Given mass of the moon is 1/81 of the mass of the earth and corresponding radius is 1/4 of the earth. If escape velocity on the earth surface is 11.2 km / s , the value of same on the surface of the moon is

Options

A. 0.14 km/s
B. 0.5 km/s
C. 2.5 km/s
D. 5 km/s

Answer

c

Explanation

`(v_(e_(2)))/(v_(e_(1)))=sqrt((M_(2))/(M_(1))xx(R_(1))/(R_(2)))=sqrt((1)/(81)xx(4)/(1))=(2)/(9)`
`v_(e_(2))=(2)/(9)xx11.2`
`=(22.4)/(9)=2.5km//s`