Question 42: Gravitation

Question

A body of mass $m$ is taken from the surface of earth to a height equal to twice the radius of earth $\left(R_e\right)$. The increase in potential energy will be $\_\_\_\_$ .
( $g$ is acceleration due to gravity at the surface of earth)

Accepted Answer

The potential energy $(U)$ of a mass $m$ at a distance $r$ from the centre of the Earth is:
$$ \mathrm{U}=-\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{~m}}{\mathrm{r}} $$
As we know that $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^2}$, so $\mathrm{GM}_{\mathrm{e}}=\mathrm{gR}_{\mathrm{e}}^2$.
$$ \mathrm{U}=-\frac{\mathrm{mgR}_{\mathrm{e}}^2}{\mathrm{r}} $$
At the surface of the Earth, the distance from the centre is $\mathrm{r}_1=\mathrm{R}_{\mathrm{e}}$ :
So, the initial gravitational potential energy of body is;
$$ \mathrm{U}_{\mathrm{i}}=-\frac{\mathrm{mgR}_{\mathrm{e}}^2}{\mathrm{R}_{\mathrm{e}}}=-\mathrm{mgR}_{\mathrm{e}} $$
The body is taken to a height $h=2 R_e$ above the surface. The distance from the centre is $r_2=R_e+h= \mathrm{R}_{\mathrm{e}}+2 \mathrm{R}_{\mathrm{e}}=3 \mathrm{R}_{\mathrm{e}}:$
$$ U_f=-\frac{m g R_e^2}{3 R_e}=-\frac{1}{3} m g R_e $$
So, the increase in potential energy is;
$$ \Delta U=U_f-U_i $$
$$\Rightarrow $$ $\Delta \mathrm{U}=\left(-\frac{1}{3} \mathrm{mgR}_{\mathrm{e}}\right)-\left(-\mathrm{mgR}_{\mathrm{e}}\right)$
$$\Rightarrow $$ $\Delta \mathrm{U}=\mathrm{mgR}_{\mathrm{e}}-\frac{1}{3} \mathrm{mgR}_{\mathrm{e}}$
$$\Rightarrow $$ $$ \Delta \mathrm{U}=\frac{2}{3} \mathrm{mgR}_{\mathrm{e}} $$
Therefore, the increase in potential energy is $\frac{2}{3} \mathrm{mgR}_{\mathrm{e}}$.
Thus, the correct option is (D).

Answer

$\frac{1}{2} m g R_e$

Answer

$\frac{3}{4} m g R_e$

Answer

$\frac{1}{4} m g R_e$

Answer

$\frac{2}{3} m g R_e$

Gravitation

A body of mass m is taken from the surface of earth to a height equal to twice the radius of earth \left(R_e\right). The increase in potential energy will be \_\_\_\_ . ( g is acceleration due to gravity at the surface of earth)

Choose the correct answer: