From 561bac75579391c14e47eaccfabdf9eda98855da Mon Sep 17 00:00:00 2001
From: Thomas Albers <thomas@thomaslabs.org>
Date: Wed, 27 Jul 2022 18:13:20 +0200
Subject: Initial commit

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+#+TITLE:Orbit
+#+SETUPFILE: ../math_options.org
+#+LATEX_HEADER: \usepackage{bm}
+#+LATEX_HEADER: \usepackage{mathtools}
+#+LATEX_HEADER: \usepackage{amssymb}
+#+LATEX_HEADER: \newcommand{\deriv}[2]{\frac{\text{d}#1}{\text{d}#2}}
+#+LATEX_HEADER: \newcommand{\unitv}[1]{\bm{\hat{e}}_#1}
+
+* 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
-- 
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