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Special relativity - why do I not change the velocity after having changed gamma?

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Star A and B are at rest relative to Earth. Star A is 27 ly from Earth, and as viewed from Earth, Star B is located beyond (behind) star A.
    (a) Star A is 27 ly from Earth. A spaceship makes a trip from Earth to star A at a speed such that the trip from Earth to Star A takes 12 years according to clocks on the spaceship. At what speed relative to earth must the spaceship travel?
    (b) Upon reaching Star A, the spaceship speeds up and departs for Star B at a speed such that the gamma factor is twice that of part A. The trip from star A to star B takes 5.0 y (spaceship time). How far in light years is star b from star A in the rest frame of Earth and the two stars?


    2. Relevant equations
    d = rt
    Δt = Δt0 / γ
    ΔL = L0γ
    γ = 1/(sqrt(1-(v/c)^2))


    3. The attempt at a solution


    So I found the first part, which is 0.914c.

    I don't completely get the second part, but this is what I did:
    v in part a = 0.914c
    I calculated gamma, which came out to be 2.46.
    So, I doubled it because that's what the question said happened.
    gamma = 4.93

    Then I calculated Δt (Earth time), which came out to:
    Δt = 5 years/(1/4.824) = 24.65 years.


    So then I had to calculate the distance from star A to star B, and I did that by using d=rt. But first, I thought that I had to re-calculate the velocity since the spaceship had "sped up," so I calculated the new speed by doing:

    1/(sqrt(1-(v/c)^2)) = 4.93
    solved for v, and found v= 0.979c.

    So then I used d=rt:
    d = (0.979c)(24.65 years) = 24.13 ly

    But the answer is 22.5 ly, and the solution says that they used 0.941c (from part A) instead. Why is this so? The gamma factor has changed because the spaceship's velocity has changed, right? So doesn't it make more sense to calculate a new velocity as well before trying to find the distance?

    Thank you for your clarification!
     
  2. jcsd
  3. May 14, 2012 #2

    BruceW

    User Avatar
    Homework Helper

    I don't think they used 0.941c because that would not give the answer 22.5 ly. Your answer seems correct to me. P.S. welcome to physicsforums! :)
     
  4. Jul 4, 2012 #3
    I got what the original poster got using the same method BUT the solution that I found is as he said, diffrent. I think they are using the length contraction formula for some reason. The solution we have is diffrent than what is in the back of the book. The back of the book matches up with the solution manual...
     
  5. Jul 4, 2012 #4
    I think the OP meant to say that they used 0.914, and that does give the answer 22.5 ly. However, my analysis agrees with the answer given by the OP and by you. The answer book must be wrong.
     
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