Kepler's first law; mathematical way of thinking I have Kepler's first law of planetary motion:
$$\text{r}=\frac{\text{p}}{1+\epsilon\cos\left(\theta\right)}\tag1.$$
Now, for $\epsilon$ I have:
$$0<\epsilon=\sqrt{1+\frac{2\cdot\eta\cdot\text{h}^2}{\mu^2}}<1\tag,$$
because it is an ellipse.
Now, for $\mu$:
$$\mu=\text{G}\cdot\text{M}\tag3,$$
and for $\eta$:
$$\eta=-\frac{\mu}{2\text{a}}\tag4,$$
where $\text{a}=\frac{\text{r}_\text{min}+\text{r}_\text{max}}{2}$.
Questions: 


*

*For the Earth's orbit around the sun, which mass ($\text{M}$) should I pick, the mass of the sun or the mass of the earth? 

*What is $\text{h}$ in the equation for $\epsilon$ and how can I find the value for it?

*What are the values of $\text{r}_\text{min}$ and $\text{r}_\text{max}$?

 A: *

*$M$ stands for the mass of the Sun.

*$h$ is the angular momentum. It is defined as $\vec h=\vec r\times\vec p$, where $\vec p$ is the linear momentum. For central forces (such as the gravitational) it is conserved and perpendicular to the plane of orbit. Therefore if you know the velocity $\vec v$ of the planet as well as its radius vector $\vec r$ at a some instant, you know the angular momentum everywhere in the orbit. You can also compute its magnitude from Kepler's second law if you know the area $A$ and period $T$ of the orbit. They are related as
$$A=\frac{hT}{2m},$$
where $m$ is the mass of the planet.

*$r_{min}$ and $r_{max}$ are minimum and maximum distance from the planet to the Sun. It can be computed to be
$$r_{\pm}=\frac{r_0}{1\pm\sqrt{1-\frac{E}{E_0}}},$$
where the sign plus (minus) refers to the minimum (maximum) distance. The constant $E$ is the mechanical energy of the planet. It is a free parameter, you can write the eccentricity of the orbit in terms of it,
$$\epsilon=\sqrt{1-\frac{E}{E_0}}.$$
The constant $E_0$ is the smallest energy a orbit can have (circular motion),
$$E_0=\frac{-G^2M^2m^3}{2h^2}.$$
Finally the constant $r_0$ is the radius of the circular orbit,
$$r_0=\frac{h^2}{GMm^2}.$$
