If the escape velocity on the earth is `11.2km-s^-1`, its value for a planet having double the radius and 8 times the mass of earth is

Options

A. `11xx 10^3m-s ^-1`
B. ` 22.4m-s ^-1`
C. `1100m-s ^-1`
D. `22.4xx10^3m-s^-1`

Answer

D

Explanation

Escape velocity ,
V_p = `sqrt (2GM_p)/(R_p)`
= `sqrt(2G8M_e)/2R_e`
= 2`sqrt (GM_e)/(R_e) = 2V_e`
`= 2xx11.2 = 22.4kms^-1`
therefore For earth surface, `V_e = 11.21kms_1`
`22.4 xx 10^3ms^-1`