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More accurate acceleration

  1. Mar 17, 2004 #1
    Hi! I'm quite new here, and I'm not sure wheter this is to go here or in the math section, so I'll just post it here since i guess nobody really care anyway.

    I've been thinking of this for a while, but I can't seem to get it right. (I'm not that good at maths but I'm learning.)

    Concider dropping a rock from a rather high altitude down to the ground. Now, using Newtonian theory, find an expression for the rocks' velocity as a function of time, that includes the fact that the gravitational attraction must become greater as the distance to Earth shrinks.

    Can somebody give me a hint?

    I know g=MG/r^2
    What confuses me is how to integrate time in this expression,
    since r=r0-gt^2/2

    Finally, I don't know why but i just like these guys:
     
  2. jcsd
  3. Mar 17, 2004 #2
    You need to set up a differential equation relating the distance from the earth to the acceleration and solve it.
     
  4. Mar 17, 2004 #3

    HallsofIvy

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    Letting r be the distance from the center of the earth,
    [tex]m\frac{d^2r}{dt^2}= \frac{-GmM}{r^2} [/tex]

    That's a non-linear differential equation but we can use the fact that t does not appear explicitely in it: Let v= dr/dt. Then (chain rule):
    [tex]\frac{d^2r}{dt^2}= \frac{dv}{dt}= \frac{dr}{dt}\frac{dv}{dr}= v\frac{dv}{dr}[/tex]
    so the differential equation becomes
    [tex]v\frac{dv}{dr}= \frac{-GM}{r^2}[/tex] and then separate:
    [tex]v dv= -GM \frac{dr}{r^2}[/tex]. Integrating:
    [tex]\frac{1}{2}v^2= \frac{GM}{r}+ C[/tex].

    (Notice that that first integral is the same as
    [tex]\frac{1}{2}v^2- GM/r= C [/tex], conservation of energy, since the first term is kinetic energy and the second potential energy.)

    That is the same as [tex]v= \frac{dr}{dt}= \sqrt{2\frac{GM}{r}+ C}[/tex] which is also integrable. You can use the fact that GM/r02= g (r0 is the radius of the earth) to simplify.
     
    Last edited: Mar 17, 2004
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