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The gravitational constant seems to be very low precision. For example, in the Wikipedia article recent measurements are given as having the significands of 6.67 and 6.69, a difference of 2 parts in 1000. I don't understand why astronomical measurements cannot be used to gain a much more accurate value. The explanation in the Wikipedia, that the force is "weak" seems like a vague answer to me.

This imprecision is a problem for me because I would like to make a simulation model of the solar system based on gravitational attraction, but with a such an imprecise constant, I don't see how I can do this to any degree of useful accuracy.

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    $\begingroup$ Look at the formula for Newtonian gravity or the formula for the Kepler problem: you need the masses of the gravitating bodies. How are you going to measure the mass of the sun or one of the planets with sufficient precision? In practice we do it the other way round: we know $G$ and the orbital parameters and from those we determine the central mass. On Earth there are other metrological problems, of course: we can't shield the weak measurement of two small masses against the strong gravity of the planet. $\endgroup$ – CuriousOne May 7 '16 at 0:47
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    $\begingroup$ G is known to a much greater precision than 2 parts per 1000. google.co.uk/… and an erratum journals.aps.org/prl/pdf/10.1103/PhysRevLett.113.039901. In part measuring G accurately has to do with trying to find out if G varies with time although measurement of orbits of planet indicate that there has been little or no change. $\endgroup$ – Farcher May 7 '16 at 7:05
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It's true that if you know the masses of, e.g. two orbiting stars $M_1$ and $M_2$, their orbital period $T$, and the distance $d$ between them, then you know $G$. And we can measure $T$ pretty well and $d$ fairly well.

But how do you think we figure out the masses of the stars? We can't just count the amount of stuff in them; we have to infer the mass from how hard they pull on other objects. So we actually determine $M$ using the known value of $G$. Since we don't know stellar masses any other way, we can't flip the measurement around to get a better value for $G$.

You might think we could calculate stellar masses directly using what we know about fusion, but that doesn't work either: a star needs to exert enough outward pressure to cancel its weight, and that weight is proportional to $G$. In other words, $G$ is an input, so it can't be an output.

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For your purpose of building a solar system dynamic model, you only need to know the product G.M for celestial bodies, not G and M separately. The product G.M is known to much higher accuracy than G itself, indeed from astronomical observations.

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    $\begingroup$ This is a comment not an answer, but I did upvote it because it is useful information for me. $\endgroup$ – Ambrose Swasey Aug 22 '19 at 19:46
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The scope of the problem: A typical textbook on physics discusses a mass of 215 kg on the surface of the earth endowed with a charge identical in magnitude to an opposing charge at earth's center so that the gravitational attraction by the entire mass of the planet is offset. In his book Six Easy Pieces, Richard Feynman explains how two grains of sand thirty meters apart would attract each other with an electrical force of three million tons without a balancing of charges, if instead of likes repelling everything attracted everything else. Still another thought experiment might make the point more succinctly.

As stated by that same Nobel prize winner in that same book, the ratio of gravitational attraction relative to the electrical repulsion between two electrons is 1/[4.17 x 10^42] : 10^-42.6+... In a standard static electricity setup involving two small masses with equal and like charges hung by insulating threads, the force [F] of repulsion is calculated to be the Coulomb constant [Ke] multiplied by the square of the charges [Q]^2, all divided by the square of the distance [L] between them: [F] = [Ke]{[Q]^2}/{[L]^2} . If the initial distance is one centimeter and that force is to be reduced to the strength of the gravitational force between those two masses by altering only [L] ,

[L] cannot be made one meter :

1/{[10^2] x [10^2]} => [F] x [10^-4] ,

[L] cannot be made one kilometer :

1/{[10^5] x [10^5]} => [F] x [10^-10] ,

[L] cannot be made one hundred million kilometers = 2/3 of the distance to the sun = .66 astronomical units :

1/{[10^13] x [10^13]} => [F] x [10^-26] ,

[L] cannot be made one hundred million times as far as .66 A U = 66 million astronomical units :

1/{[10^21] x [10^21]} => [F] x [10^-42] ,

[L] must be at least twice that far, [L] must be at least two hundred million times as far as .66 A U:

1/{[10^21.3] x [10^21.3]} => [F] x [10^-42.6] .

If that commonly exhibited, hands on, observational experiment of the electrostatic force is to be altered for observation of a force comparable to that of the gravitational force by changing only their distance of separation, those two little masses on threads must be more than 132 million astronomical units apart.

While such an experiment is unlikely to take place, if it did, at least the experimenters would not be using relatively minute masses to emulate celestially observed gravitational interactions while immersed deep in the gravity well of one of those celestial bodies, trying to obtain as good a vacuum as possible, estimating outgassing properties of materials, and/or torsion factors of fibers, and/or temperature fluctuations, and/or tidal influences while they stand in a pair of painted footprints.

Finally, there is one more thing to keep in mind. With every new publication of CODATA values, it must be said again: there exists no recognized relationship between the Newtonian gravitational constant and any of the other constants. All the constants exist in the same universe and therefore have some kind(s) of relationship(s), but experiments involving forces 10^42 times as weak as tabletop setups will be conducted as if such relationships did not exist and cannot be affecting results.

595: Cutnell, J D, & Johnson, K w, PHYSICS, (Book), Fifth Edition, copyright 2001, John Wiley & Sons, 0471 32146-X p 529

597: Feynman, Richard P, SIX EASY PIECES (Book), copyright 1963, 1989, 1995, California Institue of Technology, ISBN-13 978-0-465-02392-9 , ISBN -10 0-465-02392-4 p 29 , p 110

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    $\begingroup$ Maybe it's just me, but I find the point of this answer quite hard to understand. I don't see what it brings to the table that the other one doesn't. $\endgroup$ – Javier Dec 9 '17 at 21:48
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    $\begingroup$ I don't understand what point this answer is trying to make. It's readability also suffers because the math isn't properly marked up using mathjax, it uses nonstandard notations (why the square brackets?), numbers are written without units, and the long series of numerical calculations could have been carried out as a single line of algebra. $\endgroup$ – user4552 Dec 9 '17 at 22:19

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