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Special Relativity

  1. Aug 20, 2008 #1
    2. When travelling in space, a ship (length 1000m in its rest frame) passes an identical ship, which is at rest
    relative to Earth. The captain of the moving ship decides to measure the speed of her ship. She looks out of
    a window and uses her clock to measure the time between the moments when the head and the tail of the
    other ship pass. The result of the measurement is 1 μs. Calculate the speed of the “moving” ship relative to
    Earth.



    Attempted using both Lorentz transformations and time dilation/length contraction formulas. Tried to solve simultaneously but equation turned out to be ridiculously complex for an introductory physics course question.
     
  2. jcsd
  3. Aug 20, 2008 #2

    Doc Al

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    Staff: Mentor

    Hint: According to the captain, how far did the other ship move during that 1 μs?
     
  4. Aug 20, 2008 #3
    1000m??
     
  5. Aug 20, 2008 #4
    wouldn't that require finding the length of the captains ship from within the earths reference frame. im just confused as to how the length contraction affects this situation
     
  6. Aug 20, 2008 #5

    Doc Al

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    Nope. The distance traveled must equal the length of the ship (from head to tail) according to the captain.
    Since the captain is doing the observing, her ship is at rest. But you'll need to find the length of the other ship according to the captain.
    See above.
     
  7. Aug 20, 2008 #6
    i got a value of around 1006 metres for the length of other ship from the captains reference. to do so i had to solve a quartic equation. i used v=(length of other ship from captains reference)/(the time interval from captains reference of 1 microsecond)
    i then substituted this into L=L'/gamma where gamma is 1/sqrt(1-v^2/c^2)
    and solved for v to find the speed.

    it was something like
    v^2-(v^4/c^2) = 10^15

    does that seem right at all?
     
  8. Aug 20, 2008 #7

    Doc Al

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    Both ships have a proper (rest) length of 1000m. From the reference frame of one ship, will the other ship be longer or shorter?

    Also, until you've figured out the speed and thus gamma, how can you get a numerical answer for this length? Instead, just express it using gamma.
    Excellent.
    I think you're on the right track, but just got messed up somewhere. L' is the rest length of the ship.
    Do it over carefully and you'll get an easy equation to solve. (No quartics!)
     
  9. Aug 20, 2008 #8
    ah can you spell it out a bit more clearly. i can only seem to get the quartic equation
     
  10. Aug 20, 2008 #9
    i just can't find how i can get the velocity of the first ship, when i need the relative velocity to get gamma, and i need gamma to get the proper time and length of other ship from captains viewpoint. then i need to use these to find the velocity. it seems like its a catch 22
     
  11. Aug 20, 2008 #10

    Dick

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    The whole point of special relativity is that you can calculate by picking any inertial frame and taking it to be 'at rest'. Try solving the problem assuming the ship taking the measurement is at rest. Then there is only one gamma.
     
  12. Aug 21, 2008 #11

    Doc Al

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    There's only one speed in this problem--the relative speed of the ships, which is what you're asked to find. And that's the same speed you need to get gamma.

    Start as you did by using that equation for speed:
    All you have to do is rewrite "length of other ship from captain's reference" in terms of the ship's known rest length. (Hint: That's where gamma comes in.)

    Once you write that equation, just rearrange and solve for v. Do it step by step, showing each step if you need to.
     
  13. Aug 21, 2008 #12
    thanks for your help, i got it out
     
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