Newton made the most amazing discovery when he was able to formulate, with great accuracy the manner in which gravity worked, unfortunately he was never able to ascertain HOW gravity worked in the way that it did. This brought the rather unjust accusation that his theory of gravitation was based on the dreaded Action at a Distant (AAD) theory. AAD is dreaded because it implies that cause and effect take place instantly without any intermediate time interval in between. Needless to say, Newton was aware of this particular shortcoming in his theory, and wisely left it to future generations to figure out how gravity worked. It was this very point that Einstein leveraged to bring discredit upon Newton's theory of Universal gravity and to forward his own theory of General Relativity. In this he succeeded to a remarkable degree. But what if it were possible, using a new theory, to derive gravity without using Newton's Universal Gravitational constant OR Einstein's 'Kapp' which incidentally also uses Newton's Gravitational constant. Surely such a development would be worth a debate.
Can Gravity be derived without using Newton's Gravitational constant?
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Pinball1970
Valued Senior Member
Newton probably derived his inverse square from distance, mass and from what he know about forces.
Physics is empirical so based on measurements of bodies moving on the earth and in the sky in this case.
I honestly do not know how he measured things like distance when it came to celestial bodies, or mass for that matter.
Galileo, Halley, Newton, Copernicus, Brahe, Kepler all took lots of measurements before building the equations that described orbits.
Deriving gravity without the use of newton's Gravitational constant:
Imagine being able to derive gravity without using Newton's gravitational constant or Einstein's 'kappa' which is a constant that relates space-time to energy and momentum? Sounds impossible? Augmented Newtonian Dynamics theory, makes it possible!
According to the "Augmented Newtonian Dynamics" theory of gravity. The electron as it orbits, (yes as a particle, not as a wave-particle) the nucleus, it is constantly emitting and immediately re-absorbing 'virtual' photons. What are these virtual photons? These virtual particles, are similar in structure to real photons emitted by the electron but of such short duration <10^-16 s , that every time an electron emits a virtual particle particle, it results in the 'virtual photons' of the virtual photon aether (dark matter?) immediately forming into a line of aligned photons along the direction of propagation of the emitted 'virtual photon'. However, in the case of the emission of a virtual particle, no energy flows along the line of aligned virtual photons or line of force. Instead, it results, as has been said, into the forming of a transient line of force in the direction of propagation of the emitted virtual particle. Such a line of force transfers no energy, it merely results in a tensioning of the virtual photon field (dark matter?) for a very short time 10^-15 s. If the line of force falls on another object, then the line of force represents the shortest distance between the two objects. As a corollary to this, the object on which the transient line of force falls, immediately releases a reciprocal transient line of force that falls on the original object. So that the two are infinitesimally drawn together. That is why it could be said that gravity is always attractive. The transient nature of the line of force is also the reason that gravity is the weakest force in nature, being 10^40 times weaker than the electromagnetic force. Every electron in every atom forms transient lines of force. Therefore, although gravity is a very weak force it is also a relentless, high-density force which though weak has an irresistible cumulative effect that can stretch for immeasurable distances. Gravity is a one off effect, meaning that unlike in the production of light where trillions of identical photons are emitted every second all travelling in the same direction forming rays of light and explaining the rectilinear nature of light, virtual photons giving rise to gravity are anisometric, meaning that the only sure thing is that a virtual photon is emitted and absorbed in every orbit, but as to which position it is emitted at or the direction in which it is emitted is unpredictable and random. This is the basis of gravity according to AND theory.
Next worked examples are provided wherein gravity is derived without using Newton's Universal gravitation constant and the two results are compared. AND theory successfully predicts the same value of g as Newtonian gravity, but now with a physical mechanism rather than an abstract force.
Worked example with various elements using both Newtonian Gravitational constant and AND theory line of force:
To begin with the g of several elements will be addressed, GM/r^2:
Iron:
Using AND theory of atomic structure and lines of force to calculate the g exerted by one cubic centimetre of Iron:
Number of electrons in iron = 26
Number of atoms in 1cm3 iron = (7.34 x 10^22)
Total Number of electron in 1cm3 iron = (7.34 x 10^22) x (26) = 1.90 x 10^24
Energy in line of force = 2.7 x 10^-37 J
Using AND theory line of force to determine g of one cubic centimetre of iron:
g = (iron x line of force) = (1.90 x 10^24) x (2.76 x 10^-37) = 5.25 x 10^-13 N
Using Newtonian Gravitational constant:
Density of Iron = 7.87 gm/cm^3
Gravitational constant = G is 6.6743×10^−11 m^3 kg−1 s^−2
The Newtonian solution using g = GM/r^2
g = (7.87 x 10^-3) x (6.674 x 10^-11) = 5.25 x 10^-13 N
It is therefore demonstrated that the two methods obtain a similar result and that g according to AND lines of force, is close to Newtonian gravity.
