|WikiProject Measurement||(Rated Start-class, Mid-importance)|
|WikiProject Physics||(Rated Start-class, Low-importance)|
- 1 Is this true?
- 2 Dubious
- 3 explanation of units
- 4 effect of centripetal acceleration
- 5 constants are not italicized
- 6 "g" is subjective and varies if expressed in unit of time other than second
- 7 usage of letter/abbrev/symbol
- 8 terminology
- 9 Neither cited document contains the expression 'standard gravity'
Is this true?
The article says that "the value of G can be measured precisely", but my understanding is that, compared to other physical constants, G is actually rather difficult to measure precisely and is not known to a great degree of accuracy. Is the statement in the article just wrong, or am I misunderstanding something? Matt 01:47, 27 April 2007 (UTC).
- g is fairly easy to measure precisely, using the frequency of a long heavy pendulum or whatever. At our physics building they have an accurate value of g measured in one of the lecture theatres. 22.214.171.124 13:23, 18 May 2007 (UTC)
- I have changed the offending section to make it clear that G is rather difficult to measure precisely. Matt 01:36, 27 May 2007 (UTC)
- In fact, I have now moved that whole section to Earth's gravity. Matt 02:35, 27 May 2007 (UTC)
"For a planet R will be the planet's radius and z will be the distance from the centre of mass to the object (z ~ R) while for an accretion disk around a celestial body like a black hole R will be the distance to the black hole, and z the distance above the accretion disk." This is confusing and needs better explanation. I assume "z ~ R" means z is approximately equal to R? Why approximately? What is the geometry of an "accretion disk" and how does this relate to the formula? 126.96.36.199 (talk) 00:12, 15 February 2008 (UTC)
- The value of g0 defined above is an arbitrary midrange value on Earth...
I find it hard to believe that the value of 9.80665 is "arbitrary". Surely all those decimal places weren't plucked out of thin air? At http://www.numericana.com/answer/units.htm#g it says:
- To an actual measurement of 9.80991 m/s2 in Paris, a theoretical correction factor of 1.0003322 was applied which gives a sea-level equivalent at 45° of latitude. The result (9.80665223...) was rounded to five decimals to obtain the value officially enacted by the third CGPM, in 1901.
This explanation sounds far more likely to me, but I can't find any other sources to back it up so I haven't yet changed the article. —Preceding unsigned comment added by 188.8.131.52 (talk) 19:25, 25 April 2008 (UTC)
- The location selection (45° latitude) and the 5 decimal place rounding would both serve to render it 'arbitrary'. There's no reason to select those two values over any other, even for simplicity. Dreikin (talk) 23:18, 13 September 2008 (UTC)
The arbitrariness comes from the location chosen. The metric system is one of the glories of the French, so the location chosen must therefore be located in France - not at the equator [smallest value] nor the pole [largest]. It's the same with the initial measurement of the metre itself. 184.108.40.206 (talk) 00:31, 15 October 2008 (UTC)
- I made a small adjustment to the wording which I hope everyone will be OK with. It would be nice to specifically mention Paris, but unfortunately Paris is quite a way from 45°N and I don't know how that one resolves itself. 220.127.116.11 (talk) 18:34, 10 November 2008 (UTC).
- Sorry, of course the conversion to about 45°N would be part of the "theoretical correction factor" I guess. http://mtp.jpl.nasa.gov/notes/altitude/altitude.html says it's 45.542°. It would be interesting to know why that was chosen... 18.104.22.168 (talk) 18:45, 10 November 2008 (UTC)
explanation of units
I've never heard accelerations being described in terms of "square seconds." The phrase "per second per second" sounds a lot more meaningful. I'm having trouble even conceiving of square seconds because a number of seconds isn't a distance. Arithmetically, yes, it makes sense, because to arrive at the correct number, the seconds quantity has to be multiplied by itself. Joe (talk) 20:00, 18 May 2008 (UTC)
Both are common. In the physics community "meters per second-squared" seems far more common; consider that a newton is a "kilogram meter per second-squared" as well. I think the reference note just sounds silly; there's no special "notation" here. I'll take it out unless there's a big objection. —Preceding unsigned comment added by 22.214.171.124 (talk) 21:24, 17 June 2008 (UTC)
effect of centripetal acceleration
Nominal gravity, g0, is typically measured in the accelerating reference frame of the Earth's surface at the location where it is being measured. This is correct for most terrestrial applications. As Einstein explained with his elevator analogy, acceleration is equivalent to gravity, so it is not wrong to say that a scale measures just gravity. We should be careful, however, to say that a terrestrial scale measures g0 in its own terrestrial reference frame. It is wrong to use that measure of g0 and Newton's universal law of gravitation (in its familiar form) to calculate Earth's mass. We must either remove the centripetal acceleration of Earth's surface from the terrestrial measure of g0 or else transform the law of gravity into a terrestrial reference frame.
