Suppose a spaceship heading directly away from Earth at 0.750𝑐 can shoot a
canister at 0.500𝑐 relative to the ship. What is the velocity of the canister relative to
Earth if it is shot directly at Earth? What is the velocity of the canister relative to Earth
if it is shot directly away from Earth?
Let’s start by highlighting some of the important information given. We’re told that relative to the Earth. a spaceship is moving away from the Earth
at 0.750𝑐, where 𝑐 is the speed of light. And that the spaceship is capable of firing a canister at 0.500𝑐 in any
direction relative to the ship.
In part one, we want to solve for the velocity of the canister relative to Earth
if it is fired directly at Earth. We’ll call this speed 𝑣 sub 𝑡 for the speed of the canister as it moves toward
Earth. Then we want to solve for the velocity of the canister relative to Earth if it’s
fired directly away from Earth. We’ll call this 𝑣 sub a.
Let’s draw a diagram of the Earth and the spaceship. Relative to the Earth, the spaceship moves away from the Earth at a speed that
we’ll call 𝑣 sub 𝑠. And that speed is 0.750 times the speed of light 𝑐. We’re also told that the spaceship is able to fire a canister off at a speed of
0.500𝑐 relative to the spaceship. We’ll call that 𝑣 sub 𝑐. To move towards solving for 𝑣 sub 𝑡, the speed of the canister as it moves
toward Earth, relative to the Earth, let’s look into a relativistic velocity addition rule that’s sometimes called
Einstein velocity addition.
If we have two observers A and B where A is considered at rest and B moves
relative to A with a speed we’ll call 𝑣 and if B fires off a projectile that moves away from B with speed 𝑣 prime, then if we call 𝑢 the speed of the projectile relative to the observer at rest
A, then 𝑢 is equal to the speed of observer B, 𝑣, plus 𝑣 prime, the speed of the
projectile relative to B, all divided by one plus 𝑣 times 𝑣 prime divided by 𝑐 squared. The plus signs in this equation would change to minus signs if 𝑣 prime, the
projectile’s velocity, were opposite the direction of 𝑣.
So let’s apply this relativistic velocity addition rule to our scenario. We want to solve for 𝑣 sub 𝑡. That’s the velocity of the projectile relative to Earth when the projectile, in our case a canister, is moving directly toward the Earth. That velocity is equal to the speed of the spaceship relative to Earth, 𝑣 sub 𝑠,
minus 𝑣 sub 𝑐 divided by one minus the product 𝑣 sub 𝑠 times 𝑣 sub 𝑐 over 𝑐 squared. We’re using minus signs rather than plus signs because 𝑣 sub 𝑠 and 𝑣 sub 𝑐 are
in opposite directions from one another. These two speeds are given to us in our problem statement, so we can insert those
values into this relationship.
When we look in the denominator of the equation, we see that 𝑐, the speed of
light, cancels out, and that the numerator simplifies to 0.250𝑐. Plugging these values in on our calculator, we find that 𝑣 sub 𝑡 is equal to
0.400 times 𝑐. That’s how fast the canister appears to be moving relative to an observer on
Earth. Notice that our velocity is positive which means that for an observer on Earth,
the canister is still moving away from the Earth.
Now let’s move on to the next part where now the canister is moving away from the Earth in the same direction as
the spaceship. When we apply the relativistic velocity addition rule to this different scenario, solving for 𝑣 sub a, our equation is the same as it was before except that now instead of minus signs, we have plus signs. And again that’s because 𝑣 sub 𝑐 and 𝑣 sub 𝑠 are now acting in the same direction. Once again, we insert our values for these two velocities. We see the factors of 𝑐, the speed of light, cancel in the denominator, and our numerator sums to 1.250𝑐.
When we compute this speed, the speed of the canister measured relative to an
observer on Earth when they canister moves away from the Earth, we find a speed of 0.909 times 𝑐. Notice that this speed is less than the speed of light 𝑐 even though 𝑣 sub 𝑠
plus 𝑣 sub 𝑐 is more than the speed of light. This shows our relativistic velocity addition rule in action.