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Long Rocket

  1. Sep 5, 2010 #1
    Suppose that you are at the top of a (rigid) rocket which is half a light
    year tall. If the rocket is accelerating such that your proper acceleration
    is 1g,

    What is the proper acceleration at the bottom of the rocket?

    B =
    F c2
    c2  F

    This is one of the equations given:

    alpha B = (alpha f c^2)/(c^2 - delta alpha f)

    It seems that delta alpha f is the difference in accelerations. I can't find any equation that uses the distance. Is it OK to use the distance in this one?
  2. jcsd
  3. Sep 5, 2010 #2


    User Avatar
    Science Advisor

    Well, you could make use of the fact that any object undergoing Born rigid acceleration (such that each point has constant proper acceleration, and the length of the object in the instantaneous inertial rest frame of any point on the object doesn't change from one moment to the next) must be at rest in a Rindler coordinate system. An observer with an acceleration A has the Rindler horizon at a distance of c2/A behind them according to the article, and 1G acceleration works out to 1.03 light-years/year2, so the Rindler horizon would be 1/1.03 = 0.97 light-years behind the observer at the top of the rocket, which means it's OK for the rocket to be 0.5 light-years long. In this case the bottom would be at a distance of 0.97-0.5 = 0.47 light-years from the Rindler horizon, which means it should have a proper acceleration of 1/A = 0.47, meaning A = 1/0.47 = 2.12 light-years/year2, or about 2.12/1.03 = 2.06G.

    edit: is this a homework problem? If so I guess you'd have to use an equation that was actually discussed in the chapter in which the problem appeared...what are the meanings of the various symbols in the equation you posted?
    Last edited: Sep 5, 2010
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