Is Gravity Caused by the Motion of Particles in the Fabric of Space?

[Mr. Robin Parsons]

No time for the full responce here, not now, but I will, God willing, give you more of an answer, later, but for now, this simple point...

Originally posted by Heusdens (Ooops my error BRAD_AD23)

And FZ+ is indeed correct. You are measurign the planet's weight in terms of your gravitational field.

So there was this guy, see, he went to the top of a tower and dropped two balls, a fifty lb ball, and a five lb ball, and the gravitational attraction of the balls, on the planet's weight, wasn' t recordable/observable, and He iS FAMOUS for having done that!

Know his name??...cause you know something, HE PROVED BOTH OF YOU WRONG, waaaaaaaay back in HISTORY...

Till later...

 

[neutroncount]

Originally posted by Mr. Robin Parsons
No time for the full responce here, not now, but I will, God willing, give you more of an answer, later, but for now, this simple point...



So there was this guy, see, he went to the top of a tower and dropped two balls, a fifty lb ball, and a five lb ball, and the gravitational attraction of the balls, on the planet's weight, wasn' t recordable/observable, and He iS FAMOUS for having done that!

Know his name??...cause you know something, HE PROVED BOTH OF YOU WRONG, waaaaaaaay back in HISTORY...

Till later...

That experiment is a myth. He worked with pendulums and incline planes to realize that the specific weight of an object had no effect on its decent under the influence of gravity.

 

[Brad_Ad23]

Thank you, and Parsons, at least get the name of who you quote right.

All that would do is show the acceleration due to gravity is independent of one mass. I.E. case one of my math I included. The Earth attracts all massive objects with the same acceleration, so everything falls to it at the same dv/dt. However, not everyone has the same mass so dp/dt, or force or weight will be different. That does absolutely nothing to alter my argument in any way shape or form. It is just a side result, which as you admit is correct, otherwise you would not have posted the result.

Just in case, my resulting equation:

Take: F=dp/dt and F = GMm/r2.

Let m be the mass of the person and M the mass of the planet. dp/dt is the change in momentum per time interval (you also see this as F=ma, [mdv/dt = dp/dt, and dv/dt = a]). The mutual force of attraction between the two is F = GMm/r2.

Now, let us get the acceleration due to gravity of the earth. This case we would be looking for the weight of the person.

ma = GMm/r2. small m's go away, and we get:

a = GM/r2.

 

[FZ+]

Originally posted by Mr. Robin Parsons
No time for the full responce here, not now, but I will, God willing, give you more of an answer, later, but for now, this simple point...



So there was this guy, see, he went to the top of a tower and dropped two balls, a fifty lb ball, and a five lb ball, and the gravitational attraction of the balls, on the planet's weight, wasn' t recordable/observable, and He iS FAMOUS for having done that!

Know his name??...cause you know something, HE PROVED BOTH OF YOU WRONG, waaaaaaaay back in HISTORY...

Till later...

In fact there is a small difference due to the variation in the Earth's acceleration towards the ball. However, as you can appreciate, the mass of the ball is many orders of magnitude below that of the Earth and hence this effect is very small, often outweighed by air resistance etc, much less than any instruments by Galileo can detect.

 

[Mr. Robin Parsons]

Originally posted by FZ+
In fact there is a small difference due to the variation in the Earth's acceleration towards the ball. However, as you can appreciate, the mass of the ball is many orders of magnitude below that of the Earth and hence this effect is very small, often outweighed by air resistance etc, much less than any instruments by Galileo can detect.

I agree, it is so small because of the huge differentiation in the planet's weight, compared to the balls weights, the differential of acceleration is proportionate to the difference of the sum of the two weights, and F = ma will demonstrate that the time differential is miniscule...Really miniscule...No problemo...

Now, go to the page with the http://www.jpl.nasa.gov/earth/features/watkins.html" link, take that shape and cut it in half, use the perimeters "appearance" for the application of all of your vectors, which means, use that shape for the inhomogenous, (totally?) anisotropic, realm of wave activity.

(Please take clear note, it is NOT a "nice little {perfect?} circle")

Let's see you cancel them out now.


Oh Yes, it is required that you have a magnitude for the force, (pressurizing force at this point) well the magnitude is the speed of propagation of EMR through it's medium, solid rock.

(Don't be fooled!, there is a minimium of 26% empty space in that 'apparent' solid)


Lumpy Earth photo, (JPEG) provided by the link to the People at NASA, Thanks!


