How can one derive Newton’s gravitational $\frac{1}{r^2}$ law by considering that planets orbit the Sun in elliptical trajectories?
In fact, even before Newton, Kepler’s laws already stated that planets move around the Sun in elliptical orbits. Here we will make the following assumption: we will assume that celestial bodies attract each other according to the $\frac{1}{r^2}$ law, and we will derive the formula for the trajectory traced by a planet and verify that it is indeed an ellipse.
$$ (2a - r)^{2} = (2ea)^{2} + r^{2} - 4earcos(\pi - \theta)$$
$$ r(\theta) = \frac{a(1 - e^2)}{1 + ecos(\theta)}$$
$$ b^2 = a^2 (1-e^2)$$
$$ E= \frac{m \dot{r}^2}{2} + \frac{m\dot{\theta}^2r^2}{2} - G\frac{Mm}{r} = const \tag{1}$$
$$ L = m \dot{\theta}r^2 = const \tag{2} $$
$$ \dot{r}^2 = \frac{2E}{m} - \frac{L^2}{m^2r^2} + \frac{2GM}{r} \tag{3} $$
$$ \dot{r}^2 = (r'(\theta) \dot{\theta})^2 = \frac{L^2}{m^2r^4} (r'(\theta))^2 \tag{4} $$
$$ \frac{1}{r^4} (r'(\theta))^2 = \frac{2Em}{L^2} - \frac{1}{r^2} + 2\frac{GMm^2}{L^2r} \tag{5} $$
$$ \rho = \frac{L^2}{GMm^2}\frac{1}{r} \tag{6} $$
$$ (\frac{GMm^2}{L^2})^2(\rho'(\theta))^2 = \frac{2Em}{L^2} - (\frac{GMm^2}{L^2})^2(\rho^2 - 2\rho) \tag{7} $$
$$ (\rho'(\theta))^2 = \frac{2Em}{L^2} (\frac{L^2}{GMm^2})^2 - \rho^2 + 2\rho \tag{8} $$
$$ (\rho'(\theta))^2 = \frac{2EL^2}{G^2 M^2 m^3} - \rho^2 + 2\rho \tag{9} $$
$$ 2\rho''\rho' = - 2\rho\rho' + 2\rho' \tag{10} $$
$$ \rho'' = - \rho + 1 \tag{11} $$
$$ \rho(\theta) = 1 + ecos(\theta) \tag{12} $$
$$ r(\theta) = \frac{L^2}{GMm^2} \frac{1}{(1 + ecos(\theta))} \tag{13} $$
$$ e = \sqrt{1+ \frac{2EL^2}{G^2 M^2 m^3}} \tag{14} $$
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