A man can jump vertically to a height of `1.5 m` on the earth. Calculate the radius of a planet of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is `6.41 xx 10^(5) m`.

Answer

A, C

Explanation

`h = (u^(2))/(2g_(e))`
`:. u = sqrt(2g_(e)h)` ..(i)
For the asked planet this `u` should be equal to the escape velocity from its surface.
`:. sqrt(2g_(e)h) = sqrt(2g_(p)R_(p))`
or `(GM_(e))/(R_(e)^(2)). h = (GM_(p))/(R_(p)^(2)). R_(p)`
or `(((4)/(3) pi R_(e)^(3))rho h)/(R_(e)^(2)) = (((4)/(3) pi R_(p)^(3))rho R_(p))/(R_(p)^(2))`
or `R_(p) = sqrt(R_(e)h)`
`= sqrt((6.41 xx 10^(6))(1.5))`
`= 3.1 xx 10^(3) m`