EDIT Pssssst...this is a part of the answer to all of this riddle of gravity...

© 2003 Mr. Robin Parsons Kingston Canada

^¤Spark le and ¯Shine!

 

[Mr. Robin Parsons]

look! it changed densities!

2003-06-10

Want some other fun, try this one, the density of the surface rock on this planet is 3000 kg/m³, now, (anybody got a good supply of pre-nacent water, prefferably really really cold!) we will voyage to the Moon, with our Earth rock, and, upon landing there, we will weigh the rock, aaaaand Oh look!, the density has changed!, it is now 500 kg/m³!

Tell me, if, when I started out on the Earth, the atomic volume of the rock had been one Avogadro's number per cu mm (1 AN/mm³)*[/Color] do you think that when I arrive on the Moon, and re-weigh the rock, that that property of the rock, (1 AN/mm³) has actually changed?


*[/Color] I made up that figure, so it is just for the purpose of illustration, it is not to be seen as being the accurate one, I can (God's Grace) figure out the right one, but why bother, right now, as I only need the demonstrative ability.

 

[Brad_Ad23]

Let's see you cancel them out now.

At the center of gravity they will. I know for a fact in some earlier post of mine I said the center of gravity and the exact center of Earth need not coincide.

2003-06-10

Want some other fun, try this one, the density of the surface rock on this planet is 3000 kg/m3, now, (anybody got a good supply of pre-nacent water, prefferably really really cold!) we will voyage to the Moon, with our Earth rock, and, upon landing there, we will weigh the rock, aaaaand Oh look!, the density has changed!, it is now 500 kg/m3!

Tell me, if, when I started out on the Earth, the atomic volume of the rock had been one Avogadro's number per cu mm (1 AN/mm3)* do you think that when I arrive on the Moon, and re-weigh the rock, that that property of the rock, (1 AN/mm3) has actually changed?


* I made up that figure, so it is just for the purpose of illustration, it is not to be seen as being the accurate one, I can (God's Grace) figure out the right one, but why bother, right now, as I only need the demonstrative ability.

What the heck did you do here? You divided the density by 6 I noticed, which coincides with the rough decrease of gravity on the moon. Please tell me you are not in fact, asserting that on the moon, objects would be 1/6 as dense as on earth? Kilograms are used as a measure of weight here, but that is a misnomer that people very readily do. The actual kilogram is a measure of the amount of mass an object contains. It is a scalar quantity, meaning, its constant. Weight, is what you get when you do mass times acceleration. It is measured in Newtons, or pounds, and as people do on scales, they factor out the a and it produces a "weight" in kilograms. On the moon, the object would still have 3000kg/m³. Its "weight" would indeed be 500 kg sure, but as stated, that is a misnomer. The true definition of mass in kilograms is scalar, and invarient no matter what gravity field. The density will remain the same as a result (mainly because density is MASS per unit volume, NOT weight per unit volume).

 

[Mr. Robin Parsons]

Apparently you cannot tell what one of these is, ?, for your EDIFICATION it is called a question mark, usually used when an individual is asking a question,

Oh look! there's one now! right there, in big bold red...

Originally posted by ME!

Tell me, if, when I started out on the Earth, the atomic volume of the rock had been one Avogadro's number per cu mm (1 AN/mm3)* do you think that when I arrive on the Moon, and re-weigh the rock, that that property of the rock, (1 AN/mm3) has actually changed?

Can you see it better now? (oh look there's another one of those "thingamajigggies" that he mentioned, what was that called again??)

PS

Originally posted by Brad_AD23

At the center of gravity they will. I know for a fact in some earlier post of mine I said the center of gravity and the exact center of Earth need not coincide.

Lets keep the story straight here fellows, you keep telling me that Gravity is waning as we travel down, BY SELF CANCELATION, show me how you cancel those lines!

 

[Brad_Ad23]

You have one line pointing in one direction, and oddly enough, another pointing in the opposite. Hmmm, imagine that. It is not that hard of a concept to see for most people.

 

[heusdens]

Mr Parson.

You are an absolute genius Mr Parson, we humbly and sincerely apologize for being so wrong in this, while your brilliant mind was so right on this.

Of COURSE there is a nett force of gravity at the CENTER of gravity.
How stupid of us to assume there is not!


But.. perhaps because we are so stupid, may we still ask you a question.

If there is a nett force of gravity at the center of earth, and since a force is a vector, and we know through your brilliant mind it is not zero, all we want to know is: what is the DIRECTION of that resultant force?

