Consider a spherical planet which is rotating about its axis such that the speed of a point on its equator is v and the effective acceleration due to gravity on the equator is `1/3` of its value at the poles. What is the escape velocity for a particle at the pole of this planet.

Options

A. 3 v
B. 2 v
C. `sqrt(3)` v
D. `sqrt(2)` v

Answer

C

Explanation

We know that the escape velocity of a particle from the surface of a planet is given by
`v_(e)=sqrt((2GM)/R)=sqrt((2g R^(2))/R)" "[ :' GM =g R^(2)]`
`=sqrt(2 g R)`
where, g = acceleration due to gravity,
and R = radius of the planet
Given, the velocity at equator, `v_(E)=V`
and acceleration due to gravity at equator,
`g_(E)=1/3 g_(P)`
`:. g_(P)=3 g_(E)` ...(i)
The escape velocity of a particle at equator,
`v_(E)=sqrt(2 g_(E)R)` ...(ii)
and at poles,
`v_(P)=sqrt(2 g_(P)R)`
`=sqrt(2xx3 g_(E)R)=sqrt(3)sqrt(2 g_(E)R)`
`=sqrt(3) v_(E)`
`=sqrt(3) v`
Hence, `sqrt(3) v` is the escape velocity for a particle at the pole of this planet.