# Is the gravitational constant actually a variable

## Main Question or Discussion Point

Please don't be too cruel. I obviously have to be doing something wrong but it seems to me that the gravitational constant may be a variable based on coulombs constant and an objects charge per mass. The formula I have found would be

G = Ke x ((Q1 / M1) x (Q2 / M2))

This would give a much higher value of G for a proton and electron.
This new value for G would bring the calculated forces of gravity in line with the calculated values of coulombs law at an atomic level.

What are your thoughts on this?

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Andrew Mason
Homework Helper
Please don't be too cruel. I obviously have to be doing something wrong but it seems to me that the gravitational constant may be a variable based on coulombs constant and an objects charge per mass. The formula I have found would be

G = Ke x ((Q1 / M1) x (Q2 / M2))

This would give a much higher value of G for a proton and electron.
This new value for G would bring the calculated forces of gravity in line with the calculated values of coulombs law at an atomic level.

What are your thoughts on this?
You appear to be simply equating the gravitational to coulomb forces between an electron and a proton:

$$F_G = \frac{GM_eM_p}{r^2} = F_E = K_e\frac{Q_eQ_p}{r^2}$$

I agree that if the two forces were equal, this would be correct. However, that would not make G a variable. G would be variable only if the charge/mass ratios of the electron and proton varied. They don't.

In any event, they are not equal or even close to being equal. The ratio of the force of gravity between an electron and a proton to the coulomb force between them is about:

$$F_G/F_K = \frac{\left(GM_eM_p\right)}{\left(K_eQ_eQ_p\right)} \approx 4.4 \times 10^{-40}$$

Furthermore, gravity is always attractive. How would you explain coulomb repulsion in terms of gravity?

AM

Thank you very much for replying.

It may sound silly but could a bounce not be considered a temporary repulsion. If gravity somehow aligns the atoms similar to an electromagnet then the part touching the ground would always be the opposite polarity of the ground. But when a bounce occurs the atoms are moved around allowing for - to - or + to + interactions to occur. The object is then repelled until the atoms align again.

Can you explain why we believe G is always the same value. From all that I have read it has never been accurately measured, nor does any formula explain it. The formula I showed earlier would offer a value of 1.5 x 10^29 for G at an atomic level.

Even Keplers 3rd law G = (4pi^2 x r^3) / (T^2 x M) gives you a value of 1.5 x 10^29.

Andrew Mason
Homework Helper
It may sound silly but could a bounce not be considered a temporary repulsion. If gravity somehow aligns the atoms similar to an electromagnet then the part touching the ground would always be the opposite polarity of the ground. But when a bounce occurs the atoms are moved around allowing for - to - or + to + interactions to occur. The object is then repelled until the atoms align again.
I don't follow what you are saying. You will have to provide a better explanation of why you think gravity and the coulomb force are the same. There are many reasons why they are not: ie. the fact that the coulomb force has nothing to do with mass (eg. lightning); that the coulomb force is about 10^40 times larger than gravity; that there are two kinds of charge (+ and -) whereas there is only one kind of mass....

Can you explain why we believe G is always the same value. From all that I have read it has never been accurately measured, nor does any formula explain it. The formula I showed earlier would offer a value of 1.5 x 10^29 for G at an atomic level.
The answer is very simple: evidence. All measurements show that G is constant. The force of gravity between two masses is proportional to the product of the masses and inversely proportional to the square of their separation. These are the only factors that we have been able to find that determine the gravitational force between objects. This means that

$$F \propto Mm/R^2 => F = GMm/R^2$$

where G is the constant of proportionality.

Even Keplers 3rd law G = (4pi^2 x r^3) / (T^2 x M) gives you a value of 1.5 x 10^29.
I have no idea where you get this. The measured value is 6.67 x 10^-11.

AM

Dale
Mentor
It may sound silly but could a bounce not be considered a temporary repulsion.
No, gravity is always attractive even during a bounce. Although the object's velocity is upwards its acceleration is downwards.

