Investigating a Rocket Launch Speed Anomaly

[quasar987]

Is there something wrong with this reasoning:

First the question:

"A rocket A leave Earth at a speed v at a time where the clocks onboard and those on Earth read 0. After a time T (for the Earth's clock), a second rocket is lauched with a speed u > v. The second rocket catches up with the first one at a time t such that (t - T)u = tv, or equivalently, t = uT/(u-v)."

a) For the clocks on A, when does B leave earth?

MY ANSWER: The time between the events "A leaves" and "B leaves" is proper on earth. It is T on earth; therfor it is given on A by

$$t_A_1= T\gamma _{AT} = \frac{T}{\sqrt{1-(v/c)^2}}$$

b) For the clocks on A, when does B catches up with A?

MY ANSWER: By Lorentz,

$$t_A_2 = \frac{1}{\sqrt{1-(v/c)^2}}(t-v(vt)/c^2) = t\sqrt{1-(v/c)^2}$$

d) Deduce the speed of B as seen in frame A. Does the result agree with the law of addition of speed?

MY ANSWER: Speed of B = distance traveled in catching up with A / time took to do so

$$\Rightarrow v_{B \ in \ A} = \frac{vt_A_2}{t_A_2 - t_A_1}=\frac{vt\sqrt{1-(v/c)^2}}{t\sqrt{1-(v/c)^2} - T/\sqrt{1-(v/c)^2}} = \frac{vt}{t - T/(1-(v/c)^2)}$$
$$=\frac{uv}{(u-v)(\frac{u}{u-v}-\frac{1}{1-(v/c)^2})}$$

But according to Mapple (and to me), this is not equal to the predicted

$$v_{B \ in \ A} = \frac{u-v}{1-uv/c^2}$$

 

[learningphysics]

MY ANSWER: Speed of B = distance traveled in catching up with A / time took to do so

$$\Rightarrow v_{B \ in \ A} = \frac{vt_A_2}{t_A_2 - t_A_1}=\frac{vt\sqrt{1-(v/c)^2}}{t\sqrt{1-(v/c)^2} - T/\sqrt{1-(v/c)^2}} = \frac{vt}{t - T/(1-(v/c)^2)}$$

It should be:$$\Rightarrow v_{B \ in \ A} = \frac{vt_A_1}{t_A_2 - t_A_1}$$

If you look at it from the frame of the rocket that left first... The other rocket leaves at t1. By that time the Earth has traveled vt1. So the distance between the two rockets is vt1.

 

[quasar987]

It should be:$$\Rightarrow v_{B \ in \ A} = \frac{vt_A_1}{t_A_2 - t_A_1}$$

If you look at it from the frame of the rocket that left first... The other rocket leaves at t1. By that time the Earth has traveled vt1. So the distance between the two rockets is vt1.

Well yea...

But the time at the denominator is the time taken by the second rocket to catch up with the first. So the distance at the numerator should be the distance the second rocket traveled in catching up with the first one. And that distance is not vt1, because the first rocket continued to move during the time rocket 2 was catching up with it.

However vt2 gives the distance traveled by rocket 1 from the moment of its lauching to the moment it is catched up by rocket 2. So this has to be the distance rocket 2 has traveled to catch up with it.

Makes sense?

 

[learningphysics]

Well yea...

But the time at the denominator is the time taken by the second rocket to catch up with the first. So the distance at the numerator should be the distance the second rocket traveled in catching up with the first one. And that distance is not vt1, because the first rocket continued to move during the time rocket 2 was catching up with it.

However vt2 gives the distance traveled by rocket 1 from the moment of its lauching to the moment it is catched up by rocket 2. So this has to be the distance rocket 2 has traveled to catch up with it.

Makes sense?

Not really.

Remember we are looking at this from rocket 1's frame of reference. So rocket 1 is at rest.

t2 and t1 are the times in the frame of rocket 1 right?

Remember that rocket 1 is at rest in rocket 1's frame of reference.

Look at this in rocket 1's frame of reference:
Rocket 2 is at a distance d from rocket 1, at time t1. Then Rocket 2 moves towards Rocket 1 till it reaches rocket 1. Rocket 2 reaches rocket 1 at time t2 in rocket 1's frame of reference.

What is the distance d? Isn't this just the distance between the Earth and rocket 1 at time t1?

 

[quasar987]

That makes a hell of a lot of sense now. You are a genius m8.

 

[learningphysics]

That makes a hell of a lot of sense now. You are a genius m8.

Thanks. Glad to help!