Question #103621

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?

Expert's answer

The speeds of a spaceship and a rocket are comparable to the speed of light, so we must use the formula for adding the velocities in the relativistic case. For the case of movement along the same direction, this formula can be written as Vx=cβsp+βr1+βspβrV_x=c\frac{\beta_{sp} + \beta_{r}}{1+\beta_{sp} \beta_{r}} where βsp=Vspc\beta_{sp}=\frac{V_{sp}}{c} βr=Vrc\beta_{r}=\frac{V_{r}}{c} spaceship and rocket speed related с. We have for case (i)

Vi=c0.8+0.61+0.8⋅0.6=0.946cV_i=c\frac{0.8+0.6}{1+0.8\cdot 0.6}=0.946c . For the case (ii) we have βr=−0.6\beta_{r}=-0.6 and

Vii=c0.8−0.61−0.8⋅0.6=0.385cV_{ii}=c\frac{0.8-0.6}{1-0.8\cdot 0.6}=0.385c

Answer:  If the rocket is moving

(i) towards the earth its speed relative to the Earth is 0.946c (ii) away from the earth its speed relative to the Earth is 0.385c



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