From all that I have read it has never been accurately measured, nor does any formula explain it. The formula I showed earlier would offer a value of 1.5 x 10^29 for G at an atomic level.
You are correct, G is one of the least accurately measured universal constant with an error of about 1 part in 10^4. But you are talking about an error of 40 orders of magnitude! It is simply absurd. Life could not exist in such a universe. I don't know why you would even give this idea a second glance.

http://physics.nist.gov/cgi-bin/cuu/Value?bg

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How can coulombs law have nothing to do with mass I thought force equals mass times acceleration. It would seem to play some role. Masses all contain charged particles so I figured there should be some + and - interactions on an atomic level. They just cancel each other on a large scale.

There is a formula for an objects period of revolution t^2 = ((4pi^2 x r^3) / (G x M))

I used this formula to solve for G using an electron at the bohrs radius traveling 2.2 x 10^6 meters per second. I did not get the 6.67 x 10^-11 value of the planets. I got 1.5 x 10^29. This would explain the 10^40 difference between coulombs law and gravity.

No gravitational force has been noticed at an atomic level, and no gravitational waves have been found in the universe. Is it really that unlikely that they are the same force just with a different value for G. The speed of light is an exact measurent while the differing measurements of G allow for the possibility of an outside factor effecting it. Check out the chart on this page.

http://www.nature.com/news/2010/100823/full/4661030a.html

Dale
Mentor
How can coulombs law have nothing to do with mass
It has nothing to do with gravity, by definition.

Masses all contain charged particles so I figured there should be some + and - interactions on an atomic level. They just cancel each other on a large scale.
Sure, and those interactions dominate our "everyday" experience.

I am not saying that G is in error. At least not on a planetary level. I am saying that those slight differences at our level could show that there is a variable somewhere that may effect G greatly at smaller scales.

Andrew Mason
Homework Helper
I am not saying that G is in error. At least not on a planetary level. I am saying that those slight differences at our level could show that there is a variable somewhere that may effect G greatly at smaller scales.
If you are looking for a force that behaves like gravity, the strong nuclear force is a possibility. It is always attractive and it drops off dramatically outside of the nucleus. It also seems to be proportional to mass.

AM

Dale
Mentor
I am not saying that G is in error. At least not on a planetary level. I am saying that those slight differences at our level could show that there is a variable somewhere that may effect G greatly at smaller scales.
If at smaller scales gravity were 40 orders of magnitude stronger than it is then atomic nuclei would be much smaller than observed as would electron orbitals.

This idea is a no-starter. It doesn't explain any unexplained observation and it is contradicted by many observations. I don't know why you are even suggesting it. What is your motivation? What do you hope to accomplish?

Staff Emeritus
2019 Award
If you are looking for a force that behaves like gravity, the strong nuclear force is a possibility. It is always attractive and it drops off dramatically outside of the nucleus. It also seems to be proportional to mass.
Sorry, it can also repel, and is not proportional to mass.

Andrew Mason
Homework Helper
Sorry, it can also repel, and is not proportional to mass.
It is attractive between quarks and has an attractive effect between protons and neutrons. I wasn't aware that it was ever repulsive. Are you suggesting that a repulsive force keeps quarks separate within a proton or neutron? Can you provide a reference? Do we know how gravity operates at the sub-nuclear level?

The strong force in the nucleus is between quarks. The up and down quarks have similar masses and they seem to exert the same force on each other. How is the nuclear force not proportional to mass in that sense?

AM

Dale
Mentor
It is attractive between quarks and has an attractive effect between protons and neutrons. I wasn't aware that it was ever repulsive. Are you suggesting that a repulsive force keeps quarks separate within a proton or neutron? Can you provide a reference?
The residual strong force, the force between nucleons, is indeed repulsive at very small distances:

http://www.cartage.org.lb/en/themes/sciences/physics/NuclearPhysics/WhatisNuclear/Forces/Forces.htm [Broken]
http://www.s.u-tokyo.ac.jp/info/press/press-2007-10en.html

It also decays much faster than 1/r², which is why the EM force dominates at longer range. It also has 6 types of charge instead of 1. In addition, a graviton would be a massless, chargeless, spin-2 boson whereas the pion has mass, charge, and spin 0.

