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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}}} $

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Last updated date: 14th Apr 2024
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MVSAT 2024
Answer
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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;
r = radius
$v_e^2 = 2gh$;
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}$;
The radius is:
$ \Rightarrow r = \sqrt {\dfrac{{3gh}}{{4\pi \rho G}}} $;

Option (C) is correct.

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}}} $.

Note: Here we know the formula for escape velocity. Write the formula in terms of radius “r”. Then we equate the K.E = P.E. Here the velocity in the K.E is the same as the escape velocity, enter the relation between the kinetic energy and potential energy in terms of escape velocity in the formula for escape velocity which is written in terms of radius “r”.