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# A boy can jump to a height h from ground on earth. What should be the radius of a sphere of density $\rho$ such that on jumping on it, he escapes out of the gravitational field of the sphere?A) $\sqrt {\dfrac{{4\pi G\rho }}{{3gh}}}$;B) $\sqrt {\dfrac{{4\pi G\rho }}{{3gh}}}$;C) $\sqrt {\dfrac{{3gh}}{{4\pi G\rho }}}$D) $\sqrt {\dfrac{{3gh}}{{4\pi gh}}}$

Last updated date: 14th Apr 2024
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Hint: For the boy to escape the Earth’s gravitational field he has to achieve a maximum height. At maximum height the kinetic energy of the boy will be converted into potential energy of the boy. Apply the formula for escape velocity and apply the relation between escape velocity and the energy of the boy at max height.

Formula Used:
Escape velocity: ${v_e} = \sqrt {\dfrac{{2GM}}{r}}$;
Where,
${v_e}$= Escape velocity;
G = Gravitational Constant;
M = Mass;
$v_e^2 = 2gh$;
Where;
${v_e}$= Escape velocity;
g = Gravitational Acceleration;
h = height

Complete step by step solution:
Find out the radius of the sphere:
Apply the formula for escape velocity:
${v_e} = \sqrt {\dfrac{{2GM}}{r}}$;
To remove the root square on both the sides:
$\Rightarrow v_e^2 = \dfrac{{2GM}}{r}$;
Now we know that $Mass = Density \times Volume$;
$v_e^2 = \dfrac{{2G\rho V}}{r}$; ....(Here: $\rho$= Density; V = Volume)
Volume of Sphere is $V = \dfrac{4}{3}\pi {r^3}$; Put this in the above equation:
$v_e^2 = \dfrac{{2G\rho \dfrac{4}{3}\pi {r^3}}}{r}$;
$\Rightarrow v_e^2 = 2G\rho \dfrac{4}{3}\pi {r^2}$;
Take the rest of the variables except the radius to LHS:
$\Rightarrow \dfrac{3}{{42G\rho \pi }}v_e^2 = {r^2}$;
Now at Max height the K.E = P.E;
$\dfrac{1}{2}mv_e^2 = mgh$;
$\Rightarrow \dfrac{1}{2}v_e^2 = gh$;
The velocity will become:
$\Rightarrow v_e^2 = 2gh$;
Put the above value in the equation$\dfrac{3}{{42G\rho \pi }}v_e^2 = {r^2}$;
$\dfrac{{3 \times 2gh}}{{42G\rho \pi }} = {r^2}$;
$\Rightarrow \dfrac{{3gh}}{{4\pi \rho G}} = {r^2}$;
$\Rightarrow r = \sqrt {\dfrac{{3gh}}{{4\pi \rho G}}}$;
The radius of a sphere of density $\rho$ such that on jumping on it, he escapes out of the gravitational field of the sphere is $\sqrt {\dfrac{{3gh}}{{4\pi \rho G}}}$.