A man can jump vertically through a height 1.5 m on earth. The maximum radius of the planet of the same density as that of earth from whose gravitational field he can escape by jumping is `(R_(e )=6400 km)` :

Options

A. 6 km
B. 7 km
C. 3.1 km
D. 5 km.

Answer

C

Explanation

Here square of velocity of man who can jump 1.5 m
`=v^(2)=2gh=2xx(GM_(e ))/(R_(e)^(2)xx h)`
Square of escape velocity from planet is `v_(e )^(2)=(2GM_(p))/(R_(p))`.
Now `v_(e)^(2)=v^(2) therefore R_(p)=sqrt(R_(e )xxh)` if we substitute for velocities `v_(e )` and v and masses in terms of volume `xx` density.
Then `R_(p)=sqrt(6400xx10^(3)xx1.5)=3.1xx10^(3)m or =3.1km`.
Correct choice is (c ).