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

 

[Bandersnatch]

This is the force of attraction of the Earth at distance 1 m, correct?

No. Not only there's no force anywere in there, there's no distance of 1m either. You have plugged in the distance between Earth and Moon for R yourself, so wherefrom the idea that it's now 1m?You say you want to calculate Earth-Moon gravitational attraction. What you should be calculating then is a force. You could either use the Newton's force of gravity equation
F=GMm/R^2
or make use of the fact of Moon being on a low-eccentricity(nearly circular) orbit and calculate the centripetal force
F=mV^2/R
There's no use in comparing the two equations, since you're supplying all the values. You just plug in the numbers and get the force.

Writing an identity of GMm/R^2=mV^2/R is useful if you're missing some value and want to find it.
You can solve the above for V, R, M, or even G, if that's what you're missing. But you won't be calculating the force anymore.

What you wrote here:

GM = m*v²*r {This is not a valid identity, by the way. You left the m when it ought to have canceled out}

v²(1022² m/s = 1044484) * r (3.844*10⁸ m) = 4.015*10¹²{and here the units don't match. The unit of V^2 is (m/s)^2, not m/s}

G(6.674*10^-11) * M(5.9722*10²⁴ Kg) = 3.986*10¹²{here you wrote the unit for mass but not for G. But then again, just as previously, you don't follow with calculating units for the result on the right-hand side of the equation.}

is the calculation of the value of GM, that you then check by plugging in the numbers for these well-known constants. What is GM? Beats me. It's got no particular physical meaning, even though it's numerical value happens to be equal to GMm/R^2 where m and R are chosen to both be equal to 1(which, by the way, means it would require a black hole for the mass of Earth to be contained within 1m sphere to produce such a force). They are not the same, though, as can be shown by using unit calculus. Even though both R and m are 1, they still have their units of m and kg, so the resultant units is the Newton, as it should be for the Force of gravity, while GM has units of [m^3/s^2].
So, once again, you're not calculating any forces there.
A word of advice for solving any equations - first pick the appropriate equation. You need to understand what it means and when it's applicable. Throwing equations around without any purpose or plan is not going to work.
Next, identify what is the one thing you want to find, and isolate it one side. You know what you're looking for, so it should be easy. If it's not there, or is cancelling out, then you've got the wrong equation.
Then plug in the units only, for the variables and constants on the other side to find out whether the resultant unit matches the unit for the thing you're solving the equation for. If it doesn't(e.g., a force is not in Newtons, mass in not in kilograms), it means you borked the arithmetics, or have isolated the wrong variable for some reason.
Only then plug in the actual numbers to get the final answer. If you happen to use two or more different methods(i.e., equations) to calculate the same thing, and the results slightly don't match, then you may want to examine the assumptions that came with writing the equations, to decide which one is closer to the true value.

 

[D H]

What is GM? Beats me. It's got no particular physical meaning ...

I know you wrote that in response to to another post, so I'm quoting out of context. But I can't let that comment slide!

In the context of solar systems (and not just ours), it's GM that has huge particular physical meaning. It's G and M that are the things that have no particular physical meaning.

The Sun's standard gravitational parameter, GM_⊙, or just μ_⊙ for short, is observable via the sizes and periods of the orbits of the planets about the Sun. A planet's standard gravitational parameter μ is observable via the sizes and periods of the orbits of the planet's moons. The precision and accuracy with which these standard gravitational parameters are measured is quite phenomenal. For the Sun it's about one part in 10¹¹ (The wikipedia article on "Standard gravitational parameter" is a bit dated; it uses a four year old value with a ten-fold increase in uncertainty.)

Either way, one part in 10¹¹ or 10¹⁰ is a whole lot better than the phenomenally lousy one part in 10⁴ one would get by using G and M separately. The problem with using G and M separately is that G is one of the least precise of all the physical constants. Even though G*M might be known to many places, that G has four place accuracy means that M also has four place accuracy.

Solar system astronomers have yet another trick up their sleeves. Many of them don't use SI units. The use the solar mass as the unit of mass, the AU as the unit of length, and the sidereal year as the unit of time. In this system of units, G has a numeric value of 4*pi², exactly. Instead of a poorly-known constant, it is a defined constant. All of the uncertainty in G*M is now attributed directly to central body mass.

