Chapter 15, Problem 17A

(a)

To determine

To rank:The speeds of the probes as seen by an Earth observer from greatest to least.

Answer to Problem 17A

The speeds of the probes as seen by an Earth observer from greatest to least is $A>C>B$ .

Explanation of Solution

Given:

The given figure is shown below.

  

Formula used:

The expression for the speed of the probes as seen by an Earth observer is,

  $v=\frac{v_{\text{s}}+v_{\text{p}}}{1+\frac{v_{\text{s}}{v}_{\text{p}}}{c^2}}$

Here, $v_{\text{s}}$ is the velocity of the spaceship, $v_{\text{p}}$ is the velocity of the probe, and $c$ is the speed of the light.

Calculation:

The speed of the probes as seen by an Earth observer for case A is

  $\begin{array}{l}v=\frac{v_{\text{s}}+v_{\text{p}}}{1+\frac{v_{\text{s}}{v}_{\text{p}}}{c^2}}\\ v_{\text{A}}=\frac{\left(0.6c\right)+\left(0.6c\right)}{1+\frac{\left(0.6c\right)\left(0.6c\right)}{c^2}}\\ v_{\text{A}}=0.88c\end{array}$

The speed of the probes as seen by an Earth observer for case B is,

  $\begin{array}{l}v=\frac{v_{\text{s}}+v_{\text{p}}}{1+\frac{v_{\text{s}}{v}_{\text{p}}}{c^2}}\\ v_{\text{B}}=\frac{\left(0.6c\right)+\left(-0.4c\right)}{1+\frac{\left(0.6c\right)\left(-0.4c\right)}{c^2}}\\ v_{\text{B}}=0.26c\end{array}$

The speed of the probes as seen by an Earth observer for case C is,

  $\begin{array}{l}v=\frac{v_{\text{s}}+v_{\text{p}}}{1+\frac{v_{\text{s}}{v}_{\text{p}}}{c^2}}\\ v_{\text{C}}=\frac{\left(0.6c\right)+\left(0.4c\right)}{1+\frac{\left(0.6c\right)\left(0.4c\right)}{c^2}}\\ v_{\text{C}}=0.81c\end{array}$

By comparing the speed of the probes as seen by an Earth observer is $A>C>B$ .

Conclusion:

Thus, the speeds of the probes as seen by an Earth observer from greatest to least is $A>C>B$ .

(b)

To determine

To rank:The speed of light reflected from the departing probes as seen by the spaceship from greatest to least.

Answer to Problem 17A

The speed of light reflected from the departing probes as seen by the spaceship from greatest to least is $A=B=C$ .

Explanation of Solution

Given:

The given figure is shown below.

  

Formula used:

As per the “second postulate of special relativity”, in empty space the speed of lightin will always have the same value irrespective of the source and observer motion.

Calculation:

The speed of light is constant regardless of the speed of flashlight. It is same in all the reference frames.An observer on the spaceship will measure the speed of light as c .The speed of the flashes will not be different even if the spaceship turns around or stops.Hence, the speed of light which is reflected from the “departing probes” as observed by the spaceship from highest to least is $A=B=C$ .

Conclusion:

Thus, the speed of light reflected from the departing probes is $A=B=C$ .