Question

The escape speed of a projectile on the earth’s surface is 11.2 km 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.

Answer 1

Escape velocity of a projectile from the Earth, \(v_{esc}\) = \(11.2 \; km/s\)
Projection velocity of the projectile, \(v_p\) = \(3v_{esc}\)
Mass of the projectile = \(m\)
Velocity of the projectile far away from the Earth = \(v_f\)
Total energy of the projectile on the Earth = \(\frac{1}{2}mv^2_p - \frac{1}{2}mv^2_{esc}\)
Gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth= \(\frac{1}{2}mv^2_f\)

From the law of conservation of energy, we have
\(\frac{1}{2}mv^2_p - \frac{1}{2}mv^2_{esc}\) = \(\frac{1}{2}mv_f^2\) 

\(v_r\) = \(\sqrt{v_p ^2 - v_{esc}^2}\)

=\(\sqrt{(3v_{esc}) ^2 - (v_{esc})^2}\)

=\(\sqrt 8 v_{esc}\)

=\(\sqrt 8 \times 11.2\) = \(31.68 \; km/s\)