Question

A spaceship is moving past Earth at 0.99c. The spaceship fires two lasers. Laser A is in the same direction it is traveling, and Laser B is in the opposite direction. How fast will the light from each laser be traveling according to an observer on Earth?

Answer 1

Vx' = (Vx - u) / (1 - Vx *u / c^2) velocity transformation formula

In both cases we wish to measure the velocity in the frame of the earth which is moving at speed u = -.99 c relative to the spaceship

VA' = (c + .99c) / (1 - (-.99 c * c) / c^2) = 1.99c / 1.99 = c

VB' = (-c + .99c) / (1 - (-c * -.99c) / c^2) = .01 c / .01 = c

In both cases an observer on earth will observe the light traveling at speed c.