First slide

Gravitaional potential energy

Question

Mass of moon is (1/81) times that of earth and its radius is (1/4) as the radius of the earth' If the value of escape velocity at the surface of earth is 11 ' 2 km,/sec, its value at the surface of moon is

Easy

A

0.14km/sec
B

0.5 km/sec
C

2.5km/sec
D

5km/sec

Solution

The escape velocity is given by
$v_{e} = {(\frac{2 GM}{R})}^{1 / 2}$
For earth, $11 \cdot 2 = {(\frac{2 {GM}_{e}}{R_{e}})}^{1 / 2}$ ………..(1)
For moon, $v_{m} = {[\frac{2 G (M_{e} / 81)}{(R_{e} / 4)}]}^{1 / 2}$ ……(2)
From eq. (2),
$\begin{array}{l} v_{m} = \frac{2}{9} {[\frac{2 {GM}_{e}}{R_{e}}]}^{1 / 2} = \frac{2}{9} \times 11 \cdot 2 \\ = 2 \cdot 5 km / \sec \end{array}$

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