Proof of cause of gravity

In summary, the conversation discusses the concept of particle-wave duality and how it is caused by the motion of particles in the fabric of space. The pressure towards us from the fabric of space produces gravity, and this is the mechanism behind the acceleration due to gravity. This understanding also explains why apples fall. The conversation also mentions an article published in Electronics World, which reviews and extends the mathematical proof for the mechanism of gravity and resolves problems with general relativity. It is proposed that this model can be used to rigorously test the consequences of this physical fluid model for the fabric of space. The conversation also mentions the fixed 377 ohms impedance of the vacuum to electromagnetic energy, which suggests that the fabric of space is a non-particulate
  • #281
Notice, I said the pressure was irrelevant to the SYSTEM. Much different than saying pressure is irrelevant. Why? Because I said it was irrelevant to the system being described. Key word: system. Of course, I suppose you just didn't catch that, did you?
 
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  • #282
"If fools be silent..."

Whoever I am, I did not intend the quote to apply to anyone in particular, and certainly not as a criticism

The literal quote is in Proverbs 17:28.

Beg pardon for intruding. "...and the rest is Silence."
 
  • #283
Originally posted by Brad_Ad23
Notice, I said the pressure was irrelevant to the SYSTEM. Much different than saying pressure is irrelevant. Why? Because I said it was irrelevant to the system being described. Key word: system. Of course, I suppose you just didn't catch that, did you?

Notice in the manner in which I had quoted you, the word 'system' is there, as in "Yes", I had noticed, and Yes, it is relevant, just that, apparently, you don't think so, That Proves nothing.

Sorry r637h, had mis-read it I guess.
 
  • #284
Parsons, first of all, the system I am referring to is the truck example, not the Earth one (since that follows from the truck idea). Second of all, the website notes that it is HIGH pressure. The pressure generated by the two trucks is hardly high enough to alter properties of matter, lest the trucks start acting very strangely.

Now, notice the Earth is high pressure, and nowhere do we say pressure is irrelevant. Indeed, the balance the pressure has by representing a force directed outwards to counter gravity is quite important. I also find it rather odd that you go to all these sites, one would think that somewhere you would encounter some argument you could comprehend as to why the rest of us are right. But again, I reiterate, since you are so convinced you are correct, I look forward to reading about your works someday.
 
  • #285
And Brad_AD23 in the posting right above mine, Heusdens, the person I was originally addressing, stated...

Origianlly posted by Heusdens

So, we agree there is no nett accelartion, so no nett force.

So please, follow what is being talked about, or please, stop talking.

Thanks.
 
  • #286
But you also were referring to the truck example as well.

At any rate, the mechanics, and this may surprise you, work the exact same way regardless of the location.
 
  • #287
Originally posted by Brad_Ad23
But you also were referring to the truck example as well.

At any rate, the mechanics, and this may surprise you, work the exact same way regardless of the location.

Ahem, no I wasn't, and why would the "Mechanics of it" surprise me?

The point that I was making with Heusdens was concerning the measure of gravitational 'force' being excerted at the center.

He said "No Acceleration, no Force", I said "No acceleration, but still Force" which is where the pressure results from.

You seem to think that somehow the Earth's shell is pressurizing it's center, which would indicate that a Neutron Star's contraction would spew matter, out into the universe with tremendous force, as it would somehow be this "shell structure" you two keep inferring gravity makes, snapping into pieces as it attempted to contract, because in your world it DOES NOT contract from the center, it contracts from a shell ~4/5ths of the way down from it's surface...same as the earth's.

That is 'patently' WRONG!

EDIT a comma, like that one, there, behind , here (yuk yuk)
 
  • #288
Ok, here is some math to back this up also:

http://burro.astr.cwru.edu/Academics/Astr221/SolarSys/hydrostat.html

http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect5/subsubsection5_1_1_2.html [Broken]



Also to note, it is nuclear fission (radioactive decay) that heats up the Earth primarily, not pressure.

