14.1.8. The plane and the rocket move in the same straight line and in the same direction. Aircraft speed $\beta c$. Light pulses are emitted from the aircraft at regular intervals, which, reflected from the rocket, arrive at the aircraft at intervals $K$ times longer than the intervals of the emitted pulses. Determine the speed of the rocket relative to the plane from observations from the plane and from observations from the Ground.
Solution
Let intervals between emitted pulses (in the frame of the airplane) be $T$. We interpret the phrase "intervals $K$ times longer than the intervals of the emitted pulses" for the reflected pulses received at the airplane to be $T+KT$. Suppose the airplane and the rocket coincide at time $t=0$ when the first pulse is emitted (and immediately reflected and received). When the second pulse is emitted at time $t=T$, the rocket is $vT$ ahead of the airplane, where $v$ is the speed of the rocket relative to the airplane. Thus, the second pulse will be reflected at the rocket at time $t=T+vT/(c-v)$ and received at the airplane at time $t=T+2vT/(c-v)$. Setting $KT=2vT/(c-v)$, we obtain $v=Kc/(2+K)$. Since the speed of the ground relative to the airplane is $-\beta c$, the speed of the rocket relative to the ground is
which becomes $\beta c$ if $K=0$; that is, if intervals between reflected pulses received at the airplane equal intervals between emitted pulses, then the airplane and the rocket have the same speed.
Answer
Relative to the airplane: $\frac{Kc}{2+K}$, Relative to the ground: $\frac{[K+(2+K)\beta]c}{(2+K)+K\beta}$
Discussion
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