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

The escape velocity of a body at the surface of planet P is (1)/(sqrt(

1.	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
Answer» <P>`2.4 g_(E)` `2.8 g_(e)` `4.8 g_(e)` `3.0 g_(e)` Solution :`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)`