I had a related question in which I basically derived that equation without me knowing.
For this I did assumed that the following relation between the radius, $r$, and the true anomaly, $\theta$, is correct,
$$
r=\frac{a(1-e^2)}{1+e\cos{\theta}}
$$
with $a$ the semi-major axis and $e$ the eccentricity.
By using Kepler's second and third law you can get an expression for time (since periapsis passage) as a function of the true anomaly, which looks like this:
$$
t(\theta)=\sqrt{\frac{a^3}{\mu}}\left(2\tan^{-1}\left(\sqrt{\frac{1-e}{1+e}}\tan{\frac{\theta}{2}}\right)-\frac{e\sqrt{1-e^2}\sin{\theta}}{1+e\cos{\theta}}\right)
$$
This equations is equivalent to Kepler's equation, because the mean and eccentric anomaly are defined as followed:
$$
M=t\sqrt{\frac{\mu}{a^3}}
$$
$$
\tan{\frac{\theta}{2}}=\sqrt{\frac{1+e}{1-e}}\tan{\frac{E}{2}}
$$
A few substitutions are easy to see, namely how to get $M$ on the left hand side, by dividing by $\sqrt{\frac{a^3}{\mu}}$, and how to get the linear term of $E$, which allows use to get the following equation:
$$
M=E-\frac{e\sqrt{1-e^2}\sin{\theta}}{1+e\cos{\theta}}
$$
Proofing that the remaining term can be expressed as a product of the sine of $E$ and the eccentricity $e$ is harder, so
$$
\sin{E}=\frac{\sqrt{1-e^2}\sin{\theta}}{1+e\ \cos{\theta}}
$$
With the use of a temporary variable it is possible to proof this. This variable, lets call it $\alpha$, is defined as follows,
$$
\alpha = \tan \frac{\theta}{2} = \sqrt{\frac{1+e}{1-e}}\tan{\frac{E}{2}}.
$$
This way the trigonometry terms in the fraction become:
$$
\sin \theta = \sin \left( 2 \tan^{-1} \alpha\right) = \frac{2\alpha}{1+\alpha^2}
$$
$$
\cos \theta = \cos \left( 2 \tan^{-1} \alpha\right) = \frac{1-\alpha^2}{1+\alpha^2}
$$
Substituting these back into the fraction yields:
$$
\frac{\sqrt{1-e^2}\sin{\theta}}{1+e\ \cos{\theta}} = \frac{\sqrt{1-e^2}\frac{2\alpha}{1+\alpha^2}}{1+e\frac{1-\alpha^2}{1+\alpha^2}} = \frac{2\sqrt{1-e^2}\alpha}{1+e+(1-e)\alpha^2}
$$
Now by substituting in the expression for $\alpha$ which is an expression of $E$ gives:
$$
\frac{2\sqrt{1-e^2}\alpha}{1+e+(1-e)\alpha^2} = \frac{2\sqrt{(1+e)(1-e)}\sqrt{\frac{1+e}{1-e}}\tan{\frac{E}{2}}}{1+e+(1-e)\left(\sqrt{\frac{1+e}{1-e}}\tan{\frac{E}{2}}\right)^2} = \frac{2(1+e)\tan{\frac{E}{2}}}{(1+e)\left(1+\tan^2{\frac{E}{2}}\right)}
$$
In this last expression the dependency on the eccentrically, $e$, can be removed, which yields,
$$
\frac{2\tan{\frac{E}{2}}}{1+\tan^2{\frac{E}{2}}} = \frac{2\frac{\sin(E/2)}{\cos(E/2)}}{1+\frac{\sin^2(E/2)}{\cos^2(E/2)}} = \frac{2\frac{\sin(E/2)}{\cos(E/2)}}{\frac{\cos^2(E/2)+\sin^2(E/2)}{\cos^2(E/2)}} = 2\sin\frac{E}{2}\cos\frac{E}{2}=\sin E
$$
Therefore it can be shown that:
$$
\frac{\sqrt{1-e^2}\sin{\theta}}{1+e\ \cos{\theta}}=\sin{E}
$$
Best Answer
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.