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Law of Universal Gravitation

  1. Nov 23, 2009 #1
    Update: Thanks for the help! I'll work on it....
     
    Last edited: Nov 23, 2009
  2. jcsd
  3. Nov 23, 2009 #2
    I'm not sure what your equation is, but this is how I would derive Newton's Law

    [tex]T^2 = \frac{4\pi^2}{GM}R^3[/tex]

    In circular motion,

    [tex]F = \frac{mv^2}{R}[/tex]

    The period in circular motion is

    [tex]T = \frac{2\pi R}{v} \Rightarrow v = \frac{2\pi R}{T}[/tex]

    So, by substitution,

    [tex]F = \frac{m\left(\frac{2\pi R}{T}\right)^2}{R}=\frac{4\pi^2 mR}{T^2} [/tex]

    [tex]\Rightarrow T^2 = \frac{4\pi^2 mR}{F}[/tex]

    Substituting into Kepler's 3rd,

    [tex]\frac{4 \pi ^2 mR}{F} = \frac{4\pi^2}{GM}R^3[/tex]

    [tex]\therefore F = \frac{GMm}{R^2}[/tex]
     
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