Chapter 39, Problem 85AP

(a)

To determine

The total travel time of pulse measured by observers in S frame.

Answer to Problem 85AP

The total travel time of pulse measured by observers in S frame is $\underset{\_}{229\text{s}}$.

Explanation of Solution

Let the distance travelled by the light from spacecraft to the mirror be equal to $d_1$. From the figure, $d_1$ is equal to $d$.

Write the expression for the distance that the spacecraft travelled forward.

  $d_s=v\Delta t_s$                                                                                                                    (I)

Here, $d_s$ is the distance travelled by the spacecraft towards the mirror, $v$ is the speed of spacecraft, and $\Delta t_s$ is the time of travel of pulse from the spacecraft.

The spacecraft is travelling towards the mirror. After the reflection of pulse from the mirror it travels back to the approaching spacecraft.

Write the expression for the distance travelled by the pulse after reflection from the mirror.

  $d_2=d-v\Delta t_s$                                                                                                             (II)

Here, $d_2$ is the distance travelled by the pulse after reflection from the mirror, and $d$ is the distance travelled by the pulse to the mirror.

Write the expression for the total distance travelled by the light before and after reflection.

  $D=d_1+d_2$                                                                                                              (III)

Here, $D$ is the total distance travelled by the light.

Use expressions (I) and (II) in (III).

  $D=d+\left(d-v\Delta t_s\right)$                                                                                                  (IV)

The light travels at speed of $c$ in time $\Delta t_s$.

Write the expression for the total distance travelled by light.

  $D=c\Delta t_s$                                                                                                                   (V)

Here, $c$ is the speed of light.

Left hand side of equations (IV) and (V) are same. Therefore equate the right hand side of these equations.

  $d+\left(d-v\Delta t_s\right)=c\Delta t_s$                                                                                               (VI)

Solve equation (VI) for $\Delta t_s$.

  $\Delta t_s=\frac{2d}{c+v}$                                                                                                             (VII)

Conclusion:

Substitute $5.66\times {10}^{10}\text{m}$ for $d$, $0.650c$ for $v$, and $3.00\times {10}^8\text{m/s}$ for $c$ in equation (VII) to find $\Delta t_s$.

  $\begin{array}{c}\Delta t_s=\frac{2\left(5.66\times {10}^{10}\text{m}\right)}{c+0.650c}\\ =\frac{2\left(5.66\times {10}^{10}\text{m}\right)}{1.650\left(3.00\times {10}^8\text{m/s}\right)}\\ =229\text{s}\end{array}$

Therefore, the total travel time of pulse measured by observers in S frame is $\underset{\_}{229\text{s}}$.

(b)

To determine

The total travel time of pulse measured by the observer in the spacecraft.

Answer to Problem 85AP

The total travel time of pulse measured by the observer in the spacecraft is $\underset{\_}{174\text{s}}$.

Explanation of Solution

The observer in the spacecraft moving towards the mirror will experience a contraction in length for the distance between the mirror and the spacecraft.

Write the expression for the contracted distance between the mirror and the spacecraft.

  $L=d\sqrt{1-{\left(\frac{v}{c}\right)}^2}$                                                                                                   (VIII)

Here, $L$ is the contracted distance between the mirror and spacecraft measured by the observer in the spacecraft.

Here both mirror and pulse is moving. The speed of travel of light pulse is $c$ and the speed of travel of mirror towards the spacecraft is $v$. The total distance travelled by the pulse to the mirror and the mirror towards the spacecraft is equal to $L$.

Write the expression for the distance travelled by pulse towards the mirror measured by the observer in spacecraft.

  $L_1=c\Delta t_1$                                                                                                                  (IX)

Here, $L_1$ is the distance travelled by pulse towards the mirror measured by the observer in spacecraft, and $\Delta t_1$ is the time travel of pulse towards mirror as measured by an observer in spacecraft.

Write the expression for the distance travelled by mirror towards the spacecraft measured by the observer in spacecraft.

  $L_2=v\Delta t_1$                                                                                                                   (X)

Here, $L_2$ is the distance travelled by mirror towards the spacecraft measured by the observer in spacecraft.

Write the expression for the total distance travelled by light and mirror.

  $L=L_1+L_2$                                                                                                               (XI)

Here, $L$ is the total distance travelled by the light and the mirror.

Use expressions (IX) and (X) in (XI).

  $L=c\Delta t_1+v\Delta t_1$                                                                                                      (XII)

Solve expression (XII) to find $\Delta t_1$.

  $\Delta t_1=\frac{L}{c+v}$                                                                                                           (XIII)

Write the expression for the distance between the mirror and spacecraft when the light strikes mirror.

  $L^{\prime }=L-v\Delta t_1$                                                                                                         (XIV)

Here, $L^{\prime }$ is the distance between mirror and spacecraft when the light strikes the mirror.

The same distance $L^{\prime }$ must be travelled back by the light from mirror to reach the spacecraft.

Write the expression for the distance travelled by light after reflection from the mirror as observed by the observer in spacecraft.

  $L^{\prime }=c\Delta t_2$                                                                                                                (XV)

Equate (XV) and (XIV) and solve for $\Delta t_2$.

  $\begin{array}{c}c\Delta t_2=L-v\Delta t_1\\ \Delta t_2=\frac{L}{c}-\frac{v}{c}\Delta t_1\end{array}$                                                                                                    (XVI)

Write the expression to find the total travel time of light.

  $T=\Delta t_1+\Delta t_2$                                                                                                      (XVII)

Here, $T$ is the total time of travel by the light.

Use expressions (XVI) and (XIII) in (XVII).

  $T=\frac{L}{c+v}+\frac{L}{c}-\frac{v}{c}\Delta t_1$                                                                                          (XVIII)

Use expression (XIII) in (XVIII).

  $\begin{array}{c}T=\frac{L}{c+v}+\frac{L}{c}-\frac{v}{c}\left(\frac{L}{c+v}\right)\\ =\frac{Lc+L\left(c+v\right)-Lv}{c\left(c+v\right)}\\ =\frac{2L}{\left(c+v\right)}\end{array}$                                                                                    (XIX)

Use expression (VIII) in (XIX).

  $\begin{array}{c}T=\frac{2}{\left(c+v\right)}d\sqrt{1-{\left(\frac{v}{c}\right)}^2}\\ =\frac{2}{\left(c+v\right)}d\frac{\sqrt{c^2-v^2}}{c}\\ =\frac{2d}{c}\sqrt{\frac{c-v}{c+v}}\end{array}$                                                                                          (XX)

Conclusion:

Substitute $5.66\times {10}^{10}\text{m}$ for $d$, $3.00\times {10}^8\text{m/s}$ for $c$, and $0.650c$ for $v$ in equation (XX) to find $T$.

  $\begin{array}{c}T=\frac{2\left(5.66\times {10}^{10}\text{m}\right)}{\left(3.00\times {10}^8\text{m/s}\right)}\sqrt{\frac{c-0.650c}{c+0.650c}}\\ =\frac{2\left(5.66\times {10}^{10}\text{m}\right)}{\left(3.00\times {10}^8\text{m/s}\right)}\sqrt{\frac{0.350}{1.650}}\\ =174\text{s}\end{array}$

Therefore, the total travel time of pulse measured by the observer in the spacecraft is $\underset{\_}{174\text{s}}$.