# Talk:Acceleration due to gravity Main Article
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 Definition:  The acceleration of a ponderable object, which is near the surface of the Earth, due to the Earth's gravitational force. [d] [e]
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## Can we simplify it a bit?

This part of the article seems to repeat essentially the same equation twice:

" ... is given by:

${\vec {g}}=-G{\frac {M}{r^{2}}}{\frac {\vec {r}}{r}}$ The magnitude of the acceleration is $g=GM/r^{2}$ , with SI units of meters per second squared.

Here G is the universal gravitational constant, G = 6.67428×10−11 Nm2/kg2, ${\vec {r}}$ is the position of the test object in the field relative to the centre of mass M, and r is the magnitude (length) of ${\vec {r}}$ ."

I realize that the equation ${\vec {g}}=-G{\frac {M}{r^{2}}}{\frac {\vec {r}}{r}}$ includes all of the conventions used by physicists and mathematicians, but it simply confuses those of us who are not physicists and mathematicians. Can we not simplify it thus:

" ... is given by:

$g=GM/r^{2}$ , with SI units of meters per second squared.

G is the universal gravitational constant = 6.67428×10−11 Nm2/kg2, and r is the distance between the test object and the centre of mass M."

Those of us who are not physicists or mathematicians would find it much easier to understand if it were simplified as proposed. - Milton Beychok 03:18, 26 February 2008 (CST)

## Value of g

As far as I remember g varies by a percent or so over the earth. How can we then give so many decimals? Is there some sort of standard value?--Paul Wormer 03:30, 26 February 2008 (CST)

That is the value agreed upon by the Conférence Générale des Poids et Mesures, CGPM in 1901 as referenced in the article. I assume that it is a sea level value so it is not affected by the altitude of any locations. - Milton Beychok 04:08, 26 February 2008 (CST)
I looked around on the internet and I get the impression that gn = 9.80656 m/s2 is defined by the CGPM as the standard acceleration (a fictituous value) and that g is the local acceleration (which is the real physical value that varies by almost a percent over the globe). I read the first few sentences of the article slightly different.--Paul Wormer 09:19, 26 February 2008 (CST)
Paul, I have no objection to your re-write of the first few sentences. I just want to say that here we have a good example of the classical difference between a physcist and an engineer. Most engineers simply use g = 9.807 or even 9.8 when doing their fluid dynamics calculations and, in 99% of an engineer's work, a possible 1% error is totally insignificant. In fact, we would be very happy if the rest of our work was as good as that.
Changing the subject, I am moving the "see also" link to the "Related articles" subpage. It is my understanding that is where such links belong. - Milton Beychok 12:39, 26 February 2008 (CST)
I agree completely with your remark about accuracy, but it was not I who put all decimals of g into the article :-) --Paul Wormer 02:09, 27 February 2008 (CST)

## Notation

I'm not happy with the use of g for the exact (1/r2, non-linearized) attraction. As far as I'm aware g is used only for the linear (in height h) form of gravitation. For the time being I changed it to f, but please feel free to change it to something else.--Paul Wormer 02:17, 27 February 2008 (CST)

Paul, I like you and admire your erudition ... and I can only hope that this lengthy posting does not offend you.

• About 2 weeks ago, I wrote an article on flue gas stacks which included a section on flue gas draft (or draught) that included some equations for approximate estimation of what I call the "stack effect" or "chimney effect". The local gravitational acceleration (g = 9.807 m/s) was included in those equations and there was no article in CZ to link it to. I read Gravitation and found no mention of g at that time (although it has since been added into that article by 'Dragon' Dave McKee) ... instead Gravitation gets involved in time-space and general relativity. So I thought that once CZ attracts more engineers and more equations involving g begin to be written, we will need an article that simply defines g.
• On February 22, I posted a message in the Physics Workgroup's mailing list and asked if anyone would please write an article defining the local gravitational acceleration g in "... very simple plain English ...". You can find that message in the mailing list archive for February.
• On February 23, 'Dragon' Dave McKee created this article. Shortly thereafter, I added in the bit about gn as defined by the CGPM in 1901 ... over a century ago ... which now appears in many, many textbooks and even more technical journal articles.
• On February 26, on this Talk page,I asked that this article be simplified, and you graciously did so.

However, even now, the article seems to be focused more on $f$ than on gn. Can we not have an article that simply defines gn? After all, anyone wanting to read about $f$ can read Gravitation where it is referred to as $F$ .