 

[Mr. Robin Parsons]

Originally posted by Brad_AD23 pg 20

but the two forces still exert a pressure, in the form of an equal but opposite reaction.

 

[Mr. Robin Parsons]

Originally posted by heusdens
Mr Parson.

You are an absolute genius Mr Parson, Well thank you we humbly and sincerely apologize for being so wrong in this, Thanks again! while your brilliant mind was so right on this. Now now, flattery will get you nowheres with me...

Of COURSE there is a nett force of gravity at the CENTER of gravity.
How stupid of us to assume there is not! Good of you to admit it, but I really don't think that your all that stupid


But.. perhaps because we are so stupid, may we still ask you a question. Oh of course, but as I said, your not really stupid, you know?

If there is a nett force of gravity at the center of earth, and since a force is a vector, and we know through your brilliant mind it is not zero, all we want to know is: what is the DIRECTION of that resultant force? Well, now your just being silly, go back and re-read what I told you before last time you asked me this very same question. Cheese bud, don't you read what I post?? or are you just being your usual agrumentative self?

 

[Brad_Ad23]

Originally posted by Mr. Robin Parsons
Originally posted by Brad_AD23 pg 20

but the two forces still exert a pressure, in the form of an equal but opposite reaction.

Yes, your point? The force still has an effect even though the resultant vector of the gravity forces are zero at the point, the pressure that all the forces exert (which add up mind you, is still great because it is force divided by area. And no, it is not the resultant gravity force at the center of gravity over area. It is the sum of any of the force vectors in any direction being exerted. Yeah the pressure will be immense and come from all directions and balance out, but you still feel it squeezing you from all directions.

 

[Mr. Robin Parsons]

Originally posted by Brad_Ad23
Thank you, and Parsons, at least get the name of who you quote right.

My apologies, "To err is human, to forgive Divine"

(although, I suspect, considering the un-relenting nature of this discourse, in your case(s) (I include Heusdens in this) "Canine")

But none the less, my apologies for the slight error on my part.

Originally posted by Brad_AD23

Yes, your point?

(insert sound of POINT flying over/past his head)

But as a last mention, Nigel so sorry that I have, apparently taken over, and dominated your thread, just that, sometimes when attempting to prove that anothers "Proof", isn't, well the only way to do it is to reveal a 'Proof' that is.

As for you "other" two, I note clearly that neither of you addressed the point I made about the "plume" problem, that your responce(s) does NOT address, and actually tells that it must be behaving completely opposite to what is known, therefore illogical and inconsistant with the facts of reality as they are known.

Furthered by the simplicity that neither of you will admit that in that, (as I had told you it was) inhomogenous and anisotropic "circle", (you do know what those two words mean, don't you?*) you cannot cancel out the unequal, and therefore not opposite, lines.


*Don't bother answering, it is both a waste of time, and server space. Thanks

 

[Brad_Ad23]

Of course Parsons! You're right! IF we take any inhomogenous object we cannot have any center of gravity in it at all! Nevermind the integrals that exist for finding such things at all! It may be distributed oddly, but there will exist in it some point where all the force vectors are canceled out. As stated, the CENTER OF GRAVITY need NOT coincide with the CENTER OF THE OBJECT.

 

[Mr. Robin Parsons]

© Mr. Robin Parsons Canada 2003

13/06/2003

So Brad_AD23, let's try a simple experiment, logical and consistent.

Draw a circle r = 5, at this point we see acceleration due to gravity as being equal to 5 m per sec², and, as you have so insistently, and repetitively, told me, all of the vectors cancel to ZERO "NET" force, but there is pressure, as that is the resultant of the "Zero 'Net' force".

So Brad_AD23, as I have put in my profile, I was a mechanic, as there are lots of kinds of mechanics, Auto, Truck, Gas, Diesel, Electrical, Engineering, Robotic, Stationary, Ship, Submarine, Aerospace, Airplane, Small engine, Industrial, Spacecraft, etc. etc. I will not tell you which combinations of those I have been, but I have the experience and the learning to understand (Some/most?) mechanical things.

Now, the pressure that is exerted within that circle, r = 5, is a mechanical pressure, (from, as you two keep repeating, from "All of the weight above") and one of the features of mechanical pressures is that they are even throughout. Across the diameter of that circle the pressure is isotropic and homogenous.