 

[Bandersnatch]

In the context of solar systems (and not just ours), it's GM that has huge particular physical meaning

Cheers, D H, I wasn't aware of that(obviously).

 

[bobie]

f=G \frac{M*m}{r^2} =\frac{m* v^2 }{r}

is this law valid?

Edit: added m

 

[D H]

f=G \frac{M_1 M_2}{r^2}
f = v² / r
multiply by r² =>
f=G \frac{M*m[oon]*r^2}{r^2} =\frac{v^2 * r^2}{r} = m * v^2 * r
cancel m =>
f = GM = v_m²*r ≈ 4*10¹²

What is wrong with that?

What's wrong with that is that it's nonsense. Your units don't match. GM₁M₂/r² has units of force, or mass*length/time². v²/r has units of acceleration, or length/time². GM has units of length³/time². These are three incommensurable quantities. You cannot add, subtract, compare, or equate incommensurable quantities.

 

[bobie]

Your units don't match. .

see post #39, please , I posted in reply but is shown before yours, sorry, I must have canceled that post , is it possible to restore it , or shall re -write it?

 

[Bandersnatch]

f=G \frac{M*m}{r^2} =\frac{v^2 }{r}

is this law valid?

Check the units!

 

[Doc Al]

f=G \frac{M*m}{r^2} =\frac{v^2 }{r}

is this law valid?

The right hand side term is an acceleration, not a force. You left out the mass!

 

[bobie]

The right hand side term is an acceleration, not a force. You left out the mass!

I've done it again!

Writing an identity of GMm/R^2=mV^2/R is useful if you're missing some value and want to find it. .

I suppose that is what I was doing, finding the value of GM, the standard parameter of the Earth equals ( μ=v^2/r) the squared speed of a satellite(moon) divided by its distance. Is that correct?

 

[bobie]

No. Not only there's no force anywere in there, there's no distance of 1m either. You have plugged in the distance between Earth and Moon for R yourself, so wherefrom the idea that it's now 1m?
.

From here

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

 

[Nugatory]

From here

That was saying that if you know the force, the masses, and the distance you can calculate the value of $G$ as the unknown in $F=\frac{GMm}{r^2}$. But when you're talking about the force between Earth and the moon you know the masses, $G$, and the distance; your unknown is $F$.

 

[D H]

From here

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

The quoted post was wrong, and you were told that in post #7. G is not a force. It can't be; it doesn't have the right units.

A better way to say what phyzguy wrote is that the numerical value of G in some system of units is the numerical magnitude of the force (expressed in that system of units) between two masses of one unit mass each separated by a distance of one unit distance. With SI units, the numerical value of G in SI units is the numerical magnitude of the force expressed in Newtons between two 1 kg masses separated by by a distance of one meter.

 

[bobie]

That was saying that if you know the force, the masses, and the distance you can calculate the value of $G$ as the unknown in $F=\frac{GMm}{r^2}$. But when you're talking about the force between Earth and the moon you know the masses, $G$, and the distance; your unknown is $F$.

...or if I want to know the mass of a body I can get it from the distance and v of its satellite.

 

[bobie]

Many of them don't use SI units. The use the solar mass as the unit of mass, the AU as the unit of length, and the sidereal year as the unit of time. In this system of units, G has a numeric value of 4*pi², exactly.

I am trying to digest all the valuable information you gave me: I was trying to work this out, but while on the right side of the equation GM = v²r that value is obvious (v=2π AU/year, v²=4π²), I couldn't figure out how you can generate π on the left side.
It can only be a sort of approximation, is that correct? can you show me your result?

Thank you for your help

 

[Nugatory]

I was trying to work this out, but while on the right side of the equation GM = v²r that value is obvious (v=2π AU/year, v²=4π²), I couldn't figure out how you can generate π on the left side.

There's not much to it. You have $GM=4\pi^2$, and we've chosen units in which $M=1$, so $GM=G$ and we're left with $GM=G=4\pi^2$.