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If escape velocity on earth surface is $11.1 \mathrm{kmh}^{-1}$, 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:
$2.46 \mathrm{kmh}^{-1}$
Given, Escape velocity on the surface of the earth is given by i.e. $v_\rho=\sqrt{2 g R_e}$
Mass of the moon, $M_m=\frac{M_e}{81}$
Radius of the moon, $R_m=\frac{R_e}{4}$
$\therefore$ Escape velocity on the surface of the moon
$v_m=\sqrt{\frac{2 G M_m}{R_m}}=\sqrt{\frac{2 G\left(\frac{M e}{81}\right)}{\frac{R_e}{4}}}=\frac{2 \sqrt{2}}{9} \sqrt{\frac{G M_e}{R_e}}$
From Eq. (i),
$=\frac{2}{9} \sqrt{\frac{2 G M_e}{R_e}}=\frac{2}{9} v_e$
Escape velocity $v_e=11.1 \mathrm{~km} / \mathrm{h}$
$=\frac{2}{9} \times 11.1=2.46 \mathrm{~km} \mathrm{~h}^{-1}$
Mass of the moon, $M_m=\frac{M_e}{81}$
Radius of the moon, $R_m=\frac{R_e}{4}$
$\therefore$ Escape velocity on the surface of the moon
$v_m=\sqrt{\frac{2 G M_m}{R_m}}=\sqrt{\frac{2 G\left(\frac{M e}{81}\right)}{\frac{R_e}{4}}}=\frac{2 \sqrt{2}}{9} \sqrt{\frac{G M_e}{R_e}}$
From Eq. (i),
$=\frac{2}{9} \sqrt{\frac{2 G M_e}{R_e}}=\frac{2}{9} v_e$
Escape velocity $v_e=11.1 \mathrm{~km} / \mathrm{h}$
$=\frac{2}{9} \times 11.1=2.46 \mathrm{~km} \mathrm{~h}^{-1}$
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