Two atomic clocks are synchronized. One is put aboard a spaceship that leaves Earth at \(t=0\) at a speed of \(0.750 c .\) (a) When the spaceship has been traveling for \(48.0 \mathrm{h}\) (according to the atomic clock on board), it sends a radio signal back to Earth. When would the signal be received on Earth, according to the atomic clock on Earth? (b) When the Earth clock says that the spaceship has been gone for \(48.0 \mathrm{h},\) it sends a radio signal to the spaceship. At what time (according to the spaceship's clock) does the spaceship receive the signal?

The Lorentz factor is given by the formula: \(\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}\), where \(v\) is the speed of the spaceship and \(c\) is the speed of light. In this case, \(v = 0.750c\). Let's calculate the Lorentz factor: \(\gamma = \frac{1}{\sqrt{1 - (0.750c)^2/c^2}} = \frac{1}{\sqrt{1 - 0.5625}} = \frac{1}{\sqrt{0.4375}} \approx 1.5118\).

According to the spaceship's clock, it sends a signal to Earth after 48 hours (t' = 48 h). To find when the signal is received on Earth (t), first, we must find the dilated time using the formula \(t = t' \times \gamma\). Then, we determine the time the radio signal takes to travel from the spaceship to Earth by dividing the dilated distance traveled by the spaceship by the speed of light. t = 48 h × 1.5118 = 72.5664 h Since the spaceship is constantly moving away from Earth, the signal will take a longer time to travel back to Earth. As the spaceship is moving at 0.750c, the time taken by the signal to travel back to the Earth is: t_signal = \(\frac{72.5664 h \times 0.750c}{c} = 54.4248 h\) So, the signal will be received on Earth at: t_total = t + t_signal = 72.5664 h + 54.4248 h = 126.99 h

When the Earth's clock says the spaceship has been gone for 48 hours, it sends a signal to the spaceship. First, let's find the spaceship's position at that time. Then, let's find the time taken by the signal to reach the spaceship and add this to the 48-hour mark. Position of the spaceship after 48 hours (d) is given by: d = vt = 0.750c × 48 h = 36c h The time taken for the signal to travel to the spaceship is: t_signal_to_spaceship = \(\frac{d}{c - v} = \frac{36c h}{c - 0.750c} = \frac{36c h}{0.250c} = 144 h\) Now, let's find the time in the spaceship's frame using the time dilation formula: t'_spaceship = \(\frac{t}{\gamma} = \frac{144 h}{1.5118} \approx 95.2 h\) Thus, the spaceship will receive the signal 95.2 hours after its departure, according to the spaceship's clock.