# The meaning of the constant G

1. Sep 25, 2013

### bobie

Hi,
was it Newton the first to describe the gravitation formula?, did he do that in his Principia?
and, most of all , what is the meaning of the constant of proportionality G?

Thanks a lot

2. Sep 25, 2013

### phinds

It is exactly what the name says it is. The forces are not EXACTLY equal, they are proportional and G is the constant of proportionality, determined by experiment.

3. Sep 25, 2013

### bobie

Thanks,
Could we say that G is the force between two 1Kg-masses?

4. Sep 25, 2013

### phyzguy

We could say that G is the force in Newtons between two 1 kg masses one meter apart.

5. Sep 25, 2013

### bobie

Thanks, phyzguy, wouldn't it be better to define G in such a way?, it would be of great help to everybody

6. Sep 25, 2013

### phyzguy

What would we do differently if we defined it in such a way? In practice, the force between two 1kg masses one meter apart is immeasurably small, so to measure G we use larger masses than 1 kg that are closer together than 1 meter. We then measure the force and calculate the value of G from Newton's law of universal gravitation:
F = G*m1*m2/r^2. Since G is constant, we can use any values of m1, m2, and r and we will get the same answer. What's the difference?

7. Sep 25, 2013

### dauto

No, it wouldn't. It's not even right because G is not a force - the units don't match. You would have to say that the numeric part of G is equal to the Numeric part of the force produced by two 1kg masses 1 meter apart If we chose to use International system units. If you chose to use Imperial units you would have to say that the numeric part of G is equal to the Numeric part of the force produced by two 1slug masses 1 feet apart - not very useful since most people never even heard of the unit slug. You should try to understand the proportionality constant on its own merit, not as a force produced by a particular configuration of masses

8. Sep 25, 2013

### Philosophaie

9. Sep 25, 2013

### dauto

Right. It's not measured in Newtons, it can't be a force

10. Sep 25, 2013

### TurtleMeister

What do you mean by "the forces are not EXACTLY equal"? What are the two things that are proportional to each other?

11. Sep 25, 2013

### voko

The modern view is that the constant G (as well as many other fundamental constants) are artifacts of our system of units. We can choose units in such a way, that many fundamental constants, including G, become unity. See http://en.wikipedia.org/wiki/Planck_units

12. Sep 26, 2013

### TurtleMeister

Yes, I think we've been through this before in another thread. I remember having a hard time understanding how the proportionality could disappear by using a different system of units. I'll try to find the thread when I have more time. Thanks for the response.

13. Sep 26, 2013

### Staff: Mentor

14. Sep 26, 2013

### TurtleMeister

15. Oct 14, 2013

### bobie

16. Oct 14, 2013

### D H

Staff Emeritus
Given any one dimensioned physical constant, it's fairly easy to choose units such that the numerical value of that physical constant is one, or perhaps some other "natural" value such as 2*pi. For example, consider G, the subject of this thread. The Newtonian acceleration of a test particle subject to a gravitational field is a=GM/r2. This involves the constant G and three separate dimensions, mass, length, and time. You can pick arbitrary quantities as representing the unitary values for any two of mass, length, and time. The remaining quantity is not arbitrary if one wishes G to be numerically equal to one. For example, pick one second as the unit of time and one meter as the unit of length. With these units, G has a numerical value of one if the unit of mass is 1.498×1010 kg rather than 1 kg.

This invites a question, why stop at G? Another obvious target is the speed of light. This constant appears all over place in relativity. It's easy to choose units such that both the speed of light and the universal gravitational constant both have numeric values of one. Rhetorical question: Can we go even further? The answer is yes. From an SI (metric system) perspective, there are apparently five fundamental quantities: mass, length, time, temperature, and electrical charge. (Aside: QM adds color charge to the mix.) Choose appropriate units for energy and temperature and the Boltzmann constant has a numeric value of one. Choose appropriate units for energy and frequency and the Planck constant (or reduced Planck constant) has a numeric value of one. Choose appropriate units for charge and length and the Coulomb constant has a numeric value of one. Choosing units such that each of G, c, kB, ħ, and the Coulomb constant have numeric values of one yields the Planck units.

These choices were somewhat arbitrary. Why the Coulomb constant, for example? Particle physicists would much prefer a system of units where the electron charge has a numeric value of one. There is *no* choice of units that makes all physical constants have a numeric value of one. For example, the ratio of the mass of a proton to the mass of an electron is fundamental. You can't change that by changing units because this quantity is dimensionless. Another such dimensionless quantity is the fine structure constant α≈1/137. You can't change these two unitless quantities (along with 20 some others) precisely because they are unitless. They will have the same values regardless of choice of units.

Note that I was careful in the above to say that an appropriate choice of units makes G, c, etc. have a numeric value of one. There's a huge difference between a physical constant having a numeric value of one (but still having units) and a dimensionless physical constant. How many of our dimensions (mass, length, time, temperature, charge, ...) are truly independent? Are there *any* dimensionful quantities? These are fundamental questions regarding which physics does not yet have a definitive answer.

