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Calculation of the velocity of a Spaceship moving Relativistically

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Owen-
#1
Feb18-13, 09:35 AM
P: 40
1. The problem statement, all variables and given/known data

A spaceship travel from Earth [itex]\alpha[/itex]-Centauri (4.3 light years away) at a constant velocity. If the time elapsed onboard during the journey is 4.e years, what is the speed of the spaceship?

The spaceship is powered by a drive which works by ejecting protons behind the ship at a velocity relative to the ship of c/2. At what velocity would a person on earth observer the protons to be travelling, just as the spaceship reached constant velocity as it passed Earth on its way to [itex]\alpha[/itex]-Centauri

2. Relevant equations

[itex]\gamma=(1-\beta^2)^{-1/2}[/itex]
[itex]\beta = \frac{v}{c}[/itex]
[itex]x'=\gamma(x-vt)[/itex]
[itex]t'=\gamma(t-\frac{x \times v^2}{c^2})[/itex]


3. The attempt at a solution
[itex]x'=\gamma(4.3-vt)[/itex]
[itex]4.3=\gamma(t-\frac{x \times v^2}{c^2})[/itex]

Well essentially I couldn't see any way to cancel things with simultaneous equations, and without one other variable known I assumed it to be impossible, unless they want the answer to be in terms of something. Is this correct? Any help, even a hint in the right direction would be much appreciated. :)

Thanks in advance,
Owen.
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Owen-
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Feb18-13, 09:36 AM
P: 40
Any way to fix the typo in the title? :S
vela
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Feb18-13, 11:28 AM
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Quote Quote by Owen- View Post
1. The problem statement, all variables and given/known data

A spaceship travel from Earth [itex]\alpha[/itex]-Centauri (4.3 light years away) at a constant velocity. If the time elapsed onboard during the journey is 4.e years, what is the speed of the spaceship?
What does 4.e mean?

The spaceship is powered by a drive which works by ejecting protons behind the ship at a velocity relative to the ship of c/2. At what velocity would a person on earth observer the protons to be travelling, just as the spaceship reached constant velocity as it passed Earth on its way to [itex]\alpha[/itex]-Centauri

2. Relevant equations

[itex]\gamma=(1-\beta^2)^{-1/2}[/itex]
[itex]\beta = \frac{v}{c}[/itex]
[itex]x'=\gamma(x-vt)[/itex]
[itex]t'=\gamma(t-\frac{x \times v^2}{c^2})[/itex]


3. The attempt at a solution
[itex]x'=\gamma(4.3-vt)[/itex]
[itex]4.3=\gamma(t-\frac{x \times v^2}{c^2})[/itex]

Well essentially I couldn't see any way to cancel things with simultaneous equations, and without one other variable known I assumed it to be impossible, unless they want the answer to be in terms of something. Is this correct? Any help, even a hint in the right direction would be much appreciated. :)
Quote Quote by Owen- View Post
Any way to fix the typo in the title? :S
What typo?

Chestermiller
#4
Feb18-13, 03:58 PM
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Calculation of the velocity of a Spaceship moving Relativistically

You can do this problem using length contraction and time dilation, but the simplest and most foolproof method of doing it is to apply the LT directly.
Let S (coordinates x,t) represent the rest frame of the earth and α-centauri, and let S' (coordinates x',t') represent the rest frame of the spaceship. There are two events that are relevant:
I. Spaceship leaves earth
II. Spaceship arrives at α-centauri.

The coordinates in the S and S' frames of reference for the two events are as follows:

I. x = 0, t = 0, x' = 0, t' = 0

II. x = 4.3c, t = ?, x' = 0, t' = 4.

where x' = 0 corresponds to the location of the spaceship in its own rest frame of reference. For this problem, it is more convenient to work with the inverse LT:
[tex]x=\gamma (x'+vt')[/tex]
[tex]t=\gamma (t'+\frac{vx'}{c^2})[/tex]

You can use the first of these equations to solve part 1 for the velocity of the spaceship by substituting in the parameters for event 2 (event 1 already satisfies these equations identically).
Owen-
#5
Feb19-13, 05:25 AM
P: 40
What does 4.e mean?
sorry it should read 4.3. That might have caused some confusion.

What typo?
I'm just being stupid, there is none - Relativistically looked incorrect to me on 2nd glance :P

Chestermiller, thanks! I think I can do this now, wasn't considering needing the LT at the two distinct points - its a lot easier now that I'm looking at it properly! :)


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