Talk:Gravitational constant/Archive 2

Easy way to calculate big G without local interference
It's exceedingly obvious to me how to measure big G precisely and incontrovertibly. Send two space ships of equal and known mass into space. One of them will be placed within the pull of earth, but lined up directly under the moon. The other will be at the opposite side of earth farthest from the moon. Instruments on board the ships will all be fastened down so that the momentum of an object banging around inside the ship cannot distort the readings. There should be nothing in the ships that is not bolted firmly to the ship. When the ships are both in place, they will both be equidistant from the center of earth (not the height above land or sea level, because those both vary). The ships will then have all motion stopped such that they stay exactly in-line with the line that would join the exact center of the earth and the exact center of the moon. They are now allowed to drop to earth in complete freefall while instruments on board the ships measure their exact locations above the center of earth, using GPS and other measuring systems, and they would also measure the precise time that the space ships are at every altitude. It should be noted that the best design of these space ships would be a perfect sphere to minimize variations and differences due to wind resistance and the center of gravity in the lower atmosphere. Measuring Instruments should be in the dead center of the space crafts. There should be no humidity or rain or anything else on the day of the experiment, that is in line with the space ships' descent paths that would exert undue air resistance. Since a sphere is unwieldy to control, a secondary craft would need to carry the spheres to space before dropping them. This would help with knowing exactly how much the measuring crafts weighed since burned fuel changes the payload weight. The spheres would be reweighed on earth after a water or other safe landing, for added weight accuracy redundancy. After the data from the spheres is harvested, extremely accurate calculations can be made because of all the known variables and the large size of the objects in the experiment (space craft, the earth, the moon). Wind resistance would be minimized while the crafts are in space. The experiment will also be free of local distortions due to the gravity exerted by nearby mountains, buildings, and other objects, especially at the highest altitudes away from the earth. The experiment should ideally be repeated a total of at least 3 times to ensure accuracy and to account for the minor difference in gravity formed by bulging oceans and really tall mountain ranges. Varying conditions could be checked. Having the two space craft at opposite ends from the moon will allow for calculations based on the distance and mass of the earth and the moon, giving even greater accuracy through redundancy and two separate calculations. No measurement of big G can be completely perfect, but this is as close as we can come. The quantum/atomic measurement system used by other researchers may or may not hold true for large objects obeying "normal" laws of gravity, so the experiment above would be incontrovertible in this regard. — Preceding unsigned comment added by 70.193.36.108 (talk) 08:11, 19 May 2016 (UTC)

To the poster below: The two quantities are related by g=(GMe)÷(r^2e) what are you missing to do the calculation?

A highly precise estimate of the mass of the moon and every object in the solar system is possible by observing its size and chemical composition using astronomical spectrometry. Also, numerous spacecraft have used these masses in calculations that have helped them to fly to distant places without error. The precision of decimal places in the experiment above would far exceed the atom interferometry experiment because the mass of the objects can be measured more accurately and to a higher precision whereas atomic masses are inferred. Regardless, the atom interferometry would of course be subject to your same GM issue as the space experiment, since you would be measuring the gravity due to earth.

Nevertheless, the attraction between multiple bodies of known mass is the same concept that the Cavendish torsion balance uses, only the spacecraft idea above gets rid of most of the errors due to various frictional forces and local gravity phenomenons. — Preceding unsigned comment added by 70.199.243.40 (talk) 13:25, 19 May 2016 (UTC)


 * See WP:Discussion pages please, they exist for discussion of the article, not for general discussion of the topic. Your approach does not work - you don't measure G with it, you measure the product GM where M is the mass of earth (or moon, depending on what exactly you do). This product is known extremely well already, see the article for details. --mfb (talk) 11:17, 19 May 2016 (UTC)

Oh, and if the spacecraft experiment was performed it would give a massive improvement to the article (and the science community) since the gravitational constant could be univocally defined. That's the whole point of the article: defining the so-called constant. — Preceding unsigned comment added by 70.199.243.40 (talk) 13:34, 19 May 2016 (UTC)


 * No, it would not, see above. Also, we have literally thousands of satellites where the motion gets tracked. One of the reasons why GM is known so precisely. But we don't know the mass of Earth well enough to use it to measure G. --mfb (talk) 19:23, 19 May 2016 (UTC)

Moved from html comment
This was commented out in the article, I am moving it here for future reference:
 * The following unpublished 2011 preprint:
 * Zhiping Li, Xin Li, Quantum resonance scheme to determine the gravitational constant G (Submitted on 29 Nov 2011, revised 12 Nov 2013 ) claims a value of "6.67221937(40)e-11" (seven digits' accuracy).
 * It was rejected due to its wrong use of the Planck Constant, resulting in wrong dimensional analysis, and should not be included here without proper (published) review.

