Question:

The escape velocity from the Earth's surface is $v$. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is :

Updated On: Nov 18, 2024
  • v
  • 2 v
  • 3 v
  • 4 v
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The Correct Option is D

Solution and Explanation

$V_{e}=\sqrt{\frac{2 G M}{R}}=\sqrt{\frac{2 G \frac{4}{3} \pi R^{3} \rho}{R}}$ $ \Rightarrow V_{e} \propto R$ $\therefore \frac{V'}{V}=\frac{4 R}{R}$ $ \Rightarrow V^{'}=4 V$
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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