An astronaut travels at $0.8c$. If the time measured on their clock is 3 hours, how much time has passed on Earth?

💡 Explanation:

This question applies the concept of time dilation from Special Relativity. The time interval $(Delta t)$ in the stationary frame (Earth) is related to the proper time $(Delta t_0)$ in the moving frame (astronaut) by the formula: $Delta t = Delta t_0 / sqrt{1 - (v/c)^2}$. Given $Delta t_0 = 3$ hours and $v = 0.8c$, we calculate the Lorentz factor, $gamma = 1 / sqrt{1 - (0.8)^2}$: $gamma = 1 / sqrt{1 - 0.64} = 1 / sqrt{0.36} = 1 / 0.6 = 5/3 approx 1.667$. Therefore, $Delta t = 3 times (5/3) = 5$ hours. The observer on Earth measures a longer time interval for the event (time dilation).