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A gravity calculusy kind of quesiton

  1. Feb 25, 2006 #1
    How would i go about writing an equation for the velocity of an object released at a high altitude above the earth which takes into account the sqaure increase in acceleration with displacement (s^2).

    I posted it here because i'm aware it's more of a calculus question than physics.
     
  2. jcsd
  3. Feb 25, 2006 #2

    arildno

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    Hmm, if s is your displacement from the SURFACE, then R+s will be your distance from the centre of Earth, agreed?
    (R being the radius of the Earth)
    The adjective is calculous by the way..
     
  4. Feb 25, 2006 #3
    appreciated.....

    p.s. "The adjective is calculous by the way.." - :tongue: i know
     
  5. Feb 25, 2006 #4

    HallsofIvy

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    Gravitational force is
    [tex]\frac{-GMm}{r^2}[/tex]
    where r is the distance from the center of the earth. If your s is height above the surface of the earth, M is the mass of the earth and R is the radius of the earth, then
    [tex]\frac{d^2s}{dt^2}= \frac{-GM}{(R+s)^2}[/tex]

    That's for something falling straight down, not an orbit, of course.
    The best way to solve that differential equation is to let v= ds/dt, then not that d2s/dt2= dv/dt= (ds/dt)(dv/ds) (chain rule) = vdv/ds
    [tex]v\frac{dv}{ds}= -\frac{GM}{(R+s)^2}[/tex]
    so
    [tex]vdv= -\frac{GMds}{(R+s)^2}[/tex]
    That should be easy to integrate.
     
  6. Feb 25, 2006 #5
    damn it, i had a feeling it was going to be something that simple. Thanks a lot, you've put my mind at rest.
     
  7. Feb 25, 2006 #6

    arildno

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    Which could also be gained with somewhat fewer contortions by multiplying the 2.order diff.ex with ds/dt, and integrate wrt. time.
     
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