17. Oct 14, 2013

### bobie

They are sensible, DH, thanks.
But you already consider the electron the unit charge, without realizing it, because you are used to think in Coulombs.
Being an outsider, I can see that without esitation:
e =1 q, C = 6,24 x 10^18 e [q]

the same applies to : J = 1,5 x 10^ 33 h[/s]
if there weren't dimentions you could accept h as the unit of energy, and that would make life a lot easier for beginners and scientists alike
probably they had to modify h to h.s to compensate for Hz being 1/s so that hHz becomes J ?

Probably red tape has sedimented through centuries non rational classifications (like the name of the atomig orbitals : s, p ...) and physics could use a good overhaul and rationalization, but probably physicists are too accustomed to allow radical change?

Last edited: Oct 14, 2013
18. Oct 14, 2013

### D H

Staff Emeritus
Not at all!

First off, there's nothing in either Coulomb's law or in Maxwell's equations that assume quantized matter. There can't be for the simple reason that Coulomb's law predates quantum mechanics by over a century, and Maxwell's equations predate it by about a quarter of a century.

Secondly, the charge of an electron is not one with Planck units, where each of G, c, kB, ħ, and the Coulomb constant have a numeric value of one. The charge of an electron instead has a numerical value of about 1/11.7 in Planck units.

19. Oct 14, 2013

### bobie

I was not referring to the law but to this unit:
If you assume an electron as a unit, then the coulomb has 6.24x 10^18 elementary charges(electrons). It is exacly the same, just a change of perspective

20. Oct 14, 2013

### Imabuleva

Just to clarify, the force would have a value of 6.67*10^-11 N. Because G has a value of 6.67*10^-11 m^3*kg^-1*s^-2, G is not the actual force. It's just the constant of proportionality that relates the magnitude of the gravitational force to the geometry of the system.

21. Oct 14, 2013

### Staff: Mentor

No, the Coulomb constant has different units. Coulomb's law is:
$$F=k\frac{Q_1 Q_2}{r^2}$$
so solving for k gives
$$k=\frac{F r^2}{Q_1 Q_2}$$

So k has the same units as the expression on the right. In SI units that is

$$k \rightarrow \frac{(kg \; m/s^2) m^2}{C C}=\frac{kg \; m^3}{C^2 s^2}$$

In Gaussian units (aka electrostatic units) electric charge is measured in statcoulombs, but the statcoulomb is not a base unit, it is a derived unit where $1 statcoulomb = 1 g^{1/2}\; cm^{3/2}\; s^{-1}$. So in Gaussian units we get

$$k \rightarrow \frac{(g \; cm/s^2) cm^2}{(g^{1/2} \; cm^{3/2} \; s^{-1})^2}=1$$

So in Gaussian units k is dimensionless and equal to 1.

Last edited: Oct 14, 2013
22. Oct 14, 2013

### phyzguy

Clearly I should have said "The numerical value of G (6.67E11) is equal to the force in Newtons between two 1 kg masses 1 meter apart." I think dauto already corrected this oversight a couple of weeks ago.

23. Oct 15, 2013

### bobie

Thanks, Dalespam, that is amazingly clear! I have a few more questions:
- does that law produce an acceleration: [c]m/ s^2 like the law of gravitation?
- what is the precise value of electrostatic force Ec between two elementary charges at 1 cm distance? If I am not wrong it is roughly equal to
C/α2π => Ec=1.439958 x 10^-7 eV, is this precise enough?

- G constant does not include the /4π factor unlike Ke , why so?, does that mean that G is in effect 86.4/4π, or that propagation is different?

24. Oct 15, 2013

### Staff: Mentor

That law produces force, and then Newton's 2nd law relates that force to the acceleration.

I don't know how you came up with this, but eV is a unit of energy, not force. The force would be in newtons or dynes, depending on the system of units. In SI units:
$$F=k\frac{Q_1 Q_2}{r^2}=8.988\frac{Nm^2}{C^2}\frac{(1.6\;10^{-19}\;C)^2}{(0.01\;m)^2}=2.3\;10^{-24}\; N$$
In Gaussian units:
$$F=\frac{Q_1 Q_2}{r^2}=\frac{(4.8\;10^{-10}\;statC)^2}{(1\;cm)^2}=2.3\;10^{-19}\; Dyne$$

That is just a matter of definition. G could have been defined with a factor of 4π. The reason that you often (but not always) see that factor for k is that the factor shows up in Maxwell's equations also. If you write k with the 4π in Coulomb's law then it goes away in Maxwell's equations. That is essentially the difference between Gaussian units and Lorentz-Heaviside units:
https://en.wikipedia.org/wiki/Gaussian_units
https://en.wikipedia.org/wiki/Lorentz–Heaviside_units

25. Oct 15, 2013

### bobie

That is the formula that states that the speed of the ground state electron in H v [H1 2.1877x10^8 (C/α) equals 2π *Ecm the energy of a photon of 3.482*10^7 Hz (1.433*10^-7 eV)
Can you find the equivalent of dynes in eV?

Last edited: Oct 15, 2013