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#30725 - 05/20/09 09:52 AM Does Newton's shell theorem require repair?,
geistkiesel Offline
Junior Member

Registered: 05/20/09
Posts: 10
Loc: Arizona
I thank all the hosts and users for permitting my attempt to reach, or at least to approach scientific soundness. I have been obsessed with in Newton's Shell Theorem recently and have perceived some observed glitches that I will indicate.

The shell theorem as stated by Isaac Newton for the case of a test mass outside the surface of a thin spherical shell goes as follows: "The shell really does behave as if all the mass of the shell was concentrated at the center of the shell." I he concluded at the risk of leaping off the shoulders of some now long gone giants I am compelled to proceed.

Of all the objections to the theorem it is to me best shown how the shell integral leading to the result that the force on a test particle located a distance d from the shell center. The expression for the total force is the familiar, F = GmM/d^2. It is this expression that is referred to as "proving Newton's Shell Theorem". I am unable to discard the law of gravitation as constrained by the inverse distance squared law. To me the law says, for objects of equal mass, located at different distances from M, the object nearest m contributes a greater share of the total force on m than the object located further away. The references I have reviewed have all integrated the expression by summing the forces on rings centered on the r-axis (the m-M axis)with the plane of the circle perpendicular to the r-axis, and then adding the forces on rings from a position closest to m then in a direction retreating away down the r-axis until all rings were summed.

Bear with me on the following. Take any arbitrary ring and project it away from m to a position that reflects the ring as a mirror image of the closest ring to m. Then, draw a horizontal line linking any arbitrary dM on the closest ring with its paired mirror image dM. We are going to sum the forces associated with each mirror image pair and voila, the dM closest to m contributes a greater force on m than the paired mate on the ring farther away from m. When doing this is it not obvious that when all dM in the nearest segment of the sphere now segmented into 1/2 shell segments for the purpose of modifying the order of calculating the forces. When all dM forces in the near ½ shell segment have been counted, then all the dM forces in the farthest ½ shell have been counted.'

This method allows one to determine the exact location of the total of the combined paired forces. Briefly, finding the "gravity center" goes as follows:
Fc is the force from the dM in the closest ½ shell segment, Ff the paired mirror image force. Fc = GmdM/s^2]cos(alpha) where the cosine term effectively projects the calculated dF onto the r-axis. Fc = GmMcos(alpha)/r^2 . For just a single ring cos(&#55349;&#57084;) is a constant for the forward ring as the integration sum the forces on the rings. Ff has a cos(alpha') projection onto the r-axis also. Here &#55349;&#57084;alpha' < alpha.

Assume the test mass m = 1 and G = 1 , and M = 10 units of mass and assume the dM are each .1 and the shell dm mass is .1. Assume these dM are the first to be counted and both are on the r-axis. If Fc is located 10 units from m and Ff located 14 units from m then the forces are Fc = (1)(1)(.1)/10^2 = 10^-3. Ff = .1/14^2, or Ff = .1/196 =.0005. The total of these "point mass forces" is Fc + Ff = .001 + .0005 = .0015. Now calculate where the force is sensed to originate by m. So (.1 + .1)/ r^2 = .0015. Solving for r^2 gives us .2/r^2 = .0015. r^2 = .2/.0015 = 11.47 – the COF (center of force) of this two object system is offset from the COM in the direction of m. The COM is located at 12 unit distances.
This one calculation can be generalized to the unambiguous result that all calculations will place the COF in the 1/2 shell nearest m . The point is made, if not a bit crudely.
Another point slightly different is recognizing that Newton's shell theorem was postulated before he did all the math and physics and when the integrated force expression turned out to be, F = GmM/d^2, all of Isaac's Groupies sighed when he said [or someone said," the shell really does behave as if the mass were all concentrated at the shell center." But look at the expression. It proves nothing about concentrated mass, the development of the integral form of the expression makes no reference to the concentration of mass or its location. Isaac, however, had previously theorized that that all of this before cranking out the integral: was it luck or skill? Whatever the choice, Isaac's Theorem is seriously flawed. If it had been me, I would have claimed a state of drug addiction, or that I was severely juiced and that I had, "blown it for the m moment., but that me and my chaplain had worked out all the difficulties".

