Question

Derive the expression for escape velocity of a projectile from Earth.

Answer 1

A particle escapes from earth only it overcomes the gravitational

binding energy due to earth, and has zero energy in infinity

an Gravitational Binding energy = $\frac{-GMm}{R}$

It is projected with initial velocity $v_{esc}$ so it escapes.

$\Rightarrow$ Initial $KE=\frac{1}{2}m{v}_{esc}^2$ (m: mass of body)

From conservation of energy :-

$-\frac{GMm}{R}+\frac{1}{2}m{v}_{esc}^2=0$ $[E=0$ at $\infty ]$

$\Rightarrow \frac{GMm}{R}=\frac{1}{2}m{v}_{esc}^2\Rightarrow v_{esc}=\sqrt{\frac{2GM}{R}}$

also, as $g=\frac{GM}{R^2}$ we get $\Rightarrow v_{esc}=\sqrt{(\frac{2GM}{R^2}.R)}=\sqrt{2gR}$ (Reqd)