Next calculate the g force of gold using both AND line of force and Newtonian Gravitational constant:
Step-by-Step Calculation for g of Gold (1 cm^3)
1. Determine the number of moles in 1 cm^3 of gold
Density of gold: 19.32 g/cm^3
Molar mass of gold (Au): 196.97 g/mol
Moles of gold in 1 cm^3: moles=mass/molar mass=19.32 g/196.97 g/mol = 0.0981 moles
2. Calculate the number of atoms in 1 cm^3
Avogadro’s number: 6.022×10^23 atoms/mol
Total number of gold atoms in one cubic centimeter: = 0.0981×6.022×10^23 =5.91×10^22 atoms
Determine the total number of electrons in 1 cm^3
Gold has 79 electrons per atom
Total number of electrons in cm^3 gold: = (5.91×10^22) × 79 = 4.69×10^24 electrons
4. Compute the total energy from virtual photon interactions
Assume each virtual photon line of force contributes (2.76 x 10^-37) J
Total energy contribution using AND line of force:
g of one cubic centimetre of gold using AND line of force: ( 4.69×10^24)×(2.76 x 10^-37) = 1.28 x 10^-12 N
The g force of One cubic centimetre of gold using the virtual photon model
= 1.28 x 10^-12 N
Gravitational force of one cubic centimetre of gold using Newton's gravitational constant:
g = GM/r2
The g force of one cubic centimetre of gold using gravitational constant:
g of gold = (6.674 10^-11) x (19.32 x 10^-3) x (1) = 1.289 x 10^-12 N
(The results using AND virtual photon theory and the Newtonian gravitational constant are identical at g = 1.28 x 10^-12 N)
The results for carbon using AND theory lines of force and Newtonian gravity respectively:
g for one cubic centimetre of carbon using AND theory lines of force
Number of atoms in 1 cc of carbon = 9.07 x 10^22 atoms/cm^3
Number of electrons in carbon atom = 6
Number of electrons in 1 cc of carbon = (9.07 x 10^22) x (6) = (5.49 x 10%23)
Result of g exerted by cubic centimetre of carbon g using AND lines of force
g = (5.49 x 10^23) x (2.76 x 10^-37) = 1.52 x 10^-13 N
Result for g using Newtonian gravitational constant
Density of carbon = 2.267 cm^3
g = GM/r2
Result of g force exerted by one cubic centimetre of Carbon using gravitational constant:
g = (6.67 x 10^-11) x (2.67 x 10^-37) = 1.52 x 10^-13 N
The agreement of the two results is again very close.
Having completed a few examples it is now possible to consider larger objects like planets. While dealing with large objects like planets AND theory first calculates the volume of the object in cubic centimetres, it then determines the density per cubic centimetre of the object, finally the periodic table is looked up to get a match for the density. The g of the element is then calculated using the above methods and multiplied by the number of cubic centimetres by volume in the object. This gives the g force of the planet. To determine the acceleration due to gravity of the object. The g is divided by the radius squared (in metres) of the object. Taking the moon as the first example:
Calculating g of moon according to AND theory:
Calculating g of moon using AND theory line of force:
Volume of the moon = 2.1 x 10^25 cubic centimeters
Density of 1 cubic centimetre of the moon = 3.34 gm per cubic centimeter.
