Answer to Question #109314 in Quantum Mechanics for ss

Question

A rocket is launched from a spaceship travelling with a speed of 0.8c towards the
earth. If the speed of the rocket is 0.6c, what would be its speed as observed by an
observer on earth if the rocket is moving
(i) towards the earth (ii) away from the earth?

Answer 1

The law of velocity addition in the special theory of relativity can be written in the form

(1) $\beta_{12}=\frac{\beta_1+\beta_2}{1+\beta_1\cdot\beta_2}$ , where $\beta=\frac{V}{c}$ . In first case we have

(2) $\beta_{towards}=\frac{0.8+0.6}{1+0.8\cdot 0.6}=\frac{1.4}{1.48}=0.95$ . When the rocket is moving away from Earth the $\beta_2$ become negative.

(3) $\beta_{away}=\frac{0.8-0.6}{1-0.8\cdot 0.6}=\frac{0.2}{0.52}=0.38$

Answer: (i) when the rocket is launched from a spaceship towards the earth its speed is 0.95c as observed by an observer on earth, (ii) if the rocket starts away from the earth its speed is 0.38c