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Final velocity due to gravtiy

  1. Nov 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Assuming the Earth is at rest and alone in the universe what speed would a spaceship arriving at the surface of Earth from an infinitley large distance be if the speed at infinity= 0 and acceleration is caused only by gravity


    2. Relevant equations
    [itex]F= \frac{GmM}{x^2}[/itex]
    Also using the chain rule [itex]a=v\frac{dv}{dx}[/itex]


    3. The attempt at a solution

    well I set ma equal to the force due to gravity so

    [itex]v\frac{dv}{dx}= \frac{GM}{x^2}[/itex]

    seperating the variables gives [itex]vdv=\frac{GM}{x^2}dx[/itex]

    then

    [itex]\int vdv=GM\int x^{-2} dx[/itex]

    would the limits of integration need to be from infinity to the surface of the earth, or am i on the wrong track all together?
     
  2. jcsd
  3. Nov 1, 2013 #2

    haruspex

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    You're doing fine. Just do the integral as you suggest.
     
  4. Nov 1, 2013 #3

    TSny

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    You're on the right track. But you might need to be careful with signs. [Edit: I see haruspex beat me :smile:]
     
  5. Nov 1, 2013 #4
    Ok, then doing the indefinite integral gives [itex]v=\sqrt{\frac{2GM}{x}} + C[/itex]

    but i'm not sure about the limits, would it be [itex]\frac{1}{2}v^2=-GM\int_{\infty}^{6370}x^{-2}[/itex]

    where 6370 is the radius of the earth in km?
    Thanks
     
  6. Nov 1, 2013 #5

    haruspex

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    The limit is the radius of the Earth. What units you use are up to you, so long as you are consistent. I would think your value for G assumes m, not km.
    I think you have a sign wrong above. The minus sign should only appear after performing the integral.
     
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