Chapter 26, Problem 67P

(a)

To determine

The speed of the ship travelling according to an Earth observer.

Answer to Problem 67P

The speed of the ship travelling according to an Earth observer is $\underset{\_}{0.625c}$.

Explanation of Solution

The time elapse between the two events is $50.0\text{ns}$, the length of the ship is $12.0\text{m}$.

Let us use the symbols with primes for the astronaut’s frame and symbols without primes for the Earth’s frame. Earth measures the proper time between the events since they occur at the same location in that frame so that $\Delta t_0=\Delta t$ which is $50.0\text{ns}$. The astronaut’s measures the proper length of the ship since it is at rest in that frame so that $L^{\prime }=L_0$ which is $12.0\text{m}$.

Write the expression for the contracted length in the Earth’s frame.

    $L=\frac{L_0}{\gamma }$                                                                                                                       (I)

Here, $L$ is the contracted length in the Earth frame, $L_0$ is the proper length of the ship, $\gamma$ is the Lorentz factor.

Write the expression for the speed of the ship measured in the Earth’s frame.

    $v=\frac{L}{\Delta t}$                                                                                                                        (II)

Here, $v$ is the speed of the ship measured in the Earth’s frame, $\Delta t$ is the change in time.

Use equation (I) in equation (II).

    $v=\frac{L_0}{\gamma \Delta t}$                                                                                                                   (III)

Rearrange the above equation,

    $v\gamma =\frac{L_0}{\Delta t}$                                                                                                                  (IV)

Write the expression for the Lorentz factor.

    $\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$                                                                                                        (V)

Here, $v$ is the speed of the spaceship, $c$ is the speed of light.

Conclusion:

Substitute $12.0\text{m}$ for $L_0$, and $50.0\text{ns}$ for $\Delta t$ in equation (IV).

    $\begin{array}{c}v\gamma =\frac{12.0\text{m}/\text{s}}{50.0\text{ns}}\\ =\frac{12.0\text{m}/\text{s}}{50.0\text{ns}\times \frac{1\text{s}}{1\times {10}^9\text{ns}}}\\ =2.40\times {10}^8\text{m}/\text{s}\times \frac{1c}{3\times {10}^8\text{m}/\text{s}}\\ =0.800c\end{array}$                                                                            (VI)

Use equation (V) and (VI) in equation (IV)and solve for to $v$.

    $\begin{array}{c}v\left(\frac{1}{\sqrt{1-v^2/c^2}}\right)=0.800c\\ v^2={\left(0.800c\right)}^2\left(1-v^2/c^2\right)\\ v=\sqrt{\frac{0.640}{1.640}}c\\ =0.625c\end{array}$

Therefore, the speed of the ship travelling according to an Earth observer is $\underset{\_}{0.625c}$.

(b)

To determine

The elapsed time between light flashes in the astronaut’s frame of reference.

Answer to Problem 67P

The elapsed time between light flashes in the astronaut’s frame of reference is $\underset{\_}{64.0\text{ns}}$.

Explanation of Solution

The clock measures the proper time interval $\Delta t_0$ in the reference frame in which the clock is at rest. In any other frame, the time interval is longer.

Write the expression for the time dilation.

    $\Delta t^{\prime }=\gamma \Delta t_0$                                                                                                              (VII)

Here, $\Delta t^{\prime }$ is the time elapsed in the astronauts frame.

Use equation (V) in equation (VII),

    $\Delta t^{\prime }=\frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$                                                                                                       (VIII)

Conclusion:

Substitute $50.0\text{ns}$ for $\Delta t$ and $0.625c$ for $v$ in equation (VIII) to find $\Delta t^{\prime }$.

    $\begin{array}{c}\Delta t^{\prime }=\frac{50.0\text{ns}}{\sqrt{1-\frac{{\left(0.625c\right)}^2}{c^2}}}\\ =64.0\text{ns}\end{array}$

Therefore, the elapsed time between light flashes in the astronaut’s frame of reference is $\underset{\_}{64.0\text{ns}}$.