Answer to Question #107089 in Mechanics | Relativity for Vanessa Montoya

Question

A space explorer travels from planet α to planet µ at a speed of 0.37 c. When she is precisely halfway between
the planets, a distance of 1 light-week from each in the planets’ frame, nuclear devices are detonated on
each planet. The explosions are simultaneous in the frame of the planets. What is the difference in time of
arrival of the flashes from the explosions as observed by the space explorer?

Answer 1

The traveler will measure a time of

$$\Delta t=t\bigg(\Delta t+\frac{v\Delta x}{c^2}\bigg),$$

where

$$\Delta t=0,\\ \Delta x=1\text{ light week}\cdot c\\ t=\frac{1}{\sqrt{1-0.37^2}}=1.0764\text{ week}.$$

Therefore:

$$\Delta t'=1.0764\bigg(\frac{0.37c\cdot1\cdot c}{c^2}\bigg)=0.398\text{ week, or } 240850\text{ s}.$$