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Relativistic Doppler Effect

  1. Mar 25, 2006 #1
    This is the problem

    A shuttle is launched from a space station and travels away from it in a straight line. It rapidly accelerates and obtains a steady velocity of v = 4c/5 relative to the space station. The spaceship sends out radio signals of a frequency f. The spaceship is on a mission to dock with a rocket, which travels away from the earth at velocity u. The rocket receives the shuttle's transmission at a frequency 3f/2. Show that u of the shuttle relative to the earth is 3c/5.


    relativistic doppler effect; source moves towards the observer

    [tex] f' = f \sqrt {\frac {c+v} {c-v}} [/tex]

    velocity addition formula:

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

    My atempt

    [tex] f' = f \sqrt {\frac {c+v} {c-v}} = 3f/2 [/tex]

    hence: v = 5c/13

    this is the relative speed between the shuttle and the rocket. Therefore in frame S' of the shuttle the rocket moves with a velociy u' = 5c/13

    Now we have to transform to the Earth's frame of reference S, where v = 4c/5

    using the velocity addition formula:

    [tex] u = \frac {u' + v} {1 + v*u'/c^2} = 77/85 *c [/tex]

    which is wrong.

    Does anyone see why or can anyone give me a hint towards the right answer? That would be very helpful and much appreciated!
    Last edited: Mar 25, 2006
  2. jcsd
  3. Mar 25, 2006 #2


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    Only the last step is wrong, although it's hard to reason from the text since there's no picture showing the directions.

    Anyway, the rocket and shuttle are approaching each other while the rocket is leaving earth (so shuttle is approaching earth), so the velocity of the rocket wrt earth and the velocity of the shuttle wrt the rocket have opposite signs.
  4. Mar 26, 2006 #3
    thanks very much for this. It does make sense (although I find it easy to miss) and one get's the correct answer by altering the signs of velocity u.
    That was very helpful!!!!:smile:
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