A spaceship is moving away from Earth with a constant velocity of \(0.80 c\) with respect to Earth. The spaceship and an Earth station synchronize their clocks, setting both to zero, at an instant when the ship is near Earth. By prearrangement, when the clock on Earth reaches a reading of $1.0 \times 10^{4} \mathrm{s}$, the Earth station sends out a light signal to the spaceship. (a) In the frame of reference of the Earth station, how far must the signal travel to reach the spaceship? (b) According to an Earth observer, what is the reading of the clock on Earth when the signal is received?

First, let's calculate the Lorentz factor (γ), which will be used later for time dilation and length contraction calculations. The Lorentz factor formula is: γ = \(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\) where v is the velocity of the spaceship relative to Earth, and c is the speed of light. Here, v = 0.80c, so using this value, the Lorentz factor (γ) can be calculated as follows: γ = \(\frac{1}{\sqrt{1-\frac{(0.80 c)^2}{c^2}}}\) γ = \(\frac{1}{\sqrt{1-0.64}}\) γ = \(\frac{1}{\sqrt{0.36}}\) γ = \(\frac{1}{0.6}\) γ = 1.66667

The spaceship receives the signal after time \(t_{spaceship}\), which can be calculated using time dilation formula: \(t_{spaceship}\) = \(\frac{t_{earth}}{γ}\) Now, it is given that the Earth station sends out a light signal when clock on Earth reaches a reading of 1.0 x \(10^4\)s, therefore, \(t_{earth}\) = 1.0 x \(10^4\)s. Substituting the values: \(t_{spaceship}\) = \(\frac{1.0 \times 10^4}{1.66667}\) \(t_{spaceship} \approx 6.0 \times 10^3s\) In Earth's frame of reference, the spaceship is moving at a constant velocity of 0.80c with respect to Earth. Thus, the distance traveled by the spaceship in Earth's frame of reference can be calculated as: \(distance_{earth}\) = \(v \times t_{spaceship}\) \(distance_{earth}\) = \((0.80c) \times (6.0 \times 10^3s)\) \(distance_{earth} \approx 1.44 \times 10^{12}m\) So, the light signal must travel approximately 1.44 x \(10^{12}\) meters in the Earth's frame of reference to reach the spaceship.

We now know the distance that the light signal must travel, and the speed of light is c. Using this information, we can find the total time it takes the signal to reach the spaceship as seen by the Earth observer. \(time_{signal}\) = \(\frac{distance_{earth}}{c}\) \(time_{signal}\) = \(\frac{1.44 \times 10^{12}m}{3 \times 10^8 \frac{m}{s}}\) \(time_{signal} \approx 4.8 \times 10^3s\) Now, adding the time it takes the Earth station to send out the signal and the time it takes for the signal to reach the spaceship, we find the Earth clock reading when the signal is received: \(reading_{earth}\) = \(time_{signal} + t_{earth}\) \(reading_{earth} \approx 4.8 \times 10^3s + 1.0 \times 10^4s\) \(reading_{earth} \approx 1.48 \times 10^4s\) Thus, according to the Earth observer, the clock on Earth reads approximately 1.48 x \(10^4\) seconds when the signal is received.