Talk:Earth's gravity

From Wikipedia, the free encyclopedia

	Portal This article is within the scope of WikiProject Physics, which collaborates on articles related to physics.
B	This article has been rated as B-Class on the assessment scale.
???	This article has not yet received an importance rating within physics.
Help with this template This article has been rated but has no comments. If appropriate, please review the article and leave comments here to identify the strengths and weaknesses of the article and what work it will need.

	Earth sciences Portal Earth's gravity is part of WikiProject Geology, an attempt at creating a standardized, informative, comprehensive and easy-to-use geology resource. If you would like to participate, you can choose to edit this article, or visit the project page for more information.
B	This article has been rated as B-class on the quality scale.
High	This article has been rated as high-importance on the importance scale.

Should Pluto be removed from the list of planetary gravities? Asteron 21:22, 14 September 2006 (UTC)

[edit] Earth's gravitational field strength

Does anyone know the Earth's gravitational field strength (N/kg)in 3 decimal points? (Anonymous comment by User:213.105.217.232)

1. Definition of a Newton:
2. Divide both sides by 1 kg:
3. Cancel the kgs on the right side:

Thus N/kg is the same as m/s^2, so any value in m/s^2 in the article can be used. Standard acceleration, rounded to three decimal places, would then be 9.807 N/kg. —Ben Brockert (42) UE News 07:14, 6 May 2007 (UTC)

[edit] Gravitational field of Earth from Nasa

Someone add the picture and information from this article please.--Svetovid 20:08, 2 June 2007 (UTC)

Better pictures + complete info about the project: [1]--Svetovid 20:10, 2 June 2007 (UTC)

[edit] Table of comparative gravities: Centrifugally-adjusted value of g is incorrect (I think)

There is a lot of precision in the section Comparative gravities of the Earth, Sun, Moon and planets. First, the text above the table states that the values are for equatorial surface gravity. However, the table normalizes Earth to a value of “1” and further states that this is equivalent to an acceleration of 9.80665 m/s². It seems to me that something isn’t quite right and—at this level of precision—we’re comparing apples and oranges. The value 9.80665 m/s² takes into account the centrifugal force of Earth’s rotation. The value 9.80665 m/s² corresponds to a perfect-sphere, sea-level latitude of 50.53°. Not surprisingly, this latitude isn’t too different from that of Sèvres (Paris, which is 49° at an altitude 50 m where the BIPM is located). For many planets on Wikipedia, the surface gravity is at the equator. For still others, like Pluto, “equatorial gravity” isn’t known and it is left out. Are we to use equatorial gravity for Earth? If so, then the value certainly would not be 9.80665 m/s². It is perfectly appropriate for Wikipedia articles on individual planets to mention equatorial gravity (where available). But for a comparison table, where this sort of data isn’t always available for every planet, the values for gravity should be calculated based solely on two criteria: mass, and mean radius. If one does the same for Earth, then its mass (5.9736 × 10²⁴ kg per Wikipedia), and volumetric radius (6,371,000.8 m), and the value of G from Wikipeida (6.67428(67) × 10^–11 m³kg^–1s^–2), means that simple g = 9.8226 m/s².

At what speed will someone situated over Jupiter’s equator at a pressure altitude of 75 kPa move when overall planet rotation is added into the calculation? Do the visible cloud tops here always (or just “usually”) correspond to a pressure altitude of 75 kPa? This particular movie synched planet rotation to the visual location of the Great Red Spot as Voyager arced in.

If one instead uses the value of G and Earth’s mass as determined by the University of Washington, Measurement of Newton’s Constant Using a Torsion Balance with Angular Acceleration Feedback (109 kb PDF, here), where Earth’s mass = 5.972245(82) × 10²⁴ kg and G = 6.674215(92) × 10^–11 m³kg^–1s^–2, then g = 9.82025(13) m/s². I used the U of W values when calculating the sea-level latitude (50.53°) at which g = 9.80665 m/s².

Regardless of which data set you use for your input variables, (producing values from 9.8226 using Wikipedia data or 9.82025(13) m/s² using U of W data), when one is comparing the gravity of the planets, g for Earth is not the latitude-based, centrifugally-adjusted value of 9.80665 m/s².

To make the table consistent across all planets, damn easy to calculate, and logical, I’ve revised it with surface gravities that are non-compensated for the centrifugal effects of some arbitrarily chosen latitude. If someone prefers a different solution, whatever it is, it must apply equally to all planets and must have the proper acceleration values for that latitude. If someone proposes to use equatorial accelerations, then all planets (including Earth, which has an equatorial value of 9.78033 m/s²) must be labeled with their equatorial values. Pluto could be footnoted that its value is an exception and is not an equatorial value. Really though, lowering one’s self down into Jupiter’s clouds to a pressure altitutde of 75 kPa (some old text I just deleted) and letting one’s self “rotate” with the planet (blow with the equatorial winds) is 1) hardly a “surface gravity”, and 2) is fraught with potential errors and the need for future recalculation because it is so dependent upon the methodology used in the measurment and calculation of equatorial winds, the altitude at which they are measured, and how one calculates the overall rotational period of a big gas ball. For instance, is “equatorial wind speed” measured relative to an inertial frame of reference(?), or to the planet-wide, mean cloud-top rotational period(?), or is the equatorial wind speed relative to the rotation rate of the planet’s magnetic core?

While editing the table, I also deleted the “formula” under it for calculating surface gravity. It had only two digits of precision and required the calculation of average planet density in kilograms per cubic meter. Beyond those shortcomings, it also had nothing to do with the old table because it had no terms to adjust for the rotational speed of the planet. I replaced that formula with the actual one for calculating gravity. If the value for G changes in the future, and/or the volume or mass of a planet is revised, any editor can easily and accurately update the new table. Better yet, any editor can audit the table and others’ calculations. Greg L (my talk) 18:59, 6 February 2008 (UTC)

[edit] Conversion of Meters to Feet

The conversion from meters to feet of Standard gravity

9.80665 m/s² (32.1740 ft/s²) appears an incorrect calculation.

9.80665 m = 382.45935 in = 31.8716125 ft

32.1740 ft = 386.088 in = 9.89969 m

A difference of 3.62865 in = 0.09304 m = 0.3023875 ft

Can someone explain the differences in calculation? Galo1969X (talk) 22:51, 16 February 2008 (UTC)

• It appears you are using a conversion factor of 39 inches per meter, which is clearly incorrect. The precise conversion from meters to feet is 0.3048 m/ft (12 inches × 2.54 cm/inch). Thus, there are 100 ÷ 2.54 ≈ 39.3700787401574803149606299213 inches/m. Thus, 9.80665 m/s² ÷ 0.3048 ≈ 32.1740485564304461942257217848 ft/s². Greg L (my talk) 05:25, 22 February 2008 (UTC)