Talk:PlanetPhysics/Motion in Central Force Field

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: motion in central-force field %%% Primary Category Code: 45.50.Pk %%% Filename: MotionInCentralForceField.tex %%% Version: 6 %%% Owner: pahio %%% Author(s): pahio %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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Let us consider a body with mass $m$ in a gravitational force \htmladdnormallink{field}{http://planetphysics.us/encyclopedia/VectorField.html} exerted by the origin and directed always from the body towards the origin.\, Set the plane through the origin and the \htmladdnormallink{velocity}{http://planetphysics.us/encyclopedia/Velocity.html} \htmladdnormallink{vector}{http://planetphysics.us/encyclopedia/Vectors.html} $\vec{v}$ of the body.\, Apparently, the body is forced to move constantly in this plane, i.e. there is a question of a planar \htmladdnormallink{motion}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html}.\, We want to derive the trajectory of the body.

Equip the plane of the motion with a polar coordinate \htmladdnormallink{system}{http://planetphysics.us/encyclopedia/SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence.html} $r,\,\varphi$ and denote the \htmladdnormallink{position vector}{http://planetphysics.us/encyclopedia/PositionVector.html} of the body by $\vec{r}$.\, Then the velocity vector is \begin{align} \vec{v} \;=\; \frac{d\vec{r}}{dt} \;=\; \frac{d}{dt}(r\vec{r}^{\,0}) \;=\; \frac{dr}{dt}\vec{r}^{\,0}+r\frac{d\varphi}{dt}\vec{s}^{\,0}, \end{align} where $\vec{r}^{\,0}$ and $\vec{s}^{\,0}$ are the \htmladdnormallink{unit vectors}{http://planetphysics.us/encyclopedia/PureState.html} in the direction of $\vec{r}$ and of $\vec{r}$ rotated 90 degrees anticlockwise ($\vec{r}^{\,0} = \vec{i}\cos\varphi+\vec{j}\sin\varphi$,\, whence\, $\frac{\vec{r}^{\,0}}{dt} = (-\vec{i}\sin\varphi+\vec{j}\cos\varphi)\frac{d\varphi}{dt} = \frac{d\varphi}{dt}\vec{s}^{\,0}$).\, Thus the \htmladdnormallink{kinetic energy}{http://planetphysics.us/encyclopedia/KineticEnergy.html} of the body is $$E_k \;=\; \frac{1}{2}m\left|\frac{d\vec{r}}{dt}\right|^2 \;=\; \frac{1}{2}m\left(\!\left(\frac{dr}{dt}\right)^2\!+\!\left(r\frac{d\varphi}{dt}\right)^2\right)\!.$$ Because the gravitational force on the body is exerted along the position vector, its moment is 0 and therefore the \htmladdnormallink{angular momentum}{http://planetphysics.us/encyclopedia/MolecularOrbitals.html} $$\vec{L} \;=\; \vec{r}\!\times\!m\frac{d\vec{r}}{dt} \;=\; mr^2\frac{d\varphi}{dt}\vec{r}^{\,0}\!\times\!\vec{s}^{\,0}$$ of the body is constant; thus its \htmladdnormallink{magnitude}{http://planetphysics.us/encyclopedia/AbsoluteMagnitude.html} is a constant, $$mr^2\frac{d\varphi}{dt} \;=\; G,$$ whence \begin{align} \frac{d\varphi}{dt} \;=\; \frac{G}{mr^2}. \end{align} The central force\, $\displaystyle\vec{F} := -\frac{k}{r^2}\vec{r}^{\,0}$\, (where $k$ is a constant) has the \htmladdnormallink{scalar}{http://planetphysics.us/encyclopedia/Vectors.html} potential \, $U(r) = -\frac{k}{r}$.\, Thus the total \htmladdnormallink{energy\,}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html} $E = E_k\!+\!U(r)$ of the body, which is constant, may be written $$E \;=\; \frac{1}{2}m\!\left(\frac{dr}{dt}\right)^2\!+\frac{1}{2}mr^2\!\left(\frac{G}{mr^2}\right)^2\!-\!\frac{k}{r} \;=\; \frac{m}{2}\!\left(\frac{dr}{dt}\right)^2\!+\frac{G^2}{2mr^2}\!-\!\frac{k}{r}.$$ This equation may be revised to $$\left(\frac{dr}{dt}\right)^2\!+\frac{G^2}{m^2r^2}-\frac{2k}{mr}+\frac{k^2}{G^2} \;=\; \frac{2E}{m}+\frac{k^2}{G^2},$$ i.e. $$\left(\frac{dr}{dt}\right)^2\!+\left(\frac{k}{G}-\frac{G}{mr}\right)^2 \;=\; q^2$$ where $$q \;:=\; \sqrt{\frac{2}{m}\left(\!E\!+\!\frac{mk^2}{2G^2}\right)}$$ is a constant.\, We introduce still an auxiliary angle $\psi$ such that \begin{align} \frac{k}{G}-\frac{G}{mr} \;=\; q\cos\psi, \quad \frac{dr}{dt} \;=\; q\sin\psi. \end{align} Differentiation of the first of these equations implies $$\frac{G}{mr^2}\cdot\frac{dr}{dt} \;=\; -q\sin\psi\frac{d\psi}{dt} \;=\; -\frac{dr}{dt}\cdot\frac{d\psi}{dt},$$ whence, by (2), $$\frac{d\psi}{dt} \;=\; -\frac{G}{mr^2} \;=\; -\frac{d\varphi}{dt}.$$ This means that\, $\psi = C\!-\!\varphi$, where the constant $C$ is determined by the initial conditions.\, We can then solve $r$ from the first of the equations (3), obtaining \begin{align} r \;=\; \frac{G^2}{km\left(1-\frac{Gq}{k}\cos(C-\varphi)\right)} \;=\; \frac{p}{1-\varepsilon\cos(\varphi-C)}, \end{align} where $$p \;:=\; \frac{G^2}{km}, \quad \varepsilon \;:=\; \frac{Gq}{k}.$$\\

The result (4) shows that the trajectory of the body in the gravitational \htmladdnormallink{field}{http://planetphysics.us/encyclopedia/VectorField.html} of one point-like sink is always a conic \htmladdnormallink{section}{http://planetphysics.us/encyclopedia/IsomorphicObjectsUnderAnIsomorphism.html} whose focus contains the sink causing the \htmladdnormallink{field}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html}.\\

As for the type of the conic, the most interesting one is an ellipse.\, It occurs when\, $\varepsilon < 1$.\, This condition is easily seen to be equivalent with a negative total energy $E$ of the body.\\

One can say that any planet revolves around the Sun along an ellipse having the Sun in one of its foci --- this is \emph{Kepler's first law}.

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