A boy can jump to a height h on ground level. What should be the radius of a sphere of density d such that on jumping on it, he escapes out of the gravitational field of the sphere

The correct answer is (3gh/4πGd)⁰.⁵.

Step-by-step derivation:

Step 1: Determine the boy's jumping velocity on Earth's surface.
Let $v$ be the maximum initial velocity with which the boy can jump. When he jumps to a maximum height $h$ on the ground, where the acceleration due to gravity is $g$, his kinetic energy at the start is completely converted into potential energy at the peak of the jump:
$\frac{1}{2}m{v}^2=mgh$

Solving for $v^2$, we get:
$v^2=2gh$

Step 2: Find the escape velocity of the sphere.
Let $R$ be the radius of the sphere and $d$ be its density. The mass $M$ of the sphere is given by:
$M=\text{Volume}\times \text{Density}=\frac{4}{3}\pi R^{3}d$

The escape velocity $v_{\text{esc}}$ from the surface of this sphere of mass $M$ and radius $R$ is defined as:
$v_{\text{esc}}=\sqrt{\frac{2GM}{R}}$

Substituting the expression for mass $M$ into the escape velocity formula:
$v_{\text{esc}}=\sqrt{\frac{2G\left(\frac{4}{3}\pi R^{3}d\right)}{R}}=\sqrt{\frac{8}{3}\pi G{R}^{2}d}$

Squaring both sides of the escape velocity equation gives:
$v_{\text{esc}}^2=\frac{8}{3}\pi G{R}^{2}d$

Step 3: Relate jumping velocity to escape velocity.
For the boy to escape the gravitational field of the sphere by jumping, his jumping velocity $v$ must be at least equal to the escape velocity $v_{\text{esc}}$ of the sphere:
$v=v_{\text{esc}}\Rightarrow v^2=v_{\text{esc}}^2$

Equating the two expressions for the squared velocities:
$2gh=\frac{8}{3}\pi G{R}^{2}d$

Simplifying the equation to solve for the radius $R$:
$gh=\frac{4}{3}\pi G{R}^{2}d$

Rearranging the terms:
$R^2=\frac{3gh}{4\pi Gd}$

Taking the square root on both sides:
$R={\left(\frac{3gh}{4\pi Gd}\right)}^{0.5}$