Question

A body is projected vertically upwards from the bottom of a crater of moon of depth $\frac{R}{100}$ where R is the radius of moon with a velocity equal to the escape velocity on the surface of moon. Maximum height attained by the body from the surface of the moon is KR.The value of K is

Answer 1

BC is the crater of moon. B denotes bottom of crater.

Depth of crater = $\frac{R}{100}$

Radius of moon = R

Speed of particle at $B=V_B$

Escape velocity on surface of moon $=V_e$

$\therefore V_e=\sqrt{\frac{2GM}{R}}$

At highest point $V_A=0$

Mechanical energy is conserved in the process

Decrease in kinetic energy $=\frac{1}{2}m{V}_B^2$

Increase in $PE=U_A-U_B$

$V_A=PotentialatA=\frac{-GM}{(R+h)}$

h = height from surface

Potential at B=
$V_B=-\frac{GM}{R^3}[\frac{3}{2}{R}^2-\frac{1}{2}{r}^2]$

$=-\frac{GM}{R^3}[1.5R^2-0.5{(R-\frac{R}{100})}^2]$

$=-\frac{GM}{R}\left(1.01\right)$

$\therefore$ Increase in PE $=-\frac{GMm}{R+h}+\frac{GMm}{R}\times \left(1.01\right)$

Decrease in KE = increase in PE

$\frac{1}{2}m{V}_B^2=-\frac{GMm}{(R+h)}+\frac{GMm\times 1.01}{R}$

$\frac{GM}{R}=GM[-\frac{1}{R+h}+\frac{1.01}{R}]$

or $\frac{1}{R}=-\frac{1}{R+h}+\frac{1.01}{R}$

or $h=99R$