One could create a stellar reference frame environment (whose origin tracks Earth's orbital motion around the sun) inside a railroad car traveling due west on a horizontal track at the speed of Earth's rotation. At the equator, the train would have to go 1,674.4 km/h (465.1 m/s) relative to Earth's surface; at 60° latitude, half that fast. A scale inside the railroad car would measure Earth's gravity without the contribution of centripetal acceleration. That measure of g0 could be used with the law of gravity in its familiar form to correctly calculate Earth's mass.
The problem that I see in this article is the lack of a clear distinction between g0 and gn. Perhaps "g0" should represent the value measured in the stellar frame, while "gn" could represent the valued measured in a terrestrial frame. Surely some such convention is already established in the scientific community; does anyone here know what that convention is? Onerock (talk) 20:12, 31 August 2009 (UTC)
The denominator should use the sidereal day of 86 164.0905 seconds instead of 86 400 since inertia is relative the stars and not the Sun. David Jonsson 20:44, 11 January 2011 (UTC) — Preceding unsigned comment added by Davidjonsson (talk • contribs)
The sentence in the introduction "The acceleration of a body near the surface of the Earth is due to the combined effects of gravity and centripetal acceleration." appears to confuse "centripetal" with "centrifugal". The centripetal force (and thus the derived acceleration) is directed toward the center, while the centrifugal force is directed away from the center. Local g is lower at the equator than the poles, the rotational (pseudo)force is directed outward from the axis of rotation while the force of gravity is directed inward (down). At the equator, these forces are acting in exactly opposite directions. I'm going to change it.Enon (talk) 17:48, 12 March 2011 (UTC)
constants are not italicized
Why is g italicized in this article (as well as G)? Variables are italicized; constants are not. Do you abbreviate pi, the permitivity of free space, or Boltzman's constant? -Njsustain (talk) 19:22, 28 September 2009 (UTC)
- Well, gravity is definitely not constant. On the other hand the gravitational constant is.. constant. Do U(knome)? yes...or no 08:09, 6 December 2009 (UTC)
"g" is subjective and varies if expressed in unit of time other than second
Galileo had demonstrated that acceleration due to gravity is constant i.e. g=9.8 m/s/s or 32 ft/s/s but cognizance shows that results are not tantamount to the original experiment if unit of time is recorded other than second (e.g “g” is expressed in m/half-sec/half-sec or m/hr/hr or m/min/min). Also experimenter shouldn’t assume that it is fixed by the nature for the falling velocity of an object to be constant specifically for the duration of one second in given spacetime and should increases constantly only after every second.
