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Time Dilation Problem; Find the proper time, ts.

  1. Jul 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Alpha Centauri, the closest star to Earth, is 4.3x10^6 m away. How long would it take a spaceship to reach the star if it were travelling at 0.999c?


    2. Relevant equations

    I did get the answer as 0.2a....and the textbook also said it would take about 2 months,
    I do not understand what the a represents/stands for and how the textbook got 2 months from 2a.
    Also, I do not why I had to find tm and then to?

    3. The attempt at a solution

    ts = d/v = 4.3 x 10^16/0.999c = 1.43x10^8 s

    tm = ts/√(1-v^2/c^2
    = 1.43x10^8/√(1-(0.999c)^2/c^2
    = 3.19 x 10^9 s
    tm = 4.5a

    THE ANSWER KEy gets the same answer till the part above but then the answer key starts doing more calculations afterwards(below) which I do not get

    ts = tm√1-v^2/c^2

    = 4.5a√1-(0.999c)^2/c^2
    ts = 0.2a

    the textbook gave 0.2a but also said it will take two months to reach the star and that is what I do not understand.


    Thanks for your help.
     
    Last edited: Jul 28, 2011
  2. jcsd
  3. Jul 28, 2011 #2

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    Quick question, could you define all of your variables? a, tm, ts to etc.

    Also, is that really the full question? It doesn't state anything about reference frames.
     
  4. Jul 28, 2011 #3
    yes that is the whole question and you are right it does not tell you the frames...Im guessing you are suppose to figure out the frames yourself..


    d from earth to star= 4.3x10^16 m
    v= 0.999c=3x10^8x0.999=2.9997X10^8

    I used the distance time formula to find the ts(well I think the answer I found is ts)
    ts = 1.43x10^8 s

    In case if you do not know: ts is the proper time, the time interval between two events measured by an observer who sees the events occur at one position
    and tm is the time interval for an observer moving with a speed v relative to the sequence of events.


    And that is what confuses me... I found the right answer but i dont understand how and why

    tm=3.19x10^9s is also equal to 4.5a

    i mean what is a??! im going crazy...

    Uptill 4.5a I got but then the answer key does the next steps which makes no sense and
    says the final answer is 0.2a which is two months.

    how is 0.2a = two months?? what does a stand for and what# does it equal?
     
  5. Jul 28, 2011 #4

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    "= 3.19 x 10^9 s
    tm = 4.5a"

    Super confused about what happened here.

    Am I missing some unit of time?

    Btw, you should probobly try to put units into your solutions throughout.

    I don't think you found ts, I think you found the time in alpha centuri's reference frame. You can then use time dilation to find out how long it took in the reference frame of the ship. Or you could use length contraction to find the distance in the reference frame of the ship and solve it that way.
     
  6. Jul 28, 2011 #5
    yea sure I will do that but the book literally states in answers(unless they are wrong)

    4.5a=3.19x10^9 seconds


    i will try one more time

    thanks for your help I think I will speak to teacher about this question, maybe something is wrong with it
     
    Last edited: Jul 28, 2011
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