The escape speed of a projectile on the Earth's surface is $11.2 ext{ km/s}$. A body is projected o

Question

The escape speed of a projectile on the Earth's surface is $11.2	ext{ km/s}$. 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)

Professor PC Thomas Classes · Answer

By conservation of energy, $\frac{1}{2}m(3v_e)^2 - \frac{GMm}{R} = \frac{1}{2}mv_f^2$. Since escape speed $v_e = \sqrt{\frac{2GM}{R}}$, the initial potential energy is $-\frac{1}{2}mv_e^2$. Thus, $\frac{9}{2}mv_e^2 - \frac{1}{2}mv_e^2 = \frac{1}{2}mv_f^2$. This gives $v_f^2 = 8v_e^2$, so $v_f = \sqrt{8} 	imes 11.2 \approx 31.68	ext{ km/s} \approx 31.7	ext{ km/s}$.