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Speed of rocket as viewed from Earth

  1. May 9, 2016 #1
    1. The problem statement, all variables and given/known data
    A rocket starts at rest from the Earth, moving in the z direction so that in its own instantaneous rest frame the acceleration is always g = 9.8 m/s^2. To an observer on Earth, its position is given by

    ## z(\tau) = \frac{cosh(g\tau) - 1}{g} ##
    ## t = \frac{sinh(g\tau)}{g} ##

    where ##\tau## is the rocket's proper time. Measured in a reference frame at rest on the Earth, how far from the Earth will the rocket be in 40 years? How fast will it be going?

    2. Relevant equations

    Four-velocity:
    ## U^\mu = (\gamma, \frac{d\underline{x}}{d \tau}) = (\gamma, 0, 0, sinh(g\tau)) ##

    3. The attempt at a solution
    t = 40 years = ##1.26\times10^9 s##
    g = ##9.8m/s^2##

    ##g\tau = sinh^{-1}(gt) = sinh^{-1}((9.8)(1.26\times10^9)) = 23.93##

    ##z = \frac{cosh(g\tau)-1}{g} = \frac{cosh(23.93)-1}{9.8} = 1.26\times10^9 m##

    ## U^z = sinh(g\tau) = sinh(23.93) = 1.24\times10^{10} m/s##

    But this is greater than the speed of light so it can't be right.
     
  2. jcsd
  3. May 9, 2016 #2

    TSny

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    Your formulas for ##z## and ##t## must be assuming units where ##c = 1##. It would be a good idea to insert factors of ##c## so that you can work with meters and seconds.

    The z component of the 4-velocity is ##dz/d\tau##. But this is not the velocity you want. In the earth frame, a time increment would be ##dt##, not ##d \tau##.
     
  4. May 9, 2016 #3

    Andrew Mason

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    There seems to be an error in your calculation of ##\tau##. You have ##\tau## = 2.4 seconds (23.93/g). But for the first several months ##t \approx \tau## because the rocket speed << c.

    AM
     
  5. May 10, 2016 #4
    Yes, I forgot to say the question mentions that the equations ## z(\tau) = \frac{cosh(g\tau) - 1}{g} ## and ## t = \frac{sinh(g\tau)}{g} ## are in units where ##c=1##. So reinserting factors of ##c##, would these equations become:

    ##gt/c = sinh(g\tau /c)##
    ##z(\tau) = \frac{cosh(g\tau /c) - 1}{g/c^2} ##?

    Then for ##t=40 years=1.26 \times 10^9 s##, we have

    ##g\tau /c = sinh^{-1}(gt/c) = sinh^{-1}[\frac{(9.8)(1.26 \times 10^9)}{3 \times 10^8}]= 4.41##

    Giving a distance measured in the Earth frame:

    ## z= \frac{cosh(g\tau /c) - 1}{g/c^2} = \frac{(3 \times 10^8)^2}{9.8}[cosh(4.41) -1] = 3.69 \times 10^{17} m##.

    The speed in the Earth frame would then be:

    ##U^z = \frac{dz}{d\tau}\frac{d\tau}{dt}
    = c sinh(g\tau /c) \frac{1}{cosh(gt/c)}
    = c tanh(gt/c)
    = c tanh(4.41)
    =0.9997c##

    Is this the right way to do it?
     
  6. May 10, 2016 #5

    TSny

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    That looks correct to me.

    It might be nice to express the distance in light-years.
     
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