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- 1 Earlier comments
- 2 Wrong character
- 3 Gender specific pronouns
- 4 Counting Coefficients
- 5 References and Intro
- 6 Centrifugal Force
- 7 Broken Links
- 8 Geoid Picture
- 9 Possible image error
- 10 Sea floor
- 11 Unclear grammar
- 12 Erm... help for beginners :)
- 13 Pronunciation
- 14 Mathematical formulas
- 15 New data from Goce satellite
- 16 Rotation of the Earth
- 17 More Rotation of the Earth
- 18 Contradiction at Latitude
- 19 The work of Vaníček
- 20 Criticism of Article.
Hi, I think there's an error in the formula: you write "φ and λ are geocentric latitude and longitude respectively", but nowhere in the formula can I see any φ ...
- The problem is that Wikipedia:TeX markup permits simple TeX characters to be displayed as HTML characters, thus in a different font, which here appears to change the character, even though both are phi. Thus I changed <math>\phi</math> to <math>\phi\ </math> by adding the math text space '\ ' immediately after the math character, which forces it to be rendered as a PNG character, just like those in the formula. — Joe Kress 07:16, Apr 17, 2005 (UTC)
Gender specific pronouns
The earth has no gender, so is it necessary to call it a she ?
When you count the geopotential coefficients at the end, you get 130,317. You should probably mention what maximum degree (n) you used and why you chose that value. After all, in principle, n is infinite.
Oh, and if you're interested, here's a quick and visual definition from the project I work on (GRACE): http://www.csr.utexas.edu/grace/gravity/gravity_definition.html
- I'll try to clarify this. After all, there are new models in the works with degree=order=2160 (over 4 million coefficients), good to 1/6 degree. MFago 14:46, 5 April 2006 (UTC)
References and Intro
Was it a good idea to split the top into the short intro and a new "description" section? More closely follows other Wikipedia articles, so I made this change. Also, this article could really use more citations. Unfortunately, I haven't quite figured out the "cite" template... MFago 15:43, 5 April 2006 (UTC)
Hi, I can't find the answer to this anywhere!
Does the geoid take into account centrifugal force from the Earth's rotation? If so, would it be incorrect to refer to it as a mathematical figure of the gravity field alone? And if not, would it be incorrect to say that the mean sea level assumes the shape of the geoid?
- There is not actually "a" geoid, but rather many different variations depending on whether or not one accounts for tides, etc. I do believe that it usually does account for rotation. Note that the equation given in the article is not for the geoid itself, but for the (non-rotational part of the) potential. Computing the geoid is rather more involved. I'll see if I can relocate my reference on this (at NIMA or a related site), and perhaps add more detail to the article. Also see Physical geodesy and Geodesy. The reference text the end of Geodesy is excellent and covers most of the details. MFago 23:21, 1 June 2006 (UTC)
- Link added to above reference: No such thing as "The" EGM96 geoid, and tried to clarify a bit. Really needs more detail -- it's a complex subject, and I'm not entirely fluent. MFago 23:33, 1 June 2006 (UTC)
I was just wondering about this myself. The reason being that International Atomic Time is based on the rate of ticking of a clock sitting on the geoid. This definition removes any ambiguity due to gravitational time dilation, but i can't work out if it accounts for the dilation due to greater speed of movement of points on the equator compared to those at high latitudes. If the geoid doesn't include a centrifugal term, then it definitely doesn't. If it does, and is thus about equipotential from the point of view of a co-moving observer (as a geoid based on actual local measurements would pretty much have to be), then possibly it does, what with the equivalence principle and all that. Although i'm not at all sure about that - is there any particular reason to think time moves at the same rate at all points on an equipotential surface? No, probably not. And i don't even know what kind of geoid IAT uses. Oh god, this is confusing. I should not have to do general relativity to work out what bloody time it is! —Preceding unsigned comment added by 188.8.131.52 (talk) 18:54, 3 November 2008 (UTC)
I just saw that all of the NGA links are broken. I cannot locate any of this data on the new "enhanced" site -- it appears to be unavailable? MFago 23:52, 1 June 2006 (UTC)
- Fixed. They simply changed all *.htm to *.html, and their search engine was of no help. MFago 00:04, 2 June 2006 (UTC)
Unfortunately the picture of the Geoid (actually the geoid undulation) that used to be on this page was evidently taken from the GRACE website without proper permission. Anyone have access to another similar graphic that would be useable here? The raw data for the EGM96 geoid (less detailed than GRACE, but still illustrative) is available at: NGA EGM96 data (labeled as "Geoid Height File"). Matlab can easily make a similar plot from this data, but I do not have access to this software. Anyone able to help, or have another idea?MFago 14:36, 2 June 2006 (UTC)
