According to Anna, on Earth, Bob is on a spaceship moving at 0.8c towards Earth, and Carl, a little farther out, is on a spaceship moving a 0.9c towards Earth. According to Bob, how fast is Carl moving relative to Earth?
(a) According to Bob, how fast and in what direction is Carl moving relative to himself (Bob)?
--- I solved part a and got 0.357c and used u' = u-v divide by 1-uv/c^2.
But I confused about part b above.
Well, it seems like Bob is quite the people-watcher in space! Let's see if we can help him out.
According to Bob, Carl is on a spaceship moving at 0.9c towards Earth. Now, Bob himself is also moving towards Earth, but at a slightly slower speed of 0.8c. So, to determine Carl's speed relative to Earth according to Bob, we need to do a bit of calculation.
To find the relative velocity, we can use the relativistic velocity addition formula, which states:
v' = (v + u) / (1 + (vu / c^2))
Where v' is the relative velocity, v is Carl's velocity according to Bob (0.9c), u is Bob's velocity according to Earth (0.8c), and c is the speed of light.
Plugging in the values, we get:
v' = (0.9c + 0.8c) / (1 + ((0.9c * 0.8c) / c^2))
Simplifying, we have:
v' = (1.7c) / (1 + (0.72c^2 / c^2))
v' = 1.7c / (1 + 0.72)
v' = 1.7c / 1.72
v' ≈ 0.9884c
So according to Bob, Carl is moving at approximately 0.9884 times the speed of light towards Earth. That's quite a speedy fellow!
To solve part (a), you correctly used the relativistic velocity addition formula. However, part (b) is asking for the speed of Carl relative to Earth, according to Bob. In other words, if Bob were observing Carl's spaceship from Earth, how fast would he perceive Carl to be moving?
To solve this, we need to consider the concept of relative velocity in special relativity. To determine Carl's speed relative to Earth, as observed by Bob, we can use the relativistic velocity addition formula again.
According to Bob, Carl is already moving at a velocity of 0.9c towards Earth. So, we need to find the velocity addition between Carl's velocity relative to Bob (0.9c) and Bob's velocity relative to Earth (0.8c). This will give us Carl's velocity relative to Earth, as observed by Bob.
Using the relativistic velocity addition formula, we can substitute the values into the equation:
v' = (v + u) / (1 + (vu / c^2))
where:
v' = Carl's velocity relative to Earth, as observed by Bob
v = Bob's velocity relative to Earth (0.8c)
u = Carl's velocity relative to Bob (0.9c)
c = speed of light
Plugging in the values:
v' = (0.9c + 0.8c) / (1 + ((0.9c * 0.8c) / c^2))
= (1.7c) / (1 + (0.72 / 1))
= (1.7c) / (1.72)
≈ 0.988c
Therefore, according to Bob, Carl's spaceship is moving at a speed of approximately 0.988c relative to Earth.