This is the drawing of an object in shape of sphere, it has mass "m" and is situated somwhere in space, far away from any other objects, so far that the gravitational influence from them is negligibly small. So there we have: object, vast space around and gravitational field produced by our object. Also we know that we have no way to tell whether our object is in uniform motion or at rest,because there is nothing around that we could use as a reference.
If anyone has any comments or objections to the above scenario, feel free to post them. I will post more thoughts later if we find no flaws in the above.
A.U.

The above picture looks different from the traditional visual representation of gravitational fields where the source of gravity is always located at the very bottom of the gravitational well.

Right?

Picture is different, yes. I made up easy to visualize the concept of mass and gravitational field.It is not aimed at warping space around, therefore no gravitational well needed. It is my way to think of mass and gravitational field. It has its maximum strength at the centre, and diminishes with the distance, but is never zero. Same as radiating light from the light source, maximum at the source and goes into infinity, intensity getting less with the distance. Any way, on the graph shown the distance along X, and the intensity of gravitational field along Y, without any values attached. Bigger mass would result in curvature above that the one is shown.

=========================================================
"...It is not aimed at warping space around, therefore no gravitational well needed...".
=========================================================

Actually the image of the 'WELL' is associated with the Newtonian concept of gravity. Einsteinians, simply, commandeer it to help themselves visualize their 'WARPED SPACE'.

Furthermore, for a spherical body, the "strength of the gravitational field at the centre" is exactly equal to zero. That is the conclusion of a very famous theorem in Newton's Principia.

Yes. Its true. Two ways to fix it. One is shown below.


Another way would be to ignore the size of the object and consider only gravitational field that it is producing.


I prefere second drawing, since my future reasoning involves gravitational field and the space, outside of the object. So far OK?

I like the first picture.

One more note...

The integrated strength of a gravitational field does not decrease with distance. It remains constant and the same for all surfaces centered on its source, from the infinely small surface and up to infinity. Only the gravitational strength per unit area decreases with the square of distance as a result of dividing the same constant quantity by ever-increasing surface areas.

Would it be right to say that an object with mass "m" has the ability to attract another object with mass "m1". That ability is proportional to its mass "m" and reverse proportional to the square of the distance. I will explain the meanings. Attract means force,let letter F stand for it. F = m1 X a, where m1 is the mass of any other object and "a" is the ability of the original object to attract. So acceleration component is the ability which is made of gravitational constant times mass "m" divided by squared distance between objects. That ability exists all around our original object, regardless if there are any other objects around or not. Also that ability could be visualized as the graph on the drawing above filled with blue color, farther away from the object, less the ability to attract. Ok so far?

"Would it be right to say........"?

YES....
But Einsteinians think the above equations are only approximately correct. That is because, according to GTR, the Inverse-Square Law is only approximately valid.

Although I can't prove it, I suspect Einstein has managed somehow to smuggle in, (in his theory), Newcomb's Hypothesis which claims to account for the Advance of Planet Mercury by increasing the exponent of distance --2-- in Newton's Law of Gravitation by very tiny amount.

Yes, relativists would throw wrench in there. Let it be approximately inverse square, precise calculations are not important for now. I am going to make couple more drawings, and post it soon.
A.U.

Here is another drawing. Earth represents our first object with the ability to attract other objects. One of those "other" objects is an apple, supported by something. This apple exert a force on that support. F=m1a, where m1 is mass of the apple and "a" is the ability of Earth to attract.

Below is the drawing of a compressed spring between two plates. Spring is exerting force on the plates. In order to increase the force we would have to compress the spring more, by applying some work to it and spending some energy.

One of the ways to increase force that apple exerts on support is to increase the ability of the Earth to attract, or in formula F=ma it would be increased "a" component. Ability of the Earth to attract depends on its mass and the distance. So by doing some work and spending energy we can move Earth towards the apple thus increasing the ability of Earth to attract (this is in theory, ofcourse). Right?

The earth attracts the apple, and
the apple attracts the earth.

If one takes into account both objects at once, one certainly will be obliged to declare that Aristotle was right and Galileo was wrong. A heavy body, indeed, hits the earth surface before the lighter one, provided both are dropped one after the other from the same height. . That is because the earth falls faster towards the heavy body, and falls slower towards the less massive one.

So by dropping the two different balls from the Tower at the same time, Galileo was cheating. To test Aristotle's assertion, he should have dropped the balls in sequence.

Is there any objection to this?

Yes, its true. Heavy ball and the Earth will meet each other faster then the light ball and the Earth. But in order to pick up the difference even modren technology would probably have difficulty in conducting experiment.

Yes, he could not possibly measure a tiny
fraction of a nano-second.

But, again, he was cheating,
not in the data, but by setting up and twisting the experiment, in such a way, to make himself always
come out the winner and Aristotle the loser.

So he was a cheat!
Right?

Yes, he was a cheat. But do you think he had done it deliberately or without realizing his wrong doing? Especially without precise time keeping mechanism? Excuse him? Maybe?

I believe his 'Big Excuse' is that the precise mathematical form of the Law of Gravitation
was unknown in his time.

What is the excuse of those who believe in transformation of time mass length, but amazingly affected only "moving" object, and the same time assuring "equialency" of every reference frame? Why is it so obvious and yet so well defended?

None...I guess.

But maybe the unbelievable complixy of Einstein's Lyberinth makes a sort of an excuse. Because people usually are fascinated and hooked by obscurities and absurdities.That is why the business of Magic and Witchcraft is often succefull and good for making easy money.
Right?

Sorry...I meant 'COMPLIXITY' & 'SUCCESSFUL'.

Yes. I guess they (scientists) can play "pretend" game for many more years to come. But no matter how complex theory may appear it has to have logic in it. No logic- scrap, try again. Keep looking, keep thinking. Back many years ago the Eart was logicaly flat, until one day some things did not fit in that "logic". So flatness was "scrapped" and "roundness" introduced. Time to scrap relativity.

Re: gravity June 1 2004, 4:41 PM

AAF,
"Furthermore, for a spherical body, the "strength of the gravitational field at the centre" is exactly equal to zero. That is the conclusion of a very famous theorem in Newton's Principia."

I disagree, the gravitational field at the center of the Earth would be the inverse of the field at the surface, the extreme tension and pull on the atoms at the center causes the friction that heats the core of the Earth. As r extends out from the center the tension reverses to contraction allowing the surface of the Earth to become the attractor. The point of equilibrium "I suspect" can be approximated to the "golden mean". The point of equilibrium "golden mean or otherwise" is what prevents a body, the Earth or star, to collapse under the force of gravity irrespective of its size or mass iow Black Holes have no possibility of existing because their gravitational structure is self supporting.

Gravity is flow and fall of space.time into mass.

Force is acceleration in verse fluent pressure differential.
Energy is in verse (equal and opposite) mass by space.time constant.
Mass accelerates toward relative velocity 0; energy holds v max.
Mass holds force as attraction; energy holds force as repulsion.
Gravity is acceleration in verse fluent pressure differential.

Space drains into mass.

Peace
rwj

AAF: Furthermore, for a spherical body, the "strength of the gravitational field at the center" is exactly equal to zero. That is the conclusion of a very famous theorem in Newton's Principia.




Bob S: I disagree, the gravitational field at the center of the Earth would be the inverse of the field at the surface, the extreme tension and pull on the atoms at the center causes the friction that heats the core of the Earth. As r extends out from the center the tension reverses to contraction allowing the surface of the Earth to become the attractor. The point of equilibrium "I suspect" can be approximated to the "golden mean". The point of equilibrium "golden mean or otherwise" is what prevents a body, the Earth or star, to collapse under the force of gravity irrespective of its size or mass iow Black Holes have no possibility of existing because their gravitational structure is self supporting.




AAF: The pressure, at the center, is extremely high; but the strength of the Earth's gravitational field, at the center, must be exactly zero according to this proven and time-tested Newton's Shell Theorem

Re: gravity June 6 2009, 9:03 PM

AAF: "Furthermore, for a spherical body, the "strength of the gravitational field at the center" is exactly equal to zero. That is the conclusion of a very famous theorem in Newton's Principia."

Bob S: "I disagree, the gravitational field at the center of the Earth would be the inverse of the field at [above] the surface,..."

AAF: "The pressure, at the center, is extremely high; but the strength of the Earth's gravitational field, at the center, must be exactly zero according to this proven and time-tested Newton's Shell Theorem"

Gravity is a force of attraction (pull) not repulsion (push), an object at the center of a gravitational body, be it hollow or solid, would be subjected to expansion (tension, outward pull) not compression (pressure, inward push) because the greater mass would be above (external and spherical) to the center, the greater the external and spherical mass the greater the tension, expansion, on the center.

An object falling into a gravitational body can only fall to its center of gravitational mass because the gravitational attraction is equalized to all points outward, equalized, in this case, does not mean zero, it means equal to, or the inverse of, the gravitational potential of the body, as a whole, at its surface pulling down.

If, as you say, and I assume Newton, "The pressure, at the center, is extremely high;" then that gravity must be a point mass at the center of the body, which, if true, would mean that all bodies would have the same gravitational potential irrespective of the total mass of that body because the mass would be attracting to the center point and causing the "pressure" that you claim is on the center point.

In any experiment, the position of the observer is critical. The bulk of Newtons work required that the observer be outside of the gravitational body with the forces acting to, or toward, that body. When Newton turned his focus to the gravitational effect within the body he maintained his observational position outside of the body. When dealing with internal gravitational potential the observer must be place within the body, preferably at the center. Any dynamical gravitational forces outside of the body are not relevant to the dynamic gravitational forces within the body and, as in the present case, cause a misunderstanding of the gravitational dynamics within the body. The maths notwithstanding, on "point mass" Newton was wrong.

A body in free fall is subject to a conical gravitational "pull" to the surface, not a linear "pull" to the surface. It is only defined as linear because the fall is in a straight line but the effect is conical to the diameter of the body with the greatest pull on the center line. Once the falling body passes the surface of the gravitational body the gravitational effect begin to reverse and become inverse to the gravitational effects outside, at the surface, of the gravitational body.

Repeating a theorem over a long period of time does not mean it was "tested", it only means it was repeated; if it was wrong the first time it was used, it will still be wrong the last time it was used.

I think the problem of free falling bodies is really a density problem. The conversion from density to mass, causes serious problems in the resolution of data to the concept of gravity. Density explains why things fall and rise.

Aaron

Yes, density explains why things fall and rise...in a gravitational field. But gravity explains density, without gravity there is no density within which, things can rise or fall. That's why a less dense object rises but, does not go angular to the center of gravity and a more dense object falls but, does not fall angular to the center of gravity.

AAF: The pressure, at the center, is extremely high; but the strength of the Earth's gravitational field, at the center, must be exactly zero according to this proven and time-tested Newton's Shell Theorem





Bob S: Gravity is a force of attraction (pull) not repulsion (push), an object at the center of a gravitational body, be it hollow or solid, would be subjected to expansion (tension, outward pull) not compression (pressure, inward push) because the greater mass would be above (external and spherical) to the center, the greater the external and spherical mass the greater the tension, expansion, on the center. An object falling into a gravitational body can only fall to its center of gravitational mass because the gravitational attraction is equalized to all points outward, equalized, in this case, does not mean zero, it means equal to, or the inverse of, the gravitational potential of the body, as a whole, at its surface pulling down. If, as you say, and I assume Newton, "The pressure, at the center, is extremely high;" then that gravity must be a point mass at the center of the body, which, if true, would mean that all bodies would have the same gravitational potential irrespective of the total mass of that body because the mass would be attracting to the center point and causing the "pressure" that you claim is on the center point. In any experiment, the position of the observer is critical. The bulk of Newtons work required that the observer be outside of the gravitational body with the forces acting to, or toward, that body. When Newton turned his focus to the gravitational effect within the body he maintained his observational position outside of the body. When dealing with internal gravitational potential the observer must be place within the body, preferably at the center. Any dynamical gravitational forces outside of the body are not relevant to the dynamic gravitational forces within the body and, as in the present case, cause a misunderstanding of the gravitational dynamics within the body. The maths notwithstanding, on "point mass" Newton was wrong. A body in free fall is subject to a conical gravitational "pull" to the surface, not a linear "pull" to the surface. It is only defined as linear because the fall is in a straight line but the effect is conical to the diameter of the body with the greatest pull on the center line. Once the falling body passes the surface of the gravitational body the gravitational effect begin to reverse and become inverse to the gravitational effects outside, at the surface, of the gravitational body. Repeating a theorem over a long period of time does not mean it was "tested", it only means it was repeated; if it was wrong the first time it was used, it will still be wrong the last time it was used.





AAF: Believe it or not; Bob; when you stand half way between two super-massive bodies, the force acting on our body is precisely nil and exactly the same as being standing in the middle of Willem de Sitter's Empty Universe! And it's all thanks to Newton's Shell Theorem. Mathematically, this theorem is very simple and self-evident: Two equal masses have equal gravitational fields whose strength is equal and being canceled out at equal distances from both bodies. It's that simple. The important point to notice, here, is that opposing gravitational fields would never tear apart anybody. It's either you fall freely towards one of them; or the force resultant acting at you is zero. And it's all due to Newton's Shell Theorem. Now, why is the pressure at the center of the earth is immense and overwhelming? Why is anything placed at the center of the earth is horribly squashed and crushed? It's mostly because of the combined weight of all things on the earth's surface and below the earth's surface. For example, your own weight is transmitted in its entirety through layers upon layers of the earth's spherical body to finally reach the earth's center and play its part in squashing and crushing any miserable thing at the center of the earth. In other words, the combined weight of one half of the earth is opposing the combined weight of the other half of the earth and squeezing and squashing and crushing all things over there at the center of the earth. However, those things at the center of the earth are very dense and tough and highly resistant to any pressure brought upon them by the rest of the earth. That is why they are there in the first place. It's a kind of survival of the fittest; isn't it?

Re: gravity June 9 2009, 4:12 PM

AAF: "The pressure, at the center, is extremely high; but the strength of the Earth's gravitational field, at the center, must be exactly zero according to this proven and time-tested Newton's Shell Theorem"

Bob S: "Gravity is a force of attraction (pull) not repulsion (push), an object at the center of a gravitational body, be it hollow or solid, would be subjected to expansion (tension, outward pull) not compression (pressure, inward push) because the greater mass would be above (external and spherical) to the center, the greater the external and spherical mass the greater the tension, expansion, on the center."

AAF: "Believe it or not; Bob; when you stand half way between two super-massive bodies, the force acting on our body is precisely nil and exactly the same as being standing in the middle of Willem de Sitter's Empty Universe! And it's all thanks to Newton's Shell Theorem. Mathematically, this theorem is very simple and self-evident: Two equal masses have equal gravitational fields whose strength is equal and being canceled out at equal distances from both bodies. It's that simple."

No, it is not that simple, "canceled out" does not mean they are zero, it means they are equal. In the first place, it is clear from my above message that I am describing the gravitational effect "within" a body and you are responding about the gravitational effect "between" two bodies. In response to your point; the gravitational effect at the midway point between to equally massive bodies would be an equal gravitational pull toward each body not a "nil", or zero effect, the body would not fall toward either body because it would be in a state of equilibrium. The body in question, at the center, would also be subject to equal tidal forces.

Back to my point; I do not accept "point mass" gravity and I think I have stated the reasons why in my previous messages in this string. The gravitational effects within a body are the inverse of the gravitational effects outside of the body. The mass at the center of a gravitational body is being subjected to tidal forces that diminish, and reverse, with distance out from the center. A gravitational body is a self supporting structure that prevents it from collapsing under its own "weight".

I do not know if Newton was wrong, misinterpreted or misunderstood but "point mass" is wrong, as are Black Holes.

Bob, I agree.

Point mass is the reason that gravity does not work. Point mass does not describe the volume of the object. Thus making determinations of position with this system will always wrong.

This puts all theories that use Point mass off the table. We need to examine the motion of densities.

We only associate 4 forces with vibrations; Gluon - QCD, Photon - temp, W Boson - magnetism, Z Boson - electricity. There is no verifiable evidence of gravity. A fantasy boson call the Higg's was created to link gravity and the other bosons. This was not found. CERN broke.

In my opinion. This shows that it is still necessary to create fantasy systems to prove Zero Point Mass attraction. The things they will do to prop up the church. CERN was paid for by someone who wanted to prove the existence of the 'God' particle. That's how to market science. Sell it to the ignorant... When it was obvious that this model was incorrect. It was shut down for repairs. This is not to say that the elaborate systems did not just break. But the amount of money involved causes me to think the backer left.

So where do we go from here. The only obvious direction in the study of motion is density. The motion of density is the change in the objects temperature field, magnetic field and electric fields added to the previous density of the object. Any change in the fields will account for all motion and directions of motion.

This is fully explained at my blog
http://aaronsreality.blogspot.com

Aaron

Incompetence and ignorance explains why those posters can't explain anything correctly.

Aaron is obsessed by point mass questions because he never learned the basics, the Gauss law and its solution:

(Gauss law) . . . . . .


(solution) . . . . . .

If Aaron knew a bit of physics, he would know that this points vs density question is solved since about 3 centuries.

The same applies to Bob.
Even a beginner in physics can develop the solution of the Gauss equation near the center of the earth.
Only basic knowledge of continuum mechanics is necessary to build a sensible model of the stresses near the center of the earth, specially if the core is similar to incompressible fluid.
Calculating the central pressure under simple assumptions is quite simple, much more simpler than trying to make sense from the incoherent posts of Bob.

"The same applies to Bob.
Even a beginner in physics can develop the solution of the Gauss equation near the center of the earth."

Any theory or theorem that derives to, or propagates from, "point mass" as being at the center of a gravitational body is wrong for reasons as are previously stated by me in this string. A gravitational body "on the whole" could be considered "point mass" iff the body is considered a "point" on the whole.

Without using the concept of the point mass, how can one determine the center of mass?

Bob, sign up for this very short course:
http://web.mit.edu/professional/short-programs/courses/relativity_gravity.html?gclid=CP_l45SdgZsCFRxNagodp1l_dw

"Any theory or theorem that derives to, or propagates from, "point mass" as being at the center of a gravitational body is wrong for reasons as are previously stated by me in this string. A gravitational body "on the whole" could be considered "point mass" iff the body is considered a "point" on the whole."

In the Gauss law and its solution where do you see point masses?

Re: gravity June 11 2009, 12:29 AM

AAF: "Without using the concept of the point mass, how can one determine the center of mass?"

Point mass, center of mass and center of gravity are not the same things. Point mass, is assumed to be the center of the gravitational mass body. Center of mass, is a geometric location. Center of gravity, is a point of equilibrium of an object at rest on/in a gravitational body.

Re: gravity June 11 2009, 1:50 AM

"In the Gauss law and its solution where do you see point masses?"

Nine times at this location alone, http://en.wikipedia.org/wiki/Gauss%27s_law_for_gravitational_fields
For example, "M is the mass of the particle, which is assumed to be a point mass located at the origin." (at pg 3)

Is "M" in the equations you provided? No, but you most likely did a quick copy/paste without reading or understanding what you were trying to say.

And furthermore, if "point mass" is not part of Gauss Law then why would you use it to respond to my post, the subject is "point mass" after all.

There is absolutely no point mass in the Gauss law and its solution.
Just look at the equations with enough attention:


(Gauss law) . . . . . .

(solution) . . . . . .


In these equations you can't see any point mass, and instead you see the mass density.
Either you don't understands the basics of physics, or you are making up a fake problem.

Even the concept of point mass does make sense.
The gravitational field outside of a spherical body does not depend on its dimension, but depends only on its total mass.
This means that the outside field is exactly the same as the field of an hypothetical point mass.
That's "point" it is a convenient word when talking gravity.
It makes no harm.

