Gravitation Question 356

Question: A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is v. Due to the rotation of planet about its axis the acceleration due to gravity g at equator is 1/2 of g at poles. The escape velocity of a particle on the pole of planet in terms of v is

Options:

A) $ [v_{e}=2v] $

B) $ [v_{e}=v] $

C) $ [v_{e}=v/2] $

D) $ [v_{e}=\sqrt{3}v] $

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Answer:

Correct Answer: A

Solution:

  • $ [v=\omega R] $

    $ [g=g_{0}-{{\omega }^{2}}R] $ [$ [g=] $

    at equator, $ [g_{0}=] $ at poles] $ [\frac{g_{0}}{2}=g_{0}-{{\omega }^{2}}R;{{\omega }^{2}}R=\frac{g_{0}}{2}] $

    $ [v^{2}=\frac{g_{0}R}{2}] $

    $ [v_{e}=\sqrt{2g_{0}R}=\sqrt{4v^{2}}=2v] $



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