(The same at all points, to the center, and back out)

Now, in that circle, draw me another circle at r = 2, and here we will see the waning of the force of gravity, as both you and Heusdens keep telling me it diminishes all the way down, so here the force of acceleration due to gravity is 2 m per sec², and all of the vectors cancel out to equate to a pressure exertion. (Zero "net" force, ergo, now a pressuriz(ed)ing force)

Now, I want you to explain to everyone in this forum how the pressure that can be generated by a 2 m per sec² force, can exceed the pressure that is generated by a 5 m per sec² force, BECAUSE, very clearly, WE KNOW that the pressure is increasing, as we move towards the center of the Earth, without question.

Please use "consistent, and logical" reasoning to explain your response.


PS If you attempt to tell me, "Well (Mr.) Parsons, it sums", then I am going to have a great laugh, because if it is 'summing', then it must be at 7 m per sec² that you have dodged the plume problem, well not surprising, NOT AT ALL!

 

[Brad_Ad23]

Parsons, you have no idea about physics at all. It is a summing process (the pressure). But perhaps before we answer your question, you should answer heusdens and tell us what direction is the magnitude of the force then at the center of gravity? And also, for a more consice answer of your question, go back to one of those links I submitted sometime ago. I believe it describes a mechanism somewhere (not sure because its been awhile). At any rate the summing of the weight does occur, and the two forces meeting at one point do indeed create a pressure that is measured as force per area, and the gravitational forces do cancel out. Now, if you are so right Mr. Parsons, please submit your revolutionary work to a reputable journal and I look forward to reading it and seeing the actual mathematics behind your wacky gravity theory that has gravity vectors passing through the center of gravity like it didn't exist.

 

[Mr. Robin Parsons]

Originally posted by Brad_AD23

# 1) you should answer heusdens and tell us what direction is the magnitude of the force then at the center of gravity? # 2) And also, for a more consice answer of your question, go back to one of those links I submitted sometime ago.

# 1) Did that already, so simply follow your own advice, and go look. (I have been explicite that I would not reveal the "mechanism", simply the result, which I have, soooo)


# 2) I have, there is no explanation, that is something that both, you, and heusdens, have been rather remise on, rather you have been very repetative with things like,

GET A PHYSICS BOOK (own two)

GO BACK TO SCHOOL (still learning)

YOU DON'T KNOW WHAT NEWTON SAID (really, your right, I never met the man, but I have read some of his historical works)

If there is an explanation, cite it please, you do know where your own work is, don't you?

EDIT PS went back and fixed the error in that mis-cited post.

 

[heusdens]

What is the direction of the force of gravity at the center of gravity?

WHAT DIRECTION?

You can not escape thos question so easily.

Tell us!

 

[Mr. Robin Parsons]

Originally posted by heusdens
What is the direction of the force of gravity at the center of gravity?
WHAT DIRECTION?
You can not escape thos question so easily.

Tell us!

As I have told you, I have already posted the responce to that, it is in the previous writing.

Secondly, for someone who has offered very little, in the manner of explanations, (other then repepepepepepepetitions) I sincerley don't think you have the right to DEMAND anything from me, after all, you have never addressed the plume question, and neither have you answered to me on just how you accomplish getting the r = 2 circle to pressurize in a manner greater then the r = 5 circle has already accomplished!

(without "summing it' hence r = 7, and if you want to tell me you believe that it works as you have described it, without explaining exactly how that pressure differential is accompished, go to New York City, the have a bridge, someone wants to sell it to you, cause if you would wish to fool yourself, go right ahead, you need not my permission, simply that you will not be fooling me!...got that?)

God willing, I will cede the floor to anyone who can prove that that r = 5, r = 2 thing can be worked. (without summing it!)...otherwise simply accept that "current theory", as it was understood, has been proven to be wrong! simple as that.

 

[Mr. Robin Parsons]

nmemonics (HUH?)

© Mr. Robin Parsons 2003 (pg 15)
Think you missed that, know it actually, you missed the point which is the zero, and the energy that comes out of that/this, my point, (?) exactly!

(As Originally posted, in COLOR )



----------------

© Mr. Robin Parsons 2003 (pg 16 time 05-31-2003 10:14 PM ^{'3rd from the bottom})

By Zero Point Energy, (Principal!) the G/T cyclic, that I have mentioned, in this thread as well, it is shortwaved, and cycled back out as thermal energy, that is where the gravitational energy, culmatively, is directed.

Can you figure that one out?

 

[Brad_Ad23]

Anyways, to get back to Nigel.

