Question

Stellar system Q₁ moves away from us at a speed of 0.800c. Stellar system Q₂, which lies in the same direction in space but is closer to us, moves away from us at speed 0.400c. What multiple of c gives the speed of Q₂ as measured by an observer in the reference frame of Q₁?

Short answer

The multiple of c is 0.588.

Step-by-step solution

Describe the expression for the velocity of the particle

The relativistic velocity of the particle is given by,

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

Here, data-custom-editor="chemistry" $\mathbf{u}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{400}\mathbf{c}$ is the velocity of stellar Q₂ in reference frame, and data-custom-editor="chemistry" $\mathbf{u}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{800}\mathbf{c}$ is the velocity of the reference frame of the observer on stellar Q₁ relative to the reference frame.

Determine the multiple of c

Rearrange the equation (1) for u^'.

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

Substitute 0.800c for v, and 0.400c for u in equation (2).

$\begin{array}{rcl}{\mathrm{u}}^{\text{'}}& =& \frac{0.800\mathrm{c}-0.400\mathrm{c}}{{\displaystyle \frac{\left(0.800\mathrm{c}\right)\left(0.400\mathrm{c}\right)}{{\mathrm{c}}^2}}-1}\\ & =& \frac{0.400\mathrm{c}}{0.32-1}\\ & =& -0.588\mathrm{c}\end{array}$

Here, the negative sign indicates the opposite direction.

Therefore, the multiple of c is 0.588.