It makes also no harm to consider the earth as a point mass in most calculations of celestial mechanics. The effects of mass distribution are very often negligible. But there are situtations where they are not negligible (moon problems). For these questions, the Newton's theory of gravitation just applies equally well. Simply because the Newton's theory does not depend on any "point mass" hypothesis as it appears from the Gauss law.

This "point mass" story is not more than a cognitive problem.

Even non-spherical bodies (like a long cylinder) can be considered as a "point mass" when only the fields far from such an object are considered. (far field solution)

Here is the general solution (potential form) of the Gauss equation as a function of the distance and the spherical angles:



This solution applies to any shape.
For a spherical object, only the first term remains.
You can observe that the other terms decrease faster with distance than the spherical solution.
For example, the dipolar term (n=2) decreases 1/r² times faster that the spherical case.
Therefore, far away from any object of any shape, the field is close to the field of a "point mass".

No reason to see a problem there.

Do you really think the universe calculates these complex mathematical solutions to produce potential motion, with such speed as an analog?

Yes, you can point to every mass on an object with this, but it still uses point mass. M is every point mass of the sphere/object. The idea here is to define every point of an object as a gravitational attractor. This equation takes the centerpoint of the object and calculates outward. I am not going to dismiss Gauss's equation. It does show different masses over an object. But this does show all Point Masses over an object. These point masses are thought of as a field. But the result is a series of Point Masses.

Mass is inconsistent to the experiment of density. Changes in density change the position of the object. As the baryon spreads out; because of temp, magnetism, or electricity it decreases its density and rises. The mass of the object does not change.

Just boil some water and see density.

Archimedes described density. Eureka. We just thought Newton was cooler.

Aaron

The subject is the gravitational effects "within" a gravitational body NOT the gravitational field external to a gravitational body. In that context "point mass" refers to the center of a gravitational mass that attracts all subsequent mass to it causing pressure, the greater the accumulated mass the greater the pressure.

If you were to read the exchange between AAF and myself you might be able to stay on point.

The Gauss law applies inside as well as outside any massive body.
Did you miss that point?

The real topic is about the properties of the matter under high pressures and temperatures.

I am not so well versed in the mechanical properties of geological solids and liquids in these extreme conditions. However, I am confident that a good background in fluid mechanics and elasticity will be enough. (maybe some knowledge about MHD too for the study of magnetism)
However, I do know that cristalographic and thermodynamic data have been studied quite extensively in geological labs. Properties are typically studied up to 3000°C and 10000 bars, at least.

This kind of detailled materials knowledge combined with the simple Newton's law and geological surveys can provide quite a good understanding of what mat happen inside a body like the earth.

There is really no point about the fundamental law of gravity, except -of course- that you need data to use them properly.

You very confused about physics, it seems, Aaron:

"Do you really think the universe calculates these complex mathematical solutions to produce potential motion, with such speed as an analog? "

Mankind is part of nature, therefore -at the least- nature mirrors parts of itself in the human mind, hman theories and human calculations.

Scientific theories are not an engine for nature to work, it is an tool for humans to collect their experience in an organized way.

The "complex solutions" are not complex at all. They are simply about analysing how assymetrical field decrease with distance as compared to spherical fields.

The solutions are not complex, it is the system under study that is complex and requires more terms in the equation to be represented faithfully. The shape of the moon is more complex than a simple sphere, therefore the need for more typewriting in the equations.

The solutions are actually too simplistic to represent the nature in all its complexities. The solution displayed above deal only with a very small part of the nature. many aspects are simply not considered. Nature is much more complex.

Nature also include the "Aaron" phenomena.
This one is very difficult to understand.
Maybe because nature creates order as well as chaos.

bnon "Mankind is part of nature, therefore -at the least- nature mirrors parts of itself in the human mind, hman theories and human calculations."

Does nature mirror itself in bacteria's minds. The universe does not care what biological organism think. The universe is. It took thousands of years to figure out that objects fall and that the earth is a squashed sphere.

bnon "The solutions are actually too simplistic to represent the nature in all its complexities. The solution displayed above deal only with a very small part of the nature. many aspects are simply not considered. Nature is much more complex."

This is because of all the paradoxes that you run into trying to convert density to mass. or the strange paradoxes that are created by attractions. What about that flat universe that is required for the relativist?

This is a picture of the supermassive object in the center of the galaxy. It is ejecting material into space. And this is not that Hawkins radiation. Nor is it radiation escaping through the asymptotes. It is ejecting more than enough material to create a new star. The pic comes from the Max Planck's Institute.



http://aaronsreality.blogspot.com/2008/10/black-hole-flares.html

It is obvious that there are some topics that are off limits to cosmologist. Organic Chemistry, Chemistry, Density, Aether... An god forbid you question the axioms that produce wormholes and parallel universes.

Gravity is a zero dimension concept that does not match the evidence of the universe.

The universe is simple. The rules are continuous, in that they work the same way for all baryons. The rules are simple. Only when you quit defending other people's logic failures. The universe is simple.

Aaron
Aaron

Meaningless sentense:

"The universe is simple."

What do you mean by "simple"?
I would favor a measure of complexity/simplicity of nature by the number of bits needed to describe nature. If all planets were perfectly spherical, the universe would be much simpler than if the planets could have any shape.

I think that it is actually your point of view that is too simple.

More nonsense:

"Does nature mirror itself in bacteria's minds. The universe does not care what biological organism think. The universe is."

The surface of a lake can be mirror for the nearby trees.
Did I say that nature care about something?
Yet, you cares about these kind of stupid question, and you are part of nature.
The structure of the gene reflects/mirrors the history of life, did I say nature "cares", yet biologists care.

When we observe nature, and try to understand it, we built an image of it, that is we act like a mirror. Science is about trying to make this mirror more efficient and relevant.

Science is not about taling nonsense, like you, Aaron.
You seem embarrased by the science puzzles called "paradoxes", but you feel confortable is your useless nonsense. Is that not a big paradox?

Bob S: No, it is not that simple, "canceled out" does not mean they are zero, it means they are equal. In the first place, it is clear from my above message that I am describing the gravitational effect "within" a body and you are responding about the gravitational effect "between" two bodies. In response to your point; the gravitational effect at the midway point between to equally massive bodies would be an equal gravitational pull toward each body not a "nil", or zero effect, the body would not fall toward either body because it would be in a state of equilibrium. The body in question, at the center, would also be subject to equal tidal forces. Back to my point; I do not accept "point mass" gravity and I think I have stated the reasons why in my previous messages in this string. The gravitational effects within a body are the inverse of the gravitational effects outside of the body. The mass at the center of a gravitational body is being subjected to tidal forces that diminish, and reverse, with distance out from the center. A gravitational body is a self supporting structure that prevents it from collapsing under its own "weight". I do not know if Newton was wrong, misinterpreted or misunderstood but "point mass" is wrong, as are Black Holes.





AAF: Bob, the gravitational attraction between the two halves of a spherical body is exactly the same as the gravitational attraction between two equal bodies. Furthermore, the gravitational field is basically a field of acceleration. And so, when the gravitational fields of two equal bodies are superimposed upon each other, at each and every point of the volume between the two bodies, this equa­t­ion applies: The Acceleration Resultant of the two opposing fields (G) = The Strength of the First Field (g1) - The Strength of the Second Field (g2). At the points of equal distances from the two equal bodies, there­fore, the resultant of acceleration is always equal to zero. Use Newton's Equa­t­ion and check these results for yourself!

Sorry, but this is the correct equation:

The Acceleration Resultant of the two opposing fields (G) =
The Strength of the First Field (g1) + The Strength of the Second Field (g2)

And to be even more precise:

The Acceleration Resultant of the two fields (G) =
The Acceleration Resultant of the First field (1)+
The Acceleration Resultant of the Second fields (1)

Re: gravity June 11 2009, 1:48 PM

"The Gauss law applies inside as well as outside any massive body.
Did you miss that point?"

No Anonymous, I did not miss the point, you did, read on!

"The real topic is about the properties of the matter under high pressures and temperatures."

No, the topic is "gravity" and has been since to start of this string. A comment in this string stated that matter at the center of the Earth was under pressure and I disagreed, check my post of June 6, 2009, 11:20 am, and you will see that I am on topic. Pressure/tension and/or temperature is not a "property" of matter. The core of the Earth is under extreme tension as a result of the gravitational attraction of the mass above the core and the result is extremely high temperature causing the core to be magma or liquefied.

Newton wrongly assumed, (or was interpreted wrong) that the gravitational attraction of the Earth propagated from its center aka "point mass".

The fact that my "point" differs from Newton, Gauss el al does not mean that I "missed" the point, it means I disagree with the "point". I am doing my best to make my point and I am well within the topic/subject of this string. Thank you!

The "point" of this message board is to exchange ideas and comments on physics and/or philosophy. There is no requirement that members agree with Newton, Einstein, Gauss or You for that matter, nor is there a requirement that members have a PHD, a GED or even a name.

Re: gravity June 11 2009, 6:16 PM

AAF: "Bob, the gravitational attraction between the two halves of a spherical body is exactly the same as the gravitational attraction between two equal bodies."

Similar maybe but not exact, the attraction between two equally massive bodies is bi-directional (linear) to each body, while the attraction within a body is spherically directional (in all directions outward) which is inverse of the spherical attraction at the surface of the body,(from all directions inward).

"Furthermore, the gravitational field is basically a field of acceleration."

We have to make the distinction here between gravitational attraction and gravitational acceleration. The gravitational acceleration on a body may very well be zero, such as at the center between to bodies of equal mass but or at rest at the surface of a body, the gravitational attraction however, is not zero, it is equal in your two body system toward each body or at the center of my one body system toward the surface of that body spherically.

"And so, when the gravitational fields of two equal bodies are superimposed upon each other, at each and every point of the volume between the two bodies, this equa­t­ion applies:" [The Acceleration Resultant of the two opposing fields (G) = The Strength of the First Field (g1) - The Strength of the Second Field (g2).] font not in the original

No need to yell AAF, I can still read the fine print in my Tenure contract just fine!

No, the equation does not work for the gravitational effect "within" the body, only for the gravitational field "outside" of the body

"At the points of equal distances from the two equal bodies, there­fore, the resultant of acceleration is always equal to zero."

Yes, gravitational acceleration may very well be zero but, as I have been trying to say, the gravitational attraction is not zero! Again, the gravitational attraction between two bodies at the center point is bi-directional toward each body but, the gravitational attraction within the body is spherical toward the surface causing tension, not pressure.

"Use Newton's Equa­t­ion and check these results for yourself!"

Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood.

(an aside question for you AAF, when you check Einstein's equations, do you agree with his results?)

Bob S: Similar maybe but not exact, the attraction between two equally massive bodies is bi-directional (linear) to each body, while the attraction within a body is spherically directional (in all directions outward) which is inverse of the spherical attraction at the surface of the body,(from all directions inward). We have to make the distinction here between gravitational attraction and gravitational acceleration. The gravitational acceleration on a body may very well be zero, such as at the center between to bodies of equal mass but or at rest at the surface of a body, the gravitational attraction however, is not zero, it is equal in your two body system toward each body or at the center of my one body system toward the surface of that body spherically. No need to yell AAF, I can still read the fine print in my Tenure contract just fine! No, the equation does not work for the gravitational effect "within" the body, only for the gravitational field "outside" of the body. Yes, gravitational acceleration may very well be zero but, as I have been trying to say, the gravitational attraction is not zero! Again, the gravitational attraction between two bodies at the center point is bi-directional toward each body but, the gravitational attraction within the body is spherical toward the surface causing tension, not pressure. Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood. (an aside question for you AAF, when you check Einstein's equations, do you agree with his results?)






AAF: Okay, Bob; let's look more closely at the gravitational attraction within a spherical body! First, we assume matter is distributed uniformly or close to being distributed uniformly inside the sphere. Then we start slicing and dicing this spherical volume to equal spherical shells from the outer surface and going inward to the center. As a result, we will have a series of spherical shells: [S1, S2, S3, S4, , Sn, , Sc]. Suppose, now, the spherical body, in question, is Planet Earth; and that the spherical shells we've sliced of it are (100 miles) thick. Finally, we use Newton's Equa­t­ion to calculate the net gravitational force exerted by each spherical shell on other inner spherical shells enclosed by that specific shell. If the calculations are done correctly, the total gravitational force of this specific shell on all other inner shells will always turn out to be nil and exactly and precisely equal to zero. And this zero result remains always the same for every other shell and the shell enclosed by it. So, the final conclusion is this: outer shells have no gravitational effect on inner shells, but inner shells attract outer shells in the usual way and as if they were completely independent bodies. Have I made this fact clear, Bob? Now, let's consider very briefly an analogy that might have misled you, Bob! I presume that you like the gravitational pull to everyday mechanical pull. Say the mechanical pull as applied to a person fastened to two horses running at 180^o relative to each other. Obviously, the acceleration of that miserable person would be zero or close to zero; but he\she would be torn apart; right? That is not the way gravitational fields work. The gravitational pull is the pull of field's force, not the Archimedes lever's force. And it's called 'field's force' because it has no lever, no point of lever, no leverage, and no arm. It's just an acceleration. And when the acceleration resultant vanishes, the net field force vanishes as well. As for Einstein's Field Equa­t­ions, it's hard to say what the final results would be. It's really hard. Remember that Einstein himself, upon computing the few predictions of his General theory, suffered nervous breakdown and without the nursing care of Elsa, he would have probably lost his mind forever!

AAF and Bob;

This analogy of the spheres is still does not match evidence.

Lets look at the spheres as the different layers of the atmosphere.



http://www.williamsclass.com

Lets look at a hurricane that is traveling over the ocean. The spinning winds cause rotation in the ocean. The low pressure of the storm causes a bulge upward in the ocean. Heat and pressure are two of the main variables in this system. As the temp increases it decreases the pressure of the storm, causing an increase in intensity in the storm.


http://en.wikipedia.org/wiki/Eye_(cyclone)

The heated air is forced up the eye wall. This is an example of a temperature/pressure force on density.

There is no way that gravity can describe this system, with or without spheres.

Aaron

In this sphere of ignorance such statements can bloom:

"There is no way that gravity can describe this system (hurricane), with or without spheres."

Hurricanes can be understood from classical physics.
The Navier-Stokes equation is an excellent start:


http://en.wikipedia.org/wiki/Navier-Stokes_equations

The rotation of the earth plays a crucial role and explains at least the main features.

However, the detailled dynamical description of an hurrican is much more complex.
The Navier-Stokes equation describes only the mechanical aspect.
It is also needed to describe all the thermodynamics effect including heat exchanges and phase changes. These play an extremely important role as they are needed to understand how the mechanical power is generated.

Who said rain and heat exchange is explained by gravity?
It seems Aaron is like a baby discovering speech.

Utter stupidity:

"Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood."

Leaves me speechless.

vegetable>>Leaves me speechless.

vegetables shouldn't talk

bnon,

There are many models of hurricanes. This is the problem. No one model predicts the motion, intensity... of cyclonic systems. Ever watch the weather channel? They tend to show the best models of prediction and average the many models' points.

So If your perfect equation is correct, why are there so many models?

And why are you so afraid of what I post? Only a person in fear would resort to your behavior. Its called passive aggressive neurosis and can lead to psychotic behavior. dude, relax it is only a model. All models are science fiction until agreed upon by the group. They may or may not be correct. Just an average of what people who wear vestments agree upon.

a

Re: gravity June 13 2009, 5:02 AM

"Utter stupidity:

"Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood."

Leaves me speechless."

Really! let me try for unconsciousness:

Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood.


Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood.


Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood.


Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood.


Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood.

Re: gravity June 12 2009, 6:59 PM

AAF: "Okay, Bob;...First, we assume matter is distributed uniformly or close to being distributed uniformly inside the sphere."

That's a start.

"Then we start slicing and dicing this spherical volume to equal spherical shells from the outer surface and going inward to the center. As a result, we will have a series of spherical shells: [S1, S2, S3, S4, , Sn, , Sc]. Suppose, now, the spherical body, in question, is Planet Earth; and that the spherical shells we've sliced of it are (100 miles) thick. Finally, we use Newton's Equa­t­ion to calculate the net gravitational force exerted by each spherical shell on other inner spherical shells enclosed by that specific shell. If the calculations are done correctly, the total gravitational force of this specific shell on all other inner shells will always turn out to be nil and exactly and precisely equal to zero." [emphasis added]

That's my point A, the calculations are not done correctly because the formula is based on the gravitational force as originating from the center of the mass aka "point mass".

"And this zero result remains always the same for every other shell and the shell enclosed by it. So, the final conclusion is this: outer shells have no gravitational effect on inner shells, but inner shells attract outer shells in the usual way and as if they were completely independent bodies." [emphasis in the original]

What you are saying is, is that the inside of the shell has no gravitational attraction toward the center which is untenable, why would the inside of the shell loose its gravitational attraction and yet, the outside retain its attraction?

"Have I made this fact clear, Bob?"

You have made your "point" clear but you have not made it a "fact", AAF!

" Now, let's consider very briefly an analogy that might have misled you, Bob!"

I have not been misled, it is you who is trying to mislead me, AAF!

"I presume that you like the gravitational pull to everyday mechanical pull."

You presume wrong so I will skip your explanation.

"The gravitational pull is the pull of field's force, not the Archimedes lever's force. And it's called 'field's force' because it has no lever, no point of lever, no leverage, and no arm. It's just an acceleration."

Are you trying to tell me that gravitational attraction alone will not pull a body apart? If your answer is as vague as your answer about Einstein's equations I might suspect that you are a pseudo-relativist. If you do give me a direct answer and you would like to continue I will give you my model, if not, this statement needs repeating, Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood.

[an observation] in all the years you argued with a certain un-named relativist you never showed the frustration you have in your response to me, AAF!

Bob S: That's a start. That's my point A, the calculations are not done correctly because the formula is based on the gravitational force as originating from the center of the mass aka "point mass". What you are saying is, is that the inside of the shell has no gravitational attraction toward the center which is untenable, why would the inside of the shell lose its gravitational attraction and yet, the outside retain its attraction? You have made your "point" clear but you have not made it a "fact", AAF! I have not been misled, it is you who is trying to mislead me, AAF! You presume wrong so I will skip your explanation. Are you trying to tell me that gravitational attraction alone will not pull a body apart? If your answer is as vague as your answer about Einstein's equations I might suspect that you are a pseudo-relativist. If you do give me a direct answer and you would like to continue I will give you my model, if not, this statement needs repeating, Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood. [an observation] in all the years you argued with a certain un-named relativist you never showed the frustration you have in your response to me, AAF!






AAF: Look, Bob! Inner shells do not lose their gravitational effect on outer shells. It's the outer shells that lose their gravitational effect on the inner shells. And the main reason is that every outer shell, by definition, is surrounding the inner shells from every direction; and hence its gravitational attraction within these regions must cancel itself out. Do I feel frustrated by talking to Bob about this subject? Absolutely not! Newtonian Gravitation is a very interesting subject and I enjoy talking about it, specially with Bob.... Regarding your model, if you've already published it, then please don't show it to me; discussions won't do us any good! But if you're still developing it and fining and refining it, then please, please, please, bring it ASAP and let us take a close look at it.

"There are many models of hurricanes. This is the problem. No one model predicts the motion, intensity... of cyclonic systems. Ever watch the weather channel? They tend to show the best models of prediction and average the many models' points."

