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Twin paradox problem

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data

    A clock is placed in a satellite that orbits earth for a period of 5400 seconds. By what time interval will this clock differ from an identical clock on Earth after 1 year(assume that special relativity applies.)

    2. Relevant equations
    v(escape velocity)=11186 m/s
    possibly delta(t) = delta(t')/gamma
    gamma =1/sqrt(1-v^2/c^2)
    3. The attempt at a solution

    First I converted 1 year => 31536000 s

    after 1 year , the observers on the satelitte measured the new time of the sattilitte to be

    t(new)= 31536000 seconds + 5400 seconds = 31541400 seconds

    after 1 year , the observers measured the earth notices that the velocity of the satelitte to approach the speed of light.

    t'(new) = 31536000 seconds + 5400 seconds/gamma=31541400.000004

    the time interval between the times measured on earth and on the satelitte after 1 year = 31541400 seconds -31541400.000004=.000004 seconds

    but the back of my books says 9.6 ms. Where did I go wrong ?
     
  2. jcsd
  3. Sep 8, 2007 #2

    Doc Al

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    Staff: Mentor

    What's the speed of the satellite?
     
  4. Sep 8, 2007 #3
    It does explicity state the speed of the satellite in the problem . But I assumed the escape velocity for all objects leaving the earth is about 11186 m/s , according to the list of physical constants in the appendix of my textbooks.
     
  5. Sep 8, 2007 #4

    Doc Al

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    Does? Or does not? If not, you have to figure it out.
    What's escape velocity have to do with it? It's orbiting the earth, not escaping.
     
  6. Sep 8, 2007 #5
    It doesn't. wouldn't the velocity be v=2*Pi*(R(earth))/1 year?
     
  7. Sep 8, 2007 #6

    Doc Al

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    Only if its altitude is zero. Read this: Earth Orbits
     
  8. Sep 9, 2007 #7
    so then finding the velocity of the orbit would be pointless if we are not given the distance between the earth and sattelite
     
  9. Sep 9, 2007 #8

    learningphysics

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    Set the gravitational force equal to centripetal force... ie:

    GMm/r^2 = mv^2/r
    GM/r^2 = v^2/r

    using this equation and

    v = 2pir/5400

    You can solve for v... you have 2 equations 2 unknowns (v and r).
     
  10. Sep 9, 2007 #9
    Isn't r just the sum of the Radius of the earth + the altitude of the satelitte?
    r=R(earth) + h
     
  11. Sep 9, 2007 #10

    learningphysics

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    Yes. But you don't know the altitude.
     
  12. Sep 13, 2007 #11
    I understand how calculate the velocity but for some reason I came up with the same time interval
    I'll take you through my calculations.
    GM(earth)/r^2 = v^2/r => v=sqrt(G*M(earth)/r) = 2*Pi*r/T

    G*M(earth)/r = 4*pi^2 *r^2 /(5400s)^2 => r^3 = ((5400s)^2/(4*pi^2))*(6.67e-11)*(5.97e24 kg) => r= 6650321.521 m

    now I can find the velocity: v =2*pi*(6650321.521 m)/(5400 s) = 7738 m/s

    1 year = 31536000 seconds as measured by the person on the satellite

    the time measured by the observer on earth is gamma*proper time = 31536000 seconds*(1/(sqrt(1-(7738 m)^2/c^2) =3153600seconds

    so there is no time differences according to my calculations. What am I doing wrong ?
     
  13. Sep 13, 2007 #12

    Doc Al

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    Since the speed is so small compared to light speed, the time difference won't be much. You won't be able to find it by plugging the numbers directly into a (typical) calculator, since 1 minus a tiny number will show up as 1. Instead, use a Taylor expansion of gamma for small v/c to calculate the time difference.
     
  14. Sep 13, 2007 #13
    But are my equations and calculations are correct

    You think a TI -83 or TI-89 would approximate gamma better than a mere scientific calculator?
     
  15. Sep 13, 2007 #14

    Doc Al

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    You work looks OK to me. I don't know anything about those calculators, but I doubt they'll do any different. Do the Taylor expansion--that's the easy way.
     
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