NIF: Kepler part IV

["Rauber, Joel Phys" <RAUBERJ@mg.sdstate.edu>]

Not having seen any objections so far I continue:

> From Kepler part III, Take equation (4) and subtract equation (3) from it. 
to get.

(5) (-G(m_1+m_2))(r_2 - r_1)/(ABS(r_2-r_1)^3 + C*(r_2 - r_1) = 0

Rearrange to get

(6) G(m_1+m_2))(r_2 - r_1)/(ABS(r_2-r_1)^3 = C*(r_2 - r_1)

Equation (6) expresses an equality between vectors, therefore the magnitudes 
of the lhs and rhs equal.

So we get

(7)   G(m_1+m_2))/(ABS(r_2-r_1)^3 = C

But C = 4 pi^2/(period)^2, so substituting and performing the requisite 
multiplications give. (see, parts I and II)

(8)    Period^2 = (4 Pi^2 * ABS (r_2 - r_1)^3 / (G(m_1 + m_2))

which is Keplers 3rd law.

Joel Rauber
rauberj@mg.sdstate.edu