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Spaceship Relativity Problem

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Spaceship #1 moves with a velocity of .2c in the positive x direction of reference frame S. Spaceship #2, moving in the same direction with a speed of .6c is 3 x 10^9 m behind. At what times in reference frames S, and in the reference frame of ship #1, will 2 catch up with 1.

    2. Relevant equations



    3. The attempt at a solution

    vrelevative = .6c-.2c / (1-.6(.2)) = .455c
    Then, 10 light seconds / .455c = 22.0s

    Or...is it done classically .6-.2 = .4, making it 25s?

    For the second part, is it done like so?

    beta = 1/(sqrt (1-.2)) = 1.02
    Then t = 22/1.02 = 21.6s or if it was 25s, 25/1.02 = 24.5s

    Can someone explain to me which of these two (if either) are correct? I am not sure if it is done classically or with relativity, so can you explain why it is either.

    Thank you very much.
     
  2. jcsd
  3. Nov 7, 2012 #2

    tiny-tim

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    welcome to pf!

    hi bethany555! welcome to pf! :smile:
    yes … in any one frame, speed etc works just the way it should! :biggrin:

    (you only need those pesky gammas :grumpy: if you're having to convert measurements from another frame first :wink:)
    (btw, that's called gamma, not beta … beta is v/c :wink:)

    imo, it's always safest to use the original lorentz transformation formulas

    (rather than use short-cut contraction formulas which usually work only for rigid separations)

    in this case, you know that (in frame S) t = 25, x = … , so t' = … ? :smile:
     
  4. Nov 7, 2012 #3
    Thank you very much!

    For the second part, would it be the following --

    Vrelative = .455c (as shown above)

    L = 10c seconds (sqrt(1-.45^2)) = 8.93

    8.93/.455 = 19.6s

    I'm not sure if I'm using the correct v for all parts, would you be able to tell me if I did this correct? Thank you again so much!
     
  5. Nov 7, 2012 #4

    tiny-tim

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    no, your gamma should be for the relative speed between your two frames,

    ie between S and the frame of the 1st ship (0.2 c)

    and i really think you should use the t' = γt - γvx formula from the Lorentz transformation
     
  6. Nov 7, 2012 #5
    Thanks! So would it be 9.80cs / .455 = 21.5s?

    When I use the Lorenz equation, I get
    t' = (1/(sqrt(1-.2^2)) ) * 25 - (1/(sqrt(1-.2^2)) ) .2 * 10, I get 23.47?
     
  7. Nov 7, 2012 #6

    tiny-tim

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    what is this? :confused:
    yes, that looks ok :smile:

    (i haven't checked the actual figures)
     
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