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] $
Show Answer
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] $
