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Dilated Time [Answer Check]

  1. Jun 8, 2008 #1
    Hi I'm not sure if i am doing this correctly, can someone please check my answer! THANKS A LOT :D

    1. The problem statement, all variables and given/known data

    An Astronaut travelling at 0.90 c, with respect to Earth, measures his pulse adn finds it to be 70 beats per minute.
    a) Calculate teh time required for one pulse to occur, as measured by the astronaut.
    b) Calculate the time required for one pulse to occur, as measured by an Earth-based observer.
    c) calaculare the astronaut's pulse, as measured by an Earth-based observer.
    d) What effect, if any, would increasing the speed of the sapacecraft hav eon the astronaut's pulse as measured by teh astronaut and by the Earth observer? Why?


    2. Relevant equations

    [tex]\Delta[/tex] t = [tex]\frac{\Delta t_o}{\sqrt{1 - \frac{ v^2 }{ c^2 }}}[/tex]

    3. The attempt at a solution

    a)
    (1 min / 70 pulses) x (60 sec) = 0.85 s
    Therefore the astronaut feels a pulse every 0.85 s.

    b)
    [tex]\Delta[/tex] t = [tex]\frac{\Delta t_o}{\sqrt{1 - \frac{ v^2 }{ c^2 }}}[/tex]
    [tex]\Delta[/tex] t_o = [tex]\Delta[/tex] t x [tex]\sqrt{1 - \frac{ v^2}{c^2}}[/tex]
    [tex]\Delta[/tex] t_o = (0.85) x [tex]\sqrt{1 - \frac{ 0.90c^2}{c^2}}[/tex]
    = 6 seconds

    Therefore the time required for one pulse to occur as measured by an Earth-based observer is 6 seconds.

    c) (60 seconds / 6 seconds) = 10 beats per second
    Therefore the astronauts pulse as measured by an Earth-based observer is 10 beats per minute.

    d) As the speed of the spacecraft increases the astronaut's pulse will increase from the frame reference of the astronaut. From the frame reference of the Earth observer the pulse of the astronaut will decrease.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 8, 2008 #2

    tiny-tim

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    Hi lamerali!

    Just looking at a) and b) …

    a) is fine, of course! :smile:

    b) I don't see how you got from .85 to 6 … and you forgot to square the 0.9.
     
  4. Jun 8, 2008 #3
    Hey tiny-tim :D

    I did forget to square the 0.9 so my new answer is 0.37 s...does that sound more realistic?
     
  5. Jun 8, 2008 #4

    tiny-tim

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    Hey lamerali :smile:

    Shouldn't it be more?

    The earth observer regards the astornaut's clock as slow, so his pulse is slow, which means the pulse takes longer?
     
  6. Jun 8, 2008 #5
    isn't 0.37 seconds the time for ONE pulse to occur, his actual beat would be 162 beats per minute. if this is incorrect do you know where i am going wrong???
     
  7. Jun 8, 2008 #6

    tiny-tim

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    When the astronaut returns, he will be younger than he should be, so he will have had less pulses, so his pulses must have taken longer.

    I think your √ factor should be 1/√. :smile:
     
  8. Jun 8, 2008 #7
    tiny-tim is correct. You have the correct formula but are using it wrong. In general, [itex]\Delta t_0[/itex] represents the proper time interval, i.e. the time difference between two events as measured in a reference frame where the events occurs at the same place. In this example, this is the spaceship's frame.
     
  9. Jun 8, 2008 #8
    OHH ok so sould it be

    = (0.85) / √ (1-(0.9c^2)/c^2
    = 2 s

    therefore each pulse takes 2 seconds and his pulse is (60s / 2s) = 30 pulses per minute
     
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