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Math Help - Special relativity question

  1. #1
    MHF Contributor alexmahone's Avatar
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    Special relativity question

    A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.6c. The pursuit ship is traveling at a speed of 0.8c relative to Tatooine, in the same direction as the cruiser. What is the speed of the pursuit ship relative to the cruiser?
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    Re: Special relativity question

    I believe I answered this on another board and said that I agreed with the result you gave their. But perhaps we were both looking at this incorrectly (I'm not a physicist!). We both were doing this as \frac{u+ v}{1+ \frac{uv}{c^2}}. But relative the cruiser, Tatoine has a velocity of -.8c. So the "sum" of speeds should be \frac{-.8c+ .6c}{1+ \frac{(-.8c}(.6c)}{c^2}}= \frac{-.2c}{1- .48}= -.38c.
    Last edited by HallsofIvy; August 18th 2012 at 05:39 PM.
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  3. #3
    GJA
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    Re: Special relativity question

    Hey, guys.

    I think the numerator should be reversed:

    u=.8c is the velocity of sprinter ship relative to Tatooine and

    v=-.6c is the velocity of Tatooine relative to the cruiser ship.
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  4. #4
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    Re: Special relativity question

    If that were the case, the sprinter ship would see the cruiser ship with speed +.38c and it would not be gaining!
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    GJA
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    Re: Special relativity question

    The velocity is measured relative to the cruiser, so it should be gaining.
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  6. #6
    MHF Contributor alexmahone's Avatar
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    Re: Special relativity question

    Quote Originally Posted by HallsofIvy View Post
    I believe I answered this on another board and said that I agreed with the result you gave their. But perhaps we were both looking at this incorrectly (I'm not a physicist!). We both were doing this as \frac{u+ v}{1+ \frac{uv}{c^2}}. But relative the cruiser, Tatoine has a velocity of -.8c. So the "sum" of speeds should be \frac{-.8c+ .6c}{1+ \frac{(-.8c}(.6c)}{c^2}}= \frac{-.2c}{1- .48}= -.38c.
    Actually, Doc Al on physicsforums spotted the sign error.
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