Somebody asked:

Suppose a planet travels in an elliptical orbit around the Sun, according to Kepler's laws. What is the expected position of the planet, averaged over time?

Suppose, without loss of generality, that the ellipse has its major
axis along the *x*-axis, with the center at the origin. The
question then asks us to derive *X*, the time-averaged
*x*-position of the planet. We don't need to figure out the
time-averaged *y*-position, since by symmetry arguments, it must
be zero on average.

In order to compute the compute *X*, we must weight each
*x*-position by the time spent at that position, taken to the
limit as each time slice goes to zero. By Kepler's second law, we may
equivalently weight by the area of a wedge, extending from the Sun to
the planet's path. Let's put that in coordinate terms. Consider an
elliptical orbit, centered at the origin at *O*(0, 0), with
semi-major axis equal to 1. That is, the ends of the orbit are at
(–1, 0) and (1, 0). The Sun is, let's say, at one focus of the ellipse,
at *S*(*e*, 0), where *e* is the eccentricity of
the ellipse.

Suppose that the planet is currently at *P*(*x*,
*y*). Over an infinitesimal period of time, it will move to a
point *P*'(*x*', *y*'). Then the position
coordinate *x* is to be weighted by the area of a wedge,
extending from *S*, sweeping from *P* to *P*'.
Let's call this wedge *W*_{1}.

The key to the enterprise is recognizing that the area of
*W*_{1} is almost equal to the area of another wedge,
*W*_{2}, which is the wedge, extending from the
*origin* *O*, and sweeping from *P* to *P*'.
In particular, we may write that

area(W_{1}) = area(W_{2}) + small amount

We'll get to that small amount in a moment. Remember, again, that
*X* is equal to

the averagex-position, weighted by area(W_{1})

By the above relation, *X* is also equal to

the averagex-position, weighted by area(W_{2}),plus

the averagex-position, weighted by that small amount

Note that for every wedge *W*_{2} going to a *P*
with a positive *x*-position, there is a corresponding wedge
*W*_{2–} going to a *P*– with the corresponding
negative *x*-position. Therefore, the first quantity in the
equation above for *X* is zero! That is, *X* is simply
equal to

the averagex-position, weighted by that small amount

Now, what is that small amount? As we stated above, it is the area of
*W*_{1} minus the area of *W*_{2}. By
consulting the diagram below,

we see that this is equal to

area( SPP') – area(OPP')= area( SQP) – area(OQP')= area( SOP) – area(OSP')= ey/2 –ey'/2= e(y–y')/2

This is the "small amount" by which we must weight all the
*x*-positions. We do this simply by multiplying them together,
to yield *xe*(*y*–*y*')/2, which we'll rewrite as
(*e*/2)*x*(*y*–*y*'). This quantity must
be summed as the planet *P* traverses the entire ellipse. Since
the value of *e*/2 does not change, we'll just sum
*x*(*y*–*y*') over the entire ellipse, and multiply
*e*/2 in at the end.

As the planet moves from *P* to *P*', the quantity
*x*(*y*–*y*') can naturally be interpreted as an
area—the area of a rectangle whose width extends from the *y*-axis
to the planet's position, and whose height is the height difference
between *P* and *P*', as shown in the diagram below. Now,
when *x* is positive, the height difference *y*–*y*'
is negative, and when *x* is negative, the difference
*y*–*y*' is positive. Therefore, to be precise, we must
interpret these areas as *negative* areas.

Now observe! As the planet moves around the entire ellipse, these
small rectangles combine to cover the entire ellipse, without overlap,
and in the limit, without error either. The sum of
*x*(*y*–*y*'), over the entire ellipse, is simply
the area of the ellipse. We must be sure to interpret it as the negative
area—that is, –*A* (where *A* is the geometrical area of
the ellipse). And if the sum of *x*(*y*–*y*') over
the circle is –*A*, then the sum of
(*e*/2)*x*(*y*–*y*') is –(e/2)*A*.

All that remains now, is to divide that result by the area of the
ellipse which we averaged the *x*-position over. That area is,
again, *A*, so that the average *x*-position, weighted by
the wedge area *W*_{1}, or by time (either is equivalent
to the other) is just

–(e/2)A/A= –e/2

And this is just the position halfway between the empty focus, and the center of the ellipse. Q.E.D.

Copyright (c) 2001 Brian Tung