Question

In Fig. 37-35, three spaceships are in a chase. Relative to an x-axis in an inertial frame (say, Earth frame), their velocities are ${\mathit{v}}_{\mathbf{A}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{900}\mathit{c}$, ${\mathit{v}}_{\mathbf{B}}$, and ${\mathit{v}}_{\mathbf{c}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{900}\mathit{c}$. (a) What value of ${\mathit{v}}_{\mathbf{B}}$is required such that ships A and C approach ship B with the same speed relative to ship B, and (b) what is that relative speed?

Short answer

1. The value of $V_B$ is $0.858c$ at which ship A and C approaches ship B at same speed.
2. The relative speed of ship A or C relative to ship B will be $0.184c$.

Step-by-step solution

Given Data:

The velocity of ship A is $v_A=0.900c$.

The velocity of ship B is $v_B=0.800c$.

The velocity of ship C is $v_C=0.800c$.

A concept:

A speed having a magnitude that is a significant fraction of the speed of light. (particles) with relativistic speed:

(a) Relativistic velocity addition:

Suppose an object is moving with the velocity $u^{\text{'}}$with respect to $S^{\text{'}}$frame, which is moving with the velocity $v$with respect to frame S then the velocity of the object with respect to $u$frame is,

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

Or

$u\text{'}=\frac{u-v}{1-\frac{uv}{c^2}}$

As the ship A and ship C approaches ship B at the same speed, we can say that,

$v{\text{'}}_A=-v{\text{'}}_C$ ..… (1)

The velocity of ship A relative to ship B is,

$v{\text{'}}_A^{}=\frac{v_A-v_B}{1-\frac{v_A{v}_B}{c^2}}=\frac{c\left({\beta }_A-{\beta }_B\right)}{1-{\beta }_A{\beta }_B}$

Where, $\beta =\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.$.

Similarly, the velocity of ship C relative to ship B is

$v{\text{'}}_C^{}=\frac{v_C-v_B}{1-\frac{v_C{v}_B}{c^2}}=\frac{c\left({\beta }_C-{\beta }_B\right)}{1-{\beta }_C{\beta }_B}$

Inserting these values in equation (1),

$$\begin{gathered} \frac{c\left({\beta }_A-{\beta }_B\right)}{1-{\beta }_A{\beta }_B}=\frac{-c\left({\beta }_C-{\beta }_B\right)}{1-{\beta }_C{\beta }_B} \\ \left({\beta }_A-{\beta }_B\right)\left(1-{\beta }_C{\beta }_B\right)=\left({\beta }_B-{\beta }_C\right)\left(1-{\beta }_A{\beta }_B\right) \\ {\beta }_A-{\beta }_A{\beta }_B{\beta }_C-{\beta }_B+{\beta }_C{\beta }_B^2={\beta }_B-{\beta }_A{\beta }_B^2-{\beta }_C+{\beta }_A{\beta }_B{\beta }_C \end{gathered}$$

$$\begin{gathered} \left({\beta }_A+{\beta }_C\right){\beta }_B^2-2{\beta }_A{\beta }_B{\beta }_C-2{\beta }_B+\left({\beta }_A+{\beta }_C\right)=0 \\ \left({\beta }_A+{\beta }_C\right){\beta }_B^2-2\left({\beta }_A{\beta }_C+1\right){\beta }_B+\left({\beta }_A+{\beta }_C\right)=0 \end{gathered}$$

Inserting the values of$0.90$for${\beta }_A$and$0.80c$for ${\beta }_c$in above equation,

$$\begin{gathered} \left(0.9+0.8\right){\beta }_B^2-2\left(\left(0.8\times 0.9\right)+1\right){\beta }_B+\left(0.9+0.8\right)=0 \\ 1.7{\beta }_B^2-3.44{\beta }_B+1.7=0 \\ {\beta }_B^2-2.024{\beta }_B+1=0 \end{gathered}$$

Therefore, the expression you get is in the quadratic form, using the quadratic formula rule to get the solution of the equation.

${\beta }_B=\frac{-\left[-2.024\right]\pm \sqrt{{\left[-2.024\right]}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}$

$$\begin{gathered} {\beta }_B=\frac{2.024\pm \sqrt{4.095-4}}{2} \\ =\frac{2.024-0.308}{2} \\ =0.858 \end{gathered}$$

Hence, the speed of ship B such that ship A and ship C approaches ship B at the same speed relative to ship B is $v{\text{'}}_B=0.858c$.

(b) The speed of ship A relative to ship B:

The speed of ship A relative to ship B can be given as,

$$\begin{gathered} {v\text{'}}_A=\frac{v_A-{v\text{'}}_B}{1-\frac{v_A{v\text{'}}_B}{c^2}} \\ =\frac{0.90c-0.858c}{1-\frac{\left(0.90c\right)\left(0.858c\right)}{c^2}} \\ =0.184c \end{gathered}$$

Hence, the relative speed is $0.184c$.