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

Explanation

Given `M_(P)=10 M_(e), R_(p)=(R_(e))/10`
We know that `v_(e)=sqrt((2GM)/R)`
`:. V_(P)=sqrt((2GM_(P))/(R_(P)))=sqrt((100xx2GM_(e))/(R_(e)))=10 v_(e)`
`=10xx11=110km//s`