Find radius of such planet on which the man escapes through Jumping. The capacity of Jumping of person on earth is `1.5m` Density of planet is same as that of earth .

Explanation

For a planet: `(1)/(2) mv^(2) - (GM_(P)m)/(R_(P)) =0 rArr (1)/(2) mv^(2) = (GM_(P)m)/(R_(P))`
On earth `rarr (1)/(2) mv^(2) = m ((GM_(E))/(R_(E)^(2))) h`
`therefore (GM_(P)m)/(R_(P)) = (GM_(E)m)/(R_(E)^(2))h rArr (M_(P))/(R_(P)) = (M_(E)h)/(R_(E)^(2))`
`:.` Density (`rho`) is same `rArr (4//3piR_(P)^(3)rho)/(R_(P)) =(4//3piR_E^2hrho)/R_E^2 rArr sqrt(R_(E)h)` .