PC1672 Advanced dynamics

3.6 Central-force motion

A conservative central force is obtained from a potential which depends only on the radial distance from the centre

$\begin{displaymath}{\bf F}({\bf r})=-{{\rm d}U\over{\rm d}r}{\bf e}_r\end{displaymath}$

The equations of motion for an object moving in such a force field are three coupled nonlinear second-order differential equations. However the description of this motion can be simplified a lot if we look for constants of motion (quantities that do not change as the object moves).

A central force is parallel to the position vector ${\bf r}$ and so the torque $\hbox{\boldmath$\tau$ }={\bf r}\times{\bf F}$ on the object is zero. Since $\dot{\bf L}=\hbox{\boldmath$\tau$ }$ , the (vector) angular momentum of the object is constant. Since the angular momentum ${\bf L}=m{\bf r}\times\dot {\bf r}$ is perpendicular to ${\bf r}$ and $\dot{\bf r}$ and its direction is constant, the motion takes place in a plane perpendicular to ${\bf L}$ .

It is convenient to use plane polar coordinates and $\theta$ to describe the motion. In terms of these, the magnitude of ${\bf L}$ is

$\begin{displaymath}L=mr^2\dot\theta\end{displaymath}$

This is a constant.

By considering the area of the triangle swept out by ${\bf r}$ in a small time, we find that ${\bf r}$ sweeps out area at a rate

$\begin{displaymath}{{\rm d}A\over{\rm d}t}={1\over 2}\vert{\bf r}\times\dot{\bf r}\vert={L\over 2m}\end{displaymath}$

which is constant. This is Kepler's second law of planetary motion: the radius vector of a planet sweeps out equal areas in equal times. From a dynamical point of view we see that it is really a statement of conservation of angular momentum.

Since the force is conservative, the total energy

$\begin{displaymath}E={1\over 2}m\dot{\bf r}^2+U(r)\end{displaymath}$

is another constant of motion. The velocity can be written in terms of polar coordinates as

$\begin{displaymath}\dot{\bf r}=\dot r{\bf e}_r+r\dot\theta{\bf e}_\theta\end{displaymath}$

where ${\bf e}_r$ and ${\bf e}_\theta$ are radial and tangential unit vectors. Hence the energy can be written

$\begin{displaymath}E={1\over 2}m\dot r^2+{L^2\over 2mr^2}+U(r)\end{displaymath}$

Solving for $\dot r$ in terms of and the constants and , we get

$\begin{displaymath}\dot r=\sqrt{{2\over m}\left(E-U(r)-{L^2\over 2mr^2}\right)}\end{displaymath}$

By identifying the constants of motion, we have reduced the problem to a single first-order differential equation for

The problem now looks like one-dimensional motion in an effective potential

$\begin{displaymath}U_{\rm eff}(r)={L^2\over 2mr^2}-U(r)\end{displaymath}$

where the first term corresponds to the centrifugal force and the second to the actual force on the object.

Kepler's laws: These three laws describing the motion of the planets were first deduced empirically by Kepler from Tycho Brahe's detailed observations of the motion of Mars. They were then explained by Newton as consequences of his law of gravitation. (For more, see the NASA Kepler site and the St. Andrews history of mathematics archive.)

Textbook references

K&K 9.3, 6.3 example 6.3, 9.5
F&C 6.3, 6.4, 6.11
B&O 5.1
M 8.4-8.6
M&T 8.4-8.6

Mike Birse
17th May 2000