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authorThomas Albers <thomas@thomaslabs.org>2023-03-08 23:43:00 +0100
committerThomas Albers <thomas@thomaslabs.org>2023-03-08 23:43:00 +0100
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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