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I propose that two too-long tables in this article be removed because they obstruct the reading of the article and don't add much to the discussion. The table for "Comparative gravities in various cities around the world" was apparently generated by using a widget from Wolfram [1]. In place of this table, we could mention the gravity at a few places (fewer than in the table) and then just link that widget at the end of the essay rather than give the long table. The table for "Comparative gravities of the Earth, Sun, Moon, and planets" is poorly sourced and possibly original research. Thoughts? Isambard Kingdom (talk) 14:03, 3 March 2017 (UTC)[reply]

 Done Isambard Kingdom (talk) 13:50, 9 March 2017 (UTC)[reply]

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Quote, "The same two factors influence the direction of the effective gravity (as determined by a plumb line or as the perpendicular to the surface of water in a container). Anywhere on Earth away from the Equator or poles, effective gravity points not exactly toward the centre of the Earth, but rather perpendicular to the surface of the geoid, which, due to the flattened shape of the Earth, is somewhat toward the opposite pole. About half of the deflection is due to centrifugal force, and half because the extra mass around the Equator causes a change in the direction of the true gravitational force relative to what it would be on a spherical Earth." Does gravity involve ellipsoidal coordinates? The article Latitude#Ellipsoidal coordinates suggests this. If this is the case, then falling objects maybe adhere to a hyperbolic path, not a straight one. Also please review my question on Stack Exchange, as it deals with this exact same question. ➧datumizer  ☎  22:08, 23 August 2018 (UTC)[reply]

"Involve ellipsoidal coordinates" is a somewhat nebulous phrase. The gravity can be expressed in terms of these coordinates, but that doesn't mean that the field lines lie on a coordinate surface (unless the body is non-rotating). Also, note well, that a falling body will, in addition, experience a coriolis force, and so it won't, in general, stay in a single meridional plane.

By the way, "Talk pages are for discussing the article, not for general conversation about the article's subject" (see "talk page guidlines"). So your question doesn't really belong here. cffk (talk) 03:30, 24 August 2018 (UTC)[reply]

Please don't automatically assume I'm not asking questions with the intention of editing a Wikipedia article. ➧datumizer  ☎  22:46, 24 August 2018 (UTC)[reply]

If, as suggested in the text, the density of the Earth decreases linearly with increasing radius from a density ρ₀ at the center to ρ₁ at the surface, then the acceleration due to gravity at depth d below the surface (i.e., at ${\displaystyle r=R-d}$ ) is the following integral:

${\displaystyle g(r)=-{\frac {G\int _{0}^{r}4\pi r^{2}\rho _{r}dr}{r^{2))))$

where ${\displaystyle \rho _{r}=\rho _{0}-\left(\rho _{0}-\rho _{1}\right){\frac {r}{R))}$

Noting that ρ_r is also a function of r, the substitution must be made before the integration, and hence this integral becomes:

${\displaystyle g(r)=-{\frac {4\pi G\int _{0}^{r}r^{2}\left(\rho _{0}-\left(\rho _{0}-\rho _{1}\right){\frac {r}{R))\right)dr}{r^{2))))$

${\displaystyle g(r)=-{\frac {4\pi G\int _{0}^{r}r^{2}\rho _{0}-\left(r^{2}\rho _{0}{\frac {r}{R))\right)+\left(r^{2}\rho _{1}{\frac {r}{R))\right)dr}{r^{2))))$

${\displaystyle g(r)=-{\frac {4\pi G\int _{0}^{r}r^{2}\rho _{0}dr-4\pi G\int _{0}^{r}\left(\rho _{0}{\frac {r^{3)){R))\right)dr+4\pi G\int _{0}^{r}\left(\rho _{1}{\frac {r^{3)){R))\right)dr}{r^{2))))$

${\displaystyle g(r)=-{\frac {4\pi G[{\frac {r^{3)){3))\rho _{0}]-4\pi G[\rho _{0}{\frac {r^{4)){4R))]+4\pi G[\rho _{1}{\frac {r^{4)){4R))]}{r^{2))))$

${\displaystyle g(r)=-{\frac {4\pi }{3))G\rho _{0}r-\pi G\left(\rho _{0}-\rho _{1}\right){\frac {r^{2)){R))}$

This error was fixed by myself and RockMagnetist in May 2013, I don't know when or why it has reverted. See: https://en.wikipedia.org/w/index.php?title=Gravity_of_Earth&oldid=553887481 Please fix it again. George963 au (talk) 11:52, 16 February 2019 (UTC)[reply]

@George963 au: It took me a while to notice this post. I have fixed it. Thanks for notifying us, but it would be better if you simply fixed it yourself next time. RockMagnetist(talk) 22:19, 17 March 2019 (UTC)[reply]

@RockMagnetist: Thanks, RM. I would have fixed it, but it seems my credibility is not high enough. The last time I tried, someone promptly reversed it. (I'm not sure, but I think it might have been you!) George963 au (talk) 12:40, 6 April 2019 (UTC)[reply]

As far as I can tell, our only previous interaction was this discussion on the same subject. I don't find any edits of the article by you in the history. RockMagnetist(talk) 17:18, 6 April 2019 (UTC)[reply]

Hi, there is an error in g(r). The linear density is correct, then if we substitute it in the first equation of g(r), and we develop each term with simple math, we see there is an error: g(r) is missing the 4/3 in the second term. Incredibly in Google books, this paragraph is exactly the same in the book “Artificial Gravity: To Maintain Your Foot in the Space …” (2022) by Fouad Sabry. I wonder if it is a first edition and copied directly from Wikipedia or the wiki is copied from earlier editions from that book. IGomezLeal (talk) 08:56, 3 August 2023 (UTC)[reply]

It is better to leave the integral of g(r) to avoid this confusion and as this paragraph says the density has to be substituted inside the integral. IGomezLeal (talk) 09:02, 3 August 2023 (UTC)[reply]

I will check also the result of the integral, the order of the powers should increase. The first term on the right should be proportional to r^3 and the second r^4. IGomezLeal (talk) 09:21, 3 August 2023 (UTC)[reply]

Ok, inside the gravity integral you have forgotten the r^2 in the denominator from the gravity differential. It cancels the r^2 from the sphere volume differential, and then the result of the integral you wrote is almost correct, it is missing a 2 in the second term on the right, since the integral of r is (r^2)/2 and 4/2=2. Do you agree? IGomezLeal (talk) 09:48, 3 August 2023 (UTC)[reply]

No, I don't agree. I've put in some of the intermediate calculations above, for your reference. George963 au (talk) 14:02, 9 October 2023 (UTC)[reply]

What I'm missing on the page is a section about the history and development of the gravity of Earth. What was the Earth's gravity like before the supposed collision with Theia/Orpheus? What was it like in the Mesozoic Age (the "Dinosaur Age")? When and how did it change? Someone please make such a section. --212.186.7.232 (talk) 11:14, 17 March 2019 (UTC)[reply]

If you're talking about the value of "g" the acceleration due to gravity at the Earth's surface, then assuming no significant changes in either