B Velocity Addition Q: Ball Thrown from Moving Ship?

[NoahsArk]

Suppose a spaceship is flying past me, an observer on earth, at .86c. A traveler on the ship has a ball which is attached to a string which is 100 feet long in the ship frame. The traveler grabs the string and throws the ball out of the front window of the ship at a speed of 100 feet per second in the ship frame. Are the following calculations correct?:

1. In my frame of reference the ball is gaining distance on the ship at 50 feet per second.
2. In my frame of reference, when the string is fully extended, it is 50 feet long.

 

[Ibix]

1. In my frame of reference the speed of the ball is 50 feet per second.

The ship is doing 0.86c. The ball is going faster than that. 50 ft/s is unlikely.

2. In my frame of reference, when the string is fully extended, it is 50 feet long.

To a decent approximation, yes. From a nitpicking point of view, some care is probably needed because the string is likely to stretch and mechanical effects like that are not instantaneous - so you might need to think carefully about exactly what you mean by "when" the string is fully extended.

 

[Matterwave]

1. In my frame of reference the ball is gaining distance on the ship at 50 feet per second.

For this part of the question, use the relativistic velocity addition formula. However, 100ft/s is so small compared to 0.86c that the answer would essentially be 0.86c up to rounding.

 

[NoahsArk]

Sorry I didn't ask the first question clearly:

I didn't mean the ball's total speed in my frame of reference, I just meant how fast the ball is going with respect to the ship in my frame.

 

[FactChecker]

You see the ball being thrown at a $velocity = \Delta distance/ \Delta time$ where the traveler's measurement of both $\Delta distance$ and $\Delta time$ are distorted (in your view) by the Lorentz factor 1.96. So you see a shorter $\Delta distance$ and a longer $\Delta time$ by the Lorentz factor. The ball's speed wrt the ship should be 25 (more precisely 26.04).

EDIT: This is still missing a factor. See post #10 by @PAllen below.

 

[NoahsArk]

With a speed of .86c for the ship the Lorenz factor, I thought, was 2, which is why I thought the length of the string would be half from my perspective of what the traveler measures. That’s also why I thought the speed with respect to the ship would be half the speed than what the traveler measures.

Where did you get the speed of 25 feet per second since that is only 1/4 what the traveler measures?

Thank you.

 

[FactChecker]

Because both length and time are distorted. $(\Delta distance /\gamma(v)) / (\Delta time * \gamma(v))$

 

[NoahsArk]

I see thanks. So it would be 25 because it's half the length (50ft) divided by twice the time 2min = 25 ft/s.

 

[FactChecker]

I see thanks. So it would be 25 because it's half the length (50ft) divided by twice the time 2min = 25 ft/s.

Yes. I really should have taken the time to use better notation to indicate the ship frame measurement of distance and time.

 

[PAllen]

Simple algebra from the velocity addition formula shows that:

If u is velocity of rocket in observer rest frame, and v is velocity of ball in rocket rest frame, and if v is very small compared to u, then the difference in speed between the ball and the rocket in observer rest frame is v/γ(u)², consistent with what @FactChecker says based on a simple argument. However, to be exact, this result would be divided by (1 + uv/c²)

 

[FactChecker]

Simple algebra from the velocity addition formula shows that:

If u is velocity of rocket in observer rest frame, and v is velocity of ball in rocket rest frame, and if v is very small compared to u, then the difference in speed between the ball and the rocket in observer rest frame is v/γ(u)², consistent with what @FactChecker says based on a simple argument. However, to be exact, this result would be divided by (1 + uv/c²)

I stand corrected. I oversimplified the situation. I guess there are more complications of the relativity of simultaneity to consider. So one has to use the velocity addition formula. I guess I was close to the correct answer only because I was lucky that $u/c$ is so small in this example (as others have pointed out).