What is the escape velocity for body on the surface of planet on which the accelearation due to gravity is `(3.1)^(2)ms^(2)` and whose radius is 8100 km ?

Options

A. `2700 km s^(-1)`
B. `27.9 km s^(-1)`
C. `(27.9)/(sqrt5) kms^(-1)`
D. `27.9 sqrt5 kms^(-1)`

Answer

C

Explanation

Escape velocity of the body from the surface of the planet is `v_p = sqrt(2g_p R_p) = sqrt(2 xx (3.1)^2 xx (8100 xx 10^3) ) ms^(-1)`
` = (3.1 xx 90 xx 10 sqrt20)/(1000) kms^(-1) = (27.9)/(sqrt5) kms^(-1)`