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Homework Help: Calculation of the velocity of a Spaceship moving Relativistically

  1. Feb 18, 2013 #1
    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]\beta = \frac{v}{c}[/itex]
    [itex]t'=\gamma(t-\frac{x \times v^2}{c^2})[/itex]

    3. The attempt at a solution
    [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,
    Last edited by a moderator: Feb 18, 2013
  2. jcsd
  3. Feb 18, 2013 #2
    Any way to fix the typo in the title? :S
  4. Feb 18, 2013 #3


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    What does 4.e mean?

    What typo?
  5. Feb 18, 2013 #4
    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).
  6. Feb 19, 2013 #5
    sorry it should read 4.3. That might have caused some confusion.

    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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