Substance possessing similar density = Calcium Oxide (CaO)
CaO = (3.27 x 10^22) atoms /cm^3
CaO possesses 26 electrons per atom
CaO possesses ((3.27 x 10^22) x 26) = 8.50 x 10^23 electrons/ cm^3
Line of force = 2.76 x 10^-37 J
Therefore g calculated using AND theory = (8.50 x 10^23) x (2.76 x 10^-37) = 2.34 x 10^-13 N/cm^3
Total g of the moon according to AND theory = (2.34 x 10^-13) x (2.1 x 10^25) = 4.90 x 10^12 N
Calculating g of moon according to Newton:
Mass of the moon = 7.34767309 × 10^22 kg
Therefore g using Newton's Gravitational constant = (7.34767309 × 10^22) x (6.67 x 10^-11) = 4.90 x 10^12 N
Using AND theory line of force acceleration due to moon: (4.9 x 10^12) / (1743000)2 = 1.61 m/s^2
Using Newton's gravitational constant acceleration due to moon
= (4.90 x 10^12)/(1743000)2 = 1.61 m/s2
Conclusion:
This derivation of gravity from atomic structure, independent of the universal gravitational constant, represents a fundamental shift in understanding gravitational interactions. Rather than treating gravity as a fundamental force requiring an externally imposed constant, this approach suggests that gravity is an emergent property that emerges naturally from the intrinsic behaviour of matter at the atomic level. By linking gravitational effects to the self-regulation of electron energy through virtual photon interactions, this perspective offers a unified framework that connects microscopic quantum processes with macroscopic gravitational phenomena. This not only provides a new foundation for gravity but also reinterprets the relationship between matter, energy, and the structure of space itself.
Pinball 1970, sorry that I was not able to address you directly I ran out of space and so had to delete both a quote from your message and my response. Hence I am posting it here:
Many thanks Pinball1970 for your thoughtful comments. Absolutely, I totally agree with what you have said. The crucial information is that Newton derived gravity from what he knew about forces etc., A lot has happened since that time, which I think would have offered him clues not only to how gravity worked in the way it did but also as to why it worked in the way that it did. This is the reason that the theory is called "Augmented Newtonian Dynamics (AND) theory meaning that it is just an extension of Newtonian theory filling in missing information. Here are worked examples that should be of interest: (nb: Am always open to criticism, but the huge thing here is the close correspondence between Newtonian and AND theory)
While evaluating this theory, it should be borne in mind that the working of the nucleus is governed solely by the exchange of 'virtual particles', both to keep neutrons and protons intact through the exchange of gauge bosons of the strong force, and to hold the protons and neutrons together in the nucleus through exchange of gluons. This being so why shouldn't the atom itself be held together via exchange of virtual particles. Especially when it has been proved through the Lamb shift that such interactions do take place. Further, the proven fact that optical photons, (the only type that the electron can emit directly) are emitted at the rate of hundreds of trillions per second, something which the cloud theory of the electron in the atom cannot substantiate, is proved through the working of optical atomic clocks.
Many thanks Pinball1970 for your thoughtful comments. Absolutely, I totally agree with what you have said. The crucial information is that Newton derived gravity from what he knew about forces etc., A lot has happened since that time, which I think would have offered him clues not only to how gravity worked in the way it did but also as to why it worked in the way that it did. This is the reason that the theory is called "Augmented Newtonian Dynamics (AND) theory meaning that it is just an extension of Newtonian theory filling in missing information. Here are worked examples that should be of interest: (nb: Am always open to criticism, but the huge thing here is the close correspondence between Newtonian and AND theory)
While evaluating this theory, it should be borne in mind that the working of the nucleus is governed solely by the exchange of 'virtual particles', both to keep neutrons and protons intact through the exchange of gauge bosons of the strong force, and to hold the protons and neutrons together in the nucleus through exchange of gluons. This being so why shouldn't the atom itself be held together via exchange of virtual particles. Especially when it has been proved through the Lamb shift that such interactions do take place. Further, the proven fact that optical photons, (the only type that the electron can emit directly) are emitted at the rate of hundreds of trillions per second, something which the cloud theory of the electron in the atom cannot substantiate, is proved through the working of optical atomic clocks.
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exchemist
Valued Senior Member
Evidently Dilip James is setting density equal to M/r² in the Newtonian formula g = GM/r².
This makes no sense for at least two reasons:
1) Newton's formula describes how gravity falls off with distance, r, from the mass in question, but no such distance is specified, making it a nonsense.
2) It is also dimensionally incorrect, since density is ML⁻³, whereas M/r² is ML⁻².
This makes no sense for at least two reasons:
1) Newton's formula describes how gravity falls off with distance, r, from the mass in question, but no such distance is specified, making it a nonsense.
2) It is also dimensionally incorrect, since density is ML⁻³, whereas M/r² is ML⁻².