--dab (𒁳) 09:40, 9 June 2018 (UTC)

Quantum-relativistic formulation of gravitational constant G
A useful interpretation was added. Just used mass of proton instead of mass of electron. Алры (talk) 21:20, 4 April 2019 (UTC)
 * That is not useful, it is just completely wrong. It doesn't even have matching units. --mfb (talk) 22:32, 4 April 2019 (UTC)

Please, show me mistake if you can.Алры (talk) 23:38, 4 April 2019 (UTC)


 * The only source for your addition was a 1922 paper in a physics journal . On Wikipedia, primary sources are not enough. they must be backed up by secondary sources; textbooks or survey articles, see WP:PSTS.  So your addition was inadequately sourced.  Even if it wasn't the paper merely discussed numerical coincidences.  Numerological derivations of the gravitational constant have been discussed before and rejected, see  above.  Before adding controversial content to the article you need to get consensus on this page, see WP:CONSENSUS.  Your repeated addition of the same content after it was opposed by multiple editors is WP:EDIT WARRING and if you continue you will be blocked from editing. --ChetvornoTALK 00:52, 5 April 2019 (UTC)


 * Physical formulas must work in all unit systems. If a kilogram would have been defined to have 1% more mass Newton's law would still work, but the formulas you are trying to put in would not. The laws of physics cannot depend on our completely arbitrary choice of units. Statements like "the diameter of [whatever] expressed in meter is pi^5 times the mass of X expressed in kg" (that's the type of statement you try to add here) cannot be anything than a random coincidence.
 * To make it worse: You didn't even find such a statement. You found something that is off by half a percent and then added some nonsense about kinetic energy (with a completely arbitrary value) on top. --mfb (talk) 11:40, 5 April 2019 (UTC)


 * Besides the fringe numerology and problems with physical units, I skimmed through the paper referenced and didn't see any equations even similar to the one given by Алры. [Edit: nevermind. I see something similar, equation 11. But the expression by Алры still had wrong units.] MaxwellMolecule (talk) 13:48, 5 April 2019 (UTC)

I repeated addition of the same content because it was deleted without demonstrating where I was wrong as Mfb stated. It was not shown yet. Now Mfb made another wrong statement that my addition is a random coincidence. It is obvious for anyone who knows physics that 18th degree of fine structure constant does not allow such a speculation. Even more it explains the known difficulties with getting accuracy of G. Yet another wrong argument is about our completely arbitrary choice of units. Actually my addition was made to avoid such a problem. The attached mass of the proton is natural unit unlike G.

Chetvorno stated that my addition was inadequately sourced. It is not true. I gave primary source to show that this idea appeared on American soil just after 6 years after the fine structure constant was introduced. Now there are a lot of the contemporary secondary sources. Chetvorno also stated that I added a controversial content to the article. Please, be specific.

I don't see arguments against my addition.Алры (talk) 14:55, 5 April 2019 (UTC)


 * What are the contemporary secondary sources? MaxwellMolecule (talk) 15:54, 5 April 2019 (UTC)


 * You see the arguments, you just don't want to accept them. The content you want to add is so ridiculous that I wonder if this is deliberate trolling instead of just severe misconceptions about physics from your side. --mfb (talk) 00:15, 6 April 2019 (UTC)

There are the contemporary secondary sources in every Wikipedia. For example, let see article "Feinstrukturkonstante" in German Wikipedia in paragraph "Vergleich der Grundkräfte der Physik". There's formula
 * $$\alpha_G\ = \frac{G \, m_p^2}{\hbar \, c}\ \approx\ 5{,}9 \cdot 10^{-39},$$

where $$\alpha_G$$ is equal to $$\sqrt{3}\alpha^{18}$$ according to idea of American physicist A. C. Lunn.

Yet another example there is in Russian Wikipedia in the same article "Постоянная тонкой структуры" in paragraph "Ранние попытки". There's formula of A. C. Lunn:
 * $$\frac{G m^2}{e^2} = \frac{\alpha^{17}}{2048 \pi^6}.$$

Unfortunately, Mfb continues to show the invisible arguments like "just severe misconceptions". It looks like very personal.Алры (talk) 14:03, 6 April 2019 (UTC)


 * Don't respond to him. He's just trying to provoke you.  If you engage with him, you are just encouraging him by giving him what he wants: attention. He's harmless. --ChetvornoTALK 15:34, 6 April 2019 (UTC)

According to Chetvorno I was right from the beginning. There are no arguments against my addition yet. Please, restore it. Or, please, explain specifically why you don't want to see it. It would be interesting to know.

