Question

Inside a spacecraft travelling at a speed of 0.8c with respect to the earth, the time
interval between two events on the earth is measured to be 30 hrs. What would the
time interval be if the spacecraft was travelling at a speed of 0.9c.

Answer 1

Time intervals in the reference frame moving relative to each other are connected by the Lorentz transformation

(1) $\Delta t'= \Delta t\cdot \sqrt{1-\beta^2}$ , where $\beta=\frac{V}{c}$ . In first case the time interval on Earth is

(2) $\Delta t_E=\frac{\Delta t_1}{\sqrt{1-\beta_1^2}}$ , where $\Delta t_1=30hrs$ , and $\beta_1=0.8$ . Since the measurement of the time interval was made by the clock in the rocket, relative to which the Earth moved at a speed 0.8c. This interval measured in a rocket moving at a speed of 0.9 s will be

(3) $\Delta t_2=\Delta t_E\cdot \sqrt{1-\beta_2^2}=\Delta t_1\frac{\sqrt{1-\beta_2^2}}{\sqrt{1-\beta_1^2}}=30\frac{\sqrt{1-0.9^2}}{\sqrt{1-0.8^2}}=21.8hrs$

Answer: The time interval in the spacecraft travelling at a speed of 0.9c is 21.8hrs.