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If escape velocity on earth surface is $11.1 \mathrm{~km} / \mathrm{h}$, then find the escape velocity on moon surface. If mass of moon is $\frac{1}{81}$ times of mass of earth and radius of moon is $\frac{1}{4}$ times radius of earth.
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Verified Answer
The correct answer is:
$2.46 \mathrm{~km} / \mathrm{h}$
Given, Escape velocity on the surface of earth, $\mathrm{v}_{\mathrm{e}}=11.1 \mathrm{~km} / \mathrm{h}$
i.e.
$\begin{aligned}
& \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}_{\mathrm{e}}} \\
& \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}
\end{aligned}$
Mass of moon, $M_m=\frac{M_e}{81}$
Radius of moon, $\mathrm{R}_{\mathrm{m}}=\frac{\mathrm{R}_{\mathrm{e}}}{4}$
$\therefore$ Escape velocity on the surface of moon
$\begin{aligned}
& v_m=\sqrt{\frac{2 G M_m}{R_m}}=\sqrt{\frac{2 G \frac{M e}{81}}{\frac{R_e}{4}}} \\
& =\frac{2 \sqrt{2}}{9} \sqrt{\frac{\mathrm{GM}_{\mathrm{c}}}{\mathrm{R}_{\mathrm{c}}}}=\frac{2}{9} \sqrt{\frac{2 \mathrm{GM}_{\mathrm{c}}}{\mathrm{R}_{\mathrm{c}}}}=\frac{2}{9} \mathrm{v}_{\mathrm{e}}=\frac{2}{9} \times 11.1=2.46 \mathrm{~km} / \mathrm{h} \end{aligned}$
i.e.
$\begin{aligned}
& \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}_{\mathrm{e}}} \\
& \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}
\end{aligned}$
Mass of moon, $M_m=\frac{M_e}{81}$
Radius of moon, $\mathrm{R}_{\mathrm{m}}=\frac{\mathrm{R}_{\mathrm{e}}}{4}$
$\therefore$ Escape velocity on the surface of moon
$\begin{aligned}
& v_m=\sqrt{\frac{2 G M_m}{R_m}}=\sqrt{\frac{2 G \frac{M e}{81}}{\frac{R_e}{4}}} \\
& =\frac{2 \sqrt{2}}{9} \sqrt{\frac{\mathrm{GM}_{\mathrm{c}}}{\mathrm{R}_{\mathrm{c}}}}=\frac{2}{9} \sqrt{\frac{2 \mathrm{GM}_{\mathrm{c}}}{\mathrm{R}_{\mathrm{c}}}}=\frac{2}{9} \mathrm{v}_{\mathrm{e}}=\frac{2}{9} \times 11.1=2.46 \mathrm{~km} / \mathrm{h} \end{aligned}$
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