Question

Suppose Swarglok (heaven) is in constant motion at a speed of 0.9999c with respect to the earth. According to the earth's frame, how much time passes on the earth before one day passes on Swarglok?

Answer 1

Given:
Speed of Swarglok, v = 0.9999c
Proper time interval, ∆t = One day on Swarglok

Suppose ∆t' days pass on Earth before one day passes on Swarglok.
Now,

$$\begin{gathered} ∆t\text{'}=\frac{∆t}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ =\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(0.9999\right)}^2{c}^2}{c^2}}}} \\ =\frac{1}{0.014141782}=70.712\mathrm{days} \end{gathered}$$

Thus,
∆t' = 70.7 days