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

[knowlewj01]

Homework Statement

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=\frac{ G M1 M2 }{ a^2 }

From

M₁+M₂=\frac{4 \pi^2 a^3 }{GP^2}

Homework Equations

Centre of mass:

M₁r₁ = M₂r₂

a = r₁ + r₂

P = \frac{2\pi r}{v}

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 don't 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?

 

[ApexOfDE]

http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

Section 'Deriving K's 3rd law', I see that P = 2pir^3/2 / sqrt(GM).

Combining with centripental force, you can derive N's form.

 

[knowlewj01]

Thanks, I think i got it, it looks a bit eggy but i got there, there may be something wrong.

M = \frac{4\pi^2r^3}{GP^2}

P² = \frac{4\pi^2r^3}{GM}

\frac{4\pi^2r^2}{v^2} = \frac{4\pi^2r^3}{GM}

\frac{r^2}{v^2} = \frac{r^3}{GM}

\Rightarrow acceleration = \frac{v^2}{r}

a = centrepetal acceleration

\frac{r}{a} = \frac{r^3}{GM}

a = \frac{GM}{r^2}

\Rightarrow F= ma

F = \frac{GMm}{r^2}