The expression only says, however, "The force on m by a shell of mass M located at d." – the expression does not locate the force center!!!! Sure, the shell was located at d, but all paired force centers, dare we call them the "centers of gravity", are all off set from the COM along the r- axis in the direction of m.

I appreciate anyone who has made it this far.
I know about how we all have benefitted by standing on the shoulders of Giants, but the one I was propped up on decayed suddenly - I dropped a few ignominious feet- no surprise there, after all the dude had been dead for some 250 years, he was just dried dust and his structural integrity was found to be embarrassingly lacking, not his embarrassment , he was very, very dead, and the dead don't blush. It only took me 40 years from first exposure to the theorem before I took a long hard look.
Geistkiesel
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#30726 - 05/20/09 10:44 AM Re: Does Newton's shell theorem require repair?, [Re: geistkiesel]
Zephir Offline
Superstar

Registered: 07/01/08
Posts: 498
Originally Posted By: geistkiesel
..."The shell really does behave as if all the mass of the shell was concentrated at the center of the shell."..
This becomes true only at the sufficient distance from shell, doesn't it? Anyway, I can't see your exact problem here...

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#30731 - 05/20/09 04:11 PM Re: Does Newton's shell theorem require repair?, [Re: Zephir]
geistkiesel Offline
Junior Member

Registered: 05/20/09
Posts: 10
Loc: Arizona
Originally Posted By: Zephir
Originally Posted By: geistkiesel
..."The shell really does behave as if all the mass of the shell was concentrated at the center of the shell."..
This becomes true only at the sufficient distance from shell, doesn't it? Anyway, I can't see your exact problem here...


Originally Posted By: Zephir

Originally Posted By: geistkiesel
..."The shell really does behave as if all the mass of the shell was concentrated at the center of the shell."...
This becomes true only at the sufficient distance from shell, doesn't it? Anyway, I can't see your exact problem here...


No, Newton used the model in earth-moon calculations, even though the physical extension of both orbs is readily apparent.

I assume you are familiar with the rings on the shell locating the dM masses that are the source of the gravitational force of the thin shell mass M acting on m. After all the dM on the ring are calculated the ring moves over the surface of the sphere until the total surface of the sphere and all dM have been calculated using, F = GmM/x^2 . For the shell center located at a distance d from m have been calculated, that is after the integral has been evaluated the resulting expression for the total force on m is, F = GmM/d^2. This expression is the proof that, "the shell really does behave as if all the mass on the shell was concentrated at the shell center, the COM."

The expression is describing a force and it says nothing about the location of the force. The expression says, describes the "force of gravity by a shell of mass M with the shell center (COM) a distance d from m". Does this remind you of any proof?

There is nothing in the structure of the mathematics alluding to the location of any force, to concentration of mass. The mathematics provided a mechanism to calculate the total force on m by M. There is no calculation of the location of any combined forces.

Take three masses in a line where two of the [point] masses are a considerable distance to the third test mass. Using the force expression and calculating the forces on m singularly to get a total force, this sum can then be used to calculate the 'apparent' location of the mass contributing force on m. The force expression constrains the forces such that the closer of the two equal masses contributes a greater share of the force on m than does the more distant mass. The test mass m sees only one force. Making the calculation will result in the combined force's location off set from the COM of the two distant masses in the direction of m.

