Question

A plane trip lasts \(8.0 \mathrm{h} ;\) the average speed of the plane during the flight relative to Earth is \(220 \mathrm{m} / \mathrm{s}\). What is the time difference between an atomic clock on board the plane and one on the ground, assuming they were synchronized before the flight? (Ignore general relativistic complications due to gravity and the acceleration of the plane.)

Short answer

Answer: The time difference between the two atomic clocks is approximately 0.00767 seconds.

Step-by-step solution

Identify known values

In this problem, we are given the following values: - The average speed of the plane relative to Earth, \(v = 220\, m/s\) - The trip duration, \(t_0 = 8\, h\) We will use these values in the time dilation formula to find the time difference between the two clocks.

Convert trip duration to seconds

The time dilation formula uses seconds as its units for time, so we need to convert the trip duration from hours to seconds: \(t_0=8.0\,h \times 3600\,s/h = 28800\,s\)

Calculate the Lorentz factor

The Lorentz factor, \(\gamma\), can be found using the equation: \(\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\), where \(c\) is the speed of light \((3 \times 10^8\, m/s)\). Plug in the known values of \(v\) and \(c\): \(\gamma=\frac{1}{\sqrt{1-\frac{(220\,m/s)^2}{(3 \times 10^8\, m/s)^2}}}\) Calculate \(\gamma\): \(\gamma \approx 1.000000266\)

Calculate the time dilation

Now that we have the Lorentz factor, we can find the time dilation using the equation: \(t = \gamma t_0\) Plug in the known values of \(\gamma\) and \(t_0\): \(t = 1.000000266 \times 28800\,s\) Calculate \(t\): \(t \approx 28800.00767\,s\)

Calculate the time difference

Now that we have the time \(t\) in the moving frame (plane), we can find the time difference between the two clocks: \(\Delta t = t - t_0\) Calculate \(\Delta t\): \(\Delta t \approx 28800.00767\,s - 28800\,s\) \(\Delta t \approx 0.00767\,s\) So, the time difference between the two atomic clocks, one on the ground and the other on the plane, is approximately \(0.00767\,s\) after the 8-hour flight. Note that this value is rounded and in reality, the atomic clocks will have a much greater precision.