That's not the problem.
Hurricanes are complex systems and not easier to modelize than the weather.
The problem occurs when ignorant people express their irrelevant opinion:

"There is no way that gravity can describe this system, with or without spheres. "

As I asked you: "Who said rain and heat exchange is explained by gravity? "
If you have not understood my point, I can help you:

- the rain fall because of gravity
- the rain occurs because of water vapor cooling (temperature going closer to the dew point)
- the rain formation in itself is a very complex phenomena
- the dynamics of rain formation depends on the presence of aerosols and fine particles
- the dynamics of rain formation also depends on complex collision and coelescence phenomena
- the ascending motion of air is driven by gravity on all parts of the atmosphere
- this ascending motion is also affected by the Coriolis force
- this motion is affected also by the drag of rain
- this motion is also affected by the detailled geometry of the hurricane
- the fluid dynamics itselfs exhibits an incredible complexity, like turbulence and instabilities
- the term "T" in the Navier-Stokes equation above hides a lot of complexities that often still need to be understood
- heat exchange plays an important role, yet it is not simpler than the fluid dynamics

- yet, suppress gravity and there are no hurricanes anymore

Therefore Aaron, if you don't understand my reaction to your stupid comments, I don't care.
just show that you can do better than this:

"There is no way that gravity can describe this system, with or without spheres. "

This sentense has absolutely no meaning.
Gravity cannot describe anything, gravity is gravity and physicsts try to describe it as well as possible. I leave you to your sphere of ignorance and your illusions of knowledge.

"Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood. "

I tried, but I am still speechless.

So I try another one:

"There is no requirement that members agree with Newton, Einstein, Gauss or You for that matter, nor is there a requirement that members have a PHD, a GED or even a name."

Yes, right, there is no requirement at all.
It is even not required that a post makes sense, even the contrary: nonsense is highly appreciated.

bnon,

Let's examine your points one at a time.

- the rain fall because of gravity
Rain falls because as cold water it is more dense or less buoyant then the surrounding air. Remember Buoyancy

- the rain occurs because of water vapor cooling (temperature going closer to the dew point) Yes, that is one way of looking at density and Buoyancy

- the rain formation in itself is a very complex phenomena
Yes

- the dynamics of rain formation depends on the presence of aerosols and fine particles
Yes, the mixture of dirty water causes it to be a more dense mixture.

- the dynamics of rain formation also depends on complex collision and coelescence phenomena
Yes again, as the less dense gas collides with other gases, their temperature reduces and the gas becomes liquid. This process occurs until the liquid water is heavier than the force of the updraft.

- the ascending motion of air is driven by gravity on all parts of the atmosphere
No. The ascending motion of air is due to the changes in temperature. The air heats at the ground and rises. As it rises it cools and becomes liquid again. The liquid water is more dense then the surrounding air.


- this ascending motion is also affected by the Coriolis force
There are only 4 forces, Gluon - nuclear, W Boson - magnetism, Z Boson - electricity, and the Photon - heat. Those are the only forces that have been experimentally shown.

- this motion is affected also by the drag of rain
Of course, As rain falls it drags against the updraft. This is the friction that removes the heat from the air. Heat radiates to cold.

- this motion is also affected by the detailled geometry of the hurricane
The motion is the geometry of the cyclone. Geometry is a snapshot in time of an object.

- the fluid dynamics itselfs exhibits an incredible complexity, like turbulence and instabilities
The whole system is buoyancy. Buoyancy is only fluid dynamics. As a matter of opinion, everything is fluid dynamics.

- the term "T" in the Navier-Stokes equation above hides a lot of complexities that often still need to be understood
Like G. They change that frequently. These changes of G try to mimic the evidence. Just change the constant and you will get what you want to see.

- heat exchange plays an important role, yet it is not simpler than the fluid dynamics
Heat exchange is intrinsic to fluid dynamics. You cannot remove heat from the fluid dynamics problem.

- yet, suppress gravity and there are no hurricanes anymore
One cannot suppress gravity. Gravity is not real. Its an old model of how objects move. Evidence has force major changes in gravity. There is no supporting evidence of a graviton. Its like the higgs boson. Something that was built to try to explain interactions. But it does not.

Kepler's ellipse motion is more in line with how planets orbit the sun. This is because of the motion of the sun around the galaxy. This causes the elliptical shape of the orbit. Not gravity. Gravity cannot explain 3 body interaction. You can put up all the equations you want. If they have a G in them then they are a Zero Point Mass system and that does not exist in the universe.

The rules of the universe are simple. An atoms position in a system is based upon its density in relation to the surrounding densities and the changes in magnetism, electricity and temperature.

Aaron

There are conceptual mistakes at nearly everyone of your answers Aaron.
I cannot waste my time cleaning your glasses.
If you are interrested, chose one of your answer for more comments.
Anyone you will chose is probably wrong and I will explain you why.

I cleaned by glasses last month.

If you think there are conceptual errors in my thinking, then please destroy them. I doubt you can.

Gravity is a 400 year old erroneous concept that needs to be removed to forward science. It is like religion. You must believe. Those who believe are rewarded. Those who don't are mocked as crackpots. Remember your veggies.

It is not my problem that you and others have spent your entire lives supporting this nonsense. Basically you are defending other peoples ancient theories. Its kinda like you are defending the i Ching. That hole in the pit of your stomach is the destruction of gravity.

Aaron

Re: gravity June 13 2009, 3:19 PM

AAF: "Look, Bob! Inner shells do not lose their gravitational effect on outer shells."

I did not imply that you said that, what I said was, "What you are saying is, is that the inside of the shell has no gravitational attraction toward the center which is untenable, why would the inside of the shell loose its gravitational attraction and yet, the outside retain its attraction?"

"It's the outer shells that lose their gravitational effect on the inner shells. And the main reason is that every outer shell, by definition, is surrounding the inner shells from every direction; and hence its gravitational attraction within these regions must cancel itself out."

That statement seems to me to be the crux of our disagreement. The gravitational effect/attraction is not lost nor is it canceled out, but let's put that point aside for now.

In your post of, June 12, 2009, at 6:59 PM you made the following statement, "The gravitational pull is the pull of field's force, not the Archimedes lever's force. And it's called 'field's force' because it has no lever, no point of lever, no leverage, and no arm. It's just an acceleration." To which I asked the question "Are you trying to tell me that gravitational attraction alone will not pull a body apart?" Now, if you will please go back and review what you said about the lever force and answer my question, I can have a better understanding of your position on the subject of gravitational force and just how strong you think it is, then I will return to the point I just asked to be put aside.

Re: gravity June 13 2009, 5:15 PM

"Newton's equations work just fine for the gravitational field outside of the body but on the gravitational effects within the body he was either wrong, misinterpreted or misunderstood. "

I tried, but I am still speechless.

So I try another one:

"There is no requirement that members agree with Newton, Einstein, Gauss or You for that matter, nor is there a requirement that members have a PHD, a GED or even a name."

Yes, right, there is no requirement at all.
It is even not required that a post makes sense, even the contrary: nonsense is highly appreciated."

How very interesting, you say you are speechless yet you continue to yammer on. No Anonymous, you didn't fail, you are more unconscious than you realize.

This is from my blog athttp://aaronsreality.blogspot.com

Gravity is the bane of physics. It is the center most axiom in matter physics. The whole concept needs to be abandon if we are to advance science. A theory of motion must work at all levels; quantum, itty-bitty, tiny, normal, massive, supermassive, and ultramassive.

The first major fallacy of gravity is a concept called Zero Point Mass. This is a mass without a volume. This is not found in the universe. The main problem here is the reduction of 3 dimensional densities to 0 dimension masses. Once you reduce the density to mass you cannot return a mass to the shape of the density. So you cannot cube a zero and get anything but another zero. This breaks the commutative properties of addition.

Lets look at some of the equations and how gravity fails at a fundamental level.
1) F=ma : Force equals mass times accelleration.
We have a zero dimensional mass * a 2 dimensional vector and that does not equal a 3 dimensional field. So the main axiom of gravity fails the commutative test.

This alone should disprove gravity.

But wait there is more.

F = G(M1*M2)/r^2 : Force = (The constant of Gravity The zero dimensional mass 1 The zero dimensional mass 2)/ The 3 dimensional length between them squared.
So every object pulls every other object.

We have mass 1 pulling on mass 2 and mass 2 pulling on mass 1. F₁ = F₂. This is just wrong. The force that the moon pulls the earth is not equal to the force that the earth pulls on the moon.

A constant is what is used to fill in the gaps. When things do not work the way we want them to we just add a constant to fix the problem.

Physicist know about this problem. They created gravity waves and sphere modeling to compensate for the dimensionless mass. But gravity is still dimensionless. The dimensionless mass now creates a 3 dimensional wave.

Objects are not spherical. The Earth is not a sphere, the Sun is not a sphere. The galaxy is not a sphere.

We all know that gravity collapses under the scrutiny of the tiny. Quantum level objects do not show any signs of gravity. The Higg's Boson was 'created' in six days to solve this problem. Then he rested. They marketed the Higg's boson as the 'God' particle. Then they built a huge lab to prove the Higg's Boson's existence. When they did not find the Higg's Boson, the lab broke or the backers left. Then they created a story of how they 'created a black hole at CERN and needed to shut it down. Really, a black hole?

Gravity fails the multi-body test. You can only compute the Force between 2 objects. Any equation that uses a sum of objects fails in this way. First the two objects force is computed then the third body is computed with the resultant of the first two bodies. Then that resultant is computed with the 4 body... That is how summation works. The problem is that the distance between object 1 and 2 is not evaluated in the next iteration.

Aaron

Aaron,

There are many guys like you.
If you like, have a look at that:

http://superstruny.aspweb.cz/

Certified crank, but if you like it read it.
Personally, I had a lot of fun and I appreciated the pictures quite a lot.
I had a lot of inspirition from this crank for a few pictures I did myself.

I even don't ask you to stop your nonsense.
Asking that would mean stopping this forum!

This is a picture that illustrates why this is so:



http://www.physics.upenn.edu/courses/gladney/mathphys/java/sect5/subsubsection5_1_1_2.html

The geometrical property illustrated above implies that the total force inside a spherical shell is zero.
This occurs because of the 1/r² decrease of the gravity force.
This is of course quite difficult to check directly by an experiment.
We must trust our brain a little bit on that!

However, the 1/r² behaviour occurs also in electrostatics.
In this case an experimental check is possible, at least in principle.

To Anonymous.

This is very interesting. So in a system consisting of many concentric shells, such as Earth. The force of gravity is zero everywhere? The miners must be very lucky and oil does not have to be pumped out.

regards Gravition

AAF: Excellent illustration of the Shell Theorem, Anon!




Bob S: I did not imply that you said that, what I said was, "What you are saying is, is that the inside of the shell has no gravitational attraction toward the center which is untenable, why would the inside of the shell lose its gravitational attraction and yet, the outside retain its attraction?" That statement seems to me to be the crux of our disagreement. The gravitational effect/attraction is not lost nor is it canceled out, but let's put that point aside for now. In your post of, June 12, 2009, at 6:59 PM you made the following statement, "The gravitational pull is the pull of field's force, not the Archimedes lever's force. And it's called 'field's force' because it has no lever, no point of lever, no leverage, and no arm. It's just an acceleration." To which I asked the question "Are you trying to tell me that gravitational attraction alone will not pull a body apart?" Now, if you will please go back and review what you said about the lever force and answer my question, I can have a better understanding of your position on the subject of gravitational force and just how strong you think it is, then I will return to the point I just asked to be put aside.






AAF: Would an object placed half way between two massive objects be pulled apart? In all, except one rare and highly arranged case, the half-way object would not be pulled apart. To the contrary, this midway object would enjoy being in a free space of zero gravity! Regarding the rare case, in question, it occurs when the halfway object is less dense and very extended (e.g. red giant star) and near and half way between two dense and massive objects (e.g. two white dwarfs). In this case, even though, the difference of [g1 - g2] at each side of the middle point is not very large, one half of the midway object would accelerate towards the first object and its second half would accelerate in the opposite direction towards the second object. Assuming that it''s extended and the gravitational strength on its surface is less than the [g1 - g2] of the two massive objects, then this halfway object would lose some of its mass to these two objects and eventually become smaller and less extended. And that is it.

Stupidon:

"This is very interesting. So in a system consisting of many concentric shells, such as Earth. The force of gravity is zero everywhere? "

Use your second neuron, Aaron.
Try to understand what happens at mid distance between earth center and earth surface.
An experimenter then has half of the concentric shells below his feet and half above.
Therefore he will experience a resultant force due the shells below his feet.
The shell above his feet have an gravitational effect on him but the resultant effect is zero because attracting force from different direction cancel out.

With some training maybe you will grow a third neuron.

Anonymous,

I was just testing you. Anyway, can you plot the gravitational potential as function of the distnace from the center of Earth? Remeber that the potential and the G field must be continuos at the surface.

Regards Graviton

"I was just testing you. Anyway, can you plot the gravitational potential as function of the distnace from the center of Earth? Remeber that the potential and the G field must be continuos at the surface. "

Are you trying to look clever now?

The solution is simply g(r) = k M(r)/r² .
M(r) is a continuous function, but its derivative is discontinuous.
(so now I look clever too)
The field g(r) is therefore continous also.
The potential, obtained by integration is automatically a continuous function.

Assuming M(r) varies as r^3, the force g(r) will be a linear function of r.
If the density M'(r)/2.pi.r² is not constant, we may at least assume it is an even function of r.
In that case, up to the next order, M would vary as a r^3 + b r^5 + ...
In that case, g(r) will contain an additional term in r^3.
(now I really look smart)

Do you consider that as a relevant test?
And for which purpose?

The real exercice on this forum is to understand the people posting here.
Why do they post these stupidities (95%), or why do they answer them (5%)?
Why do they have various problems, whith Einstein, with clock synchronisation, with black holes ...
What are their geographical location and is there a cultural correlation?
Is there any link with religion or politics?
What are the age of the posters, how many retired, how many teenagers, how many game adicts, ...
These are the relevant question you can ask here.
Nothing about physics.

Dear Anonymous.

You did not pass the test.

Your formula inside of Earth does not satisfy the Gauss equation, Gauss law (Laplace with right side not equal to zero). You need to study some more and send me a new function for the potential. Keep the comments of stupidity to yourself perhaps you may need them.

Regards Graviton

Ooh stupidon,

I should have commented more.
The g(r) function I have written is the radial component of the field.
The angular component is zero, by symmetry.
Plug that into the Gauss equation, and you will see it fits.
If you need help in the calculations, post to an homework forum.

Ooh, maybe also the M(r) was not clear for you?
Don't hesitate to ask if needed.

Dear Anonymous.

I did not aks you about g(r), I only wnated to know the potential as function of r. Of course you can assume constant mass density. Why should I complicate this homework for you? Anyway, you have not answered the question of potential, so you did not pass.

If you had answered correctly I wanted to ask you a follow up question which was as follows: Since there is no g filed far from Earth the speed of light is c (vacuum). Since in the center of Earth there is no g field is the speed of light there also c (assuming a small vacuum cave there)?

But perhaps discussion with you is not that interesting, so I will not continue.

regards, Graviton.

bnon,

If you wish to learn more about my neurons. I have started writing about those few neurons I have left.

http://aaronsneuralnetworks.blogspot.com

I am working on a neural network system, and discussing what happens when neural networks fail.

You still have not addressed the issue with dimensionality.

You yammer along telling us what we can talk about. You demand that we conform to what you want to hear. I don't care what you want to hear.

It is disgusting to see you engaging people who don't think like you. You are as demanding as a religious fundamentalist. As you stated here attacking people before learning what they think. We have disparate thinking styles. If that is not to your liking then get out. Quit attacking us as crackpots. We are aware that the ideas that we discuss label us as such.

You obviously are intelligent. You have a grasp of mathematics, and common physics. Just your behavior is that of a 5 year old. I grow tire of telling you this.

"The real exercice on this forum is to understand the people posting here.
Why do they post these stupidities (95%), or why do they answer them (5%)? "

This is an open forum. We idiots need to learn to put up with people who act like obnoxious asses.

"Why do they have various problems, whith Einstein, with clock synchronisation, with black holes ... "

Because none of Einstein's babble can stand under scrutiny. E (a field) does not equal mass(a zero dimensional bs)* the variable speed of light (scalar). That is just BS. If you wish to defend it here that is fine.

"What are their geographical location and is there a cultural correlation?"

As far as I can tell there are people from all over the world here. Your behavior is keeping them from expressing their ideas.

"Is there any link with religion or politics?"
I am a independent buddhist, but that is none of your business. Belief is the only thing man will kill for. Ask a christian politician, they will tell you how they have to kill for jesus. Onward christian soldiers marching off to war.

"What are the age of the posters, how many retired, how many teenagers, how many game adicts, ... "

HOW IS THIS EVEN RELEVANT. I am a disabled epileptic, I have several thousand seizures a year. Again, you have no idea the levels of my arrogance. I have literally had to claw my way out of the abyss of constant seizures. I am surprised I still have the ability to read and write. I sit, I read and I build things. I would call that busy.

I am 40. I quit building/playing games years ago, because they are time wasters. I have 8 opteron computers most with multiprocessors, a small supercomputer. I have built an alternative model of the interactions of the universe. I don't care if the model is right or wrong, I enjoy seeing 'scientist' like you flee from such an idea. You have yet to describe any compelling reason why I should abandon the ideas I have developed.

These are the relevant question you can ask here.
Nothing about physics. "
People who demand conformity are frightened.

I like reading the discussions between AAF and Bob for example. Here is a perfect example of two people debating, and disagreeing with each other on a point by point basis, and they do this without sniping or anger.

Quit attacking us personally and work on the problems laid down. It is ok for people to be wrong. People don't need to be emotionally brutalized for being wrong.

Aaron

AAF: Well said, Aaron. ...




Anon: The real exercise on this forum is to understand the people posting here. Why do they post these stupidities (95%), or why do they answer them (5%)? Why do they have various problems, with Einstein, with clock synchronization, with black holes ... What are their geographical location and is there a cultural correlation? Is there any link with religion or politics? What are the age of the posters, how many retired, how many teenagers, how many game addicts, ... These are the relevant question you can ask here. Nothing about physics.





AAF: Good questions, Anon!
But only the question about the geographical location can be answered scientifically.
Here is the list: [1] USA: 40%, [2] Canada: 16%, [3] UK: 12%, [4] Germany: 11%, [5] Bulgaria: 5%, [6] Serbia: 2.5%, [7] Vietnam: 2%, [8] Belgium: 1.8 %, [9] Israel: 1.5%, [10] Russia: 1%, [11] Czech Republic: 1%, [12] Greece: 0.8%, [13] France: 0.8%, [14] Belarus: 0.4%, [15] Malaysia: 0.4%, [16] Australia: 0.4%, [17] Jordan: 0.3%, [18] Ukraine: 0.3%, [19] Romania: 0.2%, [20] India: 0.2%, [21] Italy: 0.2%, [22] Japan: 0.2%, [23] South Korea: 0.2%, [24] Netherlands: 0.18%, [25] South Africa: 0.17%, [26] Norway: 0.15%, and [27] The rest of the world: 1%.

AAF, and Aaron, I'll toast to that too!

Aaron,

I listed some of the ideas that pop up when I read posts here.
I agree totally with you that they are not very interresting.
I would really prefer to discuss physics.

Unfortunately I have seen too much absurdities here, too much arrogance, and too much bad faith.
Physics has become not more than a faint background.
The foreground are people shouting slogans.
None of these slogans can stand a serious discussion.
Most often these slogan have even no meaning: they cannot be translated in more precise terms.
Therefore I decided to mimick the general behaviour when expressing my own opinion.

To people talking about "math mistakes of Einstein", I answer that they are stupid.
That's the language they can understand.
Additionally I also explain them why they are wrong and why their view is unjustified.
But that's the language they don't understand.

No need to list other examples, just read the list of topics.


Aaron,

Your last post was quite sensible and convincing.
I would like so much that you discussed physics in the same way!

Anon;

It is obvious that you get common physics. What was taught to you was understood by you. You have probably gone through life being the smartest person in the room by 30 or more points. I know this is difficult. Most people cannot understand the words that come out of your mouth. It is obvious that you lost respect for most people around you because of this.