Possible image error
The geoid seems also to be used, along with satellite data, for estimating the topography of the sea floor; see Exploring the Ocean Basins with Satellite Altimeter Data, especially under "3. Explanation of the Gravity Anomaly". There may be a better home for this, but even if the main information goes elsewhere a 'see also' might be appropriate. (The gravitational anomoly is mentioned technically in Physical geodesy, but without discussion of use.) Mark— Preceding unsigned comment added by 184.108.40.206 (talk • contribs) 06:52, 27 November 2008 (UTC)
The second diagram shows the geoid as being above the ellipsoid where there is land but below it where there is an ocean. The first picture doesn't show this picture at all. Why this difference?220.127.116.11 (talk) 16:04, 20 February 2010 (UTC)
Erm... help for beginners :)
"If that perfect sphere were then covered in water, the water would not be the same height everywhere. Instead, the water level would be higher or lower depending on the particular strength of gravity in that location." - so does that mean that more gravity pulls the water-level down or does it mean that more gravity collects more water above it pushing the water level up? It may seem obvious to you, but it's not to some of us here :) Clarification would be appreciated :) Malick78 (talk) 21:37, 28 June 2010 (UTC)
I agree with Malick. Could someone who understands this properly state something like "A greater height on the geoid corresponds to a stronger local gravitational field" or vice versa? I've seen online discussions about the map from Goce in which lay-people interpret the map both ways, even after reading this Wikipedia article. BruceMcAdam (talk) 12:27, 29 June 2010 (UTC)
- I'm not an expert on this, but my guess is that the level of the water joins all points of equal gravitational potential. Thus in areas where gravity is stronger, you have to go up to a greater height to achieve the same gravitational potential as a point where the gravity is weaker. Thus the water would be higher there. Muraho (talk) 10:59, 18 August 2010 (UTC)
- I've removed that whole section. There is no direct relationship between the water level and the local strength of gravity. When traveling in a geodesic along the surface, the rate of change in the direction of the force of gravity determines whether the water surface locally follows a concave or convex curvature. When a heavy lump is ahead, the gravity vector moves a bit forward (compared to pointing to the centre of the Earth) and the water surface will curve up. Past the heavy lump, the vector will point a bit backward, and the water surface curves down. --Lambiam 09:08, 2 November 2010 (UTC)
So isn't this sentence wrong? The surface of the geoid is farther away from the center of the Earth where the gravity is weaker, and nearer where it is stronger. It contradicts both my intuition and the illustration. —Tamfang (talk) 07:27, 29 August 2013 (UTC)
How is this word pronounced? I always assumed (without really thinking about it) that it is pronounced "gee-oh-id". However, on reflection is seems more likely that it is "gee-oyd". To match ellipsoid. Is this correct? Muraho (talk) 11:00, 18 August 2010 (UTC)
- According to dictionaries it is English pronunciation: /ˈdʒiɔɪd/. --Lambiam 08:42, 2 November 2010 (UTC)
Does the inclusion of mathematical formulas in the section Spherical harmonics representation serve a purpose? Without giving the numerical values of the coefficients, they do not supply any actually useful information. Even if this could somehow be argued to serve some purpose, the right place for such detail would be the EGM96 article. --Lambiam 09:29, 2 November 2010 (UTC)
New data from Goce satellite
Scientists use a geoid model to illustrate the force of gravity on Earth from data from the Goce satellite. (BBC). Can a knowledgeable editor update this article? Dr.enh (talk) 02:41, 2 April 2011 (UTC)
Rotation of the Earth
As I understand it from [this source] the geoid is an equipotential surface based in the sum if the non-rotating gravitational potential and the centrifugal potential. This is confirmed by this sentence in the lead, '...the geoid is the equipotential surface that would coincide with the mean ocean surface of the Earth if the oceans and atmosphere were in equilibrium, at rest relative to the rotating Earth'. However, we also have this sentence, 'In that idealized situation, other influences such the rotation of the earth, winds due to solar heating, and so on have no effect'. This may suggest to some that the rotation of the Earth is not taken into account in the calculation of the geoid. I suggest that the words, 'such the rotation of the earth', are removed. Martin Hogbin (talk) 21:29, 1 April 2013 (UTC)
- I have been bold and made the change. The academic source that I added seem clear enough. Martin Hogbin (talk) 10:57, 2 April 2013 (UTC)
More Rotation of the Earth
A related observation: the first paragraph states: "The geoid, simply stated, is the shape that the surface of the oceans would take under the influence of gravity alone."
