Talk:Acceleration due to gravity

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)

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 g_n = 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)