I have had the same issues all my life. I usually took it out on teachers. It took me a while to come to this epiphany. It is not their fault. They did not understand the material enough to answer questions about it. Nor did they need to go home crying for their ignorance. Nor did I need to be kicked out of classes for asking questions. But I did become a disruption to the learning of other students. The other thing I learned is that attacking someone for their ignorance does not solve the problem. It just made me look like an ass.

Now, you are amongst people who are at least of equal intelligence. We are not so willing to be dissuaded by your usual tools. Calling me an idiot or stupid will not get me to back off an idea. Showing me a proof, illustration, equation, will cause me to think. But data will always get my attention.

So now I search the internet looking for information. I found that you tube has Graduate level classes available from Stanford, MIT, and other schools. The top physicist/scientist have place their lectures where I can watch them. I placed links on my blog to help me remember what I am working on.

Paradoxes are what my mind always finds. If a statement is incorrect I find it by the paradox it generates. Usually the author of said statement has no idea the paradox exist. 'The flat universe', this is what is generated from 2d solutions in 3d space. But we know the universe is not flat. Just look up.

People always sloganize what they do not understand. E=mC^2. It is a marketable equation. If you need money for a project, you need to communicate to people in a way that those -30 points can understand. I need 40 billion dollars to build a particle accelerator. I will use the marketing tool 'God Particle'. Now I can even get money from the religious right. I'm trying to find their god. sham-wow.

We know more about the yet to be discovered black hole. We know so much that we are certain that they lead to another universe. Where does the information go? The one that they are watching at the center of the Milky Way is not behaving like they wished. Recently, the Max Planck's Institute captured a 116 minute ejection of baryonic material from this supposed supermassive attractor. There was enough material to easily build another star. Click here for a link to this. This is not Hawkins' radiation. Stars around the black hole are not being pulled apart as describe in physics. So the evidence is starting to turn against gravity.

That picture of Eta Carina on the other thread is amazing. An asymmetric explosion. Symmetry is another disaster in physics. Most theories require symmetry to exist. We do not see symmetry in anything. Even your face is asymmetric. On average there is a 3% difference between the right side and the left side of a face.

Einstein himself, had major problems with his own work. He added ideas to perfect those ideas. But the foundation of his idea was incorrect. He knew that these issues would destroy his 'Godly' universe. That upset him. This is a person who could not look at his peers work. He sequestered himself to solve this problem and be unaffected by new ideas that were 'incorrect'.

It was absurd for people to sail west from Europe. The Church demanded a flat Earth. But the Greeks had solved this problem. They had calculated the spherical size of the Earth more than a thousand years before. I don't remember who it was.

Lets look at that identical rods problem. It is impossible to have identical rods. Lets look at 2 gold atoms far enough apart to not affect each other. They cannot be identical because of the positions of their electrons are not in the same place.

To accept as gospel, theories, is a dangerous position. History has been brutal to those who fix in their minds the absoluteness of theories.

So let's act like civil people and discuss gravity and the issues with gravity.

Aaron

Re: AAF: gravity June 14 2009, 9:49 PM

bob s: Question, can gravitation alone pull a body apart?

AAF: Question rephrased, "Would an object placed half way between two massive objects be pulled apart?"

Answer to original question, Yes, gravity alone can pull a body apart, Comet Shoemaker-Levy is a good example.

AAFs two part answer to rephrased question, part one: "In all, except one rare and highly arranged case, the half-way object would not be pulled apart."

I agree, the occurrence of two bodies pulling a body in between apart is rare, however...

Second part: "To the contrary, this midway object would enjoy being in a free space of zero gravity!"

...the midway object would not enjoy being in a "free space of zero gravity". The midway object would be located in a space of zero acceleration, not zero attraction. The gravitational attraction would be subject to the distance squared formula.

Gravity is the force of attraction. (The following assumes no forces applied to a body other than gravity)
The attraction force of gravity is recorded as acceleration.
There are two distinctly different measurements for gravity, attraction and acceleration.
Attraction for a body at rest at the surface of the Earth for example is 1G.
Acceleration is recorded as motion.
The motion of gravitational acceleration is always toward the gravitational body and the motion is recorded as "free fall" or 0Gs.
The rate of free fall recorded is variable to its (the body in free fall) distance from the gravitational source and increases as its distance to that body decreases.
A body at rest is 1G regardless of its weight.

Q. Can the attraction of gravity exist without the motion?
A. Yes, when a body comes to rest on the gravitational body the acceleration becomes zero motion. The acceleration is then recorded as weight.

Q. Can acceleration exist without attraction?
A. No, absent a gravitational attraction there is no acceleration.

Q. Can a body have weight absent acceleration?
A. No, absent gravity there is no acceleration, absent acceleration there is no weight.

Q. What happens to a body if there is no external gravitational attraction.
A. The body will be either at rest or in motion.
a. If the body is at rest, it will remain at rest.
b. If the body is in motion it will remain in constant uniform motion.

Q. If a body is absent gravitational attraction can it cause itself to move or come to rest?
A. No, a body can only move, or stop moving, by forces applied from outside of itself.

Modeling the gravitational effects within a body must begin from the center out. There are no known hollow gravitational bodies that exist nor are there any predictions that one, or more, could/would exist. If however, the model begins at the center outward as one band of bodies encircling a central body it can be shown that the outer band exerts outward tension on the center body as the center body is exerting inward attraction on the outer band. If, even if, the outer band were enlarged by adding more mass and creating distance between the band and the inner body at no place within the band will the gravitational attraction/tension on the center body become zero, acceleration would be zero. However, adding mass to the outer band would be as difficult as pulling apart a hollow sphere whose internal atmosphere had been evacuated causing a vacuum.

The following bodies assume uniform mass with the gravitational attraction of 1G.

One body, 1G attraction, 0Gs tension, 0 acceleration.
Attraction is inward pull, tension is outward pull. A body at rest is inward attraction/pull, the attraction/pull between the Earth and the Moon is outward tension/expansion.

Two bodies in contact with each other without motion or rotation.
Each body is subjected to 1G attraction and 1G tension.
Combined, 2Gs attraction, 0 tension and 0 acceleration.
The tension causes equal stress on the two bodies.

Three bodies in contact and sharing a common axis.
The end bodies are each subject to 1G attraction, 1G tension.
The center body is subjected to 2Gs attraction and 2Gs tension.
Stress on the center body is double that on the outer two bodies.
Combined, 3Gs attraction and 0 tension.


Four bodies in contact with, and surrounding, one body, 2 perpendicular axises.
Each of the four bodies is subjected to 1G attraction and 1G tension with the center body.
Each of the four outer bodies is also subjected to attraction and tension to the angular bodies at less than 1G because there is distance between them.
The center body is now subjected to 4Gs attraction and 4Gs tension.
The stress on the center body is almost 4 times that on the outer bodies
Combined, 5Gs attraction and 0 tension.

Six bodies in contact with center body and angular bodies with all sharing a common plane.
Each of the outer bodies is subjected to 3Gs attraction and 3Gs tension.
The center body is subjected to 6Gs attraction and 6Gs tension.
The stress on the center body is double that on the outer bodies.

Completing the outer sphere with six additional bodies becomes more complicated to explain than it is to understand and if you do not see my point by now, adding more bodies would be pointless. Granted, seeing my point does not mean that you agree with me. I am open to questions but if you reject my model I will not belabor my point any further. But do keep in mind my position is, that Newtons equations for concentric rings, point mass and zero gravity do not change my mind. The core of the Earth is under the stress of outward tension, which causes to core to heat, the outer shell is under the stress of inward attraction and allows the heat to dissipate, the point of equilibrium is at approximately 3/4 out on the radius which is where an object would stop falling if it were dropped toward the center of the Earth.

Aaron,

I understand and I value very much motivations and emotions.
These play a much bigger role than we could think at first sight.
Physics is not only about derivations, proof, calculations, experiments.
These are the difficult part of the job, but they cannot be accomplished by a machine.
We need motivation and we need emotions to get things done.

For me it makes a lot of sense to find that a theory is "beautiful".
This does not mean that it is right, but it motivates to learn, understand, develop.
This can of course lead sometimes to a lack of objectivity.
But the opposite can be equally true, because objectivity is needed to get the best results which can really be admired.
If a very beautiful theory fail miserably on a simple question, it should lose instantly its attraction, since it becomes like a caricature of nature.
(but a failed theory might still be useful as one can learn from that)

This is why, Aaron, it make totally sense to look at those picture that you post here and on your blog.

However, Aaron, I also urge you to look at the other side of these picture.
The main job of physics is to understand the backstage of nature.

I hope you will appreciate this simple example taken from your last post.
You said this:

" Symmetry is another disaster in physics. Most theories require symmetry to exist. We do not see symmetry in anything."

I believe this is the picture side of the story, but the backstage is totally different.
Spheres have a spherical symmetry, and yet you can assemble them to build an assymetric assemblage.

In the same way, the analysis of the most fundamental laws of physics have been discovered to be highly symmetric. Yet again, the play of these laws on a large stage does not need to show an overall symmetry, even though the details in the backstage keep all the symmetries.

I had really an emotional shock when I learned about the Noether theorem.
This theorem explicits a link between a symmetry and a conserved quantity.
(continous symmetry is involved here: like rotation by any angle, or translation)

First example.
The fundamental laws of physics are the same at any time.
This is the time-translation symmetry: move in time and you get the same physics.
The consequence of this symmetry is the energy conservation.
The Noether theorem explain how to discover the conserved quantity (energy here) by analysing mathematically the symmetry involved. Till today this symmetry has been confirmed, although it is not garnted for the eternity. Yet, in the day-to-day life, the picture is a changing world and this symmetry is far from being obvious. This symmetry is not a limitation of physics, it is instead and extraordinary discovery.

Second example.
The laws of physics are invariant by rotation, and this implies the conservation of the momentum.
However, obviously, this symmetry does not imply that all objects have a rotational symmetry!

Aaron, look also at the backstage!

Anon,

Beautiful post. Thanks. Now I can communicate with respect to you.

'First example.
The fundamental laws of physics are the same at any time.
This is the time-translation symmetry: move in time and you get the same physics.
The consequence of this symmetry is the energy conservation.
The Noether theorem explain how to discover the conserved quantity (energy here) by analysing mathematically the symmetry involved. Till today this symmetry has been confirmed, although it is not garnted for the eternity. Yet, in the day-to-day life, the picture is a changing world and this symmetry is far from being obvious. This symmetry is not a limitation of physics, it is instead and extraordinary discovery.

Second example.
The laws of physics are invariant by rotation, and this implies the conservation of the momentum.
However, obviously, this symmetry does not imply that all objects have a rotational symmetry! '

*****
The problem I have with symmetry is this. Baryons consist of 3 quarks and 3 gluons. Things with 3s have strength and stability and lack symmetry.

It is my imagination that suggest that there really isn't a quark. That would suggest that gluons are like long strands. Then a quark is like two strands intertwined and balled up gluons. A baryon has 3 gluons with 3 points where 2 gluons intertwine. This intertwining is where baryon data is stored as a Quark.

--------
http://aaronsreality.blogspot.com/2008/11/gluon-master-of-information.html

This is the postulation.

It is the Gluon that is the most important vibration in our universe. It communicates information to all other systems of vibrations as expressions. All information communication is essentially Gluon to Gluon communication. Each Gluon needs to communicate certain bits of information to other Gluons.

Gluons use certain vibrations to communicate different bits of information to other Gluons. The type of information determines the type of vibration used to communicate that information. The Bosons are the vibrations that transmit information. The Leptons and Quarks are expressions of the Gluon's information.

There are simple Gluon to Gluon communication through Quantumchromodynamics(QCD). The W+/- Boson communicates magnetism directly from Gluon to Gluon over a distance. The Photon communicates heat, pressure, and time to other Gluons through a more complicated system using both the Z Boson and the Electron. This system transmits information over great distances.

All Quarks store information for Gluons. The Strong Nuclear Force is the storage of Gluon information. Essentially a Quark is part of a Gluon and is the final expression of Gluon information.

All Leptons are responsible for secondary transmission of information for Gluons. The electron vibration loads information (described as Reiter Loading Theory) to be releasing the photon-wave as needed. The electron vibration now free of the photon-wave can accept a photon-wave and read the information from another Gluon.

Another important interaction the gluon is responsible for is the W boson communicates with dark matter. I postulate that the Gluon can have an anti-spin. This anti-spin Gluon causes expression of Dark Matter in a chain. The W boson provides magnetism information to the Dark Matter chain repelling it from the W boson. This is responsible for the bubble in Dark Matter causing Specific Baryonic Density (what was known as gravity).
---------

I also have issues with the conservation of energy stemming from pictures I see from the great smash labs. When the energy of the impact exceeds the inherit energy of the baryon we see an explosion.



Here we have vibrations dissipating over time. harmonic vibrations produce known bosons. disharmonic vibrations dissipate into the Dark Energy Aether.

Nor do I see any form of gravity in these pictures.

But this is a side thought. I suggest we start new threads on these topics.

I agree with the general concept of Noether Theorem. A system of asymmetries does create symmetric systems. A buckyball is a perfect example of this.

Aaron

Bob S: Question, can gravitation alone pull a body apart?



AAF: The question, under discussion, is: Would an object placed half way between two massive objects be pulled apart?



Bob S: Answer to original question, yes, gravity alone can pull a body apart, Comet Shoemaker-Levy is a good example.



AAF: Shoemaker's comet was torn apart by the tidal forces of Planet Jupiter. And that is very different from the tearing apart by placing an object half way between two massive bodies. And the answer, there­fore, is that in all, except one rare and highly arranged case, the half-way object would not be pulled apart.



Bob S: I agree, the occurrence of two bodies pulling a body in between apart is rare, however...



AAF: That is right, Bob! Not only that, but far from being pulled apart, this midway object would enjoy being in a free space of zero gravity.



Bob S: The midway object would not enjoy being in a "free space of zero gravity". The midway object would be located in a space of zero acceleration, not zero attraction. The gravitational attraction would be subject to the distance squared formula. Gravity is the force of attraction. (The following assumes no forces applied to a body other than gravity). The attraction force of gravity is recorded as acceleration. There are two distinctly different measurements for gravity, attraction and acceleration. Attraction for a body at rest at the surface of the Earth for example is 1G. Acceleration is recorded as motion. The motion of gravitational acceleration is always toward the gravitational body and the motion is recorded as "free fall" or 0Gs. The rate of free fall recorded is variable to its (the body in free fall) distance from the gravitational source and increases as its distance to that body decreases. A body at rest is 1G regardless of its weight. Q. Can the attraction of gravity exist without the motion?
A. Yes, when a body comes to rest on the gravitational body the acceleration becomes zero motion. The acceleration is then recorded as weight. Q. Can acceleration exist without attraction? A. No, absent a gravitational attraction there is no acceleration. Q. Can a body have weight absent acceleration? A. No, absent gravity there is no acceleration, absent acceleration there is no weight. Q. What happens to a body if there is no external gravitational attraction. A. The body will be either at rest or in motion. a. If the body is at rest, it will remain at rest. b. If the body is in motion it will remain in constant uniform motion. Q. If a body is absent gravitational attraction can it cause itself to move or come to rest? A. No, a body can only move, or stop moving, by forces applied from outside of itself. Modeling the gravitational effects within a body must begin from the center out. There are no known hollow gravitational bodies that exist nor are there any predictions that one, or more, could/would exist. If however, the model begins at the center outward as one band of bodies encircling a central body it can be shown that the outer band exerts outward tension on the center body as the center body is exerting inward attraction on the outer band. If, even if, the outer band were enlarged by adding more mass and creating distance between the band and the inner body at no place within the band will the gravitational attraction/tension on the center body become zero, acceleration would be zero. However, adding mass to the outer band would be as difficult as pulling apart a hollow sphere whose internal atmosphere had been evacuated causing a vacuum. The following bodies assume uniform mass with the gravitational attraction of 1G. One body, 1G attraction, 0Gs tension, 0 acceleration. Attraction is inward pull, tension is outward pull. A body at rest is inward attraction/pull, the attraction/pull between the Earth and the Moon is outward tension/expansion. Two bodies in contact with each other without motion or rotation. Each body is subjected to 1G attraction and 1G tension. Combined, 2Gs attraction, 0 tension and 0 acceleration. The tension causes equal stress on the two bodies. Three bodies in contact and sharing a common axis. The end bodies are each subject to 1G attraction, 1G tension. The center body is subjected to 2Gs attraction and 2Gs tension. Stress on the center body is double that on the outer two bodies. Combined, 3Gs attraction and 0 tension. Four bodies in contact with, and surrounding, one body, 2 perpendicular axises. Each of the four bodies is subjected to 1G attraction and 1G tension with the center body. Each of the four outer bodies is also subjected to attraction and tension to the angular bodies at less than 1G because there is distance between them. The center body is now subjected to 4Gs attraction and 4Gs tension. The stress on the center body is almost 4 times that on the outer bodies Combined, 5Gs attraction and 0 tension. Six bodies in contact with center body and angular bodies with all sharing a common plane. Each of the outer bodies is subjected to 3Gs attraction and 3Gs tension. The center body is subjected to 6Gs attraction and 6Gs tension. The stress on the center body is double that on the outer bodies. Completing the outer sphere with six additional bodies becomes more complicated to explain than it is to understand and if you do not see my point by now, adding more bodies would be pointless. Granted, seeing my point does not mean that you agree with me. I am open to questions but if you reject my model I will not belabor my point any further. But do keep in mind my position is, that Newtons equations for concentric rings, point mass and zero gravity do not change my mind. The core of the Earth is under the stress of outward tension, which causes to core to heat, the outer shell is under the stress of inward attraction and allows the heat to dissipate, the point of equilibrium is at approximately 3/4 out on the radius which is where an object would stop falling if it were dropped toward the center of the Earth.




AAF: That is too much, Bob! Are trying to write another Principia on this subject or what? However, I agree with all the above till I encounter this statement: "If however, the model begins at the center outward as one band of bodies encircling a central body it can be shown that the outer band exerts outward tension on the center body as the center body is exerting inward attraction on the outer band." Here, I disagree and assert the exact opposite: If the model begins at the center outward as one band of bodies encircling a central body it can be shown that the outer band exerts inward pressure on the center body as the center body is exerting outward attraction on the outer band. Now, Bob; Ive got the hunch that you arrived at the above statement as part of the counter argument against black holes; right? Well, in the past, I did the same! It worked perfectly; and if it was true, no black hole could exist. However, it turns out that the most effective instrument against black holes is none other than the variable speed of light itself. That is to say, if Einstein's Postulate of Constancy is removed, black holes cannot exist. You may have, in that case, dense objects with surface escape velocities many folds the speed of light; but they can't function as black holes. And that is because, there is no limit placed on the escape velocity and matter is able to escape from those dense objects at any time. Add to this the fact that even the tiniest angular momentum would generate, upon compression into small volumes, enormous linear velocities; and the existence of black holes ultimately becomes an absolute impossibility. In brief, black holes are the mythical crea­tures of Einstein's Relativity; and they must die out when that theory goes dead. Accordingly, Newton is right; there is no need at all for Bob's fore-mentioned central tension; correct?

Q: Would an object placed half way between two massive objects be pulled apart?

A:
It does not matter where the object is located in between two objects.
It doesn't even matter if there are two objects, one object is enough for this discussion.

Imagine the object is a physicist.
To get the head pulled away from the body, you need a difference of gravity (or force) between the head and the body. If the force is the same on the head and on the body, the acceleration will be the same, the motion will be similar and no stress will occur and the head will not be pulled apart.
If you have a difference between the force on the head and on the body, then they must separate simply because their motion wil be different. If there is an internal resistance, then the difference of the forces must overcome the internal strength.

Difference bewteen the force on head and body is, by definition, a tidal force.

