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Fall of Object

  1. Jan 25, 2006 #1
    If it takes an object at distance x from the surface of the Earth 17.5 hours to fall from space onto the surface of the Earth, how can we calculate x? The object initially has zero speed.

    I find this question hard because I've only done problems involving acceleration as a function of time, and not as a function of distance. :redface: Can anyone show me how to derive an equation of acceleration as a function of time?
  2. jcsd
  3. Jan 25, 2006 #2


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    You can use [tex] \frac{dv}{dt} = (\frac{dv}{dr})(\frac{dr}{dt}) = v \frac{dv}{dr} [/tex]
    So I think from this you can first find v(r) and then r(t) and use that to find x.
  4. Jan 25, 2006 #3
    The first thing you may want to realize is that acceleration is a function of distance between the object and the center of the earth, and that the distance is a function of time.
  5. Jan 25, 2006 #4
    I know [tex]\frac{dv}{dt} = \frac{GM}{r^2}[/tex] but I can't figure out how this may be used to obtain v as a function of r.
    How is it possible to solve the equation [tex]\frac{d^{2}r}{dt^2}=\frac{GM}{(R+r)^2}[/tex], where R is the radius of the Earth? I have not had much calculus yet, so I have not touched on solving differential equations of this difficulty.
    Last edited: Jan 25, 2006
  6. Jan 25, 2006 #5


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    You have

    [tex]\frac{dv}{dt} = \frac{GM}{r^2}[/tex]

    Now, as I said in my previous post,

    [tex] \frac{dv}{dt} = (\frac{dv}{dr})(\frac{dr}{dt}) = v \frac{dv}{dr} [/tex]

    [tex]v\frac{dv}{dr} = \frac{GM}{r^2}[/tex]

    Now, can you find v(r) from this diff equation?
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