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

Answer

D

Explanation

`V_("escape")=sqrt((2GM)/(R))`
For the Earth, `V_(e)=11kms^(-1)`
Mass of the planet = 10`M_(e)`
Radius of the planet = R/10
`thereforeU_(p)=sqrt((2GMxx10)/(R//10))=10xx11=110kms^(-1)`