The escape velocity of a body at the surface of planet P is `(1)/(sqrt(5))` times than that at surface of earth. The radius of planet is `(1)/(24)` times the radius of earth. The acceleration due to gravity for planet is

Options

A. `2.4 g_(e)`
B. `2.8 g_(e)`
C. `4.8 g_(e)`
D. `3.0 g_(e)`

Answer

C

Explanation

`V_(P) = sqrt(2g_(P) R_(P))`
`V_(e) = sqrt(2g_(e) R_(e))`
`V_(P) = (1)/(sqrt(5)) V_(e)`
`sqrt(2g_(P)R_(P)) = (1)/(sqrt(5)) sqrt(2g_(e)R_(e))`
`R_(P) = (R_(e))/(24)`
`rArr sqrt((2g_(P)R_(e))/(24)) = sqrt((2g_(e)R_(e))/(5))`
`(2g_(P)R_(e))/(24) = (2g_(e)R_(e))/(5)`
`g_(P) = (24)/(5)g_(e) = 4.8 g_(e)`