Maybe, you just don't know that the stars are made from hot hydrogen? — Preceding unsigned comment added by Алры (talk • contribs) 17:00, 6 April 2019 (UTC)


 * OK. Fringe-physics aside, lack of reliable sources is a clear-cut reason to close the discussion. Other Wikipedia articles are obviously useless as sources.MaxwellMolecule (talk) 17:12, 6 April 2019 (UTC)
 * Right, also WP:CONSENSUS is against including it. --ChetvornoTALK 17:23, 6 April 2019 (UTC)

Gravitational Constant Puzzle
Should the article assert the fundamentalism of the Gravitational constant or discuss the constant as a combination which can be assembled as a puzzle from other physical and math constants? For example:
 * $$G = ({\frac{2\pi}{137^2}})^{11} \cdot {\frac{\hbar \cdot c}{m^2}} = 6.67409850(8) \cdot 10^{-11} \ m^3 \cdot kg^{-1} \cdot s^{-2}

$$ where $$ \hbar $$ - reduced Planck constant; $$ c $$ - speed of light; $$ m = (m_n + m_p + m_e) / 2 $$ - average mass of neutron and proton + electron pair; Mikhail Vlasov Korablino (talk) 20:38, 19 February 2018 (UTC)
 * This has been discussed; see Talk:Gravitational constant/Archive 1 above. The general feeling in physics is that this is a numerical coincidence and doesn't have any physical significance, so it should not be included in the article.  --ChetvornoTALK 21:10, 19 February 2018 (UTC)


 * There is no puzzle, just bad numerology. If you play with enough constants long enough you'll always get some combinations that come close. If you replace 137 by the actual fine-structure constant the result of this expression is $$6.6356\cdot 10^{-11} \ m^3 \cdot kg^{-1} \cdot s^{-2}$$, inconsistent with G. Worry not, use tritium, because why not! 6.673906, now it is consistent with the measured value. You see how pointless this exercise is? --mfb (talk) 21:14, 19 February 2018 (UTC)

Hi Chetvorno. Yes, I agree with attributing this to numerology. My question was - should article discuss next topic: is Gravitational constant a "prime" or is it a formation of other constants. Hi Mfb. Agree with you as well except one thing. Fine Structure constant reciprocal itself is a composition of 137 and $$ \pi $$. Korablino (talk) 23:52, 19 February 2018 (UTC)


 * The inverse of the fine-structure constant is about 137.0359991. There is nothing special about 137. If something works out with 137 but not the fine structure constant, it is obviously nonsense. --mfb (talk) 10:14, 20 February 2018 (UTC)

Yes. Instead of $$ 137^2 $$ it is possible to use $$ \alpha^{-2}-\pi^2 $$. Korablino (talk) 19:05, 20 February 2018 (UTC)
 * Even more numerology. This discussion has zero chance to lead to any improvements of the article (=the purpose of talk pages), so I'll stop it now. --mfb (talk) 07:16, 21 February 2018 (UTC)

Well, in the spirit of notability, not truth, such speculation should be reported if made by someone with a reputation to lose. It may still be complete coincidence, but it's worth reporting if Dirac does it. Perhaps just getting it published in a reputable journal would be enough, you don't need Dirac-levels of credibility.

The above seems quite random (as was pointed out, alpha != 1/137), but I find this quite funny,
 * "(2\pi G)^-1 = \hbar e^(2\hbar / 3)" in "electron units"

or this,
 * G = k e^2/(20 m_p m_e)·2^{-128} vixra.org/pdf/1211.0047v6.pdf

These are both preprints, idk if either was published, but I am sure this type of paper is out there, and if based on published sources, this page certainly has room for a "numerology" section. Sure, you can always come up with this type of stuff, so it needs to pass some minimal threshold of peer review before it can even be considered amusing, but sooner or later one of these may just turn out to be meaningful. --dab (𒁳) 10:09, 9 June 2018 (UTC)

