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Question

Physics Question on Gravitation

Escape velocity from a planet is Ve V_{e} . If its mass is increased to 88 times and its radius is increased to 22 times, then the new escape velocity would be:

A

VeV_{e}

B

2ve \sqrt{2}v_{e}

C

2ve 2v_{e}

D

22ve 2\sqrt{2}v_{e}

Answer

2ve 2v_{e}

Explanation

Solution

Velocity of projection of body, at which the body goes out or gravitational field of earth or planet and never return, is called escape velocity. ve=2GMRv_{e}=\sqrt{\frac{2 G M}{R}}
where GG is gravitational constant, MM is mass of plant and ReR_{e} is radius.
When M=8M,R=2RM=8\, M,\, R=2 R,
then ve=2G(8M)(2R)ve=22GMRv_{e}=\sqrt{\frac{2 G(8 M)}{(2 R)}} v_{e}=2 \sqrt{\frac{2 G M}{R}} ve=2ve.v_{e}=2 v_{e} .
Note: Escape velocity is independent of mass of body.