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Homework Help: Twin Paradox problem

  1. May 29, 2008 #1
    [SOLVED] Twin Paradox problem

    Here is a problem about twin paradox that I can't quite figure out. I got part a, but I can't get part b. I think that the 6 years spent doing research is kinda throwing me off. Help!!

    The International Space Federation constructs a new spaceship that can travel at a speed of 0.910c. Mary, the astronaut, boards the spaceship to travel to Barnard's star, which is the second nearest star to our solar system after Alpha Centauri and is 5.98 ly away. After reaching Barnard's star, the spaceship travels slowly around the star system for 6 years doing research before returning back to Earth.
    (a) How much time does the journey take?
    11.44913698 years
    (b) How much older is her twin Frank when she returns?
    ? years
  2. jcsd
  3. May 29, 2008 #2


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    Question a is ambiguous since it does not specify in what frame...but I guess it means in the spaceship's frame.

    Since the spaceship travels slowly around the star system, the six years of research lasts that amount of time in both frames (basically the two frames are at rest relative to one another during that period). So find the time taken for the ship to get to the star as measured in Earth's frame, double that and add six years.
  4. May 29, 2008 #3
    Yep, I tried doing that, using the formula T=2L/v, added 6 to that, and got 19.1429, but the answer still wasn't right. Do you think I'm doing something wrong?
  5. May 29, 2008 #4


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    T=2L/v is the time the ship is in flight as experienced on earth. That's not the time experienced on the ship. Don't forget the time dilation factor.
  6. May 29, 2008 #5

    Doc Al

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    Looks like the right calculation to me. (You are calculating the time according to the Earth.) What did you put for the final answer to part b?

    Note: The question is a bit ambiguous. I assume they are asking: How much older is Frank than his sister.
    Last edited: May 29, 2008
  7. May 29, 2008 #6
    do you mean gamma= 1/sqrt(1-v^2/c^2)? because part 2 asks for the twin who stayed on earth's age, and my book said that the formula for that time, as measured on earth, is T=2L/v. So i figured just do that +6. I used T=2L/(gamma*v) +6 for part a's answer.

    The answer I put for part b was T=2*5.98/.91+6= 19.1429, but the homework program says it's wrong.
  8. May 29, 2008 #7

    Doc Al

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    Read the note that I added to my last post.
  9. May 29, 2008 #8
    gotcha! thank you!!:smile:
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