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

Options

A. `1.1km s^(-1)`
B. `11 km s^(-1)`
C. `110 km s^(-1)`
D. `0.11km s^(-1)`

Answer

C

Explanation

`((V_(e ))_(2))/((V_(e ))_(1))=(sqrt((2GM_(2))/(R_(2))))/(sqrt((2GM_(1))/(R_(1))))=sqrt((M_(2))/(M_(1))xx(R_(1))/(R_(2)))=sqrt(10xx10)=10`
`therefore (V_(e))_(2)=10xx11=110km//s`
So correct choice is (c )`.