Question

A particle is projected vertically upward from the surface of earth $($radius $R)$ with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of earth is

• A. $R$
• B. $2R$
• C. $3R$
• D. $4R$

Answer 1

The correct option is A $R$

Let $v$ be the speed of projection and $v_e$be the escape velocity.

As per the question,

$\frac{1}{2}m{v}^2=\frac{1}{2}\times \frac{1}{2}m{v}_e^2$

$\Rightarrow v=\frac{v_e}{\sqrt{2}}$

As we know that,

$v_e=\sqrt{\frac{2GM}{R}}$

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

Now conserving mechanical energy between the surface of the earth and the highest point (height $h$),

$U_i+K_i=U_f+K_f$

$\frac{-GMm}{R}+\frac{1}{2}m{v}^2=\frac{-GMm}{R+h}+0$

$[$ at the highest point, $K_f=0]$

$\Rightarrow \frac{-GMm}{R}+\frac{1}{2}m\frac{GM}{R}=\frac{-GMm}{R+h}$

$[$substituting $v=\sqrt{\frac{GM}{R}}]$

$\Rightarrow \frac{-GMm}{2R}=\frac{-GMm}{R+h}\Rightarrow R+h=2R$

$\therefore h=R$

Hence, option $\left(a\right)$ is correct.

$\begin{array}{c}\text{Why this question:}\\ \text{To help students the conservation of}\\ \text{mechanical energy in planet-satellite systems.}\\ \text{It is a very important concept and needs}\\ \text{to be applied frequently in problems.}\end{array}$