we have,
$\Rightarrow \Delta U = - GMm\left[ {\dfrac{{ - h}}{{R\left( {R + h} \right)}}} \right]$
Therefore on opening the bracket,
$\Rightarrow \Delta U = \dfrac{{GMmh}}{{R\left( {R + h} \right)}}$
When $h < < R$ we can neglect the $h$ in denominator and get
$\Rightarrow \Delta U = \dfrac{{GMm}}{{{R^2}}}h$
This is the change in the potential energy of the body.

Note
The gravitational potential energy of a body is its potential energy due to the gravitational force of the earth. The gravitational potential energy is taken as negative because we consider the potential energy of a body at the infinity is zero, so the potential energy increases when the body is taken further away from the earth.