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

[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 $GM=G=4\pi^2$.

 

[Imabuleva]

I'm beginning to think Bobie is trolling...

 

[bobie]

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 $GM=G=4\pi^2$.

Thanks, Nugatory,
one more question, if imabuleva allows me,
I have read that escape velocity is:
$v=\sqrt\frac{2GM}{r}$. , I suppose this is tangential escape,
what is the formula if a rocket tries to escape radially?

 

[TumblingDice]

I suppose this is tangential escape,
what is the formula if a rocket tries to escape radially?

Direction is not a factor. Tangential to radial doesn't matter as long as the path doesn't intersect the object it's escaping from.

 

[Nugatory]

$v=\sqrt\frac{2GM}{r}$. , I suppose this is tangential escape,
what is the formula if a rocket tries to escape radially?

The same. If you look at the derivation of the formula (it's a good exercise to derive it yourself) it will be pretty clear why the direction doesn't matter.