The mass of the moon is `7//81` th of earth 's mass and its radius is `1//4` th that of the earth if the escape velocity from the earth surface is 11.2 `kms^(-1)` its value for the moon will be

Options

A. `0.15 kms^(-1)`
B. `5 kms^(-1)`
C. `2.5 kms^(-1)`
D. `0.5 kms^(-1)`

Answer

C

Explanation

Here `(g_(m))/(g_(e ))=(M_(m))/(M_(e))xx(R_(e))/(R_(m))^(2)=(1)/(81)xx(4)^(2)=16/81`
`rarr g_(m)=(16)/(81)g_(e )`
`therefore v_(e)=sqrt(2g_(e)R_(e))=sqrt(2xx9.8xx(6400xx1000)=11.2 kms^(-1)`
`v_(m)=sqrt(2g_(e)R_(m))=sqrt(2xx16/81g_(e)xx1/4R_(e))`
`=2/9sqrt(2g_(e)R_(e))=2/9xx11/2=2.5 kms^(-1)`