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A body of mass ' $\mathrm{m}$ ' $\mathrm{kg}$ starts falling from a distance 3R above earth's surface. When it reaches a distance ' $R$ ' above the surface of the earth of radius ' $R$ ' and Mass ' $M$ ', then its kinetic energy is
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Verified Answer
The correct answer is:
$\frac{1}{4} \frac{\mathrm{GMm}}{\mathrm{R}}$
Initial height:
$$
\mathrm{h}=3 \mathrm{R}+\mathrm{R}=4 \mathrm{R}
$$
The potential energy of the body initially will be:
$$
\mathrm{U}_1=-\frac{1}{4} \frac{\mathrm{GMm}}{\mathrm{R}}
$$
$\therefore \quad$ At the height $\mathrm{R}$,
$$
\mathrm{h}=\mathrm{R}+\mathrm{R}=2 \mathrm{R}
$$
$\therefore \quad$ Potential energy:
$$
\mathrm{U}_2=-\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{R}}
$$
Gain in kinetic energy is equal to loss in potential energy.
$$
\begin{aligned}
\therefore \quad \mathrm{KE} & =\mathrm{U}_1-\mathrm{U}_2 \\
& =-\frac{1}{4} \frac{\mathrm{GMm}}{\mathrm{R}}-\left(-\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{R}}\right)=\frac{1}{4} \frac{\mathrm{GMm}}{\mathrm{R}}
\end{aligned}
$$
$$
\mathrm{h}=3 \mathrm{R}+\mathrm{R}=4 \mathrm{R}
$$
The potential energy of the body initially will be:
$$
\mathrm{U}_1=-\frac{1}{4} \frac{\mathrm{GMm}}{\mathrm{R}}
$$
$\therefore \quad$ At the height $\mathrm{R}$,
$$
\mathrm{h}=\mathrm{R}+\mathrm{R}=2 \mathrm{R}
$$
$\therefore \quad$ Potential energy:
$$
\mathrm{U}_2=-\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{R}}
$$
Gain in kinetic energy is equal to loss in potential energy.
$$
\begin{aligned}
\therefore \quad \mathrm{KE} & =\mathrm{U}_1-\mathrm{U}_2 \\
& =-\frac{1}{4} \frac{\mathrm{GMm}}{\mathrm{R}}-\left(-\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{R}}\right)=\frac{1}{4} \frac{\mathrm{GMm}}{\mathrm{R}}
\end{aligned}
$$
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