The general consensus is, for the following reason, that G is somehow derivative rather than fundamental, so that there is indeed some kind of composite formula (so that all candidate formulas have to be taken seriously, if they fit). The reason is this: in higher dimensional theories, including Kaluza-Klein, supergravity and string theory, the coupling coefficient will be that for the higher-dimensional version of Einstein's equations, while the 4-dimensional version of Einstein's equations will fall out of this (by a process of dimensional reduction) from the higher-dimensional field equations. The coupling coefficient for the lower-dimensional Einstein equations will be a composite formed ultimately of the higher dimensional coupling coefficient and possibly integral expressions involving the extra dimensions (e.g. the arclength of the circle in 5D Kaluza-Klein, if the extra dimension is wrapped in a circle). Note that the units don't even match. For the Einstein equations, the coupling coefficient in n-dimensions, √(Aⁿ⁻²)T/ML², where L denotes length, T time, M mass and A = [g_{μν} dx^μ dx^ν], the dimensions of the line element, which is normally taken as A = L². The coupling κ comes out of the Einstein-Hilbert action S = ∫ (1/2κ) R √|g| dⁿx, which has dimensions [S] = ML²/T. In n-dimensions [√|g| dⁿx] = √(Aⁿ), [R] = 1/A, which implies [κ] = √(Aⁿ⁻²)T/ML² = TLⁿ⁻⁴/M. For 4D, that's [κ₄] = T/M, while for (say) 5D it would be [κ₅] = TL/M. The Newton coefficient G comes out of κ₄/8π up to powers of light speed, c; but - on these dimensional grounds alone - cannot be extracted from κ₅ (nor from the coupling for any other higher-dimensional theory) directly in this way. Instead, it has to emerge as a composite of some kind. — Preceding unsigned comment added by 2603:6000:AA40:1EF2:222:69FF:FE4C:408B (talk) 18:58, 12 August 2020 (UTC)
 * Extra dimensions are just speculation for now as well. But even if they exist they don't imply a link between the gravitational constant and other constants. The macroscopic gravitational constant might be the result of a fundamental microscopic constant and calculations - but that microscopic gravitational constant can be a free parameter again. --mfb (talk) 20:59, 12 August 2020 (UTC)

The current revision is an exercise in weasel wording, "There is a good approximate 'Theory of everything' relation", "The relation can be justified for example" -- oh really, it "can" be justified? This isn't acceptable, the only thing we are interested in here is: who has proposed this in what year and in what publication, and how has the suggestion been received since? It doesn't need to be part of the mainstream consensus, but the further away it is from the mainstream, the more the text needs to focus on qualifying exactly who came up with it, when and where, and how others have reacted to it. If no experts bothered reacting to it, i.e. not even dismissively, it has no business being in the article. --dab (𒁳) 07:23, 16 April 2021 (UTC)

Variable "$h$"
In


 * $$ P^2=\frac{3\pi}{G}\frac{V}{M}\approx 10.896 \, \mathrm{ h^2 {\cdot} g {\cdot} cm^{-3} \,}\frac{V}{M}\,,$$

what is the "$h$" that is squared? — Q uantling (talk &#124; contribs) 18:34, 27 March 2023 (UTC)

In imperial units
What is the gravitational constant (6.67408 * 10^-11) converted to imperial units (in ft and lb instead of m and kg)? 212.186.0.174 (talk) 12:58, 19 May 2019 (UTC)


 * Alright, I figured out that the imperial gravitational constant is 1.0691 * 10^-9 ft^3 lb^-1 s^-2. Maybe one should add it to the page. 212.186.0.174 (talk) 16:43, 1 June 2019 (UTC)
 * No one with a sane mind uses imperial units for scientific calculations. --mfb (talk) 23:13, 1 June 2019 (UTC)


 * And I'm obviously so insane that I have to ask you "why not?" 212.186.0.174 (talk) 04:51, 2 June 2019 (UTC)
 * Science is done basically exclusively in metric because it is much easier to convert units there and because 95% of the world population uses metric everywhere. --mfb (talk) 05:06, 2 June 2019 (UTC)
 * With pounds there are two competing definitions. One is that it is a unit of mass and one is that a unit of force, and which is more common has shifted over the years (IMHO).  You'll get a different value for $G$ in imperial units depending upon which definition you are using.  — Q uantling (talk &#124; contribs) 13:57, 28 March 2023 (UTC)
 * With force pounds, $G = 3.4397 × 10−8 ft4 lbf−1 s−4$. — Q uantling (talk &#124; contribs) 16:30, 29 March 2023 (UTC)
 * Metric is the standard. Wcmead3 (talk) 04:21, 25 April 2023 (UTC)

