Question

Imagine you shine a laser beam up into the night sky at t = 0. At exactly the same time (t = 0), imagine a spaceship departs the earth heading in the same direction as the laser beam, moving at 0.9c. (That is, the ship is traveling at 90% of the speed of light - you may assume that the ship reaches this speed instantaneously). At t = 1 s, you see the leading edge of the laser beam is already 3 x 10^8 m from Earth, as expected. How far ahead of the ship is the leading edge of the laser beam at this time?

According to the theory of special relativity, the speed of light should be the same (3 x 10^8 m/s) in all reference frames, including that of the speeding ship. How can you reconcile your answer to the previous question with this idea?

Answer 1

1s of time is not 1s time with respect to spaceship.

according to time dialation equation, the equivalent of 1 s with resoect to spaceship,

t' = t*sqrt(1 - (v/c)^2)

= 1*sqrt(1 - (0.9*c/c)^2)

= 1*sqrt(1 - 0.9^2)

= 0.43589 s

distance travelled by light during this time = 3*10^8*0.43589

= 1.308*10^8 m

distance travelled by spaceship during this time = 0.9*3*10^8*0.43589

= 1.177*10^8 m

so, the leading edge of the laser with respect to spaceship = 1.308*10^8 - 1.177*10^8

= 1.31*10^7 m