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Starship special relativity problem

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A starship is moving toward a space station at half the speed of light. When it is 7.00 seconds away from reaching a space station (as measured on the ship's clocks) the starship fires a projectile toward the space station at 0.600 c (as measured by the ship's crew). When will the projectile strike the space station, as measured by the ship's clocks and also by the space station's clocks? Correct answers: -3.82 s and 3.31 s, respectively.

    2. Relevant equations

    delta t = gamma*delta t_proper
    d = rt

    3. The attempt at a solution

    I tried solving this problem from the starship's point of view. From the starship's point of view, it is stationary, the station is moving 0.5c toward it, and the projectile is moving 0.6c toward the station.

    The distance between the ship and the station @ t = -7 s can be found by
    d = rt = (0.5c)(7 s) = 3.5 Ls

    Then the time it takes for the projectile to hit the projectile to collide with the station is
    t = d / r = 3.5 Ls / (0.5c+0.6c) = 3.18s.

    But the ship's clocks started at t = -7 s, so the time read on the clocks as the projectile hits the station is t = -7 s + 3.18 s = -3.82 s, which is the correct answer.

    Now I have a few questions about this:

    Why do we say that the time at which the projectile collides with the station is t = 0?

    I'm having an awful time figuring out what is the proper time and what is the... regular time? Which is which and why?

    Could this problem be solved from the perspective of the space station? If so, how?

    Finally, how can I solve this problem by using a space time diagram like this one?

    http://i.imgur.com/BJ0tbXs.jpg?1

    What goes on the x-ct axis and what goes on the x'-ct' axis?
     
  2. jcsd
  3. Apr 30, 2013 #2
    I get -3.31s for the space station clock, not +3.31s. Is a negative or positive answer?
     
  4. Apr 30, 2013 #3
    I wanted to point out, that nothing can go past the speed of light, in any inertial frame of reference. Either there must be something wrong in your calculations and the way you got to the correct answer or the correct answer you stated is wrong.

    You don't have to. It only makes sense to do so because otherwise you'll have negative time which is much more disconcerting.

    Proper time is the time measured by a stationary frame of reference, in this case the space stations'. I think by regular time you mean the time at which it is conveniently measured? If so, then that's the spaceships'.

    It certainly can, and I think that's what this problem is supposed to entail you to do since they clearly state that it is the spaceship that's moving at .5c relative to the stationary frame of reference of the space station.
     
  5. May 1, 2013 #4
    There is nothing wrong with this calculation. All velocities relative to the observer are below c. It appears that the projectile has a relative velocity to the station of 1.1c from the ship's view but this is ok becuase relative to the ship all velocities are less than c. The velocity of the projectile to the station from the station's view must be found using Lorentz velocity transform.

    I don't think this is mentioned in the question. I might be misreading it but I think we are meant to assume that the ship's clock will read 0 when the ship reaches the station and that when the ship releases the projectile it reads -7s. These two coincidences all observers will agree on because they happen at single points in space and time. Also the station and ship agree they have a relative velocity of 0.5c. All other 'facts' are subject to transforms between frames.

    Start by finding how much time would be measured by the space station's clock as the ship approaches. Use that to find the distance between the ship and station at the start.
     
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