### Video Transcript

Suppose a spaceship heading
directly away from Earth at 0.750π can shoot a canister at 0.500π 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?

Letβs start by highlighting some of
the important information given. Weβre told that relative to the
Earth. a spaceship is moving away from the Earth at 0.750π, where π is the speed
of light. And that the spaceship is capable
of firing a canister at 0.500π 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. Weβll call this speed π£ sub π‘ 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 itβs fired directly away from
Earth. Weβll call this π£ sub π.

Letβs draw a diagram of the Earth
and the spaceship. Relative to the Earth, the
spaceship moves away from the Earth at a speed that weβll call π£ sub π . And that speed is 0.750 times the
speed of light π. Weβre also told that the spaceship
is able to fire a canister off at a speed of 0.500π relative to the spaceship. Weβll call that π£ sub π. To move towards solving for π£ sub
π‘, the speed of the canister as it moves toward Earth, relative to the Earth, letβs
look into a relativistic velocity addition rule thatβs 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 weβll call π£
and if B fires off a projectile that moves away from B with speed π£ prime, then if
we call π’ the speed of the projectile relative to the observer at rest A, then π’
is equal to the speed of observer B, π£, plus π£ prime, the speed of the projectile
relative to B, all divided by one plus π£ times π£ prime divided by π squared. The plus signs in this equation
would change to minus signs if π£ prime, the projectileβs velocity, were opposite
the direction of π£.

So letβs apply this relativistic
velocity addition rule to our scenario. We want to solve for π£ sub π‘. Thatβs 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, π£ sub π , minus π£ sub π divided by one minus
the product π£ sub π times π£ sub π over π squared. Weβre using minus signs rather than
plus signs because π£ sub π and π£ sub π 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 π, the speed of light, cancels out, and that the
numerator simplifies to 0.250π. Plugging these values in on our
calculator, we find that π£ sub π‘ is equal to 0.400 times π. Thatβs 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 letβs 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 π£ sub π, our
equation is the same as it was before except that now instead of minus signs, we
have plus signs. And again thatβs because π£ sub π
and π£ sub π are now acting in the same direction. Once again, we insert our values
for these two velocities. We see the factors of π, the speed
of light, cancel in the denominator, and our numerator sums to 1.250π.

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 π. Notice that this speed is less than
the speed of light π even though π£ sub π plus π£ sub π is more than the speed of
light. This shows our relativistic
velocity addition rule in action.