For an object to be pulled apart, the force does not matter, but tidal force matters.
The tidal force depends on:

- the variation of the force in space, also called the gradient (say dF/dr)
- the size of the object (say L)

In summary: (Tidal force) = dF/dr * L

The condition for an object to be pulled apart (in two pieces) is roghly speaking:

dF/dr * L > internal strength

Gravity wave detectors are based on similar ideas.

Actually it depends upon where in the body the head of the physicist is currently located. It commonly is located within the body, which generates a whole 'nother problem trying to get gravity to remove it. Relaxation of the sphincter is the solution.

Re: AAF, gravity June 16 2009, 9:03 PM

AAF: "However, I agree with all the above till I encounter this statement: "If however, the model begins at the center outward as one band of bodies encircling a central body it can be shown that the outer band exerts outward tension on the center body as the center body is exerting inward attraction on the outer band." Here, I disagree and assert the exact opposite: If the model begins at the center outward as one band of bodies encircling a central body it can be shown that the outer band exerts inward pressure on the center body as the center body is exerting outward attraction on the outer band."

Which is basically where our disagreement began.

I said I won't belabor the point and I won't but I do reserve the right to respond; correct?

First off, my work with the internal workings of gravitational bodies was long before my denial of Black Holes and, in fact, before I knew of Newton's hollow spheres.

Back to gravity, an analogy, the properties and nature of water can not be known from observations and measurements of the atmosphere. The same holds true for a gravitational source (body). A gravitational source (body) and its gravitational field are two distinctly different places; right? Newton worked (experimented) with a gravitational field and applied those experiments to the gravitational source (the body) and I disagree with his results. If however, I knew of a theory or an hypothesis that predicted a hollow gravitational body I would be willing to reconsider, assuming of course, I agreed with said theory and/or hypothesis. I can ponder many things AAF, but I just can not ponder a hollow gravitational body. I had hoped you could ponder the tension.

The Lagrange Points:

Bob; look closely at these points of Lagrange! Are they inspiring or what?



http://www.physics.montana.edu/faculty/cornish/lagrange.html

Re: gravity June 18 2009, 12:42 AM

AAF, "The Lagrange Points:

Bob; look closely at these points of Lagrange! Are they inspiring or what?",

Yes I am inspired, inspired to ask, why does L1 (SOHO) stay where it is?

AAF: However, I agree with all the above till I encounter this statement: "If however, the model begins at the center outward as one band of bodies encircling a central body it can be shown that the outer band exerts outward tension on the center body as the center body is exerting inward attraction on the outer band." Here, I disagree and assert the exact opposite: If the model begins at the center outward as one band of bodies encircling a central body it can be shown that the outer band exerts inward pressure on the center body as the center body is exerting outward attraction on the outer band.




Bob S: Which is basically where our disagreement began. I said I won't belabor the point and I won't but I do reserve the right to respond; correct? First off, my work with the internal workings of gravitational bodies was long before my denial of Black Holes and, in fact, before I knew of Newton's hollow spheres. Back to gravity, an analogy, the properties and nature of water can not be known from observations and measurements of the atmosphere. The same holds true for a gravitational source (body). A gravitational source (body) and its gravitational field are two distinctly different places; right? Newton worked (experimented) with a gravitational field and applied those experiments to the gravitational source (the body) and I disagree with his results. If however, I knew of a theory or an hypothesis that predicted a hollow gravitational body I would be willing to reconsider, assuming of course, I agreed with said theory and/or hypothesis. I can ponder many things AAF, but I just can not ponder a hollow gravitational body. I had hoped you could ponder the tension.




AAF: Bob; Newton's Shell theorem applies to hollow spheres and solid spheres and rings and tori and to any other shapes or forms of physical objects. Where are hollow spheres found? Mostly in gaseous clouds and nebulae in the Milky Way and other galaxies. Take, for example, the hollow ellipsoid of the Crab Nebula. Anyway, the hollow shell is the simplest to visualize and treat mathematically. In addition, any solid shell can be visualized and treated mathematically as if it were a virtual 'onion' and composed of multiple hollow shells; do you have any objection, Bob? Good! So, let me reiterate the two main conclusions of the Shell theorem: [1] Inside a hollow shell, gravitation, due to the materials of the surface of the shell, is zero everywhere within that shell. [2] A solid shell can be divided to a number of hollow shells, where gravitation, due to the materials of every shell, is zero everywhere within that specific shell.

Re: AAF, gravity June 18 2009, 6:56 PM

AAF: "Bob; Newton's Shell theorem applies to hollow spheres and solid spheres and rings and tori and to any other shapes or forms of physical objects."

Yes, I got that point, but my question was, why is the SOHO satellite parked at the L1 position in the photograph you posted? Why!...because that is the point of gravitational equilibrium between the Sun and the Earth.

Equilibrium as in: equal, not zero. Right?...Good!

Did you read the article from that picture you posted, it gave link to a derivation of Lagrange's result, at http://www.physics.montana.edu/faculty/cornish/lagrange.pdf which said, at pgs 6-7 "The presence of a positive, real root tells us that L1 and L2 are dynamically unstable. Small departures from equilibrium will grow exponentially with a e-folding time of..." maybe you should let them know that they mean "zero" not "equilibrium"... Right?...Good! Thank you!!

Bob S: Yes I am inspired, inspired to ask, why does L1 (SOHO) stay where it is?




AAF: Actually, SOHO requires periodical orbital corrections to stay there. The Lagrange Point, L1, is where the gravitational field of the sun and the gravitational field of the earth-moon system cancel each other out. And the most important question, therefore, is this: How is SOHO Orbit Location (The L1 point) Calculated?http://solar-center.stanford.edu/FAQ/QL1.html In a nutshell, Newton's Shell theorem applies to hollow spheres and solid spheres and rings and tori and to any other shapes or forms of physical objects.




Bob S: Yes, I got that point, but my question was, why is the SOHO satellite parked at the L1 position in the photograph you posted? Why!...because that is the point of gravitational equilibrium between the Sun and the Earth. Equilibrium as in: equal, not zero. Right?...Good! Did you read the article from that picture you posted, it gave link to a derivation of Lagrange's result, athttp://www.physics.montana.edu/faculty/cornish/lagrange.pdf which said, at pgs 6-7 "The presence of a positive, real root tells us that L1 and L2 are dynamically unstable. Small departures from equilibrium will grow exponentially with a e-folding time of..." maybe you should let them know that they mean "zero" not "equilibrium"... Right?...Good! Thank you!!




AAF: Good; equilibrium, here, is just a buzz word for 'zero' and for nothing else but zero! Let me quote the following: "SOHO moves around the Sun on the sunward side of Earth, where it enjoys an uninterrupted view of the Sun, by slowly orbiting around Lagrange point L1. This a spot in space where the gravitational fields of the Sun and Earth cancel each other and keep SOHO in an orbit locked in line with the two bodies.": http://www.esa.int/esaSC/120373_index_0_m.html

Re: AAF gravity June 19 2009, 6:29 PM

AAF: "Good; equilibrium, here, is just a buzz word for 'zero' and for nothing else but zero! Let me quote the following:"

Copy of email sent by me this date @ 6:21 PM,

To: Dr. Cornish, cornish@physics.montana.edu

If I may have a few minuets of your time, please.

A friend and I are in dispute over the L1 point (SOHO) in a paper you wrote. In that paper, "The Lagrange Points" (date unknown)http://www.physics.montana.edu/faculty/cornish/lagrange.pdf at the bottom of pg 6 you state "The presence of a positive, real root tells us that L1 and L2 are dynamically unstable. Small departures from equilibrium will grow exponentially with a e-folding time of..." Our dispute is over the meaning of the word "equilibrium".

One of us states, that at the L1 point the gravitational attraction between the Earth and the Sun is equal and the acceleration is zero while the other of us states that at the L1 point the gravitational attraction is zero meaning, no gravitational attraction. So the question is, is it possible that you should have used the word "zero" or "canceled" in place of the word "equilibrium"?

Thank you for your time,
bob s

Good move, Bob, even though it's highly unlikely that Prof. Cormish would reply to the above e-mail!
However, whether he responds or not, I don't acknowledge him as an authority on this subject.

So, Bob, if you're serious about getting a serious arbitrator, this is the one: http://www.jpl.nasa.gov/

And when JPL tells you that the gravitational attraction at L1 is not equal to zero, I shall surrender!

Re: AAF, gravity June 19 2009, 9:28 PM

AAF: "And when JPL tells you that the gravitational attraction at L1 is not equal to zero, I shall surrender!"

Oh! you want NASA to be the final arbitrator...OK.

http://map.gsfc.nasa.gov/mission/observatory_l2.html

From the first paragraph,
"The Lagrange Points

The Italian-French mathematician Joseph-Louis Lagrange discovered five special points in the vicinity of two orbiting masses where a third, smaller mass can orbit at a fixed distance from the larger masses. More precisely, the Lagrange Points mark positions where the gravitational pull of the two large masses precisely equals the centripetal force required to rotate with them...."

I read that as, the gravitational force is equal, with the result that the acceleration is zero, what say you? I would also add that Dr. Cornish authored the article for NASA. So I am willing to accept Dr. Cornish's explanation.

Moreover, I hope you noticed that I did not tell Dr. Cornish which position was whose, mine or yours, but I posed my question as if he had misused the word "equilibrium", which might provoke him to respond if I suggested he made a mistake. Am I being fair and equitable?

Bob S: Oh! you want NASA to be the final arbitrator...OK.http://map.gsfc.nasa.gov/mission/observatory_l2.html From the first paragraph, "The Lagrange Points
The Italian-French mathematician Joseph-Louis Lagrange discovered five special points in the vicinity of two orbiting masses where a third, smaller mass can orbit at a fixed distance from the larger masses. More precisely, the Lagrange Points mark positions where the gravitational pull of the two large masses precisely equals the centripetal force required to rotate with them...." I read that as, the gravitational force is equal, with the result that the acceleration is zero, what say you? I would also add that Dr. Cornish authored the article for NASA. So I am willing to accept Dr. Cornish's explanation. Moreover, I hope you noticed that I did not tell Dr. Cornish which position was whose, mine or yours, but I posed my question as if he had misused the word "equilibrium", which might provoke him to respond if I suggested he made a mistake. Am I being fair and equitable?





AAF: Well, NASA is the owner of SOHO and should know what life at Point L1 feel and look like. And it is not NASA per se, Bob; it's NASA's Board of Directors that should have the final word on it! Anyway, there is only one point among all the Lagrange points, which has a zero gravitational field. And that point is the one labeled as 'L1'. Finally, I don't think Dr. Cornish will respond. And the reason, I suspect, is that you made a small tactical mistake by referring to the ongoing disagreement; docs generally don't like getting involved in the controversies of others; do they?

Re: AAf, gravity June 20 2009, 3:26 PM

AAF: "Well, NASA is the owner of SOHO and should know what life at Point L1 feel and look like. And it is not NASA per se, Bob; it's NASA's Board of Directors that should have the final word on it!"

I see how it is, you can link me anywhere you want yet when that site doesn't support you, I have to go to the Board of Directors at NASA for your agreement. I doubt if the BoD at NASA knows the difference between a Lagrange Point and Cedar Point. If and when Dr. Cornish responds I'll let you know either way but, in the mean time, I have made my point quite well, which is, the Lagrange point L1 is a point of gravitational equilibrium and zero acceleration, not zero gravity.

And no! I did not make a tactical error, I posed the question as if "he" might have made a mistake so I am reasonably sure he will want to defend himself more than he would want to defend you or I. I may be wrong about Dr. Cornish answering me but I am confident as to my position on L1's gravitational equilibrium.

If it's in orbit, it's in free-fall. It will experience no inertial force. Depending upon the mass of the object, particles resting on the surface of the object at the point in question, will experience gravitational acceleration while at rest, or is it weight bring exprienced?

AAF: Happy Summer Solstice to all!





Bob S: I see how it is, you can link me anywhere you want yet when that site doesn't support you, I have to go to the Board of Directors at NASA for your agreement. I doubt if the BoD at NASA knows the difference between a Lagrange Point and Cedar Point. If and when Dr. Cornish responds I'll let you know either way but, in the mean time, I have made my point quite well, which is, the Lagrange point L1 is a point of gravitational equilibrium and zero acceleration, not zero gravity. And no! I did not make a tactical error, I posed the question as if "he" might have made a mistake so I am reasonably sure he will want to defend himself more than he would want to defend you or I. I may be wrong about Dr. Cornish answering me but I am confident as to my position on L1's gravitational equilibrium.




AAF: I'm not sure they know the difference, Bob ! But I do know that Point L1 has no gravitational field due to the sun or due to the earth. Surely, the gravitational field of Jupiter, for instance, is present at L1; but neither the sun nor the earth could have any gravitational presence at L1. Concerning SOHO, this conspicuous absence of gravitation must perpetually cost NASA money just to keep their SOHO there! Since it's impossible to make anything perform any sort of circular motion inertially and on its own, except around its own geometrical axis. SOHO therefore, would move in straight line out of L1 unless rocket boosting is used periodically to keep it there. Now, is NASA stupid? Not quite! You see, Bob; according to Kepler's laws, anything that orbits the sun, inside the earth's orbit, must have orbital speeds greater than that of the earth. And it can't be made synchronous and locked to the earth. To do so, rocket boosting is required. But rocket boosting would not do this job and with the minimum cost anywhere within the earth's orbit, except at Point L1. By choosing L1 as the parking place for its SOHO, NASA is clearly and demonstrably very smart!

Let there be a contest AAF and you get to be the judge.

For this contest we need a pit of mud.
Let this pit of mud have a breadth and width of some 20ft.
Now, let there be a rope (a very sturdy rope) suspended horizontally 4-5ft above this mud pit.
Attached to the rope, at approximately the center of the mud pit, is a small red flag.
On either side of the mud pit and holding on to the rope are two teams of people, 5 men and 5 women on each side.
The contest is a "tug of war", the winner is determined when one side pulls the rad flag to their side of the mud pit.

With it so far? Good!

As the contest begins the flag does not move, the teams respective coaches are yelling PULL...PULL...PULL! but the flag remains stationary. The team members are dug in and are beginning to sweat but the flag, although fluttering, does not move one way or the other, the crowd cheers but the flag remains stationary.
My Son turns to me and asks...what's happening Dad?
I respond...the pull of each team is canceling the pull of the other team.

You be the judge AAF, do I mean,
a.) there is no (zero) pull on the rope in spite of their efforts?
or,
b.) the pull of each team on the rope is equal?

Which shall it be AAF, a.) or b.)?

One more question:

What is the tension within the red flag attached in the middle?

Re: Anonymous, gravity June 22 2009, 2:01 AM

"What is the tension within the red flag attached in the middle?"

Typical Relativist obfuscation, avoid answering the question by misdirecting the subject. The flag is a scale of the test NOT a subject of the test, also, the degree of tension on the rope, or the flag, is not a subject of the question however, the information you seek can be gleaned from the parameters given, if read properly. Hint! the flag is attached to the rope, the rope is not attached to the flag. The "fluttering" of the flag is an indication of motion not tension.
flutter: definition
to move about or behave in an agitated aimless manner
Kind of like your participation on this board...eh Anonymous!

Think twice, bob.
Gravitation is an action at distance.
Your rope analogy is not totally relevant.
That's why the flag analogy I have suggested is useful.
Just consider the same experience with gravity instead of a rope.
It is simply because any part of the flag will experience a (near) zero force.
Only when the flag become large enough, will the stress become relevant.

Nothing to do with relativity.
Everything to do with understanding gravitational forces.
The same applies to electrostatic forces.
The principle of linear supperposition of force is an experimental observation.

Another opportunity for you: the flag is now very heavy.
The stresses within the flag are still negligible, assuming it has a small size.
Yet the forces on the two gravity (rope) side are not!

Dear Anonymous,

Why not try answering the question as posed, based on the facts given, instead of running interference for AAF so that he might avoid the answer.

Best regards,
bob s

http://www.merriam-webster.com/dictionary/cancel

cancel can:cel
Pronunciation: \?kan(t)-s?l\
Function: verb
Inflected Form(s):
canceled or cancelled; canceling or cancelling Listen to the pronunciation of cancelling \-s(?-)li?\
Etymology:
Middle English cancellen, from Anglo-French canceller, chanceller, from Late Latin cancellare, from Latin, to make like a lattice, from cancelli (plural), diminutive of cancer lattice, probably alteration of carcer prison
Date: 14th century

transitive verb
1 a: to destroy the force, effectiveness, or validity of :
b: to bring to nothingness : destroy
c: to match in force or effect: offset
d: to call off usually without expectation of conducting or performing at a later time 2 a: to mark or strike out for deletion b: omit, delete
3 a: to remove (a common divisor) from numerator and denominator
b: to remove (equivalents) on opposite sides of an equation or account


http://www.merriam-webster.com/dictionary/equal

equal
Pronunciation: \?e--kw?l\
Function: adjective
Etymology:
Middle English, from Latin aequalis, from aequus level, equal
Date: 14th century

1 a: (1): of the same measure, quantity, amount, or number as another
2): identical in mathematical value or logical denotation: equivalent
b: like in quality, nature, or status
c: like for each member of a group, class, or society
2: regarding or affecting all objects in the same way: impartial
3: free from extremes: as a: tranquil in mind or mood b: not showing variation in appearance, structure, or proportion
4 a: capable of meeting the requirements of a situation or a task
b: suitable

To cancel can mean, to equalize.
But, equal never means, to cancel.

When using the word "cancel" in reference to the gravitational forces at the L1 Lagrange point it means "equalized" gravitational force, not "zero" gravitational force.

"Why not try answering the question as posed, ..."

Simply because it is irrelevant for gravitation, it is only relevant for ropes.

Re: Anonymous, gravity June 22 2009, 2:18 PM

"Why not try answering the question as posed, ..."

"Simply because it is irrelevant for gravitation, it is only relevant for ropes."

Of course it's not relevant for gravitation you Twit! Its, (the questions) relevancy is to show the difference between "cancel" and "equal". You don't know the answer!

Anonymous is posting just to use the band width. I think he is cinci, drunk on Mogen David. https://instawine.com/cgi-bin/iw_site.cgi?mnfid=FOXCREEKLIQUORS&search=bcmnf&srec=4537&pga=search

Re: Jose, Gravity June 22 2009, 3:33 PM

"Anonymous is posting just to use the band width. I think he is cinci, drunk on Mogen David."

You may be right Jose, and I may be wrong, but I think it is AAF trying to avoid admitting I am right, in that, the gravitational force at the L1 Lagrange point is not zero, it is equalized bidirectionally to the Sun and Earth.

"You may be right Jose, and I may be wrong, ..."

You are both wrong about who I am.

"... the gravitational force at the L1 Lagrange point is not zero, it is equalized bidirectionally to the Sun and Earth."

you are right for this part of the sentense:

'... the gravitational force at the L1 Lagrange point is not zero ..."

and I know enough in audio electronics to believe this is wrong:

"... it is equalized bidirectionally to the Sun and Earth."

The truth is that the gravitational force on L1 balances the centrifugal force on L1.
You can read the whole story on wiki:http://en.wikipedia.org/wiki/Lagrangian_point
This picture from wiki is a perfect synthesis:



It is a contour plot of the effective potential.
This is the potential of the gravitational field summed to the potential equivalent to the centrifugal force.
The stationary points of the potential are the Lagrange points by definition.
On the Lagrange points there is no total force in the rotating frame of reference.

Bob, where did you see a rope there?
Bob, would a red flag on L1 be torned apart?

This assumes that the object located at L1 has very small gravitation itself. It also ignores the effects of the other planets.

Imitation is part of the learning process, even for humans.
You are on good track JR.

