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)-R omega^(2) rArr (g)/(2)=g-R omega^(2)` `R omega^(2) = (g)/(2) rArr R^(2)omega^(2)=(gR)/(2) rArr V^(2)(gR)/(2)` ….(i) `V_(e ) = sqrt(2gR)` From (i) and (ii) `V_(e ) = sqrt(2xx2V^(2))rArr V_(e )=2V` |
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