On the subject of orbits.... I was just thinking a little about details I hadn't thought about in some time and this seems clear to me:
Without assuming conservation of angular momentum or any other physical theory (including Newton's laws), and instead purely from mathematical considerations that assume a fixed central point for orbits, show that Kepler's 2nd law (equal areas in equal times) inevitably leads to the conclusion that the force of gravitation can only be directed along an instantaneous radial line connecting two points of mass, regardless of the magnitude of that acceleration. In other words, that the universal gravitation insight (that all particles of mass have an acceleration between them only along the line connecting them) is a necessary mathematical conclusion, merely from assuming Kepler's 2nd law.
Can you contribute some thoughts? I'll provide a clue a day from now, and a proof in two. The proof is short, as I see it right now.
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| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
America: Land of the Free, Because of the Brave
That's alright. We can close all our automotive plants in Canaduh... actually, that's likely.
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
America: Land of the Free, Because of the Brave
Oh yah... BC has lots of forestry... So maybe we'll cut off your supply of toilet paper too. :P
And... It's a good thing to close the American automotive plants in Canada... We're not grunt workers putting American stuff together like robots. Get Taiwan, Japan and China to do it for you.. They're making everything else.
Note: I really have no idea how much, if any, toilet paper is exported to the US. D from BC
Compute 2nd (time) derivative of area swept by orbiting planet. Separate parts and note what part must equal zero, if swept area rate is constant. (Now pick any hypothetical acceleration vector without assuming any knowledge about direction and translate it from x,y coordinates to instantaneous radial and transverse coordinate frame. Separate terms. Compare.)
x position x y position y r position radius w position angle
and that prefixing 'd' means derivative and prefixing 'd2' means the
2nd derivative.
You may recall these equivalences in mapping between an (x,y) and (r,w) position:
x = r cos w y = r sin w
From these last two, take the first and second derivatives:
dx = r * (- dw sin w) + dr * cos w = dr cos w - r dw sin w dy = r * (dw cos w) + dr * sin w = dr sin w + r dw cos w
d2x = d2r cos w - 2 dr dw sin w - d2w r sin w - r dw dw cos w = [ d2r - r (dw)^2 ] cos w - [ 2 dr dw + r d2w ] sin w d2y = d2r sin w + 2 dr dw cos w + d2w r cos w - r dw dw sin w = [ d2r - r (dw)^2 ] sin w + [ 2 dr dw + r d2w ] cos w
You can easily divide through the last pair (2nd derivatives of d2x and d2y) by (dt)^2 to refer it to time and get acceleration. Doing so will give:
ax = d2x / dt^2 = [ d2r/dt^2 - r dw/dt dw/dt ] cos w - [ 2 dr/dt dw/dt + r d2w/dt^2 ] sin w
ay = d2y / dt^2 = [ d2r/dt^2 - r dw/dt dw/dt ] sin w - [ 2 dr/dt dw/dt + r d2w/dt^2 ] cos w
By the way, while it is the case that d2x/dt^2 and d2y/dt^2 are the x and y components of acceleration on the planet, it is NOT the case that d2r/dt^2 or that d2w/dt^2 are the radial and transverse accelerations.
That's for two reasons you might want to consider: one is simply that these are apples to oranges, in that angular acceleration doesn't translate to transverse in any meaningful way, directly; but the real reason is that the basis vectors for (x,y) systems are fixed while the basis vectors for (r,w) are changing (which is why the terms that will soon come clear arise.)
So now let's look at some arbitrary acceleration vector placed on the planet at (x,y) or at (r,w). The acceleration can be broken into two easy components: 'ax' and 'ay,' which together form 'a'. Those correspond to the above ax and ay equations already given, in fact. So let's put those into radial and transverse form.
First, draw up a picture if you like, showing a planet in some position in quadrant 1 with the center being the central star (assumed unmoving.) Impress an arbitrary acceleration vector on it and draw the arrow off in some direction. From that, you can easily show the 'ax' and 'ay' portions that make it up, related to the fixed coord system. This forms a right triangle with the acceleration vector itself as the hypotenuse. Now impose another right triangle that also uses the same acceleration vector as its hypotenuse, but this time with the two sides being along the radial and the transverse directions. Different triangle, same resulting hypotenuse. To figure one from the other is just essentially a rotation of coordinate systems through some angle. In fact, this angle (if you look) is just the same angle as 'w' indicated above. This is:
ax = ar cos w - at sin w ay = ar sin w + at cos w
where 'ar' is the radial acceleration and 'at' is the transverse acceleration.
