Question

The escape velocity of a projectile on the earth's surface is $11.2km{s}^{-1}$. A body is projected out with thrice this speed. What will be the speed of the body far away from the earth, that is, at infinity? Ignore the presence of sun and other planets

Answer 1

Escape velocity from Earth is $v_{esc}=11.2km{s}^{-1}$
Projection Velocity = $v_p=3v_{esc}$
Mass of projectile m, is projected with velocity $v_f$
Total energy of projectile on Earth = $\frac{1}{2}m{v_p}^2-\frac{1}{2}m{v_{esc}}^2$
Gravitational potential energy far away from Earth is zero, so the total energy of projectile far away from the Earth will be = $\frac{1}{2}{v_f}^2$
From the law of conservation of energy,
$\begin{array}{c}\frac{1}{2}m{v_p}^2-\frac{1}{2}m{v_{esc}}^2=\frac{1}{2}m{v_f}^2\\ v_f=\sqrt{{v_p}^2-{v_{esc}}^2}=\sqrt{3{v_{esc}}^2-{v_{esc}}^2}=\sqrt{8}{v}_{esc}\\ v_f=\sqrt{8}\times 11.2=31.68km{s}^{-1}\end{array}$