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The real freefall

  1. Aug 27, 2015 #1
    We always consider the accelaration as a constant thing, while calculating the freefall problems. What if we try to calculate the real. I mean taking the accelaration GM/x^2. I tried it but i could not handle that integral(Just graduated from high school). I would be happy if you reply.
    Thank you
     
  2. jcsd
  3. Aug 27, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    There are many problems that do not. We should have some older homework problems discussing this in our homework section.

    Every satellite orbit is a related problem.
     
  4. Aug 27, 2015 #3

    Philip Wood

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    There's a trick… Assuming no tangential component to the body's motion (as would be the case if it were released from rest) then, working with radial components of force and velocity, Newton's second law gives
    [tex]\frac{GMm}{r^2}=-m\frac{dv}{dt}[/tex]
    So[tex]\int{\frac{GMm}{r^2}}dr=-m\int{\frac{dv}{dt}dr}[/tex]
    But [itex]\frac{dr}{dt}=v[/itex], so
    [tex]\int{\frac{GMm}{r^2}}dr=-m\int{v\ dv}[/tex]

    Both these integrations are easy. Either put limits in, or leave as indefinite integrals and find the value of the arbitrary constant afterwards.

    You may well now realise that the result follows immediately from energy conservation. What I did above is to establish the [itex]\frac{1}{2}m\ v^2[/itex] kinetic energy formula (and that for gravitational PE due to a spherically symmetric body) from first principles, because doing this seemed more in the spirit of your question than simply quoting energy formulae.
     
    Last edited: Aug 27, 2015
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