A planet in a distant solar systyem is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is `11kms^(-1)`, the escape velocity from the surface of the planet would be

Options

A. `1.1kms^(-1)`
B. `11kms^(-1)`
C. `110kms^(-1)`
D. `0.11kms^(-1)`

Answer

C

Explanation

(c ) ((v_e)_p)/((v_e)_e) = sqrt((2GM_p)/(R_p))/sqrt((2GM_e)/(R_e)) = sqrt((10M_e)/(M_e)xx(R_e)/(R_e//10)) =10`
`:. ((V_e)_p =10xx(v_e)_e = (10xx11 = 110km//s`