I have a few questions for you.

1. Gravity is caused by an isotropic pressure field that is shielded by mass, so you claim. That is, no matter where you are on the surface of the earth, there will be a force coming from above to push us down. We may view this as a bunch of arrows directed towards the center of the earth. My question is, what causes those arrows (in this case the pressure) to want to push down on the Earth in the first place? If we remove the earth, the vectors should still have to point towards a central point (I believe cosmologists refer to this as a hedgehog). The pressure field cannot simply revert to a no mass state, for which direction would the pressure go?


2. Regardless of the outcome of question 1, this one is alone tricky. We know that your pressure 'stuff' cannot go through mass. After all, mass is what shields us from it, and if it was able to travel through mass, it would not be able to exert the force on us. So, say we dig a tunnel down into the earth, it doesn't matter how deep. On the top of this tunnel, we put a lid, with a ball attatched to a mechanism. The tunnel is now shielded by the lid, and effectivly there should be no gravity in the tunnel now, since the pressure field cannot reach it (remember, the mass is shielding the inside of the tunnel from the force! Its as if the Earth itself was there and there was no tunnel, and if it can go through the lid, then how does it exert the force on us, and how does it not interfere with itself coming from opposite directions on the earth). So, we fire the mechanism, which let's go of the ball. According to this, the ball should remain where it is. After all, no downward force from above acting on it. It is shielded by mass from below, and by mass from above. Yet, observation time and time again shows the ball still falls. How is this so?

 

[yogi]

I would concur Brad - the mass shielding rationale leaves something out - but the equation may offer insight into a holistic explanation of gravity based upon expansion - if mass reacts to expansion then the force vectors would be isotropically convergent upon the mass if the expansion were spherically symmetrical - and the strength thereof would depend upon two factors - the amount of mass and the acceleration factor G (vol accel/per unit mass). In other words, the combination of an acceleration field G and mass leads to a spatial pressure gradient as I suggested in the gravity paradox topic on the Forum. Anyway - I also would like to hear Nigel's response to your questions.

 

[Brad_Ad23]

It could indeed offer insight, but it has physical problems. A big one is number 1. The vectors cannot be arranged like that, unless they will be so for all time. And if they were to revert back to some other form if the Earth were no longer there, which form and why were they directed to the Earth then? Number 2 is also important. I too await some explination.

 

[Nigel]

Originally posted by Brad_Ad23
Anyways, to get back to Nigel.

I have a few questions for you.

1. Gravity is caused by an isotropic pressure field that is shielded by mass, so you claim. That is, no matter where you are on the surface of the earth, there will be a force coming from above to push us down. We may view this as a bunch of arrows directed towards the center of the earth. My question is, what causes those arrows (in this case the pressure) to want to push down on the Earth in the first place? If we remove the earth, the vectors should still have to point towards a central point (I believe cosmologists refer to this as a hedgehog). The pressure field cannot simply revert to a no mass state, for which direction would the pressure go?


2. Regardless of the outcome of question 1, this one is alone tricky. We know that your pressure 'stuff' cannot go through mass. After all, mass is what shields us from it, and if it was able to travel through mass, it would not be able to exert the force on us. So, say we dig a tunnel down into the earth, it doesn't matter how deep. On the top of this tunnel, we put a lid, with a ball attatched to a mechanism. The tunnel is now shielded by the lid, and effectivly there should be no gravity in the tunnel now, since the pressure field cannot reach it (remember, the mass is shielding the inside of the tunnel from the force! Its as if the Earth itself was there and there was no tunnel, and if it can go through the lid, then how does it exert the force on us, and how does it not interfere with itself coming from opposite directions on the earth). So, we fire the mechanism, which let's go of the ball. According to this, the ball should remain where it is. After all, no downward force from above acting on it. It is shielded by mass from below, and by mass from above. Yet, observation time and time again shows the ball still falls. How is this so?

ANSWER TO QUESTION 1: The inward space pressure is produced by the conservation of volume when there is an outward motion of mass. The first time I saw the aircraft film of the 11 megaton Castle shot 2, 28 Feb 1954, I was amazed that the steam or debris at ground level moves back towards ground zero as the blast moves outward!

This is not just an afterwind due to the low pressure under a rapidly rising fireball, it is a return of air which has physically moved outward. The expansion of hot air reduces the air density near ground zero to 1 % of normal air density, and the outward-blasted air returns in a "suction" phase directed back towards ground zero. This set off my search for a similar effect in the big bang, with the fabric of space taking the place of the returning air. It is useful to have a concrete analogy in your mind to help you get through doing the maths objectively.

