91Ó°ÊÓ

We have given that a distant quasar is found to be moving away from the earth at $0.80c$.

A galaxy that is closer to the earth and follows the same line of sight is travelling away from us at $0.20c$.

We have to find the recessional speed of the quasar, as a fraction of $c$, as measured by astronomers in the other galaxy.

Assemble the frames as follows:

The $S$– Frame is the Earth.

The galaxy is the $S\text{'}$– Frame, which is moving at $0.20c$perpendicular to the $S$– Frame.

The moving quasar as measured in the $S$– Frame. According to the $S$– Frame, the velocity is $0.80c$.

Calculate the quasar's speed in relation to the galaxy:

$$\begin{gathered} u_x\text{'}=\frac{u_x-v}{1-{\displaystyle \frac{u_{x}v}{c^2}}} \\ =\frac{(0.80c)-(0.20c)}{1-{\displaystyle \frac{(0.80c)(0.20c)}{c^2}}} \\ =0.71c \end{gathered}$$

The quasar's speed in relation to the galaxy is$0.71c$.