Galaxy A is reported to be receding from us with a speed of 0.35c. Galaxy B, located in precisely the opposite direction, is also found to be receding from us at this same speed. What multiple of c gives the recessional speed an observer on Galaxy A would find for (a) our galaxy and (b) Galaxy B?

The relativistic velocity of the particle is given by,

$\mathbf{u}\mathbf{=}\frac{{\mathbf{u}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{v}}{\mathbf{1}\mathbf{+}{\displaystyle \frac{{\mathbf{u}}^{\mathbf{\text{'}}}\mathbf{v}}{{\mathbf{c}}^{\mathbf{2}}}}}$

Here, u is the velocity of the particle as measured in S, u^' is the velocity of the particle as measured in S^', v is the velocity of the S^' relative to S, and c is the velocity of the light.

If a person P is moving with speed v eastward from the person Q, then person P should observe that the person Q moving v away from person P.

In the similar way, if the Galaxy A is moving from our Galaxy with a speed of 0.35c then the observer in Galaxy A must see our Galaxy with the same speed of 0.35c or 0.35 multiple of c away from him.

Therefore the speed of an observer on Galaxy A from our galaxy is 0.35 multiples of c.

Rearrange the equation (1) as follows.

$\begin{array}{rcl}\mathrm{u}\left(1+\frac{{\mathrm{u}}^{\text{'}}\mathrm{v}}{{\mathrm{c}}^2}\right)& =& {\mathrm{u}}^{\text{'}}+\mathrm{v}\\ \mathrm{u}-\mathrm{v}& =& {\mathrm{u}}^{\text{'}}\left(1-\frac{\mathrm{uv}}{{\mathrm{c}}^2}\right)\\ \frac{\mathrm{u}\text{'}}{\mathrm{c}}& =& \frac{{\displaystyle \frac{\mathrm{u}}{\mathrm{c}}}-{\displaystyle \frac{\mathrm{v}}{\mathrm{c}}}}{1-{\displaystyle \frac{\mathrm{uv}}{{\mathrm{c}}^2}}}.......\left(2\right)\end{array}$

Substitute -0.35c for u and 0.35c for v in equation (2).

$\begin{array}{rcl}\frac{{\mathrm{u}}^{\text{'}}}{\mathrm{c}}& =& \frac{{\displaystyle \frac{-0.35\mathrm{c}}{\mathrm{c}}}-{\displaystyle \frac{0.35}{\mathrm{c}}}}{1-{\displaystyle \frac{\left(-0.35\mathrm{c}\right)\left(0.35\mathrm{c}\right)}{{\mathrm{c}}^2}}}\\ \frac{{\mathrm{u}}^{\text{'}}}{\mathrm{c}}& =& \frac{-2\left(0.35\right)}{1+{\left(0.35\right)}^2}\\ \frac{{\mathrm{u}}^{\text{'}}}{\mathrm{c}}& =& -0.62\\ {\mathrm{u}}^{\text{'}}& =& -0.62\mathrm{c}\end{array}$

Here, the negative sign indicates the direction of the galaxy B with respect to the observer in galaxy A.

Therefore, the multiple of c is 0.62.