Kepler orbit

Remarks

If the angular momentum $h = r \times r'$ is not zero, the motion lies in the plane through $0$ orthogonal to $h$. In polar coordinates $(\rho, \theta)$ in that plane, with $\theta$ measured from the periapsis, the path is the conic section $$\rho = \frac{p}{1 + e \cos \theta}, \qquad p = \frac{\norm{h}^2}{\mu},$$ with a focus at $0$ and eccentricity $e \ge 0$. It is an ellipse for $e < 1$, a parabola for $e = 1$ and a hyperbola for $e > 1$. If $h = 0$, the motion is along a line through $0$, a radial Kepler orbit.

Example: $r(t) = R (\cos \omega t, \sin \omega t, 0)$ with $R > 0$ and $\omega^2 R^3 = \mu$. Then $r'' = -\omega^2 r = -\mu\, r / R^3 = -\mu\, r / \norm{r}^3$, so the circle of radius $R$ is a Kepler orbit with $e = 0$.

Non-example: a solution of $r'' = -\mu\, r / \norm{r}^3 - \varepsilon\, r / \norm{r}^4$ with $\varepsilon \ne 0$ does not solve the equation of the Kepler problem, and in general its path is a rosette that does not close.