Comparing the Speed of Light towards and away from an Observer: Relativistic Package Delivery

Because the canister and the spaceship are moving at relativistic speeds, we must determine the speed of the canister by an Earth-bound observer using relativistic velocity addition instead of simple velocity addition.

1. Identify the knowns. $\text{v=}0.500c$; $u\prime =0\text{.}\text{750}c$
2. Identify the unknown. $u$
3. Choose the appropriate equation.
4. Plug the knowns into the equation.
   $$\begin{array}{ll}u& =& \frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{c^2}}\\ & =& \frac{0.500\text{c +}0.750c}{1+\frac{(0.500c)(0.750c)}{c^2}}\\ & =& \frac{1.250c}{1+0.375}\\ & =& 0.909c\end{array}$$

   28.34

1. Identify the knowns. $\text{v}=0.500c$; $u\prime =-0.750c$
2. Identify the unknown. $u$
3. Choose the appropriate equation. $\text{u}=\frac{\mathrm{v+u}\prime }{1+\frac{v\text{u}\prime }{c^2}}$
4. Plug the knowns into the equation.
   28.35

The minus sign indicates velocity away from the Earth (in the opposite direction from $v$), which means the canister is heading towards the Earth in part (a) and away in part (b), as expected. But relativistic velocities do not add as simply as they do classically. In part (a), the canister does approach the Earth faster, but not at the simple sum of $1.250c$. The total velocity is less than you would get classically. And in part (b), the canister moves away from the Earth at a velocity of $-0.400c$, which is faster than the $\mathrm{-0.250}c$ you would expect classically. The velocities are not even symmetric. In part (a) the canister moves $0.409c$ faster than the ship relative to the Earth, whereas in part (b) it moves $0.900c$ slower than the ship.