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Andrew Mason
Homework Helper
The residual strong force, the force between nucleons, is indeed repulsive at very small distances:

http://www.cartage.org.lb/en/themes/sciences/physics/NuclearPhysics/WhatisNuclear/Forces/Forces.htm [Broken]
http://www.s.u-tokyo.ac.jp/info/press/press-2007-10en.html

It also decays much faster than 1/r², which is why the EM force dominates at longer range. It also has 6 types of charge instead of 1. In addition, a graviton would be a massless, chargeless, spin-2 boson whereas the pion has mass, charge, and spin 0.
How does the repulsive part of the nuclear force differ from the "force" that keeps electrons in certain orbitals?

My understanding is that this is a quantum effect: no two electrons can occupy the same quantum state. Similarly, quarks cannot occupy the same quantum state so they have to keep a certain distance from each other. Is this really something we can call a force?

How would gravity behave at this level? Would quarks not still have to follow the Pauli exclusion principle if the only force keeping them together was gravity?

AM

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Staff Emeritus
2019 Award
Dale is correct, but now I am thinking about the quark force. It is definitely repulsive between two quarks of the same color charge, and it is attractive between a quark and a diquark. I'm not so sure what happens with two quarks of different colors. I think it may be attractive, but only half the strength of the repulsion of two quarks of the same color.

In any event, the color force is a very complex beast. Not nearly as simple as classical gravity.

Dale
Mentor
How does the repulsive part of the nuclear force differ from the "force" that keeps electrons in certain orbitals?

My understanding is that this is a quantum effect: no two electrons can occupy the same quantum state. Similarly, quarks cannot occupy the same quantum state so they have to keep a certain distance from each other. Is this really something we can call a force?
That is a good question. I don't know the derivation of the nuclear force (residual strong force) as shown in those diagrams. It is certainly not the same as the strong force itself which does not decay over distance, so it could very well include the Pauli exclusion principle as a part of the force.

How would gravity behave at this level? Would quarks not still have to follow the Pauli exclusion principle if the only force keeping them together was gravity?
My understanding is that at this level gravity would behave like a force mediated by a massless spin-2 boson, which the strong force is not.

gravitational constant is really a variable. i have read somewhere that a scientist has derived the value of G by the help of age of universe and range of mass and length.astime is not constant so G may not be constant.

Staff Emeritus
2019 Award
"I have read somewhere" is not an acceptable PF source.

Andrew Mason
Homework Helper
gravitational constant is really a variable. i have read somewhere that a scientist has derived the value of G by the help of age of universe and range of mass and length.astime is not constant so G may not be constant.
We don't really know whether G is constant throughout the universe or over very large distances. It is a very difficult thing to measure. We don't know whether the laws of physics, including the value of G, have always been the same in the past. That is an even more difficult thing to measure. All we can say is that there is no evidence (yet) that suggests that G is not constant over large distances and throughout the universe or over time. But in science one never rules anything out without evidence.

AM

i wonder why the value of G is 6.67*10^-11.can we derive it.

vanadium u r right.iwill soon give the name of that scientist?

We don't really know whether G is constant throughout the universe or over very large distances. It is a very difficult thing to measure. We don't know whether the laws of physics, including the value of G, have always been the same in the past. That is an even more difficult thing to measure. All we can say is that there is no evidence (yet) that suggests that G is not constant over large distances and throughout the universe or over time. But in science one never rules anything out without evidence.

AM
And purely logical evidence (which is sufficient in mathematics) is not sufficient in physics. Evidence in physics must be experimental, directly or indirectly.

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In order not to spoil He-4 production during big bang nucleosynthesis, the gravitational constant in the primordial universe can't be different from today's value by more than 9%.

Staff Emeritus