This is what the shell theorem requires as a correction. Simply calculate the forces from mirror image pairs of dM with a mirror image ring located symmetrically in the nearest ½ shell segment and the other ring in the more distant ½ shell segment. By inspection the nearest ½ shell will contribute a greater share of the total force on m than the more distant 1/2 shell segment. The 'center of force' for the shell, just call it the 'center of gravity', AKA the cg.
While the science industry continues to glorify itself from the lofty height reached when standing on the shoulders of the 'Giants', if for no other reason than this slight intrinsic error in force calculation amounts to a couple of percentage points, the complete ignorance of believing that the mass may be concentrated at the COM of the shell is witness to an aching void in lack of completeness and just plain seemed be rather ignorant regarding what should be obvious - referring to the closest dM to m contributes a greater share of the total force on m than the dM located more distant from m.

These are the two corrections for which I demand my money back.

Other somewhat irksome flaws include, 1. Treatment of the m-M system using vector analysis. If two persons hold the ends of equal length ropes where both ropes are tied to an object between the rope holders. The men pull with equal force of 50 kgf in opposite directions. Vector rules conclude the forces sum to zero, when I say the forces sum to 100 kgf. 2. There never been an observation of a stellar mass in the form of a spherical this shell, yet a universal law of gravitation is recognized and faithfully enforced, yet the law is founded in a mathematical abstraction n with absolutely no common characteristics justifying even pretending the expression has scientific value. 3. The forces outside the sphere having been calculated, now we enter the wicked witch of the west's house and are told that we could float inside the sphere without having to flap our arms – one absurd physical impossibility is now offered as justification to make real yet another event seen only by Alice when she fell down the hole.

These nit picking items are said to follow logically – scientifically – from the "proof " of the shell theorem. Why do so many commentators use the expression that the 'zero force' result felt on an object in side the shell is 'not immediately intuitive", or that it 'strains the sense of physics'.

Zephir, are you convinced, not by my incessant whining, but by the math and the physics? I just remember, a UC Berkeley PhD in Physics first lectured me on the shell theorem, but that was in another lifetime.
Where do I go to get my money back?
_________________________
Mother Nature Included Time in Her Creation so Everything Wouldn't Happen all at Once

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#30734 - 05/20/09 10:14 PM Re: Does Newton's shell theorem require repair?, [Re: geistkiesel]
Zephir Offline
Superstar

Registered: 07/01/08
Posts: 498
Originally Posted By: geistkiesel
..are you convinced, not by my incessant whining, but by the math and the physics?

Of course. Everyone, who knows at least a bit about integrall calcullus would see immediatelly, Newton's shell theorem is valid only for infinitelly thin shell.

After all, here we can find explicitelly:

Therefore, a thin shell can be treated as a point mass, provided the second object is outside the shell


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#30736 - 05/20/09 11:27 PM Re: Does Newton's shell theorem require repair?, [Re: Zephir]
geistkiesel Offline
Junior Member

Registered: 05/20/09
Posts: 10
Loc: Arizona


Originally Posted By: zephyr

Originally Posted By: geistkiesel
..are you convinced, not by my incessant whining, but by the math and the physics?

Of course. Everyone, who knows at least a bit about integral calculus would see immediately, Newton's shell theorem is valid only for infinitely thin shell.

after all, here we can find explicitly:

Therefore, a thin shell can be treated as a point mass, provided the second object is outside the shell



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What did you mean by "after all, here we can find explicitly:"?

It is not the fact of an infinitely thin shell , or even volume as is used here in the integral. The fact that the shell has extension means that the ½ shell closest to m contribute a greater share of the total force on m than the ½ shell segment farther from m.

That the theorem permit’s the "concentration of mass at a point" is of no problem, but where do you put the center of force of the shell? The physics says it cannot be at the shell center of mass (COM) as the closest ½ shell segment contributes a greater share of the total force on m than the shell segment farther from m.

Take three equal masses located on a common axis. Say that two masses, m1 and m2 are a distance of 10 units apart and m3 13 units from m1 [3 units from m2 away from m1]. The test mass m1 is looking down the common axis and can determine the direction of the force, but where is the center of the combined m2 and m3 forces located? The answer is not at the center of mass of the m2 - m3 system, that's for sure.

This location must be determined from the law of gravity, or F = GmM/r^2.

