But why an elliptical orbit

Brian Tung wrote:

I assume you mean, why planetary orbits *aren't circles*.

I couldn't tell. I still can't. I'm hoping the original poster
clarifies the question...

I drew the inference from his attempted explanations, which couldn't
distinguish between ellipses and other ovally shapes, and from the
context of the recent thread about ancient Greek astronomy. It's a
guess, though.

- Ernie http://home.comcast.net/~erniew

wrote

But Kepler's laws are so essential, you say. OK, fine.

They're essential for making a good impression on a perfect stranger. I
think I once related a story here that, as I was walking down a mountain
path near Tucson, a fellow was coming up the path and he looked at me.....
and asked, "Do you know Kepler's Laws?" Ummm... well, yes, I do! And I
proceeded to draw them out in the dirt with a stick.

"tt40" wrote in news:1129087625.368615.299390
@g47g2000cwa.googlegroups.com:

In everything I've read about planets and elliptical orbits, I can't
ever recall any author (Feynman, Newton, 'Ask an Astronomer' etc.),
explaining exactly 'why' the orbit is elliptical. Oh sure there's been
lots of mathematics to explain the orbit and how it works, but most of
the explanations don't provide a definitive statement as to why it IS
elliptical.


Until someone can do the 'why', I'm going to assume that either:

1. it's because the Sun is in a rotation of it's own and all the
planets have to scoot to 'catch' up with it. But since the Sun is
moving too, the now-receding planet's momentum carries it further away
than it expected, so it maps out a stretched circle, longer than it
thought was necessary. So it's stretching the distance on the long-arm
of the orbit and has to scoot back in again, chasing that moving Sun.


No, the motion of the Sun through the galaxy is not relevant at all. Given
the law of gravity, F=GmM/r^2 and the Newton/Galileo laws of motion, a two
body situation results in only two possible closed orbits, either a circle
or ellipse. A circle is just a special case of an ellipse. To simplify
things, you normally use the center of mass of the Sun as the origin of the
coordinate system - that way you can work the equations without having to
worry about the motion of the Sun. Other orbits are possible of course, the
parabola and hyperbola are also valid orbits but they are not closed - i.e
the planet enters and leaves the system in one pass. It is very interesting
that any orbit in a two body system can be described by one of the above
conic sections. In the real solar system things are not quite as simple
because there are many bodies and they all interact with each other via
gravity. Nevertheless, ellipses still give a pretty reasonable
approximation for the motion of the planets as the inter planet
perturbations are fairly small. It's a different story when you have to do
astro-navigation such as the dance that JPL put the Cassini space probe
through - you need a lot of precision to do what they did and are still
doing with that.

It's a lot of fun to see how Newton figured all this out - especially if
you can read Latin, there are scans of the original editions of the
Principia online he

http://burndy.mit.edu/Collections/Ba...ine/Principia/

or

2. the minute gravitational tug of the planet pulls the Sun closer (out
of its 'fixed' position relative to the planet) which increases the
mutual gravitational attraction so that they are each attracted that
little bit stronger. But since the Plant is still in orbit, it follows
the short arm of the ellipse that little bit faster or a little more
energised. It then maps out the long-arm of the ellipse and
gravitational attraction recedes just a little bit, allowing it to
'stretch that circle'.

That is almost right in a hand waving sort of way.

Klazmon.






or

(this space left blank for the correct answer)

Greg

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
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt

I (Brian Tung) wrote:
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*.

OK, it was long, and not very well written, but I think it basically has
a seed of truth to it. At some point, I'm going to try to polish it a
bit and see if it can be made understandable.

--
Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt

oriel36 wrote:
To LAL


First, a usenet lectu This is an open forum; you don't respond to
just one person, but post for all to read. If in fact you wished to
*comment* on a single person's post you use conventionalized quoting (as
I just did.)

Second, a simple request: You have morphed your identity to slip by my
killfile; I request that you maintain a consistent identity.

Chris L Peterson wrote:
On Wed, 12 Oct 2005 12:13:25 GMT, (Paul Schlyter) wrote:
OK, time to answer your question: the orbit is a conic section (the
ellipse being the most common case) because gravity is inversely
proportional to the square of the distance to the gravitating body...
snip Perhaps that is a satisfying answer to somebody already
familiar with the mathematics behind orbital dynamics, but . . . I
doubt most people find it intuitive that an inverse square
gravity law naturally leads to elliptical orbits. . . .

I always thought it is was because gravitational attraction between two
bodies was the result of two force vectors, not one.

The second smaller body has an orbital speed (angular momentum)
combined with its mass. The causes the second smaller body to pull the
larger body slightly off-center. Conversely, the larger body generates
sufficient gravitational force to still hold the smaller orbiting body
in place. As a consequence, a smaller body and larger body always orbit
a common dynamical center, offset from the true gravitational center of
gravity of the larger body.

Although a true circular orbit is theoretically possible assuming an
idealized set of initial conditions, in practice any body perturbing a
two-body orbital system will distort the idealized two-body circular
orbit. Considering the age of solar system and the density of objects
in it, the likelihood of finding any solar system object that has not
be perturbed from a true circular orbit into an elliptical orbit seems
remote.

- Canopus56

(Paul Schlyter) wrote in
:

True of course, but to a good first approximation the orbits are
ellipses (or some other kind of conic section: circle, parabola,
hyperbola).

And in the hypothetical case of

1. Newtonian mechanics being exactly valid

2. Only two bodies are present in the universe

3. The bodies are point masses.

(Paul Schlyter) wrote in :

Another hypothetical case: if gravity would have varied as the inverse
*fifth* power of the distance, all orbits would have been spirals.

Are the orbits stable if gravity varies as the inverse 4th or 6th power of
the distance?

Phew, I'm overwhelmed by the considerable and considered lengths that
respondents have made on this topic. (Of course it's tempting to be a
smart-alec and ask 'Yes, but why an ellipse?' as if to off-handedly
tilt at the signficance of mathematics in answering my question, but
that would lazy and disingenuous).

** To clarify, as was requested by some, the extended version of my
question is 'Why an ellipse and not a circle?' And thanks to those who
recognised this -- an imprecision on my part. **

It is a fascinating topic and I wish I understood (read 'could
configure my life so I had the time to learn') the maths.

Sorry that I've only had time to skim the thread, can't wait to read it
all in detail.

Greg.