Jupiter has a mass `320` times that of the earth and its readius is `11.2` times that of the earth. Determine the escape velocity from the surface of jupiter, given that the escape velocity from the surface of earth is `11.2 km s^(-1)`.

Explanation

Let `M, R` be the mass and radius of earth. Then escape velocity from earth's surface is
`upsilon_(e) = sqrt((2GM)/(R )) = 11.2 km s^(-1)`
For the jupiter, Mass `M' = 320 M , R' = 11.2 R` ltbegt Escape velocity,
`upsilon'_(e) = sqrt((2GM')/(R' )) = sqrt((2G xx 320 M)/(11.2 R))`
`= sqrt((2 GM)/(R ) xx (320)/(11.2)) = 11.2 xx sqrt((320)/(11.2))`
`= 59.86 km s^(-1)`