In essence, I think that everything after " ... different locations around the world." could be deleted since it is already in the Gravitation article. - Milton Beychok 12:44, 27 February 2008 (CST)

Milton, I commented out the two last paragraphs. I did not delete them in case somebody wants to restore them.
You give me too much credit:
• The article gravitation originates for the largest part from WP, I only added the section depending on height h (which BTW contains the term gravitational acceleration in bold together with its definition and value in three digits). Last weekend 'Dragon' Dave McKee added a value of g which was off by a factor of 10. General relativity and such comes from WP.
• The last two paragraphs (which I commented out) in the present article were written by Roger Moore in reaction to the flawed discussion by 'Dragon' Dave McKee. The Dragon drew an ellipse with the force center in the middle of the ellipse instead of in one of the foci. His discussion did not emphasize g, either. All I did was modifying the discussion of Roger three times (the first modification was on your request, the second was that I didn't like the same symbol for the exact and the linearized force, and now the third is again on your request).
• I was not aware of the history of the article, for me the contribution of the Dragon came out of the blue. I cannot find your msg in the physics mailing list, I don't know why.
--Paul Wormer 02:17, 28 February 2008 (CST)
Paul, thanks for your response and I appreciate your having commented out the last two paragraphs. Here is the url for my message in the physics mailing list: http://mail.citizendium.org/pipermail/cz-physics/2008-February/000002.html - Milton Beychok 10:41, 28 February 2008 (CST)
Milton, thank you for the link, this was clarifying. I see now that the "Dragon" only translated into Wiki some (Word/pdf) text offered by a Rumanian professor.
I still don't understand why I cannot find the physics mailing list on my own, Rumanian professors and Californian engineers are much smarter than I. I posted a help request on the forum.--Paul Wormer 11:05, 28 February 2008 (CST)
Paul,simply go to CZ:Physics Workgroup and you will see on the right-hand side "Mailing list" and underneath it "cz-physics". Just click on "cz-physics". When you get there, click on "Cz-physics Archives". - Milton Beychok 13:50, 28 February 2008 (CST)

## disambiguation pages are needed

It seems to me that you guys need a gravity (disambiguation) page, which could include something like the following:

• gravity (space-time)
• gravity (Classical)
• gravity (Earth)
• G (universal gravity constant)
• g (acceleration due to gravity) - not the letter "g"
• gn (Earth's gravity constant)

David E. Volk 11:03, 28 February 2008 (CST) and so on.

## Acceleration due to gravity

The gravitational field given is that of a point mass (or a spherical mass outside the radius of the object). The field of an oblate spheroid is not the same as that of a sphere and can cannot depend solely on the distance from the centre of the spheroid since the distribution of mass inside the spheroid (which generates the field) is important so there must be some dependence on the major and minor axes e.g. if I sit on the major axis (theta=0) and increase the minor axis there will be zero change in the gravitational field according to the article yet clearly the mass is now distributed at a greater distance from my location so the field should reduce.

I don't have time to calculate the correct field (and can't find it easily on the net) so I have edited the start of the article to correct a few things there and removed the incorrect part at the end. Some of this may want to be restored when the correct field can be added (or possibly assumptions about near sphericity explicitly stated?) but I thought it best not to leave text that is wrong remain. Roger Moore 17:16, 24 February 2008 (CST)

I've fixed a few more errors in the article which have cropped up:

• 'g' is not a constant, it is the value of the local gravitational field anywhere (not just on the Earth or other planets).
• 'g' is a vector so technically it describes both the magnitude and direction of the field but I thought it easiest just to omit 'magnitude' rather than give the details.
• the potential actually goes as $1/r$ : it is the integral of the force which goes as $1/r^{2}$ .

Roger Moore 02:48, 19 March 2008 (CDT)

## Please look at the last edit of Acceleration due to gravity

The following is a copy of talk page of Paul Wormer. --Paul Wormer 13:44, 25 March 2008 (CDT)

Paul, would you please look at last edit (by Richard Moore) of the Acceleration due to gravity article. Is it correct? And again, can it be simplified? At the very least, the parameters should be defined:

• Why not define $V_{G}$ ?
• Why not also define $G$ ?
• Why not use $m$ instead of $M$ and also define it as the Earth's mass?

That last editor assumes that all of the readers will be physicists and will understand his equation without any explanation.

