From 61b5ce20f25c5785e41574998a12c6d06eb05a5e Mon Sep 17 00:00:00 2001 From: Thomas Albers Date: Wed, 8 Mar 2023 23:43:00 +0100 Subject: Restructure build system and directory structures --- math/orbit.org | 93 ---------------------------------------------------------- 1 file changed, 93 deletions(-) delete mode 100644 math/orbit.org (limited to 'math/orbit.org') diff --git a/math/orbit.org b/math/orbit.org deleted file mode 100644 index 88079a6..0000000 --- a/math/orbit.org +++ /dev/null @@ -1,93 +0,0 @@ -#+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 -- cgit v1.2.3