Jupiter has a mass 318 times that of earth, and its radius is 11.2 times the earth's radius Estimate the escape velocity of a body from Jupiter's surface, given that the escape velocity from the earth's surface `11.2 kms^(-1)`.

Explanation

Escape velocity from the earth's surface is
`v_e =sqrt((2GM)/(R )) = 11.2 kms^(-1)`
Escape velocity from Jupiter's surface will be
`v_e' = sqrt((2GM')/(R'))`
But `M' = 318 M, R'= 11.2 R`
`v_e' = sqrt((2G(318M))/(11.2))= sqrt((2GM)/(R )xx(318)/(11.2))`
`v_e'xx sqrt((318)/(11.2))= 11.2xx sqrt((318)/(11.2))= 59.7 kms^(-1)`.