#+TITLE:Orbit
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* Disclaimer
This site as of now just a technology demonstration and its claims
should not be taken as true (even though I myself am pretty confident
they are)
* Deriving Kepler's first law from Newton's law of universal gravitation

The movement of an object with mass $m$ orbiting another body with mass $M$ is
given by Newton's law of gravitation. If $m \ll M$ it is possible to consider
the position of the larger object constant and use it as the origin of our
coordinate system. Then the following equation applies for movement of the
smaller object:

\begin{equation*}
m\ddot{\bm{r}} = - \frac{GMm}{r^3}\bm{r} \Leftrightarrow \ddot{\bm{r}} = - \frac{GM}{r^3}\bm{r}
\end{equation*}

In order to solve this differential equation we first consider the angular
momentum of or object around its orbit.

\begin{equation*}
\bm{L} = \bm{r} \times m \dot{\bm{r}}
\end{equation*}

In the abscense of external toques, because the only force acting on the object
is parallel to its position, the angular momentum is conserved.

\begin{equation*}
\deriv{\bm{L}}{t} = \dot{\bm{r}} \times m \dot{\bm{r}} + \bm{r} \times m \ddot{\bm{r}} = 0
\end{equation*}

We now multiply both sides of our equation from the right by the angular
momentum and develop the right side of the equation using vector identities.

\begin{align*}
\ddot{\bm{r}} \times \bm{L} &= -GM\frac{\bm{r} \times \left(\bm{r} \times m \dot{\bm{r}}\right)}{r^3}\\
&= - \frac{GMm}{r^3} \left(\left(\bm{r} \cdot \dot{\bm{r}}\right)\bm{r} - \left(\bm{r} \cdot \bm{r}\right)\dot{\bm{r}}\right)\\
&= GMm\left(\frac{\left(\bm{r} \cdot \bm{r}\right)\dot{\bm{r}}}{r^3} - \frac{\left(\bm{r} \cdot \dot{\bm{r}}\right)\bm{r}}{r^3}\right)\\
&= GMm\left(\frac{\dot{\bm{r}}}{r} - \frac{\left(\bm{r} \cdot \dot{\bm{r}}\right)\bm{r}}{r^3}\right)\\
&= GMm\left(\frac{1}{r} \deriv{\bm{r}}{t} + \deriv{}{t}\left(\frac{1}{r}\right)\bm{r}\right)\\
&= GMm\deriv{}{t}\left(\frac{\bm{r}}{r}\right)
\end{align*}

We observe that each side of our equation is a derivative of a quantity. We know
integrate both sides and take the integrations constant into account.

\begin{align*}
& \deriv{}{t} \left(\dot{\bm{r}} \times \bm{L}\right) = GM \deriv{}{t}\left(\frac{\bm{r}}{r}\right)\\
\Leftrightarrow \quad & \dot{\bm{r}} \times \bm{L} = GM \frac{\bm{r}}{r} + \bm{a}
\end{align*}

Our objective is now to solve the equation for $r$, so we multiply both sides by $\bm{r}$:

\begin{align*}
\dot{\bm{r}} \times \left(\bm{r} \times \dot{\bm{r}}\right) &= GM \frac{\bm{r}}{r} + \bm{a}\\
\bm{r} \cdot \left(\dot{\bm{r}} \times \left(\bm{r} \times \dot{\bm{r}}\right)\right) &= GMr + \bm{r} \cdot \bm{a}
\end{align*}

By applying a cyclic permutation of the resulting triple product and using the
known property of the scalar product we now express the equation only in terms
of the magnitudes of the vectors.

\begin{align*}
\left(\bm{r} \times \dot{\bm{r}}\right) \cdot \left(\bm{r} \times \dot{\bm{r}}\right) &= GMr + \bm{r} \cdot \bm{a}\\
\left(\frac{L}{m}\right)^2 &= GMr + \bm{r} \cdot \bm{a}\\
\left(\frac{L}{m}\right)^2 &= GMr + r a \cos\theta
\end{align*}

The last steps are to solve for $r$

\begin{align*}
r &= \left(\frac{L}{m}\right)^2 \frac{1}{GM + a \cos\theta}\\
&= \left(\frac{L}{m}\right)^2 \frac{1}{GM} \frac{1}{1 + \frac{a}{GM} \cos\theta}\\
&= \left(\frac{L}{m}\right)^2 \frac{1}{GM} \frac{1}{1 + e\cos\theta}
\end{align*}

Finally we reach our result. Objects orbiting according to Newton's Law of
Gravitation follow paths that correspond to the conic sections. Here is Kepler's
first Law a special case, where our object has a stable orbit around the larger
body.

\begin{equation*}
r &= \frac{L^2}{GM m^2} \frac{1}{1 + e\cos\theta}
\end{equation*}

* Deriving a physical interpretation of the excentricity of the orbit