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Astrophysics: Deriving Newton's Gravitational Formula from Kepler's

  1. Nov 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Dervie Newton's form of Kepler's third law.

    decrribing the orbital motion of two stars in circular orbits with masses M1 and M2, separation a, and period P

    ie.

    Obtain

    F=[tex]\frac{ G M1 M2 }{ a^2 }[/tex]

    From

    M1+M2=[tex]\frac{4 \pi^2 a^3 }{GP^2}[/tex]



    2. Relevant equations

    Centre of mass:

    M1r1 = M2r2

    a = r1 + r2

    P = [tex]\frac{2\pi r}{v}[/tex]


    3. The attempt at a solution

    [not to good at this LaTeX thing so i'll wing it]

    1: switch the (M1 + M2) For P^2

    P^2 = (4π^2 a^3)/(G(M1 + M2))

    switch the P for the term above:

    (4π^2 r^2)/v^2 = (4π^2 a^3)/(G(M1 + M2))

    π's cancel:

    r^2/v^2 = a^3 / G(M1 + M2)

    problem is here that i dont know what the r is, do i have to work out this for r1 and r2 seperatly?

    anyone done this before that could point me in the right direction?
     
  2. jcsd
  3. Nov 30, 2009 #2
  4. Dec 1, 2009 #3
    Thanks, I think i got it, it looks a bit eggy but i got there, there may be something wrong.

    M = [tex]\frac{4\pi^2r^3}{GP^2}[/tex]

    P2 = [tex]\frac{4\pi^2r^3}{GM}[/tex]

    [tex]\frac{4\pi^2r^2}{v^2}[/tex] = [tex]\frac{4\pi^2r^3}{GM}[/tex]

    [tex]\frac{r^2}{v^2}[/tex] = [tex]\frac{r^3}{GM}[/tex]

    [tex]\Rightarrow[/tex] acceleration = [tex]\frac{v^2}{r}[/tex]

    a = centrepetal acceleration

    [tex]\frac{r}{a}[/tex] = [tex]\frac{r^3}{GM}[/tex]

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

    [tex]\Rightarrow[/tex] F= ma

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