Such a surface would be very nearly a sphere. The reason that the geoid is nearly an ellipsoid is that its shape is due to gravity plus the effect of the rotating earth. The geoid entry in Encyclopedia Britanica includes the statement "This potential function describes the combined effects of the gravitational attraction of the Earth’s mass and the centrifugal repulsion caused by the rotation of the Earth about its axis."
The first paragraph of the current wiki article is therefore misleading and confusing because (1) there is no explanation of why the earth would be an ellipsoid under the influence of gravity alone, and (2) a plumb line is always under the influence of the rotating earth and would therefore not be normal to the geoid if the geoid were due to the influence of gravity alone. I would suggest that this paragraph be changed to be more accurate. "The geoid, simply stated, is the shape that the surface of the oceans would take under the combined influence of gravity and the earth's rotational velocity." Rocket Laser Man (talk) 20:21, 14 June 2013 (UTC)
- Yes, I noticed this but decided that gravity could be interpreted to include the contribution of centrifugal force. How about we have, "The geoid, simply stated, is the shape that the surface of the oceans would take under the combined influence of gravity and centrifugal force due the earth's rotation."? Martin Hogbin (talk) 00:35, 15 June 2013 (UTC)
The work of Vaníček
I notice that a 'partisan' tag has been added to the section on the work of Vaníček. Vaníček's work is reported in what, by most standards, would be considered reliable sources and thus should be included here.
- This article is supposed to be about the geoid, ideally with no undue focus on any particular author who has published research about it. I suggest removing that section if it cannot be enlarged to encompass a worldwide view. Fgnievinski (talk) 22:08, 18 June 2014 (UTC)
Criticism of Article.
My overall impression is that this is a very weak article that requires a lot of revision. For example, in the 'Description' section it states-- If the ocean surface were isopycnic (of constant density). --It is an elementary mistake to describe a two-dimensional surface as having a density. Gravity measurements are affected by the totality of rock and water densities encountered by extrapolating a plumb line from the point of measurement. Sea-water density differences are relatively trivial compared with lateral differences in rock density, which vary rapidly on a scale much smaller than the thickness of the Earth's crust. It is quite ludicrous to even mention the possibility of the geoid occupying hypothetical narrow canals cross-cutting the continents, as gravity is affected by the totality of vectors of mass both above and below the point of measurement. As the centre of gravity of the Earth - Moon system lies some distance below the surface of the Earth, satellite gravity measurements have to be constantly updated to take account of the phases of the moon and its distance from the Earth, precession etc. It is also very unlikely that the isostatic compensation depth is much more than 20 - 30 km deep. None of these caveats is addressed in the article.
The entire article should be withdrawn until someone can contribute a closer description of reality. Meanwhile it would be best to describe the geoid as a surface that is defined by gravity departure from a defined ellipsoid that has an average Earth gravity field throughout and forget about the ups and downs of the oceanic surfaces until some contributor has a basic understanding of what's going on. Geologician (talk) 21:19, 25 December 2014 (UTC)