If you can indicate, please explain how the expression for force is proof that the shell behaves as if the mass of the shell was concentrated at the COM of the shell. Wouldn't you think the imbalance of force as seen by m would off set the center of the force in the direction of the strongest applied force?
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Mother Nature Included Time in Her Creation so Everything Wouldn't Happen all at Once

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#30738 - 05/21/09 07:52 AM Re: Does Newton's shell theorem require repair?, [Re: geistkiesel]
Zephir Offline
Superstar

Registered: 07/01/08
Posts: 498
Originally Posted By: geistkiesel
The physics says it cannot be at the shell center of mass (COM) as the closest ½ shell segment contributes a greater share of the total force on m than the shell segment farther from m.
This is exactly why, we are using an average value at center. You're just reinventing light bulb.

Learn the integral calculus and the way, in which it works..

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#30806 - 05/29/09 09:03 PM Re: Does Newton's shell theorem require repair?, [Re: Zephir]
geistkiesel Offline
Junior Member

Registered: 05/20/09
Posts: 10
Loc: Arizona
Originally Posted By: Zephir
Originally Posted By: geistkiesel
The physics says it cannot be at the shell center of mass (COM) as the closest ½ shell segment contributes a greater share of the total force on m than the shell segment farther from m.
This is exactly why, we are using an average value at center. You're just reinventing light bulb.

Learn the integral calculus and the way, in which it works..


I know the calclus. The shell integral, see below, calculates frces from mdifferential mass segments on the shell. The fina result, F = GmM/d^2 says nothng about the shell behaving as if all the mass in the shell was concentated at the shell center located a d. Th statemet says, "the force on m from the total of a thin shell located at d." The shell theorem is not an "average" of forces or masses. You are in stark error when you say "we ar using an average
value at center". Average value of what? The average vlue of the mass on the shell? Your cmment maks no sense at all.

---------------------------------------------------------------------------------
This is a proof that, one may not consider that aa shell behaves as if all the mass on the shell were concentrated at the shell center", and this is, of course, is contrary to Newton's Shell Theorem attesting to the converse (see links).
http://en.wikipedia.org/wiki/Shell_theorem
http://www.physclips.unsw.edu.au/jw/NewtonShell.pdf
If contrary arguments of those responding here do not specifically go to an error in this note, the argument will be ignored and monitors who feel this one sided statement may act accordingly. I don't mind criticism or disagreement; I just want what I publish to be criticized, what I said,only.

This proof uses the system of rings containing differential masses on the shell surface, dM where the rings are centered on the r-axis linking the test mass m with the center of the shell located a distance d from m = 1. Adjust this 'standard' system to include two mirror image rings R1, located closest to m, and R2 located farthest from m. The mass on each ring is M1 = M2 = 1 (for the purposes of this demonstration).

Project the total force from each ring, F1 and F2, onto the r-axis, as the normal shell integral provides. It should be clear that m "sees" the two forces, F1 and F2, along the r-axis concentrated at both rings' centers, each located the same distance from the COM of the two rings (and parenthetically equal distance from the COM of the shell). Without losing generality, place one ring at 9 distance units, the other at 11 from m. To repeat, because the rings are mirror images each ring is the same distance from the shell center plane.

The calculations of the two forces are, F1 = 1/81, F2 = 1/121 and using F = GmM/r^2, where G =1, m = 1, M1 = M2 = 1. The forces F1 + F2 = 1/81 + 1/ 121 = 202/9801 = .020610 = F12.

This is the total force from both rings and if some symmetry is assumed let 10 distance units be the presumed COM of the two calculated forces.

Using F = GmM/r^2 again, but solving for M = F12(r^2)/Gm. Then M = .020610(100) = 2.0610 a mass 6% larger than actually available. Again, from observation, it is clear that the closest equal mass to m contributes the greater share of the total force on m and therefore [i]the COM of the two rings is not the center of force of the combined forces F1 and F2.
.