Now let's look back:
ax = d2x / dt^2 = [ d2r/dt^2 - r dw/dt dw/dt ] cos w - [ 2 dr/dt dw/dt + r d2w/dt^2 ] sin w
ay = d2y / dt^2 = [ d2r/dt^2 - r dw/dt dw/dt ] sin w - [ 2 dr/dt dw/dt + r d2w/dt^2 ] cos w
You can see the equivalence here of:
ar = d2r/dt^2 - r dw/dt dw/dt at = 2 dr/dt dw/dt + r d2w/dt^2
That's out of the way. This is the basic mathematical expression and it carries the relationship in change of basis vectors in the case of (r,w) coordinates verses the lack of such change in (x,y) coordinates. (We could arrive at this by looking at the change in basis vectors alone and taking their derivative directly, too.)
Now time to look at area swept by the planet.
We can start simple. The area of some triangle is just 1/2 base times height. We can call the base 'r', and the angle swept is still 'w'. How high? Well, the circumference of a circle is (2 pi r). For some arc, this is just (w * r), with 'w' in radians. If the arc 'w' is tiny, namely 'dw', then the arc length is 'ds' and is equal to (dw * r). So for our case, where we consider only the infinitesimal angle 'dw' to be swept, the arc (height) of our triangle is (dw * r) and the total area (dA) is:
dA = (1/2) [ dw r ] [ r ] = (1/2) r^2 dw
Let's take the derivative of that:
d2A = (1/2) [ 2 r dr dw + r^2 d2w ] = (1/2) r [ 2 dr dw + r d2w ]
Divide that through by (dt)^2 to get the 2nd derivative with respect to time and you get:
d2A/dt^2 = (1/2) r [ 2 dr dw + r d2w ] / dt^2 = (1/2) r [ 2 dr/dt dw/dt + r d2w/dt^2 ]
Now, we know that this must be set to zero, as Kepler's 2nd says the rate of area itself swept is constant. Setting this to zero suggests, of course, that (2 dr/dt dw/dt + r d2w/dt^2) must always be zero.
Hmm... Where have we seen that? Remember this:
ar = d2r/dt^2 - r dw/dt dw/dt at = 2 dr/dt dw/dt + r d2w/dt^2
Note that this is what 'at' is. So 'at' itself must be always zero. Therefore, only 'ar' can be non-zero. All acceleration must be, if Kepler's 2nd is true, radial along the position line connecting two points of mass.
Now note that none of this assumes anything about the physical laws that describe the form of that acceleration. This doesn't deal with
f = m a
or,
f = G m1 m2 / r^2
It's just about accelerating influences and what Kepler's suggests about just that aspect alone. And there is a deeper question that should be prodding your mind at this point about what should be the case regarding any object under any arbitrary acceleration that is always directed towards some one fixed point.
Thanks Jonathan, that was elegant. The amazing part of this is that Kepler wasn't aware of differential equations, they weren't into this at that time. As far as I know. It took them another century or so.
As I understand it, during Kepler's time, algebra was still being made mathematically rigorous. Calculus was for Leibnitz and Newton and I think Newton's earliest paper was maybe in the late 1660's timeframe. Kepler died in 1630.
One of the nifty things about Kepler is that he understood that his laws weren't satisfying. He wrote, paraphrasing from German, that they were merely "a smaller pile of dung to replace a larger one." They didn't explain, they just described. And he knew the difference and wanted more.
So far as I'm aware of, right now, Newton didn't use the logic I did, either. He was actually able to show that Kepler's 2nd was a consequence of the laws he developed. But that's going the other way, really.
What I think I've shown here is that one could inevitably arrive at the insightful conclusion that all matter attracts all other matter from Kepler's 2nd law alone. Newton himself had the tools and yet arrived as his own insights about the universality of gravitation from the ideas of others and himself, but without finding this particular chain of logic.
I looked on the web and I've been able to find articles there on related ideas but not this particular reasoning. For example, there is one called "Analytical Proof of Newton's Force Laws." (Google it.) But if you read it I think you'll see on page 11 that the author comes at this in a completely different direction and assumes that transverse acceleration is zero and proceeds. Yet there is no such need for assumption because it can be shown to be zero, since mathematical analysis predicts something quite generally that is then found to be true by physical observation interpreted through other, more basic reasoning.
I happen to like this particular approach, because it also says something about acceleration generally, that is directed towards or away from a single point (or may even be of zero magnitude.) Look at the case of a straight line and a particle following along it at a constant velocity as it sweeps out areas relative to a point not on that line. You can mark out a 'height' of each triangle formed that remains the same regardless of which triangle, and it is obvious that the base of each triangle will be the same given the velocity is constant and we pick out fixed time periods. So all areas must be the same. Here, no acceleration. In the case of elliptical orbits, the acceleration varies strongly with r^-2, yet even here equal areas are swept in equal times __because__ the acceleration is always radial. This is a much more general result than most folks realize. And it is interesting because of that.
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