How to derive inverse square relation in Newton's Law of Gravitation from Kepler's laws? I read Feynman's Lost lecture a while ago. In that spirit, I'm trying to do a more simplified version of coming up with the inverse square relation.
Using Kepler's 2nd,
$ A_1 = \frac{1}{2}r_1^2\theta_1 $
$ A_2 = \frac{1}{2}r_2^2\theta_2 $
$ A_1 = \frac{1}{2}r_1^2\omega_1 t $
$ A_2 = \frac{1}{2}r_2^2\omega_2 t$
$ m\omega_1 A_1 = \frac{1}{2}mr_1^2\omega_1^2 $
$ m\omega_2A_2 = \frac{1}{2}mr_2^2\omega_2^2 $
$ mv_1 A_1 = \frac{1}{2}r_1^2F_1 $
$ mv_2A_2 = \frac{1}{2}r_2^2F_2 $
Dividing one by the other, I get
$\frac{F_1}{F_2} = \frac{v_1}{v_2}\frac{r_2^2}{r_1^2}$
which is almost there, except for the $\frac{v_1}{v_2}$
Any suggestions on how to proceed from here?
 A: Consider a circular orbit (Kepler's first law tells us this is possible, as circles are particular cases of ellipses). By Kepler's second law, the speed $v$ is constant along the orbit. We can obtain its dependence on $r$ using Kepler's third law: $T^2\propto r^3$. The result is $v^2\propto 1/r$. 
For the orbit to be circular, the force should satisfy
\begin{equation}
  F(r)=m\frac{v^2}{r}\propto\frac{1}{r^2}.
\end{equation}

We can also find the direction of the force from Kepler's laws! We work in two dimensions because Kepler's first law tell us that the orbits stay in a plane. The acceleration in radial coordinates is
\begin{equation}
  \vec{a} =
  (\ddot{r}-r\dot{\theta}^2)\hat{r} +
  (r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta}.
\end{equation}
Notice that the $\hat{\theta}$ component of the acceleration is just $\frac{1}{r}\frac{d}{dt}(r^2\dot{\theta})$, and that $r^2\dot{\theta}$ is the areal velocity, which is constant by Kepler's second law. Therefore, the acceleration has the direction of $\hat{r}$, and so does the force. The latter should then be of the form
\begin{equation}
  \vec{F}(\vec{r})=-F(r)\hat{r},
\end{equation}
where the dependence only on $r$ and not $\theta$ is a consequence of the isotropy of space.
