Example 28.3 from College Physics, 28.4 Relativistic Addition of Velocities

Showing that the Speed of Light towards an Observer is Constant (in a Vacuum): The Speed of Light is the Speed of Light

Suppose a spaceship heading directly towards the Earth at half the speed of light sends a signal to us on a laser-produced beam of light. Given that the light leaves the ship at speed $c$ as observed from the ship, calculate the speed at which it approaches the Earth.

A spacecraft is heading towards earth v equals zero point five zero zero times c. A laser beam from the ship travels towards the Earth with velocity c as shown by a vector. A second spaceship traveling away from the Earth. The velocity of the second ship and second laser are the same as the first, but in the opposite direction.

Figure 28.16

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Because the light and the spaceship are moving at relativistic speeds, we cannot use simple velocity addition. Instead, we can determine the speed at which the light approaches the Earth using relativistic velocity addition.
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Solution
1. Identify the knowns. $v= 0.500 c$ ; $u' = c$
2. Identify the unknown. $u$
3. Choose the appropriate equation. $u = \frac{v+u'}{1 + \frac{v u'}{c^{2}}}$
4. Plug the knowns into the equation.
  $\begin{array}{l} u & = & \frac{v+u'}{1 + \frac{v u'}{c^{2}}} \\ = & \frac{0.500 c + c}{1 + \frac{(0.500 c) (c)}{c^{2}}} \\ = & \frac{(0.500 + 1) c}{1 + \frac{0.500 c^{2}}{c^{2}}} \\ = & \frac{1.500 c}{1 + 0.500} \\ = & \frac{1.500 c}{1.500} \\ = & c \end{array}$
  28.33
Discussion
Relativistic velocity addition gives the correct result. Light leaves the ship at speed $c$ and approaches the Earth at speed $c$ . The speed of light is independent of the relative motion of source and observer, whether the observer is on the ship or Earth-bound.

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