And, to note also, no matter what, the sum of the gravitational forces at the center is zero. But as we have been trying to say, this does not mean that nothing is happening. Zero net force is not the same as absence of force. A simple analogy:

Take the Earth and divide it into two equal parts. Seperate it some distance and put yourself at the center between the two hemispheres. Since the gravitational attraction cancels out, you won't move. BUT, stay there for a few minutes and the two hemisphere will come together and crush you with a pressure proportional to the force of attraction between them and your surface area. Does that perhaps better clairify how no net force can still exist in the presence of pressure?

also: I am not quite sure where you get a 4/5th from nor this shell pressurization. And as for the truck stuff, perhaps a contextual misread. At any rate, the above situation dealing with the force of gravity and pressure is physics. A simple bit of research will show that.
 
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  • #289
There's no way to be totally silent: It's against human nature.

I only want to say (for now) that I find the dialog (multilog?)
in this thread absolutely fascinating. Mirable dictu! I think I'm actually learning something.

Best regards to All.
 
  • #290
Originally posted by Brad_Ad23
Also to note, it is nuclear fission (radioactive decay) that heats up the Earth primarily, not pressure.

And, to note also, no matter what, the sum of the gravitational forces at the center is zero. But as we have been trying to say, this does not mean that nothing is happening. Zero net force is not the same as absence of force. A simple analogy:

Take the Earth and divide it into two equal parts. Seperate it some distance and put yourself at the center between the two hemispheres. Since the gravitational attraction cancels out, you won't move. BUT, stay there for a few minutes and the two hemisphere will come together and crush you with a pressure proportional to the force of attraction between them and your surface area. Does that perhaps better clairify how no net force can still exist in the presence of pressure?

also: I am not quite sure where you get a 4/5th from nor this shell pressurization. And as for the truck stuff, perhaps a contextual misread. At any rate, the above situation dealing with the force of gravity and pressure is physics. A simple bit of research will show that.

According to your source for the pressure, they 'average out' the density, hence it is not looking for from where the pressure is generated, nor does it give a result in concordance with what I have/had, the Access Science site's graph told me between 3, and 4 Megabars of pressure at the center.

Your is 1.7x10E6 atmospheres, which, when I convert from Megabars to atmospheres, I get @ 3 megabar = 2,960,769.5079428 atmosphere [standard] and @ 4 megabar = 3,947,692.6772571 atmosphere [standard], so I would respectfully suggest that your site is simplfieing the math just a little bit much.

The part of your post that I have emboldened is self-contradicting as you first tell me the gravitational attractions cancel, then, that they will crush me.

Your problem is a simply one you keep treating gravity as if it can be dealt with as a 'discrete' force, "pieces and parts" of what acutally acts in a wholistic manner.

Where do you see that pressure coming from?, arising from?

The "4/5ths" was explained, in one of my posts, some pages back.
 
  • #291
Originally posted by Mr. Robin Parsons
Ahem, no I wasn't, and why would the "Mechanics of it" surprise me?

The point that I was making with Heusdens was concerning the measure of gravitational 'force' being excerted at the center.

He said "No Acceleration, no Force", I said "No acceleration, but still Force" which is where the pressure results from.

You seem to think that somehow the Earth's shell is pressurizing it's center, which would indicate that a Neutron Star's contraction would spew matter, out into the universe with tremendous force, as it would somehow be this "shell structure" you two keep inferring gravity makes, snapping into pieces as it attempted to contract, because in your world it DOES NOT contract from the center, it contracts from a shell ~4/5ths of the way down from it's surface...same as the earth's.

That is 'patently' WRONG!

EDIT a comma, like that one, there, behind , here (yuk yuk)

Yes Mr Parson. And you have an ABSOLUTE TALENT for not understanding what someone else is telling you.

The discussion was not about wether or not there is the force of gravity at the center, but wehter there is a NETT RESULTANT FORCE.
And there isn't since all forces balance there, and hence no gravitational acceleration.

Else I didn't claim, and especially this pressure thing, I never claimed there not be a lot of pressure there.
 
  • #292
Please stop wasting server space!

2003-06-08

Lets try it this way Guys, F = ma, so, when I hold a golf ball in my hand, balanced forces, no acceleration, I "drop the ball" (so to speak) and gravity continues to act upon the mass that was previously in a 'balanced holding', by accelerating it towards, and into, the ground.

Gravity continues to act, holding the ball securely to the face of the planet.

