# Homework Help: Astrophysics: Deriving Newton's Gravitational Formula from Kepler's

1. Nov 30, 2009

### knowlewj01

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

From

M1+M2=$$\frac{4 \pi^2 a^3 }{GP^2}$$

2. Relevant equations

Centre of mass:

M1r1 = M2r2

a = r1 + r2

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

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. Nov 30, 2009

3. Dec 1, 2009

### 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 = $$\frac{GMm}{r^2}$$