Video Transcript
Suppose the speed of light were only 3000 meters per second. A jet fighter moving toward a target on the ground at 800 meters per second shoots bullets, each having a muzzle velocity of 1000 meters per second. What are the bulletsβ velocity relative to the target?
In this problem statement weβre told to imagine that π the speed of light is 3000 meters per second and that a jet fighter is approaching a target on the ground at 800 meters per second, which weβll call π£ sub π, and that the jet fighter fires bullets which leave the jet fighter with a muzzle velocity of 1000 meters per second, which weβll call π£ sub π.
Knowing all this, we want to solve for the bulletsβ velocity relative to the target, what weβll call π£ sub π‘. We can begin by drawing a diagram of our scenario. In this scenario, we have our target, the box off to the right, and the jet fighter approaching at a speed π£ sub π and then firing bullets at the target which move relative to the jet with speed π£ sub π.
We want to know the velocity of the bullets relative to the target, π£ sub π‘. And remember, weβre imagining that the speed of light π is just 3000 meters per second. To solve for π£ sub π‘, weβll perform a velocity addition where our velocities are high enough that this addition is relativistic; that is, it takes into account the principles of relativity.
When we add two velocities in this equation, π’ and π’ prime relativistically, that means that instead of simply adding them like we would classically, we now have a denominator, one plus the product of those two velocities weβre adding divided by the speed of light squared.
If we apply this relationship to our particular scenario, then π£ sub π‘, the speed of the bullets relative to the target, equals π£ sub π plus π£ sub π divided by one plus the product of π£ sub π and π£ sub π divided by π squared. Since we know all three of these values, we can plug them in now.
We substitute 800 meters per second for π£ sub π, 1000 meters per second for π£ sub π, and 3000 meters per second for π. When we calculate π£ sub π‘, we find a result of 1.65 kilometers per second. This is how fast the bullets would approach the target under relativistic velocity addition. Notice that this is slower than if we had just added the velocities of the jet and the bullets classically.