PlanetPhysics/Motion in Central Force Field

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

Equip the plane of the motion with a polar coordinate [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|system]] $r,\phi$ and denote the [[../PositionVector/|position vector]] of the body by $\overrightarrow{r}$.\, Then the velocity vector is

$\begin{array}{c}\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\frac{d}{dt}(r{\overrightarrow{r}}^0)=\frac{dr}{dt}{\overrightarrow{r}}^0+r\frac{d\phi }{dt}{\overrightarrow{s}}^0,\end{array}$

where ${\overrightarrow{r}}^0$ and ${\overrightarrow{s}}^0$ are the [[../PureState/|unit vectors]] in the direction of $\overrightarrow{r}$ and of $\overrightarrow{r}$ rotated 90 degrees anticlockwise (${\overrightarrow{r}}^0=\overrightarrow{i}\cos \phi +\overrightarrow{j}\sin \phi$,\, whence\, </math>\frac{\vec{r}^{\,0}}{dt} = (-\vec{i}\sin\varphi+\vec{j}\cos\varphi)\frac{d\varphi}{dt} = \frac{d\varphi}{dt}\vec{s}^{\,0}$).Thusthe[[../KineticEnergy/|kineticenergy]]ofthebodyis<math>E_k=\frac{1}{2}m{\left|\frac{d\overrightarrow{r}}{dt}\right|}^2=\frac{1}{2}m({\left(\frac{dr}{dt}\right)}^2+{\left(r\frac{d\phi }{dt}\right)}^2).$ Because the gravitational force on the body is exerted along the position vector, its moment is 0 and therefore the [[../MolecularOrbitals/|angular momentum]] $\overrightarrow{L}=\overrightarrow{r}\times m\frac{d\overrightarrow{r}}{dt}=m{r}^2\frac{d\phi }{dt}{\overrightarrow{r}}^0\times {\overrightarrow{s}}^0$ of the body is constant; thus its [[../AbsoluteMagnitude/|magnitude]] is a constant, $m{r}^2\frac{d\phi }{dt}=G,$ whence

$\begin{array}{c}\frac{d\phi }{dt}=\frac{G}{m{r}^2}.\end{array}$

The central force\, $\overrightarrow{F}:=-\frac{k}{r^2}{\overrightarrow{r}}^0$\, (where $k$ is a constant) has the [[../Vectors/|scalar]] potential \, $U(r)=-\frac{k}{r}$.\, Thus the total [[../CosmologicalConstant/|energy\,]] $E=E_k+U(r)$ of the body, which is constant, may be written $E=\frac{1}{2}m{\left(\frac{dr}{dt}\right)}^2+\frac{1}{2}m{r}^2{\left(\frac{G}{m{r}^2}\right)}^2-\frac{k}{r}=\frac{m}{2}{\left(\frac{dr}{dt}\right)}^2+\frac{G^2}{2m{r}^2}-\frac{k}{r}.$ This equation may be revised to ${\left(\frac{dr}{dt}\right)}^2+\frac{G^2}{m^2{r}^2}-\frac{2k}{mr}+\frac{k^2}{G^2}=\frac{2E}{m}+\frac{k^2}{G^2},$ i.e. ${\left(\frac{dr}{dt}\right)}^2+{(\frac{k}{G}-\frac{G}{mr})}^2=q^2$ where $q:=\sqrt{\frac{2}{m}(E+\frac{m{k}^2}{2G^2})}$ is a constant.\, We introduce still an auxiliary angle $\psi$ such that

$\begin{array}{c}\frac{k}{G}-\frac{G}{mr}=q\cos \psi ,\frac{dr}{dt}=q\sin \psi .\end{array}$

Differentiation of the first of these equations implies $\frac{G}{m{r}^2}\cdot \frac{dr}{dt}=-q\sin \psi \frac{d\psi }{dt}=-\frac{dr}{dt}\cdot \frac{d\psi }{dt},$ whence, by (2), $\frac{d\psi }{dt}=-\frac{G}{m{r}^2}=-\frac{d\phi }{dt}.$ This means that\, $\psi =C-\phi$, where the constant $C$ is determined by the initial conditions.\, We can then solve $r$ from the first of the equations (3), obtaining

$\begin{array}{c}r=\frac{G^2}{km(1-\frac{Gq}{k}\cos (C-\phi ))}=\frac{p}{1-\epsilon \cos (\phi -C)},\end{array}$

where $p:=\frac{G^2}{km},\epsilon :=\frac{Gq}{k}.$\\

The result (4) shows that the trajectory of the body in the gravitational [[../VectorField/|field]] of one point-like sink is always a conic [[../IsomorphicObjectsUnderAnIsomorphism/|section]] whose focus contains the sink causing the [[../CosmologicalConstant/|field]].\\

As for the type of the conic, the most interesting one is an ellipse.\, It occurs when\, $\epsilon <1$.\, 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 Kepler's first law .