What is the significance of the constant G in Newton's gravitation formula?

[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

 

[phinds]

... what is the meaning of the constant of proportionality G?

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.

 

[bobie]

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

 

[phyzguy]

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

 

[bobie]

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

 

[phyzguy]

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

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?

 

[dauto]

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

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

 

[Philosophaie]

Quote by dauto:
No, it wouldn't. It's not even right because G is not a force - the units don't match.

The units are

G = 6.67384 E-11 m^3/kg/s^2 = 6.67384E-11 N(m/kg)^2

from

https://en.wikipedia.org/wiki/Gravitational_constant

 

[dauto]

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

 

[TurtleMeister]

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.

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

 

[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

 

[TurtleMeister]

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

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.

 

[Dale]

Maybe you mean this thread. See post 17 in particular

https://www.physicsforums.com/showthread.php?t=709247

 

[TurtleMeister]

Maybe you mean this thread. See post 17 in particular

https://www.physicsforums.com/showthread.php?t=709247

That's not the one I was thinking of, but it looks very interesting. And it does seem to address my question. Thanks

 

[bobie]

Maybe you mean this thread. See post 17 in particular

https://www.physicsforums.com/showthread.php?t=709247

Thanks, Dalespam, that is very interesting , it would be wonderful to dispose of dimensions. I posed a similar question here:
https://www.physicsforums.com/showthread.php?t=715997
Does the same apply to Coulomb constant?has it got the same dimensions?

 

[D H]

I remember having a hard time understanding how the proportionality could disappear by using a different system of units.

Given anyone 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, k_B, 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.

Thanks, Dalespam, that is very interesting , it would be wonderful to dispose of dimensions.

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.

 

[bobie]

Particle physicists would much prefer a system of units where the electron charge has a numeric value of one.

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?

 

[D H]

But you already consider the electron the unit charge, without realizing it, because you are used to think in Coulombs.

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, k_B, 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.

 

[bobie]

First off, there's nothing in either Coulomb's law or in Maxwell's equations that assume quantized matter. .

I was not referring to the law but to this unit:

...elementary positive charge. This charge has a measured value of approximately 1.602176565(35)×10−19 coulombs.[

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

 

[Imabuleva]

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

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.

 

[Dale]

Does the same apply to Coulomb constant?has it got the same dimensions?

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.

 

[phyzguy]

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.

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.

 

[bobie]

No, the Coulomb constant has different units.
... in Gaussian units k is dimensionless and equal to 1.

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 E_c between two elementary charges at 1 cm distance? If I am not wrong it is roughly equal to
C/α2π => E_c=1.439958 x 10^-7 eV, is this precise enough?

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

 

[Dale]

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?

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

- what is the precise value of electrostatic force E_c between two elementary charges at 1 cm distance? If I am not wrong it is roughly equal to
C/α2π => E_c=1.439958 x 10^-7 eV, is this precise enough?

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:
$$F=\frac{Q_1 Q_2}{r^2}=\frac{(4.8\;10^{-10}\;statC)^2}{(1\;cm)^2}=2.3\;10^{-19}\; Dyne$$

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

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

 

[bobie]

I don't know how you came up with this, but eV is a unit of energy, not force. ]

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 *E_cm 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?

 

[phyzguy]

Can you find the equivalent of dynes in eV?

Bobie,

You seem to have some conceptual confusion on the meaning of units. For each physical quantity, there is a corresponding unit. Dynes is a unit of force, and eV is a unit of energy. There is no equivalence between these two different things. It's like asking "How many apples correspond to one bicycle?"

 

[bobie]

Dynes is a unit of force, and eV is a unit of energy. There is no equivalence between these two different things.

I surely can't express myself properly:
The force of the proton gives the electron the energy to stay in the orbit and this energy must be equal to v^2/r , correct? that energy that pulls the electron is in eV, correct?

 

[Nugatory]

I surely can't express myself properly:
The force of the proton gives the electron the energy to stay in the orbit and this energy must be equal to v^2/r , correct? that energy that pulls the electron is in eV, correct?

No. $v^2/r$ is an acceleration.

 

[phyzguy]

I surely can't express myself properly:
The force of the proton gives the electron the energy to stay in the orbit and this energy must be equal to v^2/r , correct? that energy that pulls the electron is in eV, correct?

Electrons don't really orbit, so let's take a better example of the Earth orbiting the sun. There is a certain force F = G*Msun*Mearth/Rsun-earth^2, which keeps the Earth in it's orbit. The Earth also has a certain energy, which would be the sum of its potential and kinetic energy, so you could say there is a correspondence between the force on the Earth and its energy. But this correspondence only applies to that specific case. The correspondence would be different for Mars, for example. Does this help?

 

[bobie]

No. $v^2/r$ is an acceleration.

I studied sime time ago that to keep something in orbit
F_cm/r² = v^2/r
am I wrong?
in H₁(Bohr's) radius = 0.529177*10^-8
in order to calculate v^2/r we must divide the electrostatic force at 1cm by r²
If this is correct how do we express that force?

 

[Imabuleva]

You can use Newton's second law for columb attraction, an assumption of quantized angular momentum, and a classical energy balance of kinetic and potential energy to derive the Bohr radius, namely:

Newton's second law:
ƩF = m*v² / r = k*q² / r^2

Quantized angular momentum:
m*v*r = n*h / 2*∏

Energy balance:
0.5*m*v² - k*q*q / r = E

Solve for r, v, and E in the case of atomic Hydrogen and you get the Bohr radius, velocity, and energy of the electron. Electrons don't really orbit in any classical sense, though. That's why you have to assume quantized angular momentum.

 

[Dale]

bobie, you need to wait a while before you will be ready for quantum mechanics. And I recommend against these semi classical models.

 

[bobie]

You can use Newton's second law
ƩF = m*v² / r = k*q² / r^2
Solve for r, v, and E in the case of atomic Hydrogen .

I know the radius_
m(9.1*10^-28 g) * v²(2.18*10⁸² c/s = 4.786*10¹⁶⁾ / r (0.529*10^-8 c) =
k*q^{2/ r2 =} 8.4 x 10^-4 (??) = F
this is the force at distance r, correct?
multiplying by r²: m*v²*r = F_cm

8.23*10^-3* r² = 2.3*10^-19 (??)

this should be the electrostatic force at 1cm, correct? but what is the unit? the dyne or the the dyne*cm the erg, the work it does at 1 cm?
whatever the unit, if we convert it to Hz do we get F_cm = 3.4818*10⁷ = v/2π?
could we express the same relation in dynes?

 

[bobie]

bobie, you need to wait a while before you will be ready for quantum mechanics. And I recommend against these semi classical models.

I suppose gravitation can do without QM, can you help me calculate the earth-moon attraction?

GM = m*v²*r
v²(1022² m/s = 1044484) * r (3.844*10⁸ m) = 4.015*10¹²

G(6.674*10^-11) * M(5.9722*10²⁴ Kg) = 3.986*10¹²
This is the force of attraction of the Earth at distance 1 m, correct?
I took the data from wiki but the results do not match

 

[Dale]

I suppose gravitation can do without QM, can you help me calculate the earth-moon attraction?

GM = m*v²*r
v²(1022² m/s = 1044484) * r (3.844*10⁸ m) = 4.015*10¹²

G(6.674*10^-11) * M(5.9722*10²⁴ Kg) = 3.986*10¹²
This is the force of attraction of the Earth at distance 1 m, correct?
I took the data from wiki but the results do not match

An attraction is a force, so the correct formula is:
$$f=G \frac{M_1 M_2}{r^2}$$

Putting those in I get f = 2.05E20 N