Ernie Wright wrote:
But to see why this is so, you can't escape the math that describes how
gravity works. It's not explainable as scooting, wobbling, sloshing, or
anything like that. An ellipse is simply what happens for a broad set
of distance and velocity settings.

Ernie, I'm going to try an explanation, and you tell me how far off I
get. It's not going to be a rigorous explanation, but I'll try to make
it specific to the ellipse. Warning: This may use no complicated math,
but it is *long*.

I suppose many of you have seen those Chinese juggling toys, where the
juggler holds something like a jump rope in two hands, and a spindle
travels back and forth on top of the rope. Imagine attaching the two
handles to a plank, so that their separation is fixed. In that case,
the spindle, as it moves back and forth, must travel in an ellipse (that
is, relative to the two handles), with the foci at the two handles.
This follows from one of the standard definitions of an ellipse.

Suppose that at some point in time, the two handles look like this (and
use a fixed-width font for the diagrams in this post)--

A


B

with the rope trailing down between them. If we let the spindle come to
rest on the rope, it will hang something like this, with P representing
the spindle for reasons that will soon become clear (if they aren't
already).

A
\
\
\ B
\ /
\ /
P

Observe two things about this diagram. One, the distance AP is just
twice the distance BP. That, of course, is just a result of the way I
designed the diagram; at different orientations or different rope
lengths, it would be a different ratio. What is not just an accident
of the way I designed the diagram is the fact that the rope lengths AP
and BP both make identical angles with the vertical.

This would not be true in general for any spindle path; if, for
instance, the half of the rope attached to A were ordinary and the half
attached to B were made out of phone cord, the orbit would not be
elliptical, and very likely AP would be closer to vertical than BP.

One consequence of the angles being equal is that the tension on the
ropes AP and BP are equal; they both pull the same amount, even though
AP is longer than BP. If they didn't, then the sideways pull would
be greater to one side, and the spindle wouldn't yet have come to rest.
If the spindle weighs 10 ounces, say, then handle A exerts a pull of a
bit more than 5 pounds, and handle B exerts an identical pull--the
extra bit being due to the fact that AP and BP aren't quite vertical,
so that part of the pulls from the handles are sideways and cancel each
other out.

Now, let's send the spindle swinging a little.

A
\
\
\ B
\ /
\ /
- P -

If the spindle is slowed by friction with the rope (as it always is in
the real world), it will come to rest again at the same point P, which
is another indication that the rope tensions from A and B balance out
and the force is directed upward.

If we are to apply this to the situation with the Sun and a planet,
however, we must put the Sun at either A or B, and the force cannot be
directed upward. It has to be directed to wherever the Sun is.

Let's suppose the Sun is at A. Then the force on P (now the planet) has
to come from A. For that to happen, there must be a second force on the
planet, which "pushes" the force vector, so that instead of pointing up,
it points toward A. Since the planet isn't actually veering from an
elliptical path, such a force must be sideways--to the left. In other
words, if the planet is revolving counterclockwise, it must be slowing
down in its orbit.

If, on the other hand, the Sun is at B, the second force that we must
add in order to make the total force point toward B is a sideways force
to the right. For a counterclockwise-revolving planet, the planet must
be speeding up.

What's important to notice is that in either case--the force applied by
a Sun at A or that applied by a Sun at B--those two forces are the same
for an identical small amount of motion around point P. It would not
be for any other orbit but an elliptical one. As a result, the change
in motion experienced by the planet is the same whether the Sun is at A
or at B.

At this crucial point, we bring up Kepler's second law, which for bodies
of fixed mass is equivalent to the conservation of angular momentum. We
have assumed, in equating the force at A and the force at B, that the
motion of P is the same in either case. But Kepler's second law says
that it isn't. Kepler's law says that the planet must sweep out equal
areas in equal times.

For a given small amount of motion around point P, the planet sweeps out
twice the area for a Sun at A as it does for a Sun at B. This stems
from both the fact that AP is twice BP, and that AP and BP make equal
angles with the vertical. If, as Kepler's second law states, the planet
sweeps out equal areas in equal times, then it must sweep out half the
area in half the time.

That is to say, if the Sun is at B, the planet completes a bit of motion
in half the time it would take if the Sun were at A; it must move twice
as fast. As a result, the change in motion is twice as much, and since
it takes place in half the time, the acceleration is four times as
great. (For instance, it takes four times the acceleration to change
from -20 km/s to 20 km/s in half a second as it does to go from -10 km/s
to 10 km/s in a full second.)

These ratios hold only because AP is twice BP. If AP were k times BP,
we would complete, for a Sun at B, k times the change in motion in 1/k
of the time, as we would for a Sun at A, yielding k^2 times the
acceleration. Or, equivalently, the acceleration for the Sun at A would
be 1/k^2 of the acceleration for the Sun at B.

But A and B are entirely interchangeable. That means that for two
symmetrically placed points on the orbit, if the ratio of the distances
is k, the ratio of the accelerations (and hence, forces) must be 1/k^2.
There are many ways that force can vary with respect to distance that
exhibit this property, but the simplest one by far is the one that says
it is always proportional to the inverse-square. (Otherwise, something
funny would happen as the planet passed the minor axis of the ellipse.)
If that's right--and this is the biggest leap I'll ask people to make--
then an ellipse comes as a result of an inverse-square force.

For the other direction, I will resort to determinism: If a force can
yield a certain result given some initial conditions, it must always do
so. In that case, if an inverse-square force can yield an elliptical
orbit, then given the same initial conditions, it must always do so.

--
Brian Tung
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