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Relatvitiy - Spaceship traveling into Galaxy

  1. Jan 26, 2009 #1
    1.The radius of our galaxy is approx. 3x10^20 m. A spaceship sets out to cross the galaxy in 25 years, as measured on board the ship. With what uniform speed does the spaceship need to travel? How long wold the trip take, as measured by a timepiece stationed on Earth?



    2. v=d/t



    3. So far I have converted the 25 years into seconds using proportions of 1 yr = 31556952 to get 788923800 seconds in 25 years. I found the velocity using v=d/t. d = radius of galaxy and t = 788923800 seconds

    I am not sure if I am doing this right, and what do I have to do find out how long this trip would take? I don't think my distance is correct

    thanks
     
  2. jcsd
  3. Jan 26, 2009 #2

    nrqed

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    You can't use that equation with the values given since the "d" is given in the frame of the galaxy and the "t" is in the frame of the spaceship. You need to use all values in one frame. You could express the time it takes in the frame of the galaxy in terms of the time in the frame of the spaceship. If it takes 25 yrs in the frame of the ship, how long does it take in the frame of the galaxy? (you can't get a number but you can find an expression in terms of the 25 years and of the speed). Once you have that time "t'", then you may use v = d/t'. And you will be able to solve for the speed v.
     
    Last edited: Jan 26, 2009
  4. Jan 26, 2009 #3

    Delphi51

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    Oops, nrgd has given a far better answer! I fear I missed something - not surprising with relativity.
     
  5. Jan 26, 2009 #4
    I knew there was something wrong with using the radius as my d. thanks for that, but I'm still unsure of how to attack the problem
     
  6. Jan 26, 2009 #5

    nrqed

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    If it takes 25 years in the ship's frame, it will take (gamma times 25) years as measured in the galaxy's frame.
     
  7. Jan 27, 2009 #6
    I got through a little bi t of the problem. I learned that I had to convert the radius of the galaxy into light years and use it as the length.

    Converting 3x10^20m to lights years is -> 31712.47 light years. I have to use the formula L(v) = L_0 * sqrt(1-v^2/c^2) Now I have two lengths in this equation, I have one of the length (31712.47 lt yrs) what should be the other length? I am trying to solve for velocity from that equation ( L(v) = L_0 * sqrt(1-v^2/c^2) )
     
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