Nabeshin, your argument works the other way too.
For a circular orbit, a = v^2/r (that's universal, nothing to do with gravitation). Plug in
v=\omega r = \frac{2\pi r}{T}
and obtain
a = \frac{4 \pi^2 r}{T^2}
On the other hand, from Kepler's law, T^2 ~ r^3, so
F = ma \propto \frac{4 \pi^2...