Bob S: Let there be a contest AAF and you get to be the judge. For this contest we need a pit of mud. Let this pit of mud have a breadth and width of some 20ft.
Now, let there be a rope (a very sturdy rope) suspended horizontally 4-5ft above this mud pit. Attached to the rope, at approximately the center of the mud pit, is a small red flag. On either side of the mud pit and holding on to the rope are two teams of people, 5 men and 5 women on each side. The contest is a "tug of war", the winner is determined when one side pulls the red flag to their side of the mud pit. With it so far? Good! As the contest begins the flag does not move, the teams respective coaches are yelling PULL...PULL...PULL! but the flag remains stationary. The team members are dug in and are beginning to sweat but the flag, although fluttering, does not move one way or the other, the crowd cheers but the flag remains stationary. My Son turns to me and asks...what's happening Dad? I respond...the pull of each team is canceling the pull of the other team. You be the judge AAF, do I mean, a.) there is no (zero) pull on the rope in spite of their efforts? or, b.) the pull of each team on the rope is equal? Which shall it be AAF, a.) or b.)?






AAF: I bet the 5-Men Team shall win quickly, and very likely will drag and throw the other team into the mud pit. But, for the sake of argument, let's assume that the five men are midgets and the five women are giants! Well, in such a case, the pulling forces are opposite and equal. And the ground under the feet of each team is the main lever point and unmovable. Recall that Archimedes once said: "Give me a lever point and I shall move the earth". Fortunately for SOHO, there is nothing that can function even remotely as a lever point at the Lagrange Point, L1. There, the gravitational field of the sun just cancels and annihilates the gravitational field of the earth, and vice versa. We have to conclude, there­fore, that when the gravitational acceleration disappears, the gravitational field disappears as well. Now, this is the question: Just few thousands of miles beyond Lagrange L1, the sun appears to have less mass, and the orbital velocity is slightly less than that of the earth and capable of making SOHO locked and permanently parked in the earth's gravitational shadow. So, why didn't NASA choose that orbit for its solar satellite? That is the question, Bob. Anyway, it's probably much easier to calculate the gravitational cancellation point for the [earth & the moon] than for the [sun & the earth + the moon)]: http://www.physicsforums.com/showthread.php?t=145633 Finally, it's the Big Question to my Co-defender (Anon): Could it be Monseigneur Lagrange had chosen the right point beyond the gravitational cancellation point of (the earth & the sun) and called it L1? Well, in that case, the gravitational cancellation point, under discussion, would be unnamed; and obviously I would gain the right to have it named after me and to be called 'AAF Point'; do I have the right to have it named after me, Bob?

It's true, Anon!

The AAF Point A1 is closer to Earth than the Lagrange Point L1:

"There is one point where the two gravitational fields truly cancel out, 260,000 kilometres from Earth in the direction of the sun - much closer than any of the five Lagrangian points. At this true cancellation point, the trajectory of any orbiting body would briefly become a straight line.":http://www.newscientist.com/article/mg20127010.600-trip-to-lagrange.html



Just to spread the word for Opera...

Re: AAF, gravity June 22 2009, 8:11 PM

AAF, when I initiated my disagreement with you on June 1 2004, 4:41 PM in this string I had a reasonable expectation that you would at least understand, if not agree, what my position was but now, when I read this response;

"AAF: I bet the 5-Men Team shall win quickly, and very likely will drag and throw the other team into the mud pit. But, for the sake of argument, let's assume that the five men are midgets and the five women are giants!"

When in fact this is what I said;
"On either side of the mud pit and holding on to the rope are two teams of people, 5 men and 5 women on each side."
I have to say...what was I thinking??? I made it quite clear that each team was composed of 5 men and 5 women. The truth is, you have no idea what I have been trying to tell you, you seem more interested in what you have to say rather than what you read. And your buddy Anonymous, whether he be you, Stanly16, Cincirob or even ol'Max, has no clue as to what the disagreement was about!

And you say, "for the sake of argument,". NO! The argument is; from gravity June 22 2009, @ 8:11 PM
"AAF,
"Furthermore, for a spherical body, the "strength of the gravitational field at the centre" is exactly equal to zero. That is the conclusion of a very famous theorem in Newton's Principia."

I disagree, the gravitational field at the center of the Earth would be the inverse of the field at the surface, the extreme tension and pull on the atoms at the center causes the friction that heats the core of the Earth. As r extends out from the center the tension reverses to contraction allowing the surface of the Earth to become the attractor. The point of equilibrium "I suspect" can be approximated to the "golden mean". The point of equilibrium "golden mean or otherwise" is what prevents a body, the Earth or star, to collapse under the force of gravity irrespective of its size or mass iow Black Holes have no possibility of existing because their gravitational structure is self supporting."

I saw some days back that we were in a dead lock of disagreement and I said I would no longer belabor my point but you kept up, and now, I've changed my mind and will continue to belabor my point quite simply because I know I am right which is, the gravitational force on the center of the Earth is tension not zero, tension as in outward stress concentric to the spherical mass on the Earth.

You take your friends, real or imagined, car pool it to a Library and study up on how to use a dictionary! You might just learn that it is zero that is the buzz word for equilibrium not vice versa, zero is, on the other hand, the word that denotes acceleration. Otherwise en-that I'm cool.

Re: AAF, gravity June 22 2009, 8:11 PM

AAF: "Well, in that case, the gravitational cancellation point, under discussion, would be unnamed; and obviously I would gain the right to have it named after me and to be called 'AAF Point'; do I have the right to have it named after me, Bob?"

No, you do not have the right, just because you have calculated a different distance for the L1 Lagrange point does not change the metrics of that point only the distance at which the metrics apply, gravitational attraction equal, gravitational acceleration zero. Hell, you even said it yourself on June 11 2009, @ 6:16 PM
"At the points of equal distances from the two equal bodies, there­fore, the resultant of acceleration is always equal to zero. Use Newton's Equa­t­ion and check these results for yourself!"
Trouble is, you were so busy shouting at me you did not realize you were agreeing with me.

bob:

And your buddy Anonymous, whether he be you, Stanly16, Cincirob or even ol'Max, has no clue as to what the disagreement was about!
...
I disagree, the gravitational field at the center of the Earth would be the inverse of the field at the surface, the extreme tension and pull on the atoms at the center causes the friction that heats the core of the Earth.
____________________

Total nonsense:

- you guessed the wrong names
- the physics is wrong
- you missed my point with the flag which explains clearly why your physics is wrong

With a rope, you can tear the flag (or the rope) in two because the pull is propagated by the rope to sides of the flag.
By gravity you cannot tear the flag (if the flag is small enough): the two teams replaced by two massive bodies will change the physics completely.

The gravity force on the left corner of the flag will be zero because it is the resultant of the attraction from the two massive bodies acting oppositely. The same will apply for the right corner of the flag, approximatively because the forces change with distance and the balance will be slightly perturbed. Therefore the stress within the flag will be negligible.

As I told you the stress is, roughly speaking: dF/dr * (size of the flag)

The stress at the center of the earth is propagated by the rocks (or the liquid) and is caused by the weigth of the upper parts. The stress is not caused by the gravity in the center, but by the gravity away from the center.

Trying the Socratic Method or playing the fool, Bob?

Bob S: AAF, when I initiated my disagreement with you on June 1 2004, 4:41 PM in this string I had a reasonable expectation that you would at least understand, if not agree, what my position was but now, when I read this response; "AAF: I bet the 5-Men Team shall win quickly, and very likely will drag and throw the other team into the mud pit. But, for the sake of argument, let's assume that the five men are midgets and the five women are giants!" When in fact this is what I said; "On either side of the mud pit and holding on to the rope are two teams of people, 5 men and 5 women on each side." I have to say...what was I thinking??? I made it quite clear that each team was composed of 5 men and 5 women. The truth is, you have no idea what I have been trying to tell you, you seem more interested in what you have to say rather than what you read. And your buddy Anonymous, whether he be you, Stanly16, Cincirob or even ol'Max, has no clue as to what the disagreement was about! And you say, "for the sake of argument,". NO! The argument is; from gravity June 22 2009, @ 8:11 PM "AAF, "Furthermore, for a spherical body, the "strength of the gravitational field at the centre" is exactly equal to zero. That is the conclusion of a very famous theorem in Newton's Principia." I disagree, the gravitational field at the center of the Earth would be the inverse of the field at the surface, the extreme tension and pull on the atoms at the center causes the friction that heats the core of the Earth. As r extends out from the center the tension reverses to contraction allowing the surface of the Earth to become the attractor. The point of equilibrium "I suspect" can be approximated to the "golden mean". The point of equilibrium "golden mean or otherwise" is what prevents a body, the Earth or star, to collapse under the force of gravity irrespective of its size or mass iow Black Holes have no possibility of existing because their gravitational structure is self supporting." I saw some days back that we were in a dead lock of disagreement and I said I would no longer belabor my point but you kept up, and now, I've changed my mind and will continue to belabor my point quite simply because I know I am right which is, the gravitational force on the center of the Earth is tension not zero, tension as in outward stress concentric to the spherical mass on the Earth. You take your friends, real or imagined, car pool it to a Library and study up on how to use a dictionary! You might just learn that it is zero that is the buzz word for equilibrium not vice versa, zero is, on the other hand, the word that denotes acceleration. Otherwise en-that I'm cool.






AAF: I like to see 5 men on one side and 5 women on the other side; what is wrong with that? Besides, this rearrangement would not affect or alter the main point of your scenario in any way; correct? So let me now update what I said about the Lagrangian Point, L1: The [gs MINUS ge] at L1 is positive and greater than zero. I thought Monseigneur Lagrange was interested in computing the zero-gravitation point, but he was in fact interested only in finding the right orbits for a third body around two massive bodies. It follows, therefore, that NASA just took Lagrange's calculations out of his books and used 'em for its SOHO; and it wasn't quite smart as I first thought! However, the gravitational cancellation point, under discussion, is still unnamed by the International Astronomical Union; so, Bob; do I have the right to have it named after me?





Jose: Sure, I think you should have a galaxy named after you. Who cares about piddling stars?





Bob S: No, you do not have the right, just because you have calculated a different distance for the L1 Lagrange point does not change the metrics of that point only the distance at which the metrics apply, gravitational attraction equal, gravitational acceleration zero. Hell, you even said it yourself on June 11 2009, @ 6:16 PM "At the points of equal distances from the two equal bodies, there­fore, the resultant of acceleration is always equal to zero. Use Newton's Equa­t­ion and check these results for yourself!" Trouble is, you were so busy shouting at me you did not realize you were agreeing with me.





AAF: Thank you very much for your support, Jose. Look, Bob; let's follow the lead of Shoemaker and Levvy and rename the AAF Point! The new name of the gravitational cancellation point between two massive bodies now is the AAF-Bob Point. That is the name we should submit together for consideration by the above International naming body; OK? Notice also that I didn't calculate the new position of the AAF-Bob Point; the New Scientist did it.

The AAF-Bob Point: New Scientist

Re: AAF, gravity June 23 2009, 6:40 PM

AAF, "Notice also that I didn't calculate the new position of the AAF-Bob Point; the New Scientist did it."

If we are to take notice of something, then you should take notice of this site,
http://www.esa.int/esaSC/SEMM17XJD1E_index_0.html
where it says,
"Lagrange points" [which includes L1] "are locations in space where gravitational forces and the orbital motion of a body balance each other."
Notice also that "balance" is synonymous with "equilibrium" while, and notice this, "equilibrium" is not synonymous with "zero". So, unless you know of a secret passage way for gravity to get around your "zero" gravity point, two bodies, such as the Earth-Sun have no gravitational attraction, right AAF?

Should you not fully understand what I am saying, take notice of an earlier statement of yours on June 09, 2009, @ 4:12 PM,
AAF: "Believe it or not; Bob; when you stand half way between two super-massive bodies, the force acting on our body is precisely nil and exactly the same as being standing in the middle of Willem de Sitter's Empty Universe!"
You do know that "nil" means "nothing" which is synonymous with "zero" and not, (I repeat, not!) synonymous with "balance" nor "equilibrium". I have no idea where you came up with the idea that "half way between two super-massive bodies" the gravitational attraction is "nil-canceled-zero", but the concept is even more preposterous than time dilation, right AAF?

Popular science sentenses, like this one ...

"Lagrange points ... are locations in space where gravitational forces and the orbital motion of a body balance each other"

... can only lead to endless and useless discussions.
Forces can be "balanced": this happens when their resultant is zero.
A force cannot be in balance with a motion, or at least not without some further definition.

The given reference (http://www.esa.int/esaSC/SEMM17XJD1E_index_0.html ) is not bad, but it is not a good basis for a discussion, it is only a basis for popular science.

It is no surprise therefore that the discussion here is running like a chicken without head.
After all, what is now the real topic?
It will now become a discussion about the bob's guy dislike for this sentense:

"... half way between two super-massive bodies, the force acting on our body is precisely nil ..."

And it looks like a very very serious question for him.
He compares this topic to his crusade against time dilation in SR!
The phylosophycal Gods will go to war on this one!

Really debating like that is just killing the time.
Even in his bathroom a space engineer would not take it seriously.
Near an L point, a spacecraft experiences nearly no gravitational force.
What truly is all this fuss about?

Re: Anonymous, gravity June 25 2009, 4:15 AM

Anonymous: "Really debating like that is just killing the time.
Even in his bathroom a space engineer would not take it seriously.
Near an L point, a spacecraft experiences "nearly no" gravitational force.
What truly is all this fuss about?"

You really are a Ditz Anon., you think "nearly no" is a better description of the gravitational forces at L1 than "balanced"?

And, "What truly is all this fuss about?". I told you you had no idea of the subject at hand. Two Ditzy remarks in the same paragraph makes you a Double Ditz (DD). I'd sell you a clue DD but I doubt you'd know what to do with it! Ha...ha...hahahahaha! OH GOD!... with you around DD, I need a pee bucket by my computer.

(note to God) "OH GOD" was an expression, not a request. I Gotta be careful what I pray for.

Oh! what the heck, I'll give you two clues for free,
"The given reference http://www.esa.int/esaSC/SEMM17XJD1E_index_0.html ) is not bad, but it is not a good basis for a discussion, it is only a basis for popular science."
"esa" (from the link) stands for the (European Space Agency)! And it is your Goombah, AAF, that posted a link to "esa" in the first place so he felt is was a reliable source.

I said "nearly no" simply because the field is not constant.

If you move a little bit away from the L point, the gravity becomes non-zero and changes with position. This is important because a finite object will experience slighly different forces on its different part. This cause small internal forces. It is possible to measure gradient of a gravitational field and this has been done. Gravitational wave detectors are based on this principle and this explains their size.

Concerning the ESA reference, you should keep in mind that ESA does publish educational material. The supposed audience of this web page are kids between 12 and 16, I guess.

For the rest of you post, I could not understand it.
As I told you already before, I learned english in "Gravitation" by JA Wheeler first, and then in many other physics books. I am not able to understand Mikey mouse.

Re: Anonymous (aka DD) gravity June 25 2009, 1:14 PM

DD: I said "nearly no" simply because the field is not constant.

If you move a little bit away from the L point, the gravity becomes non-zero"

The gravitational attraction at a Lagrange Point (at any Lagrange Point) is NOT, (NEVER-EVER) ZERO!!!
IF, the gravitational attraction at the L1 Lagrange point were to be ZERO there would be NO gravitational attraction between the two bodies in question.

If there is zero gravity between two bodies it is because of distance and there is no longer a Lagrange point because, there must be gravitational attraction for there to be a Lagrange point in the first place.

"The gravitational attraction at a Lagrange Point (at any Lagrange Point) is NOT, (NEVER-EVER) ZERO!!! "

You are right, Bob.

That's what I explained in a previous post: it is a balance between gravitational forces and the centrifugal force that defines the L points. When I said "nearly zero", I wanted to say that this balance (grav.+centif.) is zero in one point and close to zero in th vicinity.

Because of that, a body can be at rest in a L point.
In this case, this body will experience "nearly no" internal stresses.
"Nearly" of course because the finite extend of this body will imply some stress anyway.
By the way, this stress, near a equilibrium point does not vary as the size, but it varies as size².

Finally note that only two of the L points are stable: those that are not aligned with on the sun-earth axis. If you look at the lines of force near the instable points you will see they look like hyperbolas while they look like ellipses near the stable points. The unstable points are sometimes called X points and the others O points. The perturbation of orbits near X points is one of my favorite topics.

Sorry for that:

"Because of that, a body can be at rest in a L point. "

I should have said:

"Because of that, a body can be at rest in a L point, in the rotating frame of reference. "

Very sorry to downgrade this forum even further with such badly written posts.
Nearly as bad as chat.

Bob S: If we are to take notice of something, then you should take notice of this site, http://www.esa.int/esaSC/SEMM17XJD1E_index_0.html where it says, "Lagrange points" [which includes L1] "are locations in space where gravitational forces and the orbital motion of a body balance each other." Notice also that "balance" is synonymous with "equilibrium" while, and notice this, "equilibrium" is not synonymous with "zero". So, unless you know of a secret passage way for gravity to get around your "zero" gravity point, two bodies, such as the Earth-Sun have no gravitational attraction, right AAF? Should you not fully understand what I am saying, take notice of an earlier statement of yours on June 09, 2009, @ 4:12 PM, AAF: "Believe it or not; Bob; when you stand half way between two super-massive bodies, the force acting on our body is precisely nil and exactly the same as being standing in the middle of Willem de Sitter's Empty Universe!" You do know that "nil" means "nothing" which is synonymous with "zero" and not, (I repeat, not!) synonymous with "balance" nor "equilibrium". I have no idea where you came up with the idea that "half way between two super-massive bodies" the gravitational attraction is "nil-canceled-zero", but the concept is even more preposterous than time dilation, right AAF?






AAF: Bob, your attention for one minute, please! All previous references about Lagrange's Points, including the above, are now made outdated and somewhat obsolete by this recent article by this man: Derek Bolton as a response to these incorrect remarks by this other man: Stuart Clark Here is the list of the new corrections: (1) Lagrange's Point L1 and the Cancellation Point of gravity between two massive bodies do not occupy the same location, but two different locations, where the former is always located farther away from the less massive body and closer to the more massive body; while the latter is always farther away from the more massive body and closer to the less massive body. (2) For every pair of two equal bodies, no matter how big or how small they are, the Cancellation Point of gravitation and the Center of Mass of the two bodies are always located at the same location in space and half way between the two massive bodies. (3) From (2), it follows at once that gravity at the center of a spherical body is always equal to zero. (4) At Lagrange's Point L1, the difference between [the strength of the sun's gravitational field & the strength of the earth's gravitational field] is greater than zero and slightly less than the strength of the sun's gravitational field at the distance of the earth's orbit. Derek Bolton, in his for-mentioned paper, stated that the difference is equal to the strength of the sun's gravitational field at the distance of the earth's orbit. That is an obvious mistake on his part. Since it's quite clear that if he was right, then any object, at Lagrange's Point L1, would drift away very quickly, under its own angular momentum, towards the earth's orbit.

Bob:

Unless you're too grief-stricken by the untimely departure of Michael Jackson to care, allow me to clarify and qualify Remark #1 and make it more precise!




Lagrange's Point L1 and the Cancellation Point of gravity between two massive bodies do not occupy the same location, but two different locations, where the former is always located farther away from the less massive body and closer to the more massive body (than the latter); while the latter is always farther away from the more massive body and closer to the less massive body (than the former).

Notice, please:


[1] Any object, at Lagrange's Point L1, drifts away very quickly towards higher orbits, when its linear velocity is higher than the orbital velocity of L1.


[2] Any object, at Lagrange's Point L1, drifts away very quickly towards lower orbits. when its linear velocity is less than the orbital velocity of L1.

Re: Anonymous, gravity June 25 2009, 4:12 PM

bob s: "The gravitational attraction at a Lagrange Point (at any Lagrange Point) is NOT, (NEVER-EVER) ZERO!!!"