From that we can deduce that, the acceleration has ceased, but gravity is still excerting a force as to hold the ball to the face of the planet. We can measure that force, easily, by simply inserting a scale between the interface of contact, and we now see that the energy of gravity, being opposed by a greater resistance, generates a pressure we call "Mass".

Gravity is still working, it has never stopped, not since the formation of the planet, it is inceassent, it is continous, and persistant, throughout time.

So in the equation F = ma, we need a repair, in our thinking, actually, because when gravity acts to hold a piece of mass/matter to it's gravitational face, we can easily surmise the action of a force, even though we see no acceleration, we still know that the effort/attempt of/to accelerate, exerts force called pressure, or weight/mass, so we need to see that in the Equation the acceleration value can only be seen as a 1 (one). (with NO qualifiers like m/sec2)


The reasoning is simply that, now we see that F = m(1), or F = mass, because we know that the fact of "mass" is as a result of gravities attempt at accelerating something, in a place where the opposition to that acceleration, is greater then the force applied, and we weigh that 'force', on a scale, and call it pressure, from a mass.

The force is the weight, the weight is a measure of a force, the force that is being measured is still gravity, just that like the two trucks at opposition, there is an attampt at acceleration, that generates a pressure, measurable with a scale, as either a pressure, or a mass, although we wouldn't normally recognize a reading "that way" as a mass, it is, effectively, the same thing.

(because you need "mass", to make pressure)

So we now see that gravity is a pressurizing force, a wave of energy that attempts to accelerate mass that is oppositionally resistive to that attempt at acceleration, the result of which is pressure(ization, of mass)

Try that one....
 
  • #293
Mr Parsons: But F = ma (or more accurately F d/dt(mv)) is applied only on the resultant force on the object. In the case of the object, there is not resultant force because the sum of the force of gravity "down" and the electromagnetic reaction force up is zero. It's not opposed by greater resistance, because then we would have an acceleration upwards.
 
  • #294
Originally posted by FZ+
Mr Parsons: But F = ma (or more accurately F d/dt(mv)) is applied only on the resultant force on the object. In the case of the object, there is not resultant force because the sum of the force of gravity "down" and the electromagnetic reaction force up is zero. It's not opposed by greater resistance, because then we would have an acceleration upwards.

Static resistance, the sum of the forces is not zero, the measurable force is the weight of the mass, hence it's ability to exert pressure, as it so clearly does, like a six ton rock.

EDIT ...Like a six ton Rock, which, BTW, weighs in at one ton, on the Moon, because the Gravitationally attempted acceleration excerts 1/6th (less) the force, upon the rock.
 
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  • #295
No sir. When you are weighing, you are in fact weighing the difference in the two force vectors by their effect on both ends in compressing the scales. When you weigh the person, you are simultaneously weighing the planet in terms of the person's gravitational field.
 
  • #296
The Net (please for the love of physics learn what the word net means when applied to force!) force cancels out..ie. it is zero. This does not by any stretch of any good imagination mean gravity is not present. As has been stated numerous times by everyone but you, zero net force/acceleration can and does exist when two even forces balance out, but the two forces still exert a pressure, in the form of an equal but opposite reaction. Hence you are squished, but you don't move from one side to another.
 
  • #297
Originally posted by FZ+
No sir. When you are weighing, you are in fact weighing the difference in the two force vectors by their effect on both ends in compressing the scales. When you weigh the person, you are simultaneously weighing the planet in terms of the person's gravitational field.

Ahem, you are 'weighing' the planet's gravitational attraction, as it acts upon the person.

If you remove the weight from the face of the planet, the planet does NOT come flying out at you, (a slight decompression may happen) as it affords no vector of "force", it affords a vector of resistance to force.

Brad_AD23, the 'zero net force' you would wish me to believe in, is wrong, simply on the fact of the energy, that is gravity, is still acting upon the matter, that is what is compressing it. Nothing else can.

Originally posted by Brad_AD23

As has been stated numerous times by everyone but you, zero net force/acceleration can and does exist when two even forces balance out, but the two forces still exert a pressure, in the form of an equal but opposite reaction. Hence you are squished, but you don't move from one side to another.

Not from what you and heusdens have been stating on these pages, as you keep trying to tell me that the center is at zero, less gravity there, according to BOTH of you, then above it, and the simplicity is that it is the Force, of the acceleration, due to the passage of gravitational energy, waveform, that is causing the compression of matter at that center.