Using the values of M1 and M2 used in determining the forces F1 and F2, and calculating r from the combined forces, F12 = F1 + F2, r^2 = 2/.02081 = 97.040, or r = 9.851 a distance off set from the COM of M1 and M2 in the direction of m. The number 2 is used in place of M in the general statement GmM/r^2, or "M1 = 1" + "M2 = 1".
Any mirror image rings will always provide a center of force as discussed here being located in the shell half nearest the test mass. The expression F = GmM/d^2 states only the force of mass M on m from a spherical shell located a distance d from m.

Symmetry arguments do not distribute mass to be other than the actual distribution with the resultant generated forces.
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Mother Nature Included Time in Her Creation so Everything Wouldn't Happen all at Once

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#30829 - 06/02/09 04:12 AM Re: Does Newton's shell theorem require repair?, [Re: geistkiesel]
geistkiesel Offline
Junior Member

Registered: 05/20/09
Posts: 10
Loc: Arizona
Originally Posted By: zephir
The physics says it cannot be at the shell center of mass (COM) as the closest ½ shell segment contributes a greater share of the total force on m than the shell segment farther from m.This is exactly why, we are using an average value at center. You're just reinventing light bulb.



No, there is npo av eraging her. The force F GmM/d^2 is used as a proof that the shell acts as if all the force were concentrated at the center of the shell. The expression merely locates he shell cnter at d from m and the expression says nothing about about locating the center of the combined forces acting on m.

If the shell theorem was meely a convenient averaging process don't you think this matter would be mentioned?

Zephir, the shell theorem is described as a "proof" that , the test mass m may consider the mass of the shell located at the shell center.

My next post will indcate the error in calculating earth-moon interactions using the "shell theorem".

I didn't reinvent this light bulb published by Isaac Newton, I am merely poining out the error in perception of what exactly the theorem does state and how some historical misperceptions may be corrected for - if anyone finds a different result or intention I suggest you reevaluate your analysis of this matter.
_________________________
Mother Nature Included Time in Her Creation so Everything Wouldn't Happen all at Once

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#30835 - 06/04/09 01:47 AM Re: Does Newton's shell theorem require repair?, [Re: geistkiesel]
Zephir Offline
Superstar

Registered: 07/01/08
Posts: 498
Originally Posted By: geistkiesel
..one may not consider that aa shell behaves as if all the mass on the shell were concentrated at the shell center..
Why not? This is simply a geometrical approximation, which becomes more and more relevant, as the distance increases. At the very large distance scale such shell would behave like pin-point object with respect to gravity spreading. Try to draw image bellow at scale and you'll see, why Newton could use it for derivation of gravity force between Earth and Moon without excessive simplification.



Originally Posted By: geistkiesel
..the shell theorem is described as a "proof" that ..
Nope, it's a geometrical approximation (..a quite trivial and apparent, IMO..), which Newton has used for estimation of gravity force in spherical geometry - no less, no more. In more detailed analysis we should use a more exact approximation and Mr. Newton realized it too a quite well. For more exact derivation of gravity force near sphere please check the article here and here.

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#30841 - 06/04/09 01:20 PM Re: Does Newton's shell theorem require repair?, [Re: Zephir]
geistkiesel Offline
Junior Member

Registered: 05/20/09
Posts: 10
Loc: Arizona
Originally Posted By: Zephir
Originally Posted By: geistkiesel
“One may not consider that aa shell behaves as if all the mass on the shell were concentrated at the shell center.”


Why not? This is simply a geometrical approximation, which becomes more and more relevant, as the distance increases. At the very large distance scale such shell would behave like pin-point object with respect to gravity spreading. Try to draw image below at scale and you'll see, why Newton could use it for derivation of gravity force between Earth and Moon without excessive simplification.