Sign of the force
In a comment, writes "the text defines F as the attractive force. No negative sign is needed for this case where the force is defined to be attractive". I agree that no negative sign is needed when the force is defined to be attractive. However, my experience (and maybe yours too?) is that Gmm/r^2 gravitational force is almost always written with a negative sign. I don't think that we are doing the reader any favors by going counter to this convention. I would rather go with convention by giving the minus sign in the equation, and adjust the text so that it works with the equation. Might something along those lines work? — Q uantling (talk &#124; contribs) 21:46, 27 March 2023 (UTC)


 * As far as I recall, most books don't write the minus sign. See, for instance MacDougal. And Feynman, Ohanian and Hewitt don't do it either in their basic books. - DVdm (talk) 22:12, 27 March 2023 (UTC)


 * Ohanian & Ruffini Gravitation and Spacetime has the minus sign. Halliday and Resnick does not.  — Q uantling (talk &#124; contribs) 22:25, 27 March 2023 (UTC)
 * I have Ohanian's Physics, vol 1, Second Edition here before me, chapter 9, section 9.1, eq (1), page 212. No minus sign. Just the (standard) non-vectorial form. - DVdm (talk) 22:35, 27 March 2023 (UTC)
 * Textbooks are what matter when it comes to Wikipedia, but I can't help venting anyway. Gravitational potential energy is negative, $$-\frac{GMm}{r}$$; we can at least agree on that, right?  And force is the negative gradient of the potential.  And $$-\frac{d}{dr} \left(-\frac{GMm}{r}\right) = -\frac{GMm}{r^2}$$, with the minus sign.  Sure, calling the force as positive but in the opposite direction works, but it will confuse the student who is learning the canonical ways of dealing with potential energy, conserved forces, Hamiltonians, etc.  — Q uantling (talk &#124; contribs) 01:42, 28 March 2023 (UTC)
 * My view is that defining the force so that it is positive in the direction of r increasing is more mathematically logical. The equation then requires a minus sign. It is, however, a more difficult concept for readers who are fairly new to the concepts of the article. I can see that there is an argument in favour of defining the force as attractive, so as to avoid having a minus sign in the equation, given that the article is about the gravitational constant rather than the Newtonian law of gravitation.
 * However, I'd like to say that the article as a whole is quite messy. TowardsTheLight (talk) 12:49, 28 March 2023 (UTC)
 * Apologies for reverting the text, but the edit changed the meaning in a way that was incorrect. The text had been written in a way that meant that the force F was positive, so the formula does not have a minus sign. TowardsTheLight (talk) 12:41, 28 March 2023 (UTC)
 * What if we insert "magnitude" into the text? Because the formula doesn't give the force vector, that might be appropriate even without the debate about the sign.  Using "magnitude" justifies the use of the positive sign to people like me who would otherwise be supporting the negative sign.
 * According to Newton's law of universal gravitation, the magnitude of the attractive force ($F$) between two point-like bodies is directly proportional to the product of their masses, $m_{1}$ and $m_{2}$, and inversely proportional to the square of the distance, $r$, between their centers of mass:
 * Thoughts? — Q uantling (talk &#124; contribs) 13:05, 28 March 2023 (UTC)
 * I would be happy to have either a reference to the magnitude of the force, or the current wording. TowardsTheLight (talk) 15:35, 28 March 2023 (UTC)
 * Perhaps "strength" would be more accessible than "magnitude"; when talking about force, they mean pretty much the same thing.
 * According to Newton's law of universal gravitation, the strength of the attractive force ($F$) between two point-like bodies is directly proportional to the product of their masses, $m_{1}$ and $m_{2}$, and inversely proportional to the square of the distance, $r$, between their centers of mass:
 * Is that better?  are you using "reference" in a strict sense, meaning that you'd want the modification to be within  markup, or would the change I am proposing meet with sufficient approval?  — Q uantling (talk &#124; contribs) 16:18, 29 March 2023 (UTC)
 * That would be fine. Personally, I would prefer magnitude to strength, because magnitude is something that novices would understand as a word, even if they do not understand its mathematical meaning, while people with a proper understanding of the subject would understand magnitude in its full meaning. But either strength or magnitude would be fine. And I meant "reference" in a general sense, not as a source of information. TowardsTheLight (talk) 19:24, 29 March 2023 (UTC)
 * I edited the article and went with "magnitude". If you revert, please say why here.  Thank you — Q uantling (talk &#124; contribs) 19:17, 30 March 2023 (UTC)
 * That's absolutely fine. The statement is correct in all all respects. TowardsTheLight (talk) 19:39, 1 April 2023 (UTC)
 * "Magnitude" is fine. Wcmead3 (talk) Wcmead3 (talk) 04:12, 25 April 2023 (UTC)