Eccentricity and Keplerian orbit

Question

A particle is moving in elliptical orbit, with ecccentricity $e$. Let $r(t)$ be the distance of the particle from one focus. Seeing this as a perturbed circular motion, one find that $r(t)=\frac{1}{2}(1-e\cos(t))+O(e^2)$. Now every book or article says that this is a known fact, and moreover some authors say $e$ is the eccentricity while some others say it's eccentric anomaly (https://en.wikipedia.org/wiki/Eccentric_anomaly). How can one arrives to this expression? Thanks in advance.

Answer 1

The "exact" normalized expression for the polar equation of conic section curve with the Sun placed at the origin is $$r(t)=\frac{r_0}{1+e \cos(t)}.$$

(here, you have taken $r_0=\frac12$). See for that first Kepler's law in

https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion

Thus, consider expansion $$\frac{1}{1+\epsilon}\approx 1-\epsilon+\epsilon^2-...$$ and stop at order $2$.