A planet in a distant solar systyem is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is `11kms^(-1)`, the escape velocity from the surface of the planet would be

Options

A. 110km/s
B. 11km/s
C. `1.1km//s`
D. `0.11km//s`

Answer

A

Explanation

Let `V_(P)` and `V_(e)` be the escape velocities from the planet and the earth.
Then `(V_(P))/(V_(e))=sqrt((GM_(P))/(R_(P))xx(R_(e))/(GM_(e)))=sqrt((M_(P))/(M_(e))xx(R_(e))/(R_(P)))`
`therefore (V_(P))/(11)=sqrt(10xx10)`
`therefore V_(P)=11xx10=110 km//s`.