First slide

Gravitaional potential energy

Question

The acceleration due to gravity on the surface of the moon is $\frac{1}{6}$ that on the surface of earth and the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and moon will be :

Moderate

A

$\frac{\sqrt{6}}{2}$
B

$\sqrt{24}$
C

3
D

$\frac{\sqrt{3}}{2}$

Solution

$\frac{{(v_{e})}_{E a r t h}}{{(v_{e})}_{M o o n}} = \sqrt{(\frac{g_{E a r t h}}{g_{M o o n}}) \cdot (\frac{R_{E a r t h}}{R_{M o o n}})} = \sqrt{24}$

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