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Relativistic acceleration with integration problem

  1. May 5, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider a space ship that accelerates so that the passengers feel and acceleration equal to that of gravity of the earth’s surface, g. If the space ship undergoes this acceleration for a time T, show that the final velocity is given by:
    V=c[1+(c/gT)^2]^(-1/2)


    2. Relevant equations

    F=(gamma^3)ma where gamma= [1-(v/c)^2]^-1/2 (the Lorentz factor)

    3. The attempt at a solution
    Since the passengers always feel the acceleration of gravity, you don’t actually feel acceleration you feel the force mg. So at any time the passengers must feel mg so:

    mg=(gamma^3)ma
    a=g/(gamma^3)

    I converted gamma into the function and tried to integrate to obtain V but that’s where I got stuck .

    a=g[1-(v/c)^2]^3/2

    I trying to integrate with respect to dt but v it self is a dx/dt so I am wierded out by that. I went to the professor and he gave me a hint in which I have to convert adt into another thing using the chain rule (whatever that means).
    Somehow I think I have to integrate with respect to dv and have my limit be gT instead of T but iam not sure how to get to there.

    Any help is greatly appreciated I have my test this week and I need to get a A :(.
     
    Last edited: May 5, 2010
  2. jcsd
  3. May 5, 2010 #2

    tiny-tim

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    Hi Gravitino22! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Your professor is probably thinking of something like
    ∫ f(v) dt = ∫ [f(v) / (dv/dt)] dv/dt dt = … :wink:
     
  4. May 5, 2010 #3

    vela

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    Use the definition of acceleration is to express the LHS in terms of v and t.
     
  5. May 5, 2010 #4
    Thats what i was trying

    a=dv/dt then pass dt to the RHS to integrate and which also has a v=dx/dt which is where iam confused.

    And for tiny-tim's response

    Iam not seeing where your going with that...

    =f(v)dtdv? double integration?
     
  6. May 5, 2010 #5

    ideasrule

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    a=dv/dt, so you can write dv/dt=g[1-(v/c)^2]^3/2. You can integrate that to get a relationship between v and t. It's true that v=dx/dt, but since you're not asked to find x as a function of t, this equation is irrelevant.
     
  7. May 5, 2010 #6

    vela

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    The differential equation is separable, so you can get all the v's on one side and t on the other and then integrate each side.
     
  8. May 5, 2010 #7
    Thanks vela,

    I was trying that approach earlier but i cant solve that nasty integral, going to keep trying
     
  9. May 5, 2010 #8

    vela

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    Use the trig substitution v/c=sin θ. Then don't make the really stupid mistake I did, and the answer (for the integral) will pop right out.
     
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