The escape speed from jupiter is approximately `59.5 kms^(-1)` and its radius is about `12` times that to earth. From this we may estimate the mean density of jupiter to be about (radius of earth =escape speed from the earth is `11.2 kgm^(-1)`

Options

A. `5` times that of earth
B. `0.2` times that of the earth
C. `2.5` times that of the earth
D. `0.4` times that of earth

Answer

A

Explanation

`V_(e)=sqrt((2GM)/R)=(sqrt((2G4/3piR_(epsilon)^(3)rho_(epsilon))/(R_(epsilon))))/(sqrt((2G4/3piR_(j)^(3)delta_(j))/(R_(j))))`
`rArr V_(e)alphaRsqrt(rho)`