Earth diameter 12756 km reduce to 1

Moon Diameter 3476 km reduced to 3476/12156 = .28

The earth-moon mean distance 384403 / 12156 = 31.622

Moon orbit eccentricity - .0549

Earth constant gravity 980.665 cm/sec^2

Moon 162 cm/sec^2

These numbers will suffer under any approximation

The solar system is not, in general, orbitally planar - the solar system entities all move in a helical trajectory with the sun the axis of the helix. As such interplanetary activity would appear to deserve treatment as the physics is best described - why approximate?

See Flanders take on the speed of gravity forces - effectively infinite.

In a state of pure speculation could Mercury’s anomalous orbit error be corrected with a proper law of gravity applied? Relativity theory and classic physics models both corrected the problem - I suspect fudging on the math and physics by some, but this is only a speculation.

http://metaresearch.org/cosmology/speed_of_gravity.asp
[img]http://www.aetherwavetheory.info/images/physics/gravity/shell_theorem.gif[/img]


Well at least you agree with me that the shell center is not the center of mass attraction between a test particle and the shell. What do you object to by using an accurate system , as opposed to, an 'approximation'? I understand completely your position here and I understand we are, or may be, nit picking rhetorically to a certain degree.



Originally Posted By: geistkiesel
..the shell theorem is described as a "proof" that ..
Nope, it's a geometrical approximation (..a quite trivial and apparent, IMO..), which Newton has used for estimation of gravity force in spherical geometry - no less, no more. In more detailed analysis we should use a more exact approximation and Mr. Newton realized it too a quite well. For more exact derivation of gravity force near sphere please check the article here and here. [/quote]
These are the concluding remarks of your links - the language is certainly not in terms of "approximations” or in terms of “averaging”.

FIRST LINK - "This is the desired goal, to show that the force from a thin spherical shell is exactly the same force as if the entire mass M were concentrated at the center of the sphere! Physically, this is a very important result because any spherically symmetric mass distribution can be build up as a series of such shells. This proves that the force from any spherically symmetric mass distribution on a mass outside its radius is the same as if the total mass were a point mass concentrated at the center of the sphere."

I say that the force expression, the result F = gMm/D^2
of the integration in the link merely states the forece on a test mass by a sphere located at a distance D from M.

SECOND LINK - "The net gravitational force on a point mass inside a spherical shell of mass is identically zero! Physically, this is a very important result because any spherically symmetric mass distribution outside the position of the test mass m can be build up as a series of such shells. This proves that the force from any spherically symmetric mass distribution on a mass inside its radius is zero. If a given mass m is inside a spherically symmetric distribution of mass, that part of the mass outside its radius does not contribute to the net force on it. "

As I claim the integration is not complete and we must all recognize that an actual "shell" is an unobserved abstraction only and is being used to describe proof of a flawed system - flawed in the sense that I have been describing.
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#30842 - 06/04/09 02:14 PM Re: Does Newton's shell theorem require repair?, [Re: geistkiesel]
Zephir Offline
Superstar

Registered: 07/01/08
Posts: 498
Originally Posted By: geistkiesel
shell center is not the center of mass attraction between a test particle and the shell
It isn't. It's not even of infinitely small thickness.

Does it make another problem in your understanding of integration? If not, why not?

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#31377 - 08/02/09 04:45 AM Re: Does Newton's shell theorem require repair?, [Re: Zephir]
geistkiesel Offline
Junior Member

Registered: 05/20/09
Posts: 10
Loc: Arizona
Originally Posted By: Zephir
Originally Posted By: geistkiesel
shell center is not the center of mass attraction between a test particle and the shell
It isn't. It's not even of infinitely small thickness.

Does it make another problem in your understanding of integration? If not, why not?

I never had a problem with integration. There is, however, a problem with the orientation of the various differential masses in the "rings". The center of the shell sees all the differential volumes equally. The projected areas of each differential volumes are equal as seen by the COM of the shell.