ANSWER TO QUESTION 2: X-rays go through your hand, which is nearly empty. Gravity is the weakest force known in the universe. The amount of shielding is exceedingly small. We do not shield the Earth to any appreciable extent, nor does your tunnel lid. It takes a mass as great as the Earth to cause the shielding which makes all objects fall down with the same acceleration.

I think that Mr Parsons raised a point that a sheet of steel should have a different weight standing on end than it would have if flat. This doesn't measurably happen because (1) the steel is not doing the shielding that causes the acceleration towards the earth, that is being done by the Earth shielding the steel, and that shield is irrespective of the orientation of the shield, and (2) the gravitational mass points in the steel (fundamental particles) are very small, and even in a large amount of material, they will not be exactly behind one another.

If you had a big enough shield, then you could indeed do what you say. A lid with the mass of the Earth placed over a tunnel would have a decent effect on the space pressure and hence "gravity" in the tunnel.

What I would like to add is that, looking back at all this, I can see why people like Newton were sensible to not say what they had, mathematically proved until pushed. I am not a clever person in terms of exam grades, but I put a lot of hard work into the gravity mechanism, and thought that it was worth publication. However, if you work on anything like this, anything basic, you are censored out. You are ridiculed. If you then hit back, you have a personality problem. Anything you do is not tolerated, and you are called intolerant if you push. Science is not fun.

 

[Mr. Robin Parsons]

Goodness, I used "Common Knowledge" to demonstrate that what Brad, and Heusdens, were saying was the manner of operation, was wrong!

I thought science was fun, because anyone who can understand/learn/explore/discover/advance it, can succeed at it!

(but you need to be able to prove it/&yourself)

 

[Brad_Ad23]

You still left a big gaping hole there Nigel. The vectors of the pressure that are supposedly being directed towards the mass have to have some reason for going that way. And what of when the mass is no longer at that point in space? How do these vectors readjust themselves to in effect return to this isotropic pressure field?


And 2 still leads to a begging question: If the shielding is so weak, and thus it can traverse through the earth, why are we not being pushed off the Earth from a mysterious force that comes out of the ground? I am not referrign to the normal force either. Or does it eventually just stop? And if that is the case, where? Certainly not the non-existant center of gravity (according to Mr. Parsons). So I guess my question still remains, how does it operate inside a massive body. And for that matter in orbits? Orbits are determined by the gravitational attraction of the body. But if it is the result of some shielding, and not the geometry of spacetime being warped, why is it that orbits exist? It should seem that any object in orbit should fall immediately towards the object it is orbiting in that case, since there is a force now pushing it towards the object but no shielding in any other direction in its orbit.

Also, why is it that if say the moon is at one point in its orbit, the side of the Earth opposite the moon will receive a stronger (not noticably stronger mind you) pull down? If they are already shielded from the rest of the earth, the moon should not make any difference.

 

[Mr. Robin Parsons]

Originally posted by Brad_AD23

(snip) Certainly not the non-existant center of gravity (according to Mr. Parsons) (snip).

Really? when/where did I state that there was no center of gravity??

 

[Brad_Ad23]

Well by definition the center of gravity of a body is where the internal gravitational effects are zero, but ok. They can't cancel out at the center of gravity according to you. There.

 

[Mr. Robin Parsons]

Brad_AD23 wasn't 'nuthin' new from you...

2003-06-18

Originally posted by Yogi...https://www.physicsforums.com/showthread.php?threadid=2435&perpage=&pagenumber=1"

(snip)the first time I encounted this notion was in Ted Harrisons textbook "Cosmology" in connection with black holes - the idea being that space flows in toward black holes at a speed c at the event horizon ergo, light cannot escape since it cannot travel upstream at greater than c. Tom Martin has developed the mathematical theory for a number of experiments(snip)

Originally posted by Yogi
(snip)In other words, the combination of an acceleration field G and mass leads to a spatial pressure gradient as I suggested in the gravity paradox topic on the Forum. Anyway - I also would like to hear(snip)

Oh!, I get it now, your all trying to limelight

LIMELIGHTERS is what you are!

Tell me, how does YOUR theory of gravity make it compress to the center??


(Hahahahahahahhehehehehehehehehhohohohohohohohohohohhuhuhuhuhuhuhuyuckyuckyucknyucknyucknyuck ad infinitum...)