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

Given Data

• Escape speed \(v_e = 11.2\) km/s = \(1.12 \times 10^4\) m/s
• Launch speed \(v_i = 3v_e = 33.6\) km/s

Conservation of Energy

Total mechanical energy conserved (no atmosphere, ignore other bodies):

$$E_i = E_f$$ $$\frac{1}{2}mv_i^2 - \frac{GMm}{R} = \frac{1}{2}mv_\infty^2 + 0$$

Escape speed definition: \(\frac{1}{2}mv_e^2 = \frac{GM}{R}\)

$$\frac{1}{2}v_i^2 - \frac{1}{2}v_e^2 = \frac{1}{2}v_\infty^2$$ $$v_\infty^2 = v_i^2 - v_e^2$$

Calculation

$$v_\infty^2 = (3v_e)^2 - v_e^2 = 9v_e^2 - v_e^2 = 8v_e^2$$ $$v_\infty = \sqrt{8} \, v_e = 2\sqrt{2} \, v_e$$

Numerical value:

$$v_\infty = \sqrt{8} \times 11.2 = 2.828 \times 11.2 = 31.7 \, \text{km/s}$$

Final Speed Far Away

\(v_\infty = \textbf{31.7 km/s}\)

Physical Insight

• Escape speed: \(v_\infty = 0\) km/s
• 3× escape: excess kinetic energy \(\frac{1}{2}m(9v_e^2 - v_e^2) = 4mv_e^2\) converts to asymptotic speed
• \(\sqrt{8} \approx 2.83\), so retains ~94% of launch speed far away

Energy Breakdown

State	KE	PE	Total E
Surface	\(\frac{9}{2}mv_e^2\)	\(-mv_e^2\)	\(\frac{7}{2}mv_e^2\)
Infinity	\(4mv_e^2\)	0	\(\frac{7}{2}mv_e^2\)