Kepler's problem

The motion of the sun and the earth is an example of a ``two-body problem''. This is a relatively simple problem that can be solved analytically (it interesting to notice here, however, that adding a new object to the problem, the moon for instance, make it completely intractable. This is the famous ``three body problem''). We can assume that, to a good approximation, the sun is stationary and is a convenient origin of our coordinate system. This is equivalent to changing to a ``center of mass'' coordinate system, where most of the mass is concentrated in the sun. The problem can be reduced to an equivalent one body problem involving an object of reduced mass $\mu$ given by

$$\mu=\frac{mM}{m+M}$$

Since the mass of the earth is $m=5.99\times 10^{24}$ kg and the mass of the sun is $M=1.99\times 10^{30}$ kg we find that for most practical purposes, the reduced mass of the earth-sun system is that of the earth. Hence, in the following we are going to consider the problem of a single particle of mass $m$ moving about a fixed center of force, which we take as the origin of the coordinate system. The gravitational force on the particle m is given by

$${\mathbf F}=-\frac{GMm}{r^3}{\mathbf r},$$

where the vector ${\mathbf r}$ is directed from $M$ to $m$, and $G$ is the gravitation constant

$$G=6.67\times 10^{-11} \frac{m^3}{kg.s^2}$$

The negative sign implies that the gravitational force is attractive, and decreases with the separation $r$. The gravitational force is a ``central force'': its magnitude depends on the separation between the particles and its direction is along the line that connects them. The assumption is that the motion is confined to the $xy$ plane. The angular momentum ${\mathbf L}$ lies on the third direction $z$ and is a constant of motion, i.e. it is conserved:

$$L_z=({\mathbf r}\times m{\mathbf v})_z=m(xv_y-yv_x)=\mathrm{const.}$$

An additional constant of motion ins the total energy $E$ given by

$$E=\frac{1}{2}mv^2-\frac{GmM}{r}$$

If we fix the coordinate system in the sun, the equation of motion is

$$m\frac{d^2{\mathbf r}}{dt^2}=-\frac{mMG}{r^3}{\mathbf r}$$

For computational purposes it is convenient to write it down in cartesian components:

$$F_x=-\frac{GMm}{r^2}\cos{\theta}=-\frac{GMm}{r^3}x,$$ (28)

$$F_y=-\frac{GMm}{r^2}\sin{\theta}=-\frac{GMm}{r^3}y.$$ (29)

Hence, the equations of motions in cartesian coordinates are:

$$\frac{d^2x}{dt}=-\frac{GM}{r^3}x,$$ (30)

$$\frac{d^2y}{dt}=-\frac{GM}{r^3}y,$$ (31)

where $r^2=x^2+y^2$. These are coupled differential equations, since each differential equation contains both $x$ and $y$.