# motion in central-force field

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

Equip the plane of the motion with a polar coordinate system $r,\,\varphi$ and denote the position vector of the body by $\vec{r}$.  Then the velocity vector is

 $\displaystyle\vec{v}\;=\;\frac{d\vec{r}}{dt}\;=\;\frac{d}{dt}(r\vec{r}^{\,0})% \;=\;\frac{dr}{dt}\vec{r}^{\,0}+r\frac{d\varphi}{dt}\vec{s}^{\,0},$ (1)

where $\vec{r}^{\,0}$ and $\vec{s}^{\,0}$ are the unit vectors  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 kinetic energy 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 angular momentum

 $\vec{L}\;=\;\vec{r}\!\times\!m\frac{d\vec{r}}{dt}\;=\;mr^{2}\frac{d\varphi}{dt% }\vec{r}^{\,0}\!\times\!\vec{s}^{\,0}$

of the body is constant; thus its magnitude is a constant,

 $mr^{2}\frac{d\varphi}{dt}\;=\;G,$

whence

 $\displaystyle\frac{d\varphi}{dt}\;=\;\frac{G}{mr^{2}}.$ (2)

The central force  $\displaystyle\vec{F}:=-\frac{k}{r^{2}}\vec{r}^{\,0}$  (where $k$ is a constant) has the scalar potential  $U(r)=-\frac{k}{r}$.  Thus the total energy  $E=E_{k}\!+\!U(r)$ of the body, which is constant, may be written

 $E\;=\;\frac{1}{2}m\!\left(\frac{dr}{dt}\right)^{2}\!+\frac{1}{2}mr^{2}\!\left(% \frac{G}{mr^{2}}\right)^{2}\!-\!\frac{k}{r}\;=\;\frac{m}{2}\!\left(\frac{dr}{% dt}\right)^{2}\!+\frac{G^{2}}{2mr^{2}}\!-\!\frac{k}{r}.$

This equation may be revised to

 $\left(\frac{dr}{dt}\right)^{2}\!+\frac{G^{2}}{m^{2}r^{2}}-\frac{2k}{mr}+\frac{% k^{2}}{G^{2}}\;=\;\frac{2E}{m}+\frac{k^{2}}{G^{2}},$

i.e.

 $\left(\frac{dr}{dt}\right)^{2}\!+\left(\frac{k}{G}-\frac{G}{mr}\right)^{2}\;=% \;q^{2}$

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

 $\displaystyle\frac{k}{G}-\frac{G}{mr}\;=\;q\cos\psi,\quad\frac{dr}{dt}\;=\;q% \sin\psi.$ (3)
 $\frac{G}{mr^{2}}\cdot\frac{dr}{dt}\;=\;-q\sin\psi\frac{d\psi}{dt}\;=\;-\frac{% dr}{dt}\cdot\frac{d\psi}{dt},$

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

 $\displaystyle r\;=\;\frac{G^{2}}{km\left(1-\frac{Gq}{k}\cos(C-\varphi)\right)}% \;=\;\frac{p}{1-\varepsilon\cos(\varphi-C)},$ (4)

where

 $p\;:=\;\frac{G^{2}}{km},\quad\varepsilon\;:=\;\frac{Gq}{k}.$

By the http://planetmath.org/node/11724parent entry, the result (4) shows that the trajectory of the body in the gravitational field (http://planetmath.org/VectorField) of one point-like sink is always a conic section  whose focus the sink causing the field.

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.

## References

• 1 Я. Б. Зельдович &  А. Д. Мышкис: Элементы  прикладной  математики.  Издательство  ‘‘Наука’’.  Москва (1976).
Title motion in central-force field MotionInCentralforceField 2013-03-22 18:52:41 2013-03-22 18:52:41 pahio (2872) pahio (2872) 14 pahio (2872) Derivation msc 15A72 msc 51N20 Kepler’s first law CommonEquationOfConics PropertiesOfEllipse