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

Updated On: Sep 3, 2024
  • $\frac {\sqrt {6}} {2}$
  • $\frac{1}{\sqrt{24}}$
  • $3$
  • $\frac {\sqrt {3}}{2}$
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The Correct Option is B

Solution and Explanation

The correct option is (B): \(\frac{1}{\sqrt{24}}\)
The escape velocity of an object is given by the equation \(v_{e}=\sqrt{\frac{2 G M}{R}}\)
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 
\(v_{e}=\sqrt{\frac{2 \frac{1}{6} G M}{\frac{1}{4} R}}\)
\(=\frac{1}{\sqrt{24}} \sqrt{\frac{2 G M}{R}}\)
Hence, the ratio of escape velocity on moon and earth will be \(\frac{1}{\sqrt{24}}\).
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Concepts Used:

Escape Speed

Escape speed is the minimum speed, which is required by the object to escape from the gravitational influence of a plannet. Escape speed for Earth’s surface is 11,186 m/sec. 

The formula for escape speed is given below:

ve = (2GM / r)1/2 

where ,

ve = Escape Velocity 

G = Universal Gravitational Constant 

M = Mass of the body to be escaped from 

r = Distance from the centre of the mass