While on board a star ship, you intercept signals from four shuttle craft that are moving either directly toward or directly away from you. The signals have the same proper frequency f_o. The speed and direction (both relative to you) of the shuttle craft are (a) 0.3c toward, (b) 0.6c toward, (c) 0.3c away, and (d) 0.6c away. Rank the shuttle craft according to the frequency you receive, greatest first.

In astronomy applications, the velocities of galaxies are estimated using Doppler shifts. Doppler shift is the difference between the observed and proper wavelength of light.The wavelength measured in the rest frame of the source is called proper wavelength ${\lambda }_0$. And the detected wavelength $\lambda$is related to the proper wavelength as

$\lambda ={\lambda }_0\sqrt{\frac{1+\beta }{1-\beta }}$

Where,

$\beta$is the speed parameter($v/c$). The wavelength ad frequency is related by

$\lambda =\frac{c}{f}$

Inserting this in above expression, we get

$$\begin{gathered} \frac{c}{f}=\frac{c}{f_0}\sqrt{\frac{1+\beta }{1-\beta }} \\ f=f_0\sqrt{\frac{1+\beta }{1-\beta }} \end{gathered}$$

For part (a), a shuttle moves at 0.3c towards the star ship. As the separation decreases, the speed parameter will have opposite signs. The frequency observed from this shuttle is

$$\begin{gathered} f_a=f_0\sqrt{\frac{1+\beta }{1-\beta }} \\ =f_0\sqrt{\frac{1+0.3}{1-0.3}} \\ =f_0\sqrt{\frac{1.3}{0.7}} \\ =1.36f_0 \end{gathered}$$

For part (b), the shuttle moves at 0.6c towards the star ship. As the separationdecreases, the speed parameter will have opposite signs. The frequency observed from this shuttle is

$$\begin{gathered} f_b=f_0\sqrt{\frac{1+\beta }{1-\beta }} \\ =f_0\sqrt{\frac{1+0.6}{1-0.6}} \\ =f_0\sqrt{\frac{1.6}{0.4}} \\ =2f_0 \end{gathered}$$

For part (c), the shuttle moves at 0.3c away from the star ship. The frequency observed from this shuttle is

$$\begin{gathered} f_c=f_0\sqrt{\frac{1-\beta }{1+\beta }} \\ =f_0\sqrt{\frac{1-0.3}{1+0.3}} \\ =f_0\sqrt{\frac{0.7}{1.3}} \\ =0.73f_0 \end{gathered}$$

For part (d), the shuttle moves at 0.6c away from the star ship. The frequency observed from this shuttle is

$$\begin{gathered} f_d=f_0\sqrt{\frac{1-\beta }{1+\beta }} \\ =f_0\sqrt{\frac{1-0.6}{1+0.6}} \\ =f_0\sqrt{\frac{0.4}{1.6}} \\ =0.5f_0 \end{gathered}$$

The ranking of frequency received from the shuttles is $f_b>f_a>f_c>f_d$.