The escape velocity on the surface of the earth is 11.2 `kms^(-1)`. If mass and radius of a planet is 4 and 2 tims respectively than that of the earth, what is the escape velocity from the planet?

Options

A. `11.2kms^(-1)`
B. `1.112kms^(-1)`
C. `15.8kms^(-1)`
D. `22.4kms^(-1)`

Answer

C

Explanation

`V_(e_(1))=sqrt((2GM)/(R )), V_(e_(2))=sqrt((2G4M)/(2R))=sqrt((4GM)/(R ))`
`therefore V_(e_(2))=sqrt(2)[(2GM)/(R )]=sqrt(2)xx V_(e_(1))`
`= 1.414xx11.2=15.8` km/s