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 $\overrightarrow{v}$ of the body.  Apparently, the body is forced to move constantly in this plane, i.e. there is a question of a planar motion.  We want to derive the trajectory of the body.

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

$\overrightarrow{v}={\displaystyle \frac{d\overrightarrow{r}}{dt}}={\displaystyle \frac{d}{dt}}(r{\overrightarrow{r}}^{\mathrm{\hspace{0.17em}0}})={\displaystyle \frac{dr}{dt}}{\overrightarrow{r}}^{\mathrm{\hspace{0.17em}0}}+r{\displaystyle \frac{d\phi }{dt}}{\overrightarrow{s}}^{\mathrm{\hspace{0.17em}0}},$(1)

where ${\overrightarrow{r}}^{\mathrm{\hspace{0.17em}0}}$ and ${\overrightarrow{s}}^{\mathrm{\hspace{0.17em}0}}$ are the unit vectors in the direction of $\overrightarrow{r}$ and of $\overrightarrow{r}$ rotated 90 degrees anticlockwise (${\overrightarrow{r}}^{\mathrm{\hspace{0.17em}0}}=\overrightarrow{i}\cos \phi +\overrightarrow{j}\sin \phi$,  whence  $\frac{{\overrightarrow{r}}^{\mathrm{\hspace{0.17em}0}}}{dt}=(-\overrightarrow{i}\sin \phi +\overrightarrow{j}\cos \phi )\frac{d\phi }{dt}=\frac{d\phi }{dt}{\overrightarrow{s}}^{\mathrm{\hspace{0.17em}0}}$).  Thus the kinetic energy of the body is

$$E_k=\frac{1}{2}m{\left|\frac{d\overrightarrow{r}}{dt}\right|}^2=\frac{1}{2}m\left({\left(\frac{dr}{dt}\right)}^2+{\left(r\frac{d\phi }{dt}\right)}^2\right).$$

Because the gravitational force on the body is exerted along the position vector, its moment is 0 and therefore the angular momentum

$$\overrightarrow{L}=\overrightarrow{r}\times m\frac{d\overrightarrow{r}}{dt}=m{r}^2\frac{d\phi }{dt}{\overrightarrow{r}}^{\mathrm{\hspace{0.17em}0}}\times {\overrightarrow{s}}^{\mathrm{\hspace{0.17em}0}}$$

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

$$m{r}^2\frac{d\phi }{dt}=G,$$

whence

${\displaystyle \frac{d\phi }{dt}}={\displaystyle \frac{G}{m{r}^2}}.$(2)

The central force  $\overrightarrow{F}:=-{\displaystyle \frac{k}{r^2}}{\overrightarrow{r}}^{\mathrm{\hspace{0.17em}0}}$  (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

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

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.

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

where

$$q:=\sqrt{\frac{2}{m}\left(E+\frac{m{k}^2}{2G^2}\right)}$$

is a constant.  We introduce still an auxiliary angle $\psi$ such that

${\displaystyle \frac{k}{G}}-{\displaystyle \frac{G}{mr}}=q\cos \psi ,{\displaystyle \frac{dr}{dt}}=q\sin \psi .$(3)

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

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

where

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

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.

As for the of the conic, the most interesting one is an ellipse.  It occurs, by thehttp://planetmath.org/node/11724parent entry, when  .  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.