PlanetPhysics/Motion in Central Force Field

Let us consider a body with mass $$m$$ in a gravitational force field 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 $$\begin{matrix} \vec{v} \;=\; \frac{d\vec{r}}{dt} \;=\; \frac{d}{dt}(r\vec{r}^{\,0}) \;=\; \frac{dr}{dt}\vec{r}^{\,0}+r\frac{d\varphi}{dt}\vec{s}^{\,0}, \end{matrix}$$ 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 $$\begin{matrix} \frac{d\varphi}{dt} \;=\; \frac{G}{mr^2}. \end{matrix}$$ The central force\, $$\vec{F} := -\frac{k}{r^2}\vec{r}^{\,0}$$\, (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 $$\left(\frac{dr}{dt}\right)^2\!+\frac{G^2}{m^2r^2}-\frac{2k}{mr}+\frac{k^2}{G^2} \;=\; \frac{2E}{m}+\frac{k^2}{G^2},$$ 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 $$\begin{matrix} \frac{k}{G}-\frac{G}{mr} \;=\; q\cos\psi, \quad \frac{dr}{dt} \;=\; q\sin\psi. \end{matrix}$$ 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 $$\begin{matrix} r \;=\; \frac{G^2}{km\left(1-\frac{Gq}{k}\cos(C-\varphi)\right)} \;=\; \frac{p}{1-\varepsilon\cos(\varphi-C)}, \end{matrix}$$ where $$p \;:=\; \frac{G^2}{km}, \quad \varepsilon \;:=\; \frac{Gq}{k}.$$\\

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

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 Kepler's first law.