Chapter 4, Problem 4.6P

(a)

To determine

The time measured by the clock on the earth.

Answer to Problem 4.6P

The time measured by the clock on the earth is $5\text{yr}$.

Explanation of Solution

Write the expression for time measured by the clock on the earth.

    $\Delta t=\frac{d}{u}$        (1)

Here, $d$ is the distance to $\alpha$ centauri, $u$ is the speed of the star ship.

Conclusion:

Substitute $4\text{ly}$ for $d$, $0.8c$ for $u$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (1).

    $\begin{array}{c}\Delta t=\frac{4\text{ly}}{0.8\left(3\times {10}^8\text{m}/\text{s}\right)}\\ =\frac{\left(4\text{ly}\times \frac{9.5\times {10}^{15}\text{m}}{1\text{ly}}\right)}{0.8\left(3\times {10}^8\text{m}/\text{s}\right)}\\ =15.83\times {10}^7\text{s}\times \frac{1\text{yr}}{3.14\times {10}^7\text{s}}\\ =5\text{yr}\end{array}$

Thus, the time measured by the clock on the earth is $5\text{yr}$.

(b)

To determine

The rest time measured by the starship pilot.

Answer to Problem 4.6P

The rest time measured by the starship pilot is $3\text{yr}$.

Explanation of Solution

Write the expression for rest time measured by the starship pilot.

    $\Delta t_{\text{rest}}=\Delta t_{\text{moving}}\sqrt{1-\frac{u^2}{c^2}}$        (2)

Here, $\Delta t_{\text{moving}}$ is the time measured by the clock on the earth, $u$ is the speed of the star ship, $c$ is the speed of the light.

Conclusion:

Substitute $5\text{yr}$ for $\Delta t_{\text{moving}}$, $0.8c$ for $u$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (2).

    $\begin{array}{c}\Delta t_{\text{rest}}=\left(5\text{yr}\right)\sqrt{1-{\left(\frac{0.8\left(3\times {10}^8\text{m}/\text{s}\right)}{3\times {10}^8\text{m}/\text{s}}\right)}^2}\\ =\left(5\text{yr}\right)\left(0.6\right)\\ =3\text{yr}\end{array}$

Thus, the rest time measured by the starship pilot is $3\text{yr}$.

(c)

To determine

The distance between the earth and $\alpha$ centauri measured by the starship pilot.

Answer to Problem 4.6P

The distance between the earth and $\alpha$ centauri measured by the starship pilot is $2.4\text{ly}$.

Explanation of Solution

Write the expression for distance between the earth and $\alpha$ centauri measured by the starship pilot.

    $L_{\text{moving}}=L_{\text{rest}}\sqrt{1-\frac{u^2}{c^2}}$        (3)

Here, $L_{\text{moving}}$ is the distance between the earth and $\alpha$ centauri measured by the starship pilot, $L_{\text{rest}}$ is the distance measured from earth.

Conclusion:

Substitute $4\text{ly}$ for $L_{\text{rest}}$, $0.8c$ for $u$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (3).

    $\begin{array}{c}{L}_{\text{moving}}=\left(4\text{ly}\right)\sqrt{1-{\left(\frac{0.8\left(3\times {10}^8\text{m}/\text{s}\right)}{3\times {10}^8\text{m}/\text{s}}\right)}^2}\\ =\left(4\text{ly}\right)\left(0.6\right)\\ =2.4\text{ly}\end{array}$

Thus, the distance between the earth and $\alpha$ centauri measured by the starship pilot is $2.4\text{ly}$.

(d)

To determine

The time interval between the reception of one of the radio signal and the reception of next signal aboard the starship.

Answer to Problem 4.6P

The time interval between the reception of one of the radio signal and the reception of next signal aboard the starship is $18\text{months}$.

Explanation of Solution

Write the expression for time interval between the arrival of two light signal.

    $\Delta t_{star}=\frac{\Delta t_{\text{rest}}}{\sqrt{1-\frac{u^2}{c^2}}}\left(1+\frac{u}{c}\cos \theta \right)$        (4)

Here, $\theta$ is the angle between the two signals.

Conclusion:

Substitute $6\text{months}$ for $\Delta t_{\text{rest}}$, $0°$ for $\theta$ $0.8c$ for $u$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (4).

    $\begin{array}{c}\Delta t=\frac{\left(6\text{months}\right)}{\sqrt{1-{\left(\frac{0.8\left(3\times {10}^8\text{m}/\text{s}\right)}{3\times {10}^8\text{m}/\text{s}}\right)}^2}}\left(1+\frac{0.8\left(3\times {10}^8\text{m}/\text{s}\right)}{3\times {10}^8\text{m}/\text{s}}\left(\cos 0\right)\right)\\ =\frac{\left(6\text{months}\right)}{0.6}\left(1.8\right)\\ =18\text{months}\end{array}$

Thus, the time interval between the reception of one of the radio signal and the reception of next signal aboard the starship is $18\text{months}$.

(e)

To determine

The time interval between the reception of one of the radio signal and the reception of next signal on the earth.

Answer to Problem 4.6P

The time interval between the reception of one of the radio signal and the reception of next signal on the earth is $20\text{months}$.

Explanation of Solution

Write the expression for time interval between the reception of one of the radio signal and the reception of next signal on the earth.

    $\Delta t_{\text{earth}}=\frac{\Delta t_{\text{clock}}}{\left(1+\frac{u}{c}\cos \theta \right)}\sqrt{1-\frac{u^2}{c^2}}$        (5)

Conclusion:

Substitute $6\text{months}$ for $\Delta t_{\text{rest}}$, $0$ for $\theta$ $0.8c$ for $u$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (5).

    $\begin{array}{c}\Delta t_{\text{earth}}=\frac{\left(6\text{months}\right)}{\left(1+\frac{0.8\left(3\times {10}^8\text{m}/\text{s}\right)}{3\times {10}^8\text{m}/\text{s}}\left(\cos 0\right)\right)}\sqrt{1-{\left(\frac{0.8\left(3\times {10}^8\text{m}/\text{s}\right)}{3\times {10}^8\text{m}/\text{s}}\right)}^2}\\ =\frac{\left(6\text{months}\right)}{1.8}\left(0.6\right)\\ =20\text{months}\end{array}$

Thus, the time interval between the reception of one of the radio signal and the reception of next signal on the earth is $20\text{months}$.

(f)

To determine

The wavelength of the starship’s receiver.

Answer to Problem 4.6P

The wavelength of the starship’s receiver is $45\text{cm}$.

Explanation of Solution

Write the expression for wavelength of the starship’s receiver.

    ${\lambda }_{\text{star}}={\lambda }_{\text{earth}}\sqrt{\frac{1+\frac{u}{c}}{1-\frac{u}{c}}}$        (6)

Here, ${\lambda }_{\text{earth}}$ is the wavelength of the signal on the earth.

Conclusion:

Substitute $15\text{cm}$ for ${\lambda }_{\text{earth}}$, $0.8c$ for $u$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (6).

    $\begin{array}{c}{\lambda }_{\text{star}}=\left(15\text{cm}\right)\sqrt{\frac{1+\left(\frac{0.8\left(3\times {10}^8\text{m}/\text{s}\right)}{3\times {10}^8\text{m}/\text{s}}\right)}{1-\left(\frac{0.8\left(3\times {10}^8\text{m}/\text{s}\right)}{3\times {10}^8\text{m}/\text{s}}\right)}}\\ =\left(15\text{cm}\right)\left(3\right)\\ =45\text{cm}\end{array}$

Thus, the wavelength of the starship’s receiver is $45\text{cm}$.