Anon.: "You are right, Bob."

Thank you.

Anon.: "That's what I explained in a previous post:"

I wish you had been more clear as to the date of that post.

Anon.: "it is a balance between gravitational forces and the centrifugal force that defines the L points. When I said "nearly zero", I wanted to say that this balance (grav.+centif.) is zero in one point and close to zero in the vicinity."

By "(grav.+centif.)" I assume you mean "acceleration".

Anon.: "Finally note that only two of the L points are stable: those that are not aligned with on the sun-earth axis. If you look at the lines of force near the instable points you will see they look like hyperbolas while they look like ellipses near the stable points. The unstable points are sometimes called X points and the others O points. The perturbation of orbits near X points is one of my favorite topics."

I have no doubt that the instability of the X points would be a topic of future interest but for now I only wish to establish the gravitational effects at the L1 Lagrange point Sun/Earth system which from my point of view is:
1.) The gravitational attraction is equal.
2.) The gravitational acceleration is zero.


Re: Anonymous, gravity June 25 2009, 4:15 PM

Anon.: "Very sorry to downgrade this forum even further with such badly written posts."

Your posts are not "badly written", they are broken English. I have little doubt that you have as much trouble understanding me as I do you.

This is the date of the post: June 22 2009, 4:15 PM .
I admit it is not very clear.

Re: Anonymous, gravity June 26 2009, 5:02 PM

Anon.: "This is the date of the post: June 22 2009, 4:15 PM .
I admit it is not very clear."

That! is the sum of your reply to my previous post, Ok then.

Re: Anonymous, gravity June 22 2009, 4:15 PM

bobs: "... the gravitational force at the L1 Lagrange point is not zero, it is equalized bidirectionally to the Sun and Earth."

Anon.: "you are right for this part of the sentense:

'... the gravitational force at the L1 Lagrange point is not zero ..."

True enough.

Anon.: "and I know enough in audio electronics to believe this is wrong:

"... it is equalized bidirectionally to the Sun and Earth."

"equalized bidirectionally" means equal in two directions, toward the Sun and toward the Earth. Look to the picture you posted, the gravitational lines equalize at the L1 point and the two blue pointers, one pointing toward the Sun and the other pointing toward the Earth. What could possibly be "wrong" with that sentence?

Anon.: "The truth is that the gravitational force on L1 balances the centrifugal force on L1."

I agree completely, so where is you think we may still disagree?

Re: AAF, gravity June 25 2009, 10:15 PM

AAF,
"Bob:

Unless you're too grief-stricken by the untimely departure of Michael Jackson to care, allow me to clarify and qualify Remark #1 and make it more precise!"

There is no need to clarify or qualify your Remark #1. The location of the Lagrange points is not at issue here. The present issue is, the gravitational effects at the L1 Lagrange point. The original subject of disagreement is the gravitational effects at the core of the Earth, right AAF?

Do you recall making this statement on June 01, 2004, @ 4:41 PM? I do.
AAF,
"Furthermore, for a spherical body, the "strength of the gravitational field at the centre" is exactly equal to zero. That is the conclusion of a very famous theorem in Newton's Principia."

And I responded to you on June 06, 2009, @ 11:20 AM,
bob s:
"I disagree, the gravitational field at the center of the Earth would be the inverse of the field at the surface, the extreme tension and pull on the atoms at the center causes the friction that heats the core of the Earth. As r extends out from the center the tension reverses to contraction allowing the surface of the Earth to become the attractor. The point of equilibrium "I suspect" can be approximated to the "golden mean". The point of equilibrium "golden mean or otherwise" is what prevents a body, the Earth or star, to collapse under the force of gravity irrespective of its size or mass iow Black Holes have no possibility of existing because their gravitational structure is self supporting."

The gravitational effects at the centre of the Earth is not zero (as in nil) it is tension (as in the inverse of the gravitational effects at the surface of the Earth), right AAF?

And the gravitational effects at the L1 Lagrange point is not zero, (as in nil) it is equal, (as in bidirectional), with one direction toward the Sun and the other direction toward the Earth, right AAF?

I do not know who first proposed the cockamamie idea of zero (as in nil) effects for the centre of the Earth and zero (as in nil) for the L1 Lagrange point but they were/are wrong, right AAF?

If someone has calculated a different positions for those point that's great, but, it doesn't change the gravitational effects for those points only the positions of those points which, as I have already said, is not an issue being currently discussed, right AAF?

Anon: You are right, Bob. That's what I explained in a previous post: it is a balance between gravitational forces and the centrifugal force that defines the L points. When I said "nearly zero", I wanted to say that this balance (grav.+centif.) is zero in one point and close to zero in th vicinity. Because of that, a body can be at rest in a L point. In this case, this body will experience "nearly no" internal stresses. "Nearly" of course because the finite extend of this body will imply some stress anyway. By the way, this stress, near a equilibrium point does not vary as the size, but it varies as size². Finally note that only two of the L points are stable: those that are not aligned with on the sun-earth axis. If you look at the lines of force near the instable points you will see they look like hyperbolas while they look like ellipses near the stable points. The unstable points are sometimes called X points and the others O points. The perturbation of orbits near X points is one of my favorite topics.




AAF: Your grasp of this subject is quite good, Anon; I wish you had done the same with Relativity! But let me emphasize one important point, in this regard. The balance between the gravitational force and the centrifugal force is a general characteristic of all closed orbits, not just Lagrange's Points: Orbits




Bob S: You really are a Ditz Anon., you think "nearly no" is a better description of the gravitational forces at L1 than "balanced"? And, "What truly is all this fuss about?". I told you you had no idea of the subject at hand. Two Ditzy remarks in the same paragraph makes you a Double Ditz (DD). I'd sell you a clue DD but I doubt you'd know what to do with it! Ha...ha...hahahahaha! OH GOD!... with you around DD, I need a pee bucket by my computer. (note to God) "OH GOD" was an expression, not a request. I Gotta be careful what I pray for. Oh! what the heck, I'll give you two clues for free, "The given referencehttp://www.esa.int/esaSC/SEMM17XJD1E_index_0.html is not bad, but it is not a good basis for a discussion, it is only a basis for popular science." "esa" (from the link) stands for the (European Space Agency)! And it is your Goombah, AAF, that posted a link to "esa" in the first place so he felt is was a reliable source.




Anon: Very sorry to downgrade this forum even further with such badly written posts.



Bob S: Your posts are not "badly written", they are broken English. I have little doubt that you have as much trouble understanding me as I do you.



Anon: Really debating like that is just killing the time. Even in his bathroom a space engineer would not take it seriously. Near an L point, a spacecraft experiences "nearly no" gravitational force. What truly is all this fuss about?




AAF: 'Goombah Bono'; go easy on Anon! His writing is very typical of science writing in general. That is how most manuscripts look like when first received by Physical Review. And that is the main reason why scientific journals employ copy editors in the first place: One more reason why the world needs copyeditors

Anon.: "and I know enough in audio electronics to believe this is wrong:

"... it is equalized bidirectionally to the Sun and Earth."

Bob: "equalized bidirectionally" means equal in two directions, toward the Sun and toward the Earth. Look to the picture you posted, the gravitational lines equalize at the L1 point and the two blue pointers, one pointing toward the Sun and the other pointing toward the Earth. What could possibly be "wrong" with that sentence?
_________________________________________________________________

Here is the caption of the picture on wiki:

"A contour plot of the effective potential of a two-body system (the Sun and Earth here) due to gravity and the centrifugal force as viewed from the rotating frame of reference in which Sun and Earth remain stationary. Objects revolving with the same orbital period as the Earth will begin to move according to the contour lines showing equipotential surfaces. The arrows indicate the gradients of the potential around the five Lagrange points downhill toward or away from them, but at the points themselves these forces are balanced."

_________________________________________________________________

Near the L1 point, on this picture, you see indeed two arrows in opposite directions.
These arrows do not represent the attraction by the sun and by the earth.
One of the arrows represents the total force on an object when it is a little bit on the side of the sun.
The other arrow represents the same thing for an object a little bit on the side of the earth.
The size of these forces increases whith the distance away from the L1 point.
On the L1 point, this force is zero.
The arrow on the left and on the right can be equal or can be different depending on the chosen distances from the L1 point.

Note also that the "total force" means: attration by the sun + attraction by the earth + centrifugal force.

I said that the sentense was wrong, because I did not see any meaning for this sentense.
My understanding of English may be weak as well as my writing.
But more certainly, I really belief that it is very difficult to talk physics without formulas or at least without reference to formulas.
The caption of the picture on wiki illustrates that well.
Look at the words "effective potential": both words have a definite mathematical translation.
They could have presented another picture for the "effective force" which sum the gravitation forces and the centrifugal force. But the potential is easier to draw and to read.

Re: Anonymous, gravity June 27 2009, 4:36 PM

Anon.: "Here is the caption of the picture on wiki:"

"A contour plot of the effective potential of a two-body system (the Sun and Earth here) due to gravity and the centrifugal force as viewed from the rotating frame of reference in which Sun and Earth remain stationary. Objects revolving with the same orbital period as the Earth will begin to move according to the contour lines showing equipotential surfaces. The arrows indicate the gradients of the potential around the five Lagrange points downhill toward or away from them, but at the points themselves these forces are balanced."

Did you miss the part that read "...but at the points themselves these forces are balanced."
"balanced" as in equalized, "points themselves" as in L1, L2, L3, L4 and L5.

And did you read the full article which read,
"The Lagrange points mark positions where the combined gravitational pull of the two large masses provides precisely the centripetal force required to rotate with them."

"Centripetal force" as in equalizing the centrifugal force.

And further,
"the gravitational fields of two massive bodies combined with the centrifugal force are in balance at the Lagrangian points,"

"Balance" same as above.

And further,
"he found five specific fixed points where the third body experiences zero net force"

"Zero net force" as in equalized gravitational force.

And still further,
"This is due to Newton's second law (F = dp/dt), where p = mv (p the momentum, m the mass, and v the velocity) is invariant if force and position are scaled by the same factor. A body at a Lagrangian point orbits with the same period as the two massive bodies in the circular case, implying that it has the same ratio of gravitational force to radial distance as they do."

"Same ratio of gravitational force" means the centripetal is equal to the centrifugal force. If the gravitational force were zero (as in nil) the body at L1 would not orbit with the two massive bodies.

Anon.: "I said that the sentense was wrong, because I did not see any meaning for this sentense."

Hopefully now you do.

Anon.: "But more certainly, I really belief that it is very difficult to talk physics without formulas or at least without reference to formulas.
The caption of the picture on wiki illustrates that well.
Look at the words "effective potential": both words have a definite mathematical translation."

The maths are in the article. It take words to describe the maths, if you don't understand the words how do you expect to understand the maths? Take Newton's second law for example (F = dp/dt) "F" force, can never go to zero, "never" (as in not ever!) can you understand?

"... It take words to describe the maths, if you don't understand the words how do you expect to understand the maths? Take Newton's second law for example (F = dp/dt) "F" force, can never go to zero, "never" (as in not ever!) can you understand?"

Why do you think that naming "F" as "force" explains anything more?
And yes, I do not understand what you mean with "can never go to zero".
And further why would you add such comments when F=dp/dt contains all you need to know?
However, i can admit that a few words are sometimes necessary, at least to explain how symbols have to be measured using instruments.

Re: Anonymous, gravity June 28 2009, 6:38 AM

bob s: "... It take words to describe the maths, if you don't understand the words how do you expect to understand the maths? Take Newton's second law for example (F = dp/dt) "F" force, can never go to zero, "never" (as in not ever!) can you understand?"

Anon.: "Why do you think that naming "F" as "force" explains anything more? And yes, I do not understand what you mean with "can never go to zero". And further why would you add such comments when F=dp/dt contains all you need to know? However, i can admit that a few words are sometimes necessary, at least to explain how symbols have to be measured using instruments."

This is a joke, Right? You have got to be putting me on! You want me to explain to you how, from this equation (F = dp/dt) F (force) can never (as in not ever) go to zero! It's like I said Anon, it take words to describe the maths, if you don't understand the words how do you expect to understand the maths and obviously you don't understand words or maths.

How about you use Newton's second law (F = dp/dt) and show me, mathematically wise, how F can go to zero.

Bob, AAF, JR and bnon

Because of that, a body can be at rest in a L point.
In this case, this body will experience "nearly no" internal stresses.
"Nearly" of course because the finite extend of this body will imply some stress anyway.
By the way, this stress, near a equilibrium point does not vary as the size, but it varies as size².

---

SOHO is at L1. The idea here is that the mass of the sun and earth cause an area of 'nearly no' stresses. I strongly disagree.



SOHO uses error correction to maintain its position. The L point has no real effect on the position of the spacecraft.

The reason is that the L point is really a peak between two great densities. The dense sun creates W Bosons that push back the dark matter. The earth also pushes back the dark matter. The area known as L1 is between these two and is a peak.

There are tremendous forces acting upon SOHO. First is temp, second is magnetism, third is electrical(not so much). Just because gravity is not a force does not suggest that there are 'nearly no stresses' affecting this craft.

Looking at the evidence. Soho is using fuel faster than anticipated. It is sliding down the peak. This means there is no stable point called L

Aaron

Aaron L points June 28 2009, 9:17 AM

Aaron: "Looking at the evidence. Soho is using fuel faster than anticipated. It is sliding down the peak. This means there is no stable point called L"

SOHO's life expectancy was two years and was extended to 11 years so I doubt its fuel usage was faster than anticipated.

Mission lifetime
"SOHO was designed for a nominal mission lifetime of two years. In 1997 the mission was extended until 2003 because of its spectacular success. In 2002, a second extension of another four years was granted, that is, through March 2007. This will allow SOHO to cover a complete 11-year solar cycle."
(source)http://sohowww.nascom.nasa.gov/about/about.html

Bob,

I cannot find the article that discussed this. I retract my comment until I can back it up.

Do you know which spacecraft are in L points? I use to get this info from google. Now I can't find it. Frustrating.

a

Bob,

I remember that this was not an article that I read. It was a thought I had. I found my notes. My memory sucks.

I postulated that if L points did exist as described they would be filled with debris. We would see these area of debris around all the planets in various amounts based on the density of the planet. There would be more debris at these points then in any other area of the solar system. They would just collect junk. I doubt a spacecraft would survive the debris.

These points are clean. This is the proof that natural debris does not collect in these L points. We would see the debris with telescopes. Thus the L points have no specific difference than any other point in the solar system.

So I will partially remove my previous comment until natural debris is found in these points. Just the SOHO part.

a

Aaron, L points June 29 2009, 7:28 PM

Aaron, "Do you know which spacecraft are in L points? I use to get this info from google. Now I can't find it. Frustrating."

First, the link I gave in my previous post was for the SOHO home page.
Second as to your question, As far as I know there are geosynchronous satellites parked in the Earth/Moon L1 points. As to the Sun Earth/Moon L1 point I think SOHO is still there.


Aaron, Oops June 29 2009, 7:39 PM

Aaron: "I postulated that if L points did exist as described they would be filled with debris. We would see these area of debris around all the planets in various amounts based on the density of the planet."

Debris have been found at other Sun/Planets Planet/Moon L4-5 points.
The following is from the NASA site,
(source)http://map.gsfc.nasa.gov/mission/observatory_l2.html
"The L4 and L5 points are home to stable orbits so long as the mass ratio between the two large masses exceeds 24.96. This condition is satisfied for both the Earth-Sun and Earth-Moon systems, and for many other pairs of bodies in the solar system. Objects found orbiting at the L4 and L5 points are often called Trojans after the three large asteroids Agamemnon, Achilles and Hector that orbit in the L4 and L5 points of the Jupiter-Sun system. (According to Homer, Hector was the Trojan champion slain by Achilles during King Agamemnon's siege of Troy). There are hundreds of Trojan Asteroids in the solar system. Most orbit with Jupiter, but others orbit with Mars. In addition, several of Saturn's moons have Trojan companions. No large asteroids have been found at the Trojan points of the Earth-Moon or Earth-Sun systems. However, in 1956 the Polish astronomer Kordylewski discovered large concentrations of dust at the Trojan points of the Earth-Moon system. Recently, the DIRBE instrument on the COBE satellite confirmed earlier IRAS observations of a dust ring following the Earth's orbit around the Sun. The existence of this ring is closely related to the Trojan points, but the story is complicated by the effects of radiation pressure on the dust grains."

Hope the link works for you.

Well,

This is from wiki,http://en.wikipedia.org/wiki/Kordylewski_cloud
'The existence of the Kordylewski clouds is still under dispute.[2] The Japanese Hiten space probe, which passed through the libration points to detect trapped dust particles, did not find an obvious increase in dust levels above the density in surrounding space.'

Now here is the info in the Hiten space probe.http://en.wikipedia.org/wiki/Hiten
'The primary mission was concluded on March 30, 1991 and the follow-on mission was started. On April 24, 1991 Hiten left Earth orbit and went to the Moon using Belbrunos route. On October 2, 1991 Hiten reached the Moon at the prescribed distance. After which, it was put into a looping orbit which passed through the L4 and L5 Lagrange points to look for trapped dust particles. No obvious increase was found by the Munich Dust Counter (MDC). After two months in lunar orbit, the spacecraft's orbit was decaying, so the last of Hitens fuel was used to crash it into the lunar surface on April 10, 1993.'

From your previous post.
'Recently, the DIRBE instrument on the COBE satellite confirmed earlier IRAS observations of a dust ring following the Earth's orbit around the Sun.'

This would confirm what I have described as the Earth pushing the dark matter away. This pressurization would cause a ring around the sun just outside the earth's orbit.

There are many sites that contain papers that cost money to read. I cannot afford this. It irks me to find science is following money not data or analysis. This is why the wiki references.

I have spent many a years reading about spacecraft. I have 'Space Mission Analysis and Design' by Larson and Wertz. I got it for $20 on amazon used. It is an amazing book. In there they describe L points. That was the first time I read about them, it shocked me.

I have an idea how these L points would work in my theory, without gravity. But I want to think on it for a day. These Greeks could put a Trojan horse in my theory. But I doubt. This idea I have seems to fit the model. If the Trojans are all in L4 and L5 then I can explain them. L4 is the planet pushing the dark matter away causing a lip. That lip would be enough to continue to push objects. L5 would be where the dark matter is returning to normal from the encounter with the planet. Here the objects would be pushed by the returning dark matter. L1 would be more difficult to describe in this model. L3 is in dispute. L2 would just be moved into the magnetosphere tail.

As for SOHO in L1. It just seems to me that they got the angular velocity correct to maintain a stable orbit around the sun. They make an excuse as to why the craft is not in L1 but in a 'Halo orbit' around L1. They cannot perch the craft on L1 because of solar radiation interference. But if L1 is like a inverted bowl with an object perched on top then the craft would be above the elliptic. This inverted bowl is actually a fair description how this craft would work in my model. Pushing causes inverted bowls. Attraction causes regular bowls.http://sohowww.nascom.nasa.gov/explore/faq.html

Thanks for making me think. I'll get back on this.
a

p.s. I published my first paper here. "Disproof of Gravity".

Aaron, Scratching head June 29 2009, 10:51 PM

Aaron: "I published my first paper here. "Disproof of Gravity"."

Yes, I saw that, congratulations!

Aaron: "Thanks for making me think. I'll get back on this."

Ok, but be advised that I have little dispute over the existence of Liberation points. The dispute I have been involved in is primarily the gravitational effects on a body at the L1 position. I maintain that the gravitational attraction at L1 (Sun Earth/Moon) is equal bidirectionally toward the Sun and the Earth and gravitational acceleration is zero. Others maintain that the gravitational attraction is zero. Their concept is based on "point mass" which, even if Newton's second law predicts, I reject for gravitational effects "within" a body. However, if point mass is considered as the whole of the body (as used in a gravitational field), as opposed to the center of the body, I agree with Newton.