Hence that waveform, energy, that is gravity, must still be acting right down to the center. Otherwise NO COMPRESSION.
 
  • #298
look!, the past, it's catching up!

So Brad to recap yours and Heusdens arguements, "a brief history of time" (Sorry Stephan Hawking)
Originally posted by Heusdens pg 17

Right at that point, we imagine to have a whole bunch of vectors, in different directions and different magnitudes. Now the application of some math allow you to pick pairs of vectors with same magnitue and opposing directions, which sum to a resultant vector of zero magnitude, and the "proof" includes that this can be done in such a way that all the vectors can be removed. Resulting in this zero gravity force at the center.

and from page 16...

I don't have a massive pressure problem. I already explained what caused the pressure. It's because all the mass that DOES have weight ABOVE the center of gravity, that is pushing there.

There is absolutely no problem with that.

and from page 15...

It can be proven very easily however that right at the center, all forces of gravity of all the mass of earth, cancel out.
Now take any part of the Earth mass, and calculate this mass and it's distance to the center of gravity. That will enable you to calculate the force of gravity from that part of Earth's mass. Which is not zero. But at the exact opposite direction, we will find an equal mass at equal distance. This follows the fact that the density of Earth at any specific depth is more or less the same, and the Earth is a round spherical object. So we can deconstruct all of Earth's mass into small parts (as small as you want) and calculate the force of gravity from all parts. Since we always can find for every force vector an equal and exactly opposite directed force vector, this means that the resulting force of gravity exactly results in a force of zero.

And you Brad_Ad23

Originally posted by Brad_AD23 pg 14

Ok Parsons, listen up carefully. The gravitational force is directed downwards. To the center of gravity. At this point, the center of gravity, this is where all the little vectors of force meet up. At this point, there is no attraction, since there is no distance to go to get to the center of gravity. But, what else is going on? Hmm, well we have all this weight above the point. Yeah, that is directed downward as well. But what else is going on? We have pressure. Yep, Force divided by area. What else? As you go down deeper, you get more pressure, because there is more weight above you. Check. So anything else? Well, it would appear that if gravity's strength decreases, should there not be less pressure? Nope. The pressure exerted on any region down there, is a result of the weight above it. Not due to the gravitational force at that depth. It is also balanced as per the 3rd law of motion--An equal, but opposite, reaction for every action. The pressure from above pushes down, and the matter being acted upon pushes back with equal force.

So if you would like to change your stories, again, please, do so, you'll only make yourselves look, well, you know "that word" don't you!

(More so then you already are??)
 
  • #299
But why would I change my story when it is true? As you go down deeper you feel less and less gravity acting on you. The resultant gravity force does indeed become weaker. However, as you do down you also have more mass above you. The more mass above you, the more force it contributes in the opposite direction as that below you, the resultant force thus becomes weaker. The effect this has, is the same as the analogy with the planets crushing you. You have the combined forces acting in two opposite directions. The resultant force vector is zero, however the resultant pressure is not. Why? Because of the laws of motion, and mechanics. You have an equal amount of gravity itself acting around both sides, and an enourmous weight all around. All this mass is trying to pull itself together (since that is what gravity does...it causes stuff to contract). But, at the center, it is equally distributed in all directions. The resultant is no gravity vector. You are correct when saying the gravity energy is still there, yes, it does not go away. However when I have a force of 5 Newtons acting on something from the left, from the right, and from up and down, that particle will not move. What it will feel however, is the resultant force per unit area that the vectors exert on it, also known as pressure. So even though the forces cancel out to the same, the pressure is still there. And remember, pressure from all directions will be equal as well (equal depths same pressure when in equillibrium, which at the center there is).

However, that you say energy compresses matter, that is rather disconcerting to know. Energy in of itself cannot exert a force. Energy may be imparted into something, which MAY cause it to exert a stronger, or weaker force.

