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Special theory of relativity: velocity of airplane

  1. Jun 7, 2014 #1
    1. The problem statement, all variables and given/known data
    If a clock in an airplane is found to slow down by 5 parts in 1013, (i.e. Δt/Δto = 1 + 5x10-13), at what speed is the airplane travelling? (Hint: You may need to use the binomial expansion for γ.)

    To be honest, I'm really confused about what this question's telling me. What's in the brackets?

    2. Relevant equations
    γ = 1/√(1-v2/c2)
    That's the equation I used.

    3. The attempt at a solution
    When I rearranged the equation to find the velocity, I got v = √((1-(1/γ)2)(c2))

    I subbed in 1 + 1.5x10-13 for γ but I keep getting 0 as my velocity and the answer's supposed to be 300 m/s. Am I even doing this right? Thanks in advance! I appreciate any help you can give me!
     
  2. jcsd
  3. Jun 7, 2014 #2
    1st why did you sub 1 + 1.5x10-13 for γ instead of 1 + 5x10-13, as described in the problem.

    2nd yes your method is correct but either you're misusing the calculator or the calculator can't handle the calculation. Yes, calculators don't always provide correct answers. That's why the problem said "You may need to use the binomial expansion for γ" by what they mean a Taylor expansion.
     
  4. Jun 7, 2014 #3
    Oops! Sorry! I meant to write 5x10-13. I'm afraid I don't really know what a Taylor expansion is... How does this work?
     
  5. Jun 7, 2014 #4
    Taylor's expansion is very important and useful but it takes more than a few lines to explain it. You should definitely learn it, but for right now I would try using a better calculator. Use the calculator that comes with Windows. It's very good.
     
  6. Jun 8, 2014 #5

    vela

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    The binomial expansion says that
    $$(1+x)^m = 1 + \frac{m}{1!} x + \frac{m(m-1)}{2!}x^2 + \frac{m(m-1)(m-2)}{3!}x^3 + \cdots.$$ In particular, for ##m=-1/2##, you get
    $$(1+x)^{-1/2} \cong 1 - \frac 12 x.$$ Therefore, you have
    $$\gamma = \left[ 1 + \left(-\frac{v^2}{c^2}\right)\right]^{-1/2} \cong 1 + \frac 12 \frac{v^2}{c^2}.$$
     
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