A rocket ship leaves earth at a speed of $\frac{\mathbf{3}}{\mathbf{5}}\mathbf{c}$. When a clock on the rocket says has elapsed, the rocket ship sends a light signal back to earth.

(a) According to earth clocks, when was the signal sent?

(b) According to earth clocks, how long after the rocket left did the signal arrive back on earth?

(c) According to the rocket observer, how long after the rocket left did the signal arrive back on earth?

Given data:

The speed of a rocket is$v=\frac{3}{5}c$ .

The elapsed time is t=1hr .

a)

As the rocket clock runs slow, write the expression for the reading of the earth clock.

$t^{\text{'}}=\gamma t$

${\mathit{t}}^{\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}}$ …… (1)

Substitute the value of in the above expression.

$$\begin{gathered} t^{,}=\frac{1}{\sqrt{1-{\displaystyle \frac{{\left({\displaystyle \frac{3}{5}}c\right)}^2}{c^2}}}}\left(1hr\right) \\ {t}^{,}=\frac{1}{\sqrt{1-{\displaystyle \frac{9}{25}}}} \\ {t}^{,}=\frac{1}{\sqrt{{\displaystyle \frac{16}{25}}}} \end{gathered}$$

On further solving,

$$\begin{gathered} t^{,}=\sqrt{\frac{25}{16}} \\ {t}^{,}=\frac{5}{4}hr \\ {t}^{,}=1\mathrm{hr}15\min \end{gathered}$$

Therefore, the time of sending a signal after take-off is 1hr 15 min.

(b)

Write the relation between distance, speed, and time.

$d=v{t}^{\text{'}}$

Substitute the value of vand$t^{\text{'}}$in the above expression.

$$\begin{gathered} d=\left(\frac{3}{5}c\right)\times \left(\frac{5}{4}hr\right) \\ d=\frac{3}{4}\mathrm{light}\mathrm{hours} \end{gathered}$$

As the rocket already travels 1 hr 15 min after departure, the light signal reaches the earth 2hrs after take-off.

Therefore, the time (according to earth clocks) taken by the light signal to reach the earth is 2hrs .

(c)

Substitute the value of the time (rocket) taken by a light signal reaches the earth, i.e.,t=2hrs in equation (1).

$$\begin{gathered} t^{\text{'}}=\frac{1}{\sqrt{1-{\displaystyle \frac{\left({\displaystyle \frac{3}{5}}c\right)}{c^2}}}}\left(2hrs\right) \\ {t}^{\text{'}}=\frac{1}{\sqrt{{\displaystyle \frac{25-9}{25}}}}\left(2hrs\right) \end{gathered}$$

On further solving,

$$\begin{gathered} t^{\text{'}}=\sqrt{\frac{25}{16}}\times \left(2\mathrm{hrs}\right) \\ {t}^{\text{'}}=\frac{5}{4}\times \left(2\mathrm{hrs}\right) \\ {t}^{\text{'}}=\frac{5}{2}\mathrm{hrs} \\ {\mathrm{t}}^{\text{'}}=2\mathrm{hrs}30\min \end{gathered}$$

Therefore, the time (according to the rocket observer) taken by the light signal to reach the earth is 2hrs 30 min .