The moon has a mass of `(1)/(81)` that of the earth and radius of `(1)/(4)` that of the earth. The escape speed from the surface of the earth is 11.2 km`//`s. The escape speed from surface of the moon is-

Options

A. `1.25 km s^(-1)`
B. `2.49 km s^(-1)`
C. `3.7 km s^(-1)`
D. `5.6 km s^(-1)`

Answer

B

Explanation

On the earth , the escape speed is `v_e = sqrt((2GM_e)/(R_e))`
On the moon , the escape speed is `v_m = sqrt((2GM_m)/(R_m))`
`therefore (v_m)/(v_e) = sqrt(( (M_m)/(M_e) ) xx ( (R_e)/(R_m) ) ) = sqrt( (1/81) xx (4/1) ) = 2/9`
or ` v_m = 2/9 xx v_e = 2/9 xx 11.2 kms^(-1) = 2.49 kms^(-1)`