A man can jump up to a height of 1.5 m on the earth's surface. What should be the radius of a planet having the same average density as that of the earth so that the man can come out of the gravitational field of that planet in one jump ? Radius of the earth is 6400 km.

Explanation

The man can jump up to a height h on the earth.s surface.
Hence, his initial kinetic energy =his potential energy at height h.
Thus `1/2 mv^2=mgh` [m = mass of the man v=initial velocity]
or, `v=sqrt(2gh)" "……..(1)`
The man can jump on the surface of any planet with this velocity v. if v equals the escape velocity from a planet , the man can go out of the gravitational pull of that planet. Let the acceleration due to gravity for a planet be `g^.`, its radius `R^.` and average density `rho`, then
`g^.=4/3 piGR^.rho`
As `4/3piG and rho` are constants `g^.prop R^.` .
Hence `(g^.)/(g) =(R^.)/R or g^.=g*(R^.)/R" "..........(2)`
[R=radius of the earth]
Escape velocity for the planet is `v=sqrt(2g^.R^.)" ".......(3)`
From (1) and (3) ,
`sqrt(2gh)=sqrt(2g^.R^.) or, gh=g^.R^.`
`or, g^.=g*h/(R^.)" "...........(4) `
From (2) and (4)
`(gR^.)/R=(gh)/(R^.)`
`or, R^.=sqrt(hR)=sqrt(1.5xx(6400xx10^3))=3098m =3.098 km`.