I know from reading your paper that you also reject "point mass" but from my point of view your W boson is just a different name for gravity. Quite frankly, I have no desire to learn a whole new lexicon just to describe what it well known to me as gravity. Without gravity your buoyancy concept would not work, your helium filled balloon not withstanding.

Bob,

Thanks for reading my paper.

Yes, I disapprove of gravity. Mass has no volume. It cannot be described as shells or waves. Shoot it does not even fit basic mathematics. Gravity states that always 2 masses will attract to each other.

To Newton's credit, he was unaware of helium.

As for the W Boson, I describe this as magnetism. It does push back dark matter. That would look like attraction if you were unaware of things like Helium. I understand how Newton generated this idea. Attraction is easy to evaluate. But it fails at a fundamental level. Helium is not the only example of density. It is just the most obvious.

I disagree with all concepts of gravity. Including the L points. L4 and L5 may work but for very different reasons. Even NASA suggest that the L point is an inverted bowl. This would evaluate to pushing not pulling.

Density dude. Density

a

"SOHO is at L1. The idea here is that the mass of the sun and earth cause an area of 'nearly no' stresses. I strongly disagree.

SOHO uses error correction to maintain its position."
_______________________________________________
Aaron,

When I said "nearly no stresses", this did not imply "no drift".
The stress refers to an internal stress in the structure of the spacecraft, caused by (non homogeneous) gravitational and inertial forces.

A drift may need corrections.
This is especially necessary because the L1 point is an instable equilibrium.
This means that any deviation from the equilibrium point could lead to a loss of SOHO.
Actually the forces in the vicity of L1 are like those experienced by an object falling on a saddle (see picture).

In contrast, the L4 and L5 are stable equilibrium.
There, a spacecraft could stay like in a well.
On these points however, drift corrections might be necessary too.
This would be the case if the speed of the spacecraft with respect to the L point would be too high. In this case it could escape the well definitively. The corrections could stabilize the spacecraft in the potential well for ever if not disturbed by other celestial bodies.

________________________________________
Saddle point: unstable equilibrium:

Bob,
Great Mathematical pic. Which program did you use to create that. I learned on Mathematica. I now have Wmaxima. I just have not taken the time to learn it. It is probably similar to Mathematica. Also what equation did you use to create that result.

But again there is that pushing. Attraction cannot create this. Pulling causes regular bowls. This is an inverted bowl or saddle shape.

I drew on your pic. I hope you don't mind. This is an example of pushing dark matter away with a magnetosphere. The Sun's magnetosphere causes the general shape of the dark matter. It is similar to this pic.
M104


This is what I imagine when the sun radiates energy. The cloud around the star is similar to our Oort cloud and the other ice cloud. Not enough coffee.

Now let's imagine L1 in front of the Earth's motion around the Sun. The process causes a lip in the dark matter.
From your pic.

The yellow lines represent the labeled magnetospheres pressing against one another. The pink line represents the direction of travel of the Earth.

This is a pushing process.

Grateful Dead on the stereo. Doing physics. It's going to be a great day.
a

Aaron, I did not provide those pics. Anonymous did.

It labeled my posting as Bob.

Or most likely is that I typed in Bob's name in the name box. Human error.

Anon, Sorry not enough coffee.

a

Aaron, Lexicon June 30 2009, 12:54 AM

Aaron,
"Bob,

Thanks for reading my paper."

Ya know Aaron, I was just going back over that paper you published, you know..to see if I might reconsider on some of your points and I realized that the paper had no acknowledgments, There are people here, I may not be one of them, but they have given you insights and help with the formulation of that paper, don't you think you should have acknowledged the help of your friends here on the forum...I'm not saying--I'm just saying!

You are correct. I have spent so much energy in the complication of, am I capable of actually describing these advanced topics. This is just fear.

I do need to acknowledge my friends here who helped me define these ideas. I will add acknowledgements to all future papers. I am barely learning to write semi-professional papers. I feel like an infant in a world of adults.

Thanks for the error correction

But I will only acknowledge people who use their correct names.

I am sending another paper in a few minutes called 'Planets are ordered by density'

a

AAF: Unless you're too grief-stricken by the untimely departure of Michael Jackson to care, allow me, Bob, to clarify and qualify Remark #1 and make it more precise!




Bob S: There is no need to clarify or qualify your Remark #1. The location of the Lagrange points is not at issue here. The present issue is, the gravitational effects at the L1 Lagrange point. The original subject of disagreement is the gravitational effects at the core of the Earth, right AAF? Do you recall making this statement on June 01, 2004, @ 4:41 PM? I do. AAF, "Furthermore, for a spherical body, the "strength of the gravitational field at the centre" is exactly equal to zero. That is the conclusion of a very famous theorem in Newton's Principia." And I responded to you on June 06, 2009, @ 11:20 AM, bob s: "I disagree, the gravitational field at the center of the Earth would be the inverse of the field at the surface, the extreme tension and pull on the atoms at the center causes the friction that heats the core of the Earth. As r extends out from the center the tension reverses to contraction allowing the surface of the Earth to become the attractor. The point of equilibrium "I suspect" can be approximated to the "golden mean". The point of equilibrium "golden mean or otherwise" is what prevents a body, the Earth or star, to collapse under the force of gravity irrespective of its size or mass iow Black Holes have no possibility of existing because their gravitational structure is self supporting." The gravitational effects at the centre of the Earth is not zero (as in nil) it is tension (as in the inverse of the gravitational effects at the surface of the Earth), right AAF? And the gravitational effects at the L1 Lagrange point is not zero, (as in nil) it is equal, (as in bidirectional), with one direction toward the Sun and the other direction toward the Earth, right AAF? I do not know who first proposed the cockamamie idea of zero (as in nil) effects for the centre of the Earth and zero (as in nil) for the L1 Lagrange point but they were/are wrong, right AAF? If someone has calculated a different positions for those point that's great, but, it doesn't change the gravitational effects for those points only the positions of those points which, as I have already said, is not an issue being currently discussed, right AAF?





AAF: Yes; the strength of the gravitational field at the center of a spherical body is exactly equal to zero. You can't argue with Newton against that! Newton is not a weakling like Einstein. And his arguments are always decisive and self-evident. Newton has concluded that gravity at the center is nil; and he is right. Now, Bob, would explain what exactly you mean by this phrase "the inverse of the field at the surface"? As I understand it, it means that the strength of the gravitational has its maximum value at the surface and drops to zero at infinity. And so, if its strength at the center is the reverse of its strength at the surface, then its strength at the center must be zero; right, Bob? And there is no tension; there is only inward pressure at the center of the earth. You don't wrestle with Newton against that; do you? Regarding Lagrange's Points, the strength of the gravitational field at Lagrange's L1 and Lagrange's L2 is the same and slightly less than the strength of the sun's gravitational field at the earth's orbit. And you don't want to wrestle with Lagrange against that; do you? Remember this, Bob; and be prepared! Arguing with Einstein & his followers is most of the time like playing in the Midget professional wrestling But arguing with Newton & Lagrange is always like wrestling with Stone Cold Steve Austin

AAF and Bob

I think I have a solid argument against Newton and Lagrange. I will publish more papers that strengthen this argument. I just finished a paper on the structure of baryons.

a

Re: AAF, gravity June 30 2009, 6:58 PM

AAF: "Unless you're too grief-stricken by the untimely departure of Michael Jackson to care, allow me, Bob, to clarify and qualify Remark #1 and make it more precise!

Bob S: There is no need to clarify or qualify your Remark #1. The location of the Lagrange points is not at issue here. The present issue is, the gravitational effects at the L1 Lagrange point. The original subject of disagreement is the gravitational effects at the core of the Earth, right AAF? Do you recall making this statement on June 01, 2004, @ 4:41 PM? I do. AAF, "Furthermore, for a spherical body, the "strength of the gravitational field at the centre" is exactly equal to zero. That is the conclusion of a very famous theorem in Newton's Principia." And I responded to you on June 06, 2009, @ 11:20 AM, bob s: "I disagree, the gravitational field at the center of the Earth would be the inverse of the field at the surface, the extreme tension and pull on the atoms at the center causes the friction that heats the core of the Earth. As r extends out from the center the tension reverses to contraction allowing the surface of the Earth to become the attractor. The point of equilibrium "I suspect" can be approximated to the "golden mean". The point of equilibrium "golden mean or otherwise" is what prevents a body, the Earth or star, to collapse under the force of gravity irrespective of its size or mass iow Black Holes have no possibility of existing because their gravitational structure is self supporting." The gravitational effects at the centre of the Earth is not zero (as in nil) it is tension (as in the inverse of the gravitational effects at the surface of the Earth), right AAF? And the gravitational effects at the L1 Lagrange point is not zero, (as in nil) it is equal, (as in bidirectional), with one direction toward the Sun and the other direction toward the Earth, right AAF? I do not know who first proposed the cockamamie idea of zero (as in nil) effects for the centre of the Earth and zero (as in nil) for the L1 Lagrange point but they were/are wrong, right AAF? If someone has calculated a different positions for those point that's great, but, it doesn't change the gravitational effects for those points only the positions of those points which, as I have already said, is not an issue being currently discussed, right AAF?

AAF: Yes; the strength of the gravitational field at the center of a spherical body is exactly equal to zero. You can't argue with Newton against that! Newton is not a weakling like Einstein. And his arguments are always decisive and self-evident. Newton has concluded that gravity at the center is nil; and he is right. Now, Bob, would explain what exactly you mean by this phrase "the inverse of the field at the surface"? As I understand it, it means that the strength of the gravitational has its maximum value at the surface and drops to zero at infinity. And so, if its strength at the center is the reverse of its strength at the surface, then its strength at the center must be zero; right, Bob? And there is no tension; there is only inward pressure at the center of the earth. You don't wrestle with Newton against that; do you? Regarding Lagrange's Points, the strength of the gravitational field at Lagrange's L1 and Lagrange's L2 is the same and slightly less than the strength of the sun's gravitational field at the earth's orbit. And you don't want to wrestle with Lagrange against that; do you? Remember this, Bob; and be prepared! Arguing with Einstein & his followers is most of the time like playing in the Midget professional wrestling But arguing with Newton & Lagrange is always like wrestling with Stone Cold Steve Austin"

Get real AAF, I have already addressed every one of those issues beginning with my first message on this string on June 06, 2009 @ 11:20 AM and I stand behind every thing I have said since then. As to Steve Austin, bring it on, but make sure he can read and knows how to use a Dictionary.

Now, unless you have something new to add I suggest you go out and play with Cincirob, just be in before dark!

(one more thing, if Steve does come tell him to bring the Undertaker, his services will be needed.)

The only reason I would talk about wrestlers or michael jackson is if they were used in a terminal velocity experiment or better yet used in a particle accelerator.

Again I think I can argue that density is how objects move. Gravity has shown that it is an unacceptable model of motion.

a

"Now, Bob, would explain what exactly you mean by this phrase "the inverse of the field at the surface"? "

AAF, this is Bob's opinion.
He has no obligation to explain and even less to write formulas.
This forum is only about expressing opinion.
Even units of measurements need not to be detailled, only opinions matter.
If the field at the surface is in kg*m/s², it really does not matter that the inverse of that is s²/kg/m.

Why do you want to discuss opinions, even on physics, if the game is about telling anything?
Why do you want precision, if anybody telling Eisnstein a weakling is a genius while those asking for justifications are called fools here.

Re: gravity July 1 2009, 2:58 AM

AAF: "Now, Bob, would [you] explain what exactly you mean by this phrase "the inverse of the field at the surface"?"

First, what I said was "...the gravitational field at the center of the Earth would be the inverse of the field at the surface,".

Second, no word or combination of words, in that sentence can be inferred to mean that gravity is zero at infinity as no such place or condition exists.

Third, The boundaries are clearly defined; center of Earth and surface of Earth.

Fourth, the sentence should be self-explanatory.

Fifth, in spite of the fact that the sentence should be self-explanatory I did try and explain to more fully several times in this string therefor, explaining the sentence again would be redundant. Redundancy is not required on this board.

AAF, "this is Bob's opinion."

Sixth, My opinion is based on the fact that gravitational bodies attract, do you disagree?

AAF: "He has no obligation to explain and even less to write formulas."

Seventh, I disagree, there is an obligation to provide explanation when asked whereas, providing formulas is optional.

Eighth, I fulfilled my obligation to explain, several times in this string and I did provide a formula at least once in this string.

Anonymous: "This forum is only about expressing opinion."

Ninth, the purpose of this forum is clearly stated on the index page and reads;
"This resource is for the use of visitors to the site in the discussion of problems and opinions on physics and philosophy. Please feel free to post your observations on these and related subjects."

Anonymous: "Even units of measurements need not to be detailled, only opinions matter."

Tenth, units of measurements are standardized and a reasonable expectation exists that participants to this forum should have a working knowledge of same.

Anonymous: "If the field at the surface is in kg*m/s², it really does not matter that the inverse of that is s²/kg/m."

Eleventh, I fully explained the inverse gravitational effect at the center of the Earth previously in this string.

Twelfth, it's really quite simple, if the gravitational attraction at the surface of the Earth is 1G down the inverse gravitational attraction at the center is 1G up. If you feel an explanation is needed then read my messages beginning on June 07, 2009 @ 12:05 PM. If you had had any real interest in my opinion (which is based on fact) then you would have already read those messages. When you joined the discussion it was your responsibility to understand the issues and subject of the discussion.

Anonymous: "Why do you want to discuss opinions, even on physics, if the game is about telling anything?"

Your question does not make sense.

Anonymous: "Why do you want precision, if anybody telling Eisnstein a weakling is a genius while those asking for justifications are called fools here."

A person asking for justification is not a fool but, a person thinking Einstein a genius is! Ingenious, I would agree with!

I did not ask for more than this:

"Twelfth, it's really quite simple, if the gravitational attraction at the surface of the Earth is 1G down the inverse gravitational attraction at the center is 1G up."

Notice that in maths the inverse of x is 1/x, and the opposite is -x.
Note also that you have not specified the direction of gravity at the center of the earth.
Does it point in your direction, or in mine?

Re: Anonymous, gravity July 1 2009, 1:05 PM

Anonymous: "I did not ask for more than this:"

bob s: "Twelfth, it's really quite simple, if the gravitational attraction at the surface of the Earth is 1G down the inverse gravitational attraction at the center is 1G up."

First, I may, if I choose, respond to your message in part or in whole, I chose to respond to the whole of it.

Second, my obligation to explain ends with having to explain the difference between "down" and "up" both of which are clearly stated in the quoted paragraph!

Third, redundancy here is optional, not required, I gave you the location for the beginning of my full explanation in my previous message on July 01, 2009 @ 10:21 AM.

Anon.: "Notice that in maths the inverse of x is 1/x, and the opposite is -x."

Yes, I noticed. Did you notice I said "inverse gravitational effect"? Also, I am not required to give definitions for commonly used words; "inverse effect" does not mean "inverse maths".

Anon.: "Note also that you have not specified the direction of gravity at the center of the earth."

Yes I did, and it was clearly stated in the sentence you quoted in this message.

Anon.: "Does it point in your direction, or in mine?"

Yours is a fragmented sentence without defining what the word "it" is referring to. Also, it does not appear relevant to the subject at hand.

The power of the mathematical language: going straight to the point.

My point is simply: F(0)=0

Where F(r) is the gravitational force (vector) at distance r from the center of a spherical body.

Re: Anonymous, gravity July 1 2009, 5:00 PM

Anon.: "The power of the mathematical language: going straight to the point.

My point is simply: F(0)=0

Where F(r) is the gravitational force (vector) at distance r from the center of a spherical body."

You failed to respond to my question, which was;
"My opinion is based on the fact that gravitational bodies attract, do you disagree?"
Failure to respond to my inquiries nulls any responsibility I may have to respond to yours! Moreover, your point is not relevant to the issue.

Bob,

"My opinion is based on the fact that gravitational bodies attract, do you disagree?"

I disagree.

There are many reasons I disagree. Dimensionless points in space don't represent natural objects. An oblong odd shaped asteroid cannot be described by gravity. A barycenter needs to be evaluated. Barycenters are again a 0 dimensional representation of an object by evaluating the two edges and determining the midpoint. This is not an accurate representation of the object.

Attraction based on mass is not the determinate of motion of the object. The sun is a great radiator of energy. It does not attract. It burns material and expels radiation and light. It ejects this radiation constantly. There are huge bursts of radiation that create space weather.

The sun also has huge magnetic fields. These magnetic fields pressurize the surrounding dark matter, leaving a bubble of baryonic material inside. The shape of this extends to the ice clouds. This radiation emitted from the sun extends at least 50,000 au. [1]

Black bodies are radiators.
a

Attraction
July 1 2009, 8:42 PM

Bob,
"My opinion is based on the fact that gravitational bodies attract, do you disagree?"

Aaron: "I disagree.
Attraction based on mass is not the determinate of motion of the object."

My opinion is based on a two body system ie "bodies" (in the plural).
Your disagreement is based on a one body system ie "the object" (in the singular).
So, I agree! the motion of your "object" can not be based on attraction. However, although I agree with the basis of your disagreement, your disagreement does not negate my opinion.

Since the rest of your message does not apply to the subject at hand, I chose to not respond.

Bob,

"My opinion is based on a two body system ie "bodies" (in the plural).
Your disagreement is based on a one body system ie "the object" (in the singular).

So, I agree! the motion of your "object" can not be based on attraction. However, although I agree with the basis of your disagreement, your disagreement does not negate my opinion."

"My opinion is based on the fact that gravitational bodies attract, do you disagree?"
+++++++++

This is what you asked. I am only answering the question. I know it was not posed to me, but I figured that this was an open forum so if I responded, you would understand. You don't have to like it but it is powerful enough to stop the dream of gravity or attraction at a distance.

(1)

(2)

These equations are not just for one body systems. Each baryon uses this method to evaluate its position. So by default, if each baryon is evaluating itself to the information it is receiving from other baryons, then that must be a many body system. Not just a 2 body system.

These equations are the relation between the baryon and the information that other baryons communicate to it. It also describes the relation between it and the dark matter.

So this is a many body system. It is scalable to any size. I show that each subsystem (Magnetism, Electricity, Temperature) are separate and can effect the other. Also each proven force is accounted for and utilized at the same moment. No need for frames of references. This is an analog system.

These equations show that in the field if one side of the baryon is hotter than the other it will move away from the heat source.

This is a 4 dimensional field with the time aspect built into the system. A photon which transmits heat information also communicates pressure. A heat source like a broiler release the same photons. If the receiving baryon is 5 inches away then it will cook the baryon. Where as if the receiving baryon is 10 feet away it will not cook.

Remember that gravity can only explain 2 body interactions. It cannot resolve anything but attraction in the same iteration. It cannot explain how temperature, magnetism, or electricity affect an objects motion.

In conclusion this is a correct evaluation of baryonic communication. This communication results in pressure, temperature, motion and many other effects.

I will have a paper soon that expresses this information in a more detailed way. I have explained this at my blog, but as anon stated, it is difficult to read from bottom to top. So this paper will be a top down explanation of this equation and its implications.

I am watching videos on Density Function Theory and algorithm development from MIT. From this I should be able to develop a working computer model analysis of this system. Just a little programming problem.

Most of us here agree that gravity does not work in one way or another. You are arguing with AAF as to the correct way gravity should be evaluated. Why is it so hard to evaluate a system that if proven correct could replace the arcane concept of attraction? Or did I answer my own question.

I would like you to read and disprove my paper 'Disproof of Gravity'. Just beat the hell out of it. Please, take it point by point and destroy it. Show your friends how you destroyed it. I would love to quit thinking about this. So if anyone can disprove my theory. I would love it. Then I could go about my pathetic life.

a