As falling velocity of object changes/ increases at every fraction of second in given spacetime therefore acceleration is produced at every fraction of second rather than per second. Thus the confirmed value of” g” which is equal to 9.8 m/s/s or 32 ft/s/s is contentious, not viable and should not be considered as “fait accompli” albeit rewarding.126.96.36.199 (talk) 17:20, 31 May 2010 (UTC)khattak#1-420
- The OP raised this at WP:RD/S#Question about the standard value of "g"?. Gravitational acceleration is (close to) a constant and is not subjective. Time passes continuously and the OP needs to learn that the normal expression of acceleration g=9.8 m/s/s does not mean anyone thinks nature increases the speed in jumps of +9.8m/s every second. The OP can express the same acceleration as g=
58835280m/min/min which is correct. However it is nonsense to attack the common use of the second as time unit as "contentious, not viable etc." Cuddlyable3 (talk) 21:09, 31 May 2010 (UTC)
- There is no need for your confused experimenter to assume measurements every second. g=0.0000098 m/s per microsecond or 0.0000000098 m/s per nanosecond. Dbfirs 07:23, 1 June 2010 (UTC)
An equation of instantaneous velocity for falling mass is v = g x t in which a magnitude of 9.8 m/s/s [acceleration] can't be preceded in the very first infinitesimal falling of an object as "g" also incepts from zero [besides "v" and "t"] and should increase gradually to its constant value of 9.8 m/s/s.
Here is how "v" and "t" are aberrational in the equation v = gt even if it is presumed that "g" is constant.
As "v" is directly proportional to "t" in aforementioned equation therefore both ["v" and "t"] should start from zero and after sometime "v" should overcome "t" due to their gradual increasement but there is no chance at all for the numerical value of "v" to meet with numerical value of "t" in the very first development of mass falling as "v" is always ahead of "t" at 9.8 times "t" [9.8xt] in the equation.
Neither I'm lambasting nor my intent is to prevail over the contest rather consider this onus through abstruse reasoning that these are just state of the art STANDARDS work method/ procedure /guidance and shouldn't be occulted peremptorily. Thus, if I have understood all above correctly, there may be grounds and opportunities for neoteric computerized brains to make such praxises (process) more amenable to the mother nature. 188.8.131.52 (talk) 01:41, 28 December 2010 (UTC) khattak#1-420
- You seem to have a basic misunderstanding about the nature of acceleration. It does not start from zero, it is constant right from the instant when the body is released. The equation v = gt is independent of the units chosen, though the numerical values will depend on the units chosen, and the units of g and t must be consistent, and together they will determine the units of v. Dbfirs 08:35, 28 December 2010 (UTC)
usage of letter/abbrev/symbol
So, is the proper written term not G, or gees, or even g, but g? Or is it g0?
- (In Edward James Olmos's voice:) "...Now you're on your attack run. You launch your missiles. So you gotta jam that pedal into the firewall and hold a six g0 turn for ten seconds or you die. Ten, nine, eight, seven, six, five, four-- (clang) This was only three g0s, Starbuck, not six. I'm sorry, it's a tough one but you're staying home." 184.108.40.206 (talk) 20:44, 30 September 2013 (UTC)
I don't understand why an acceleration rate is referenced with the same term as the force creating it (i.e. standard GRAVITY). I have never heard it used that way, and I get after my students whenever they use gravity to refer to the ACCELERATION due to gravity. (Just as bad as when they say it's -9.81 m/s^2. NO!)220.127.116.11 (talk) 16:15, 1 October 2013 (UTC)Amy
Neither cited document contains the expression 'standard gravity'
Neither cited document contains the expression 'standard gravity'. Instead, each document, citing the declaration of the 3rd CGPM, 1901, refers to 'standard weight' and 'standard acceleration due to gravity' viz:
• Declaration on the unit of mass and on the definition of weight; conventional value of gn (CR, 70)
Taking into account the decision of the Comité International des Poids et Mesures of 15 October 1887, according to which the kilogram has been defined as unit of mass;
Taking into account the decision contained in the sanction of the prototypes of the Metric System, unanimously accepted by the Conférence Générale des Poids et Mesures on 26 September 1889;
Considering the necessity to put an end to the ambiguity which in current practice still exists on the meaning of the word weight, used sometimes for mass, sometimes for mechanical force;
The Conference declares
1. The kilogram is the unit of mass; it is equal to the mass of the international prototype of the kilogram;
2. The word “weight” denotes a quantity of the same nature as a “force”: the weight of a body is the product of its mass and the acceleration due to gravity; in particular, the standard weight of a body is the product of its mass and the standard acceleration due to gravity;3. The value adopted in the International Service of Weights and Measures for the standard acceleration due to gravity is 980.665 cm/s2, value already stated in the laws of some countries.