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

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. The resulting a we get should be roughly 9.8 meters per square second. Plug this into the F = ma equation and we get the weight of the person in Earth's gravity field. But, remember, for every actions there is an equal but opposite reaction. In this case, the changes in momentum must be equal obviously. But one has a large Mass and the other has a small mass. For the momentums to equal, the acceleration, a, ( remember, m dv/dt = dp/dt, and dv/dt = a) one body must have a smaller change in velocity than the other. That body is the earth. This means it is being acted upon by a tiny gravity field. And who's is that? Why the person's! Instead of canceling out the small m, cancel out the big one and find the acceleration. This is the weight of Earth in your gravitational field, which as FZ+ pointed out, you are also weighing. To make this simpler refer to Newton's 3rd law. If you were not exerting equal magnitudes of force, then either A)you would fly into the earth, or B)the Earth would fly into you.


Also, I add that resistence to force is not a vector. You can have a resistive force, or a force of resistance and that will be a vector, but a resistence in of itself to force is no vector.
 
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  • #300
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html#isq

In order for me to see some consistancy here, it was necessary for me to see the logic and develpement of a model of apprehension, based on preceding work done.

Like understanding Euclid's psotulates and then finding his fifth had moved to a different realm. Girolma Saccheri leads us into further thinking.

The logic of this approach must be geometrical defined. Is the metric understood here? I am in need of some education, so if anyone can help, I might help you extend your own knowlegde (logically and geometrically).

Sol
 
  • #301
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...
 
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  • #302
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.
 
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  • #303
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.
 
  • #304
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.
 
  • #305
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" [Broken] 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 ¯Shi[COLOR=#aoaoao]n[/COLOR][COLOR=#eoeoeo]e[/COLOR]!
 
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  • #306
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)* 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.
 
  • #307
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).
 
  • #308
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!
 
  • #309
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.
 
  • #310
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?
 
  • #311
Originally posted by Brad_AD23 pg 20

but the two forces still exert a pressure, in the form of an equal but opposite reaction.
 
  • #312
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?
 
  • #313
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.
 
  • #314
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
 
Last edited:
  • #315
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.
 
<h2>What is gravity?</h2><p>Gravity is a natural phenomenon by which all objects with mass are brought towards one another. It is a fundamental force of nature that is responsible for the motion of planets, stars, and galaxies.</p><h2>What causes gravity?</h2><p>The current accepted theory is that gravity is caused by the curvature of spacetime. The presence of mass or energy warps the fabric of spacetime, creating a gravitational field that pulls objects towards each other.</p><h2>How was the cause of gravity discovered?</h2><p>The concept of gravity has been studied and theorized by scientists for centuries. The most famous theory is Newton's law of universal gravitation, which was proposed in the 17th century. However, the understanding of gravity has evolved with the development of Einstein's theory of general relativity in the early 20th century.</p><h2>Can the cause of gravity be proven?</h2><p>While the concept of gravity has been proven through various experiments and observations, the exact cause is still a topic of debate and ongoing research. Theories such as general relativity and quantum gravity attempt to explain the cause of gravity, but there is still much to be discovered and understood.</p><h2>How does the cause of gravity affect our daily lives?</h2><p>The cause of gravity is essential in understanding the motion of objects and the behavior of the universe. It allows us to predict and explain phenomena such as planetary orbits, tides, and the formation of galaxies. Without gravity, life as we know it would not exist.</p>

What is gravity?

Gravity is a natural phenomenon by which all objects with mass are brought towards one another. It is a fundamental force of nature that is responsible for the motion of planets, stars, and galaxies.

What causes gravity?

The current accepted theory is that gravity is caused by the curvature of spacetime. The presence of mass or energy warps the fabric of spacetime, creating a gravitational field that pulls objects towards each other.

How was the cause of gravity discovered?

The concept of gravity has been studied and theorized by scientists for centuries. The most famous theory is Newton's law of universal gravitation, which was proposed in the 17th century. However, the understanding of gravity has evolved with the development of Einstein's theory of general relativity in the early 20th century.

Can the cause of gravity be proven?

While the concept of gravity has been proven through various experiments and observations, the exact cause is still a topic of debate and ongoing research. Theories such as general relativity and quantum gravity attempt to explain the cause of gravity, but there is still much to be discovered and understood.

How does the cause of gravity affect our daily lives?

The cause of gravity is essential in understanding the motion of objects and the behavior of the universe. It allows us to predict and explain phenomena such as planetary orbits, tides, and the formation of galaxies. Without gravity, life as we know it would not exist.

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