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Question

Physics Question on Escape Speed

The acceleration due to gravity on the surface of the moon is 1/61/6 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

A

62\frac {\sqrt {6}} {2}

B

124\frac{1}{\sqrt{24}}

C

33

D

32\frac {\sqrt {3}}{2}

Answer

124\frac{1}{\sqrt{24}}

Explanation

Solution

The correct option is (B): 124\frac{1}{\sqrt{24}}
The escape velocity of an object is given by the equation ve=2GMRv_{e}=\sqrt{\frac{2 G M}{R}}.
where GG is the gravitational constant, MM is the mass of the planet, and RR is the distance you're at from the centre of the planet.
In this case, the acceleration due to gravity on the surface of the moon is one sixth that on the surface of earth's and the diameter of the moon is one fourth of that of earth.
Therefore, the escape velocity in the moon will be
ve=216GM14Rv_{e}=\sqrt{\frac{2 \frac{1}{6} G M}{\frac{1}{4} R}}
=1242GMR=\frac{1}{\sqrt{24}} \sqrt{\frac{2 G M}{R}}.
Hence, the ratio of escape velocity on moon and earth will be 124\frac{1}{\sqrt{24}}.