What rubbish! If you use your eyes exchemist and your brain, you will find that the great thing is that there is no reference to mass at all in the AND (Augmented Newtonian Dynamics Theory) solution for gravity, it is merely the value for line of force, multiplied by the number of electrons in a cubic centimetre of the substance. Nothing to do with mass or r^2.
exchemist
Valued Senior Member
You wrote:
QUOTE
Using Newtonian Gravitational constant:
Density of Iron = 7.87 gm/cm^3
Gravitational constant = G is 6.6743×10^−11 m^3 kg−1 s^−2
The Newtonian solution using g = GM/r^2
g = (7.87 x 10^-3) x (6.674 x 10^-11) = 5.25 x 10^-13 N
UNQUOTE
You are multiplying density by G and setting that equal to g.
Since g=GM/r², it follows you are setting density equal to M/r².
Not meaning to be rude, but a very cursory reading and a somewhat dense interpretation. There are two solution in every example. One of the solutions is using AND theory to derive gravity and the other solution is using Newton's gravitational constant to derive gravity (g) (note differs from substance to substance) using Newtons Universal gravitational constant of G = 6.6 x 10^-11 approx. This same Universal gravitational constant constant is not used at all in the AND theory solution. Yes exactly no distance is used because it is calculating the g of one cubic centimetre, if you like I can include the 1 and square it.
exchemist
Valued Senior Member
The passage of yours that I quoted claims to show a result from Newtonian gravitation. It does not mention your AND theory.
And it is algebraically wrong.
Nevertheless, correct when used in this context, whichever notation is used. It would be (Gm/1^2) = g that;s it!
exchemist
Valued Senior Member
Can't be. The units don't stack up.
g has (SI) units of m/sec².
G x density has units of (m³/kg-sec²) x (kg/m³) = 1/sec².
So it's wrong.
exchemist
Valued Senior Member
Tell Newton you can’t do algebra?
exchemist
Valued Senior Member
His equation is fine. It is your substituting density for M/r² in Newton's formula that is wrong. I have explained this.
You are an idiot.
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In science the only idiots are those who raise objections for the sake of raising objections and nothing else. But enough of that, I will ignore your posts in future, so don't bother answering to this post. However, in the interests of the general discussion, I feel it is important enough to explain my point of view.
If the only utility of Newton's law of Universal gravity were to determine the gravitational attraction between two bodies, it would leave a lot of questions unanswered. For instance what is the gravitational force exerted by a single body? Surely this is the reason for mass existing in the equation in the first place; in order that bodies possessing different masses would exert different gravitational force? This being so the equation g = Gm/r^2 seems to be the most appropriate to determine the gravitational force exerted by a single body. One way to do this is to check this equation with respect to the gravitational force exerted by the earth.
Given that the gravitational force g = Gm/r^2 . Taking the density of earth (approx.) to be 5.5 gm/cm^3. The first step in order to solve the equation using the metric (MKS) system is to convert gm to kg. Therefore : (0.0055) x (6.67 x 10^-11)/1m = 3.6 x 10^-13 N/cm^3.
To get the total gravitational force exerted by the earth it is necessary to multiply the volume of the earth in cm^3 by (3.6 x 10^-13 N) Volume of earth in cm^3 = 1.09 x 10^27 cm^3. Therefore total gravitational force exerted by the earth = (1.09 x 10^27) x (3.6 x 10^-13) = 3.92 x 10^14 N, and the acceleration due to that force would be (3.92 x 10^14)/((6.37 x 10^6)^2) = 9.6 m/s^2 which is close to the actual value of gravitation due to acceleration 9.8 m/s^2
Similarly, if we use the AND system: Number of electrons in one cm^3 of earth considering earth has a density of 5.5 gm/cm^3 and 55 mol. Then there are approx. 5.9 x 10^22 atoms cm^3 taking an average of 20 electrons per atom gives 1.18 x 10^24 electrons per cm^3. g = (1.18 x 10^24) x (2.76 x 10^-37) = 3.25 x 10^-13 N and the total gravity exerted by earth = (3.5 x 10^-13) x (1.09 x ^27) = 9.4. Total gravity exerted by earth = 9.4 m/s^2. Not quite accurate but acceptable in my opinion.
Нужно различать собственную гравитацию тела(вопрос: как выглядит гравитация при наличии одного единственного объекта), которая впрочем возможна только у составных массивных объектов, и гравитацию между разными объектами. Хотя, по сути своей, у них одна и та же природа.
Pinball1970
Valued Senior Member
Gravity is not a force in GR
