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

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

$\displaystyle E_k \;=\; \frac{1}{2}m\left\vert\frac{d\vec{r}}{dt}\right\vert^2
...
...!\left(\frac{dr}{dt}\right)^2\!+\!\left(r\frac{d\varphi}{dt}\right)^2\right)\!.$

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

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

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

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

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

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

This equation may be revised to

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

i.e.

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

where

$\displaystyle q \;:=\; \sqrt{\frac{2}{m}\left(\!E\!+\!\frac{mk^2}{2G^2}\right)}$

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)

Differentiation of the first of these equations implies

$\displaystyle \frac{G}{mr^2}\cdot\frac{dr}{dt} \;=\; -q\sin\psi\frac{d\psi}{dt} \;=\; -\frac{dr}{dt}\cdot\frac{d\psi}{dt},$

whence, by (2),

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

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

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


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.



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As of this snapshot date, this entry was owned by pahio.