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Relativity Question

  1. May 9, 2005 #1
    Here is the question:

    An armada of spaceships that is 1.00 ly long (in its rest frame) moves with speed .800c relative to a ground station in frame S. A messenger travels from the rear of the armada to the front with a speed of .950c relative to S. How long does the trip take as measured in:

    (a) the messenger's rest frame?
    (b) the armada's rest frame?
    (c) an observer's point of view in frame S?


    I am having a very hard time with relativity right now... Do I use the relativistic velocity equation?

    [tex]u = \frac {u^{\prime} + v}{1+u^{\prime} v/c^2} \mbox { (relativistic velocity)}[/tex]

    Or any other equations like:

    [tex]\triangle t = \gamma \triangle t_0[/tex]
    [tex]L = \frac {L_0}{\gamma}[/tex]

    The thing that would help me most and I would be most grateful for is if someone would answer one of the questions in detail offering reasoning for each step of the problem, and offer some hints on solving the other ones. Answers to compare with might be helpful too.

    Thanks in advance.

    -Mark
     
  2. jcsd
  3. May 9, 2005 #2
    help

    would anyone please help?
     
  4. May 9, 2005 #3

    James R

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    An armada of spaceships that is 1.00 ly long (in its rest frame) moves with speed .800c relative to a ground station in frame S. A messenger travels from the rear of the armada to the front with a speed of .950c relative to S. How long does the trip take as measured in:

    (a) the messenger's rest frame?
    (b) the armada's rest frame?
    (c) an observer's point of view in frame S?

    All the normal non-relativistic equations work provided you take all quantities in the same frame of reference.

    Part (a)

    In the messenger's frame, we can use s=vt to work out the trip time. But before we can do that, we need to know how far the armada moves as its front moves to the messenger (who is stationary in this frame), and how fast the armada moves in the same frame.

    The speed of the armada relative to the messenger is given by the velocity addition formula.

    [tex]w = \frac{v - u}{1 - uv/c^2}[/tex]

    where v is the messenger's speed relative to S, u is the armada's speed relative to S, and w is the armada's speed relative to the messenger.

    We get

    [tex]w = \frac{0.950c - 0.800c}{1 - (0.950)(0.800)} = 0.625c[/tex]

    The length of the armada L in the messenger's frame, is:

    [tex]L = L_0 / \gamma[/tex]

    where

    [tex]\gamma = \frac{1}{\sqrt{1-(w/c)^2}} = 1.28[/tex]

    So

    [tex]L = 1.00 ly / 1.28 = 0.78 ly[/tex]

    The time taken, in this frame, is:

    [tex]t = L/w = 0.78 ly / (0.625 ly/yr) = 1.25 years[/tex]

    Part (b)

    We could follow the same procedure as in part (a) here. You should try that. But, there's another way to do this - use the time dilation formula between the messenger's frame and the armada's frame.

    The messenger's elapsed time is 1.25 years, and the messenger is moving relative to the armada, so the armada measures a longer time, by a factor of [itex]\gamma[/itex] (given above).

    The time taken is 1.60 years, in this frame.

    I'll leave you to do part (c).
     
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