The escape velocity of the surface of earth is 11.2 km/s. What would be the escape velocity of the surface of another planet of the same  mass but `(1^(th))/4` times the radius of the earth ?

Options

A. 44.8 km/s
B. 22.4 km/s
C. 5.6 km/s
D. 11.2 km/s

Answer

B

Explanation

`implies` For surface of earth.
`v_e=sqrt((2GM_e)/R_e)= 11.2 (km)/s`
For planet
`v_p = (sqrt(2GM_p))/R_p`
mass of planet `M_p=M_e`
Radius of planet `R_p = R_e/4`
`v_p = sqrt((2GM_e)/(R_e/4)) `
`=2 sqrt((2GM_e)/R_e)`
`=2(v_e)`
`= 1 (11.2 km//s) `
` =22.4km//s `