Question

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

• A. $\frac{\sqrt{6}}{2}$
• B. $\frac{1}{\sqrt{24}}$
• C. $3$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (B)

The escape velocity of an object is given by the equation $v_e=\sqrt{\frac{2GM}{R}}$.
where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ 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
$v_e=\sqrt{\frac{2\frac{1}{6}GM}{\frac{1}{4}R}}$
$=\frac{1}{\sqrt{24}}\sqrt{\frac{2GM}{R}}$.
Hence, the ratio of escape velocity on moon and earth will be $\frac{1}{\sqrt{24}}$.