Would it be incorrect to simply replace his equation with:

$\quad g\equiv {\frac {Gm_{e}}{r_{e}^{2}}}$ where $G$ is the universal gravitational constant, $m_{e}$ is Earth's mass and $r_{e}$ is the Earth's radius

which is as written in the article Gravitation#Gravitational potential. Or am I completely incorrect? Regards, - Milton Beychok 18:30, 24 March 2008 (CDT)

Milton, I wrote the section Gravitation#Gravitational potential and I believe it to be correct. Roger (not Richard) Moore made the point that acceleration is a vector, which is why he wrote fat g. This is true, but in the approximation that the Earth is a non-spinning, perfect, homogeneous sphere the vector character is not so relevant, because then g is a vector with one component only (directed to the center of the sphere).
Altogether the article now suffers a severe case of Wikipeditis. It started out completely wrong with an erroneous Kepler orbit, and then different people (including myself) tried to clean it up as politely as possible. By polite I mean trying to save as much as possible of the work of the previous author. Your points are well-taken, VG, G, and M must be defined. There is some duplication in the article that should be removed as well. Maybe I will write a new version from scratch tomorrow, discarding politeness. The only thing I don't understand is, why is there a standard value with 6 decimal figures, this is unphysical as the variation in the value is in the third decimal figure. Is the standard perhaps meant to enable checking of computations?--Paul Wormer 21:06, 24 March 2008 (CDT)
Paul, thanks for offering to rewrite the article. I don't care if g is expressed to 6 decimal figures or 3 decimal figures. As I have said before, g is used a great deal by engineers dealing with fluid dynamics and we need a straightforward article about g which is clearly written in understandable language that we can link to when writing articles. In other words, an article that can be understood by reasonably intelligent people who are not advanced physicists. I look forward to seeing your rewrite. Regards, - Milton Beychok 22:18, 24 March 2008 (CDT)

End copy from talk page of Paul Wormer

I think the article can do both. Yes, the introductory section, and perhaps the first section after that, should clearly and simply explain the basic, 'simple' form. Then later sections could tackle the more advanced issues. J. Noel Chiappa 13:58, 25 March 2008 (CDT)

## Removed material

The following content was commented out, and was removed; putting a copy here so people have easy access to it:

In the sciences, the term acceleration due to gravity refers to a quantity g describing the strength of the local gravitational field. The quantity has dimension of acceleration, i.e., m/s2 (length per time squared) whence its name.

In the article on gravitation it is shown that for a relatively small altitude h above the surface of a large, homogeneous, massive sphere (such as a planet) Newton's gravitational potential V is to a good approximation linear in h: V(h) = g h, where g is the acceleration due to gravity. This aproximation relies on h << Rsphere (where Rsphere is the radius of the sphere). The exact gravitational potential is not linear, but is inversely proportional to the distance, r, from the centre of the Earth:

$V_{G}={\frac {GM}{r}}$ .

On Earth, the term standard acceleration due to gravity refers to the value of 9.80656 m/s2 and is denoted as gn. That value was agreed upon by the 3rd General Conference on Weights and Measures (Conférence Générale des Poids et Mesures, CGPM) in 1901. The actual value of acceleration due to gravity varies somewhat over the surface of the Earth; g is referred to as the local gravitational acceleration .

Any object of mass m near the Earth (for which the altitude h << REarth) is subject to a force m g in the downward direction that causes an acceleration of magnitude gn toward the surface of the earth. This value serves as an excellent approximation for the local acceleration due to gravitation at the surface of the earth, although it is not exact and the actual acceleration g varies slightly between different locations around the world.

More generally, the acceleration due to gravity refers to the magnitude of the force on some test object due to the mass of another object. Under Newtonian gravity the gravitational field strength, due to a spherically symmetric object of mass M is given by:

$f=G{\frac {M}{r^{2}}}.$ The magnitude of the acceleration f is expressed in SI units of meters per second squared. Here G is the universal gravitational constant G = 6.67428×10−11 Nm2/kg2  and $r$ is the distance from the test object to the centre of mass of the Earth and M is the mass of the Earth.

In physics, it is common to see acceleration as a vector, with an absolute value (magnitude, length) f and a direction from the test object toward the center of mass of the Earth (antiparallel to the position vector of the test object), hence as a vector the acceleration is:

${\vec {f}}=-G{\frac {M}{r^{2}}}{\vec {e}}_{r}\quad {\hbox{with}}\quad {\vec {e}}_{r}\equiv {\frac {\vec {r}}{r}}.$ 