Question

A spaceship is traveling away from Earth at \(0.87 c .\) The astronauts report home by radio every \(12 \mathrm{h}\) (by their own clocks). At what interval are the reports sent to Earth, according to Earth clocks?

Short answer

Based on the concept of time dilation in special relativity, the time interval between the astronauts' reports as observed by people on Earth is approximately 21.57 hours.

Step-by-step solution

Understand the concept of time dilation

Time dilation is a result of Albert Einstein's theory of special relativity. It states that the time experienced by an observer in a relatively moving reference frame will be different from the time experienced by a stationary observer in their own reference frame. The equation that relates the time intervals in both reference frames is given as: \(T_e = \dfrac{T_s}{\sqrt{1 - \dfrac{v^2}{c^2}}}\) Here, \(T_e\) - Time interval on Earth \(T_s\) - Time interval on the spaceship \(v\) - Velocity of the moving reference frame (spaceship) \(c\) - Speed of light

Plug in given values

We are given that the astronauts send their reports every 12 hours (by their own clock), and the spaceship is moving away from Earth at a speed of 0.87c (where 'c' is the speed of light). We can plug in these values into the equation for time dilation: \(T_e = \dfrac{12 \text{ h}}{\sqrt{1 - \dfrac{(0.87c)^2}{c^2}}}\)

Simplify the equation

Now we can simplify the equation by addressing the term inside the square root and removing the speed of light 'c' from the equation: \(T_e = \dfrac{12 \text{ h}}{\sqrt{1 - \dfrac{(0.87)^2}{1}}}\)

Solve for the time interval on Earth

Calculate the time interval on Earth by solving for \(T_e\): \(T_e = \dfrac{12 \text{ h}}{\sqrt{1 - (0.87)^2}}\) We find that \(T_e \approx 21.57 \text{ h}\) (rounded to two decimal places).

Conclusion

According to Earth clocks, the astronauts' reports are sent to Earth every 21.57 hours, which is time dilation's effect on the time interval between the reports.