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Calculating height achieved under changing gravitional field

  1. Feb 21, 2005 #1
    For some reason, this is alluding me at the moment. We know the gravitional acceleration equation is g = GM/r^2, integrate that in respect to r to yield -GM/r

    I thought I could use

    ?Y = VoT + .5AT^2

    and subsitute GM(-1/Ro + 1/R) into A

    For some reason this is not working, is my line of reasoning correct?
  2. jcsd
  3. Feb 21, 2005 #2

    Doc Al

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    Staff: Mentor

    Not exactly:
    [tex]g = dv/dt = - GM/r^2[/tex]
    [tex]v dv/dr = - GM/r^2[/tex]
    Now you can integrate with respect to r:
    [tex]v^2/2 = GM/r + C[/tex]

    That's only good for uniformly accelerated motion.
    No. What problem are you trying to solve? The height of a projectile as a function of time? That's not so simple.
  4. Feb 22, 2005 #3
    Thank you very much! :)
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