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Question

Physics Question on Gravitation

Escape velocity for 'a projectile at Earth's surface is v,. A body is projected from earth's surface with velocity 2ve2v_e. The velocity of the body when it is at infinite distance from the centre of the Earth is

A

vev_e

B

2ve2 \, v_e

C

2ve\sqrt{2} \, v_e

D

3ve\sqrt{3} \, v_e

Answer

3ve\sqrt{3} \, v_e

Explanation

Solution

Let vv be the speed of the body when it is at infinite distance from the centre of the Earth and u be speed of projection of the body from the earth's surface
According to law of conservation of mechanical energy,
12mu2GMmR=12mv20\frac{1}{2}mu^{2} - \frac{GMm}{R} = \frac{1}{2}mv^{2} - 0
or, v2=u22GMRv^{2} =u^{2} - \frac{2GM}{R}
v2=u2ve2(ve=2GMR)v^{2} = u^{2} -v^{2}_{e} \left( \because \:\: v_e = \sqrt{\frac{2GM}{R}} \right)
v=(2ve)2ve2v = \sqrt{\left(2v_{e}\right)^{2 } - v_{e}^{2}}
=3ve= \sqrt{3} v_{e}