InterviewSolution
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InterviewSolution
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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 planet in terms of `V`.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 `g_(e)=g_(p)-Romega^(2)impliesg/2=g-Romega^(2)` `Romega^(2)=g/2impliesR^(2)omega^(2)=(gR)/2` `V^(2)=(gR)/2.....(1)` `V_(e)=sqrt(2gR).....(2)` From `(1)` and `(2)` `V_(e)=sqrt(2xx2V^(2))impliesV_(e)=2V` |
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