*Michael Fowler, UVa.*

Of course, Kepler’s Laws originated from observations of the
solar system, but

The standard approach in analyzing planetary motion is to
use (*r*, *q
*) coordinates, where *r* is the distance from the origin—which we take
to be the center of the Sun—and *q * is the angle
between the *x*-axis and the line from the origin to the point in
question.

In the picture above, in which I have greatly exaggerated
the ellipticity of the orbit, suppose the planet goes from *A* to *B*,
a distance D*s*, in a short time D*t*, so its speed in
orbit is D*s*/D*t*. Notice that the velocity can be resolved into
vector components in the radial direction D*r*/D*t*
(in this case negative) and in the direction perpendicular to the radius, *r*D*q*/D*t*. The short line *BC* in the diagram above
is perpendicular to *SB* (*S* being the center of the Sun), and
therefore becoming perpendicular to *SC* as well in the limit of *AB*
becoming an infinitesimally small distance.

In the limit of small D*s*, then, we have _{} where the angular
velocity _{}.

We shall consider Kepler’s Second Law (that the planet sweeps out equal areas in equal times) first, because it has a simple physical interpretation.

Looking at the above picture, in the time D*t*
during which the planet moves from *A* to *B*, the area swept out is
the approximately triangular area *ABS*, where *S* is the center of
the Sun. For the distance *AB* sufficiently small, this area tends to that
of the long thin triangle *BSC*, which has a base of length *r*D*q*
and a height *r*. Using area of a
triangle = ½ base´height,
it follows immediately that

_{}

Now, the angular momentum *L* of the planet in its
orbit is given by

_{}

so *the rate of sweeping out of
area is proportional to the angular momentum, and equal to L/2m*.

*the rate of change of angular momentum is equal to the
torque of the forces acting on the system*.
For the planet orbiting the Sun, this torque is *zero*: the only
force acting is gravity, and that force acts in a line from the planet towards
the center of the Sun, and therefore has zero torque—no leverage—about that
central point, so the planet’s angular momentum about that point must remain
constant. Consequently, the rate at which area is swept out is *also*
constant, and Kepler’s Second Law follows: *equal areas are swept out in
equal times*.

We’re now ready for Kepler’s First Law: *each planet moves in an elliptical orbit with the Sun at one focus of
the ellipse. *

Let us begin by reviewing some basic facts about ellipses.

An ellipse is essentially a circle scaled shorter in one
direction, in (*x*, *y*) coordinates it is described by the equation

_{},

a circle being given
by *a* = *b*. The lengths *a*
and *b* are termed the *semimajor axis* and the *semiminor axis*
respectively.

An ellipse has two foci, shown *F*_{1} and *F*_{2}
on the diagram, which have the optical property that if a point source of light
is placed at *F*_{1}, and the ellipse is a mirror, it will
reflect—and therefore *focus*—all the light to *F*_{2}. One way to *draw* an ellipse is to take
a piece of string of length 2*a*, tie one end to *F*_{1} and
the other to *F*_{2}, and hold the string taut with a pencil
point. Anywhere this happens on a flat piece of paper is a point on the
ellipse. In other words, the ellipse is the set of points *P* such that *PF*_{1}
+ *PF*_{2} = 2*a*.

The *ellipticity *of the ellipse *e* is defined by
writing the distance from the center of the ellipse to a focus *OF*_{1}
= *ea*.

In fact, in analyzing planetary motion, it is more natural
to take the origin of coordinates at the center of the Sun rather than the
center of the elliptical orbit. It is
also more convenient to take (*r*, *q *) coordinates instead
of (*x*, *y*) coordinates, because the strength of the gravitational
force depends only on *r*. Therefore,
the relevant equation describing a planetary orbit is the (*r*, *q *)
equation with the origin at one focus.
For an ellipse of semi major axis
*a* and eccentricity *e* the equation is:

_{}.

It is not difficult to prove that this is equivalent to the
traditional equation in terms of *x*, *y* presented above.

Note: I’m including the calculus derivation of the elliptic orbit, not to be found in the textbooks, just so you can see that it’s calculus, not magic, that gives this result. This is an optional section, and will not appear on any exams.

We now back up to Kepler’s *First* Law: proof that the
orbit is in fact an ellipse if the gravitational force is inverse square. As usual, we begin with *F* = *ma*, in
vector form. The force is *GMm*/*r*^{2} in a radial inward
direction. But what is the acceleration? Is it just *d*^{2}*r*/*dt*^{2}? Well, no, because if the planet’s moving in a
circular orbit it’s still accelerating inwards at *r**w*^{2}
(same as *v*^{2}/*r*) even though *r* is not changing at
all. The total acceleration is the *sum*, so *ma* = *F* becomes:

_{}

This isn’t ready to integrate yet, because *w*
varies too. But since the angular momentum *L* is constant, *L* = *mr*^{2}*w*,
we can get rid of *w* in the equation to give:

_{}

This equation can be integrated, using two very unobvious
tricks, figured out by hindsight. The
first is to change go from the variable *r* to its inverse, *u* = 1/*r*.
The other is to use the constancy of angular momentum to change the variable *t*
to *q*.

Anyway,

_{}

so

_{}.

Therefore

_{}

and similarly

_{}.

Substituting in the equation of motion gives:

_{}

This equation is easy to solve! The solution is

_{}

where *A* is a constant of integration, determined by
the initial conditions.

This is equivalent to
the standard (*r*, *q *) equation of an ellipse of semi major axis *a*
and eccentricity *e*, with the origin at one focus, which is:

_{}.

The time it takes a planet to make one complete orbit around
the Sun *T* (one planet year) is related to the semi-major axis a of its
elliptic orbit by

_{}.

We have already shown how this can be proved for circular orbits, however, since we have gone to the trouble of deriving the formula for an elliptic orbit, we add here the(optional) proof for that more general case.

The area of an ellipse is *pab*, and the rate of
sweeping out of area is *L*/2*m*, so the time *T* for a complete orbit is evidently

* _{}*.

Putting the equation

_{}

in the standard form

_{},

we find

_{}.

Now, the top point *B* of the semi-minor axis of the
ellipse (see diagram above) must be exactly *a* from *F*_{1} (visualize the string *F*_{1}*BF*_{2}),
so using Pythagoras’ theorem for the triangle *F*_{1}*OB* we
find

_{}.

Using the two equations above, the square of the orbital time

_{}

We have established, then, that the time for one orbit
depends *only* on the semimajor axis of the orbit: *it does not depend
on how elliptic the orbit is*!