Law 3

We'll show this is true only for circular orbits. To derive this for general elliptical orbits requires more advanced math than we're using at present.

So let's apply ${\bf F}~=~ m{\bf a}$ to a circular orbit. The acceleration points towards the sun and has a magnitude $\omega^2 R$. The gravitational force points in the same direction. Setting $ma$ equal to 1.1 we have

$$F ~=~ G {Mm\over R^2} ~=~ ma ~=~ m \omega^2 R$$ (1.10)

Solving for $\omega$

$$\omega^2 ~=~ G {M\over R^3}$$ (1.11)

The period $T$ is related to $\omega$ through

$$\omega ~=~ {2\pi\over T}$$ (1.12)

Substituting for $T$ we finally have

$$T^2 ~=~ {4\pi^2 \over GM}R^3$$ (1.13)

Well this is Kepler's third law with the constant of proportionality explicity calculated.