However, the test mass sees an entirely different view that is not accounted for in the shell integral. Starting with the shell segment nearest m (and farthest away) m sees exactly the same as the COM of the shell. As the rings develop and begin to move up the shell surface the differental masses begin to rotate relative to m
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#31378 - 08/02/09 05:03 AM Re: Does Newton's shell theorem require repair?, [Re: geistkiesel]
geistkiesel Offline
Junior Member

Registered: 05/20/09
Posts: 10
Loc: Arizona
Zephir,

From the link we have a major overhaul to consider.

http://www.metaresearch.org/cosmology/speed_of_gravity.asp

Van Flandern tells us that a large bulk of the astronomy/cosmological industry has considered for 60+ years, more if Newton's fears are considered, that the speed of gravity forces is at least Vg > 10^10*c, or as some say, instantaneous. This is fairly easy to swallow as if the Vg of the force of gravity were equal to c, chaos would reign as a simple analysis shows. A Fg originating on the sun for instance would arrive at Jupiter in 20 minutes or so. JUpiter would respond with a return reaction force arriving at the sun 20 minutes later. The orbits of al.l the planets as you can see whould become very chaotic in the struggle to maintain conservation of angular momentum . In any event observation results indicate that Vg conservatively is Vg > 10^10*c. Hence we can dispense with Newton's nightmare of 'action at a distance' through space. This scenario requires a jettisoning of the 'vector' nature of the universal law of gravitation. The force between two masses confronting each other are, as a non-scientific minded friend, "just is".

This rapid force exchange, with motion an intrinsic element in all of this, doesn't allow a knee jerk subtraction of so called 'off axis' forces as is considered by the shell theorem.
Any thoughts here?

Speculation being a a major elemnt in the theory of free form of thought and analysis, I opine that gravity, the force that is, is a manifestaion of conservation of angular momentum, but in a modified form of what we learned in school.
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#31379 - 08/02/09 05:07 AM Re: Does Newton's shell theorem require repair?, [Re: geistkiesel]
geistkiesel Offline
Junior Member

Registered: 05/20/09
Posts: 10
Loc: Arizona
This section should be added to the reply immediately before the one preceeding this post.


However, the test mass sees an entirely different view that is not accounted for in the shell integral. Starting with the shell segment nearest m (and farthest away) m sees exactly the same as the COM of the shell. As the rings develop and begin to move up the shell surface the differental masses begin to rotate relative to m until the line of sight to the shell is tangent at which point m is looking at the thinest cross section of width times thickness. One cannot fix the view b y claiming ever smaller shell differential segments as the problem is merely amplified by the increase in differential segments.

At the tangent point mentioned the half of the differential volume center of the square (measured from the closest 1/2 volume to m contributes more mass force than the 1/2 square segment farther away. Hence, treating each differential volumke as a 'point' as the mathem atics tells us can be done introduces another intrinsic error in the integration results. Therefore, using the integral as generaso instead of using ifferential ;ly desribed requires a fix to account for the rotating volumes. This can get hairy but substituting diferential spheres for the rectangular volumes provides an ideally uniform total of differential volumes that now differ in force on m by the locations of the rings and a simple handling of the spheres where it is known that the 1/2 spnhere segments closest to m [provide a greater share of the force on m than the segment in farthest 1/2 segments. I suppose some overall average of the differential spheres handled as point masses with the correction added will serve our purposes here.
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#31380 - 08/02/09 06:03 AM Re: Does Newton's shell theorem require repair?, [Re: geistkiesel]
geistkiesel Offline
Junior Member

Registered: 05/20/09
Posts: 10
Loc: Arizona
The shell model above in post 30734 should be modified by adding a mirror image ring in the 1/2 segment farthest from m and the integration then will be able to calculate and compare the massforces from each ring on m. Clealry the mass force center, call it the center of gravity is located in the 1/2 shell segment closest to m.

Returning to Flandern http://www.metaresearch.org/cosmology/speed_of_gravity.asp
and giving the rapid force creation of gravitational attraction which Einstein would never have accepted willingly, serious scrutiny of matters outside the general lines of midstream astronomy is long over due.
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Mother Nature Included Time in Her Creation so Everything Wouldn't Happen all at Once

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