The acceleration due to gravity on the surface of the moon is one-sixth that on the earth. The radius of the moon is about one-fourth tha tof the earth. If `v_(e)` is the escape velocity on the surfaceo f the earth, then escape velocity on the surface of the moon will be

Options

A. `(4v_(e))/(6)`
B. `(v_(e))/(24)`
C. `v_(e)sqrt((4)/(6))`
D. `(v_(e))/(sqrt(24))`

Answer

d

Explanation

`(v_(m))/(v_(e))=sqrt((M_(m))/(M_(e)).(R_(e))/(R_(m)))`
`therefore" "(v_(m))/(v_(e))=sqrt((g_(m))/(g_(e))xx(R_(m))/(R_(e)))`
`"Now, "g_(m)=(g_(e))/(6)" and "R_(m)=(R_(e))/(4)`
`therefore" "(v_(m))/(v_(e))=(1)/(sqrt(6xx4)).`