 # The escape speed of a projectile on the earth’s surface is 11.2 km s–1. Question:

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

Solution:

Escape velocity of a projectile from the Earth, $v_{\mathrm{esc}}=11.2 \mathrm{~km} / \mathrm{s}$

Projection velocity of the projectile, $v_{p}=3 v_{e s c}$

Mass of the projectile $=m$

Velocity of the projectile far away from the Earth $=v_{f}$

Total energy of the projectile on the Earth $=\frac{1}{2} m v_{p}^{2}-\frac{1}{2} m v_{\mathrm{cc}}^{2}$

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth $=\frac{1}{2} m v_{f}^{2}$

From the law of conservation of energy, we have

$\frac{1}{2} m v_{\mathrm{p}}^{2}-\frac{1}{2} m v_{\mathrm{cc}}^{2}=\frac{1}{2} m v_{\mathrm{f}}^{2}$

$v_{\mathrm{f}}=\sqrt{v_{\mathrm{p}}^{2}-v_{\mathrm{esc}}^{2}}$

$=\sqrt{\left(3 v_{\mathrm{esc}}\right)^{2}-\left(v_{\mathrm{ec}}\right)^{2}}$

$=\sqrt{8} v_{\mathrm{esc}}$

$=\sqrt{8} \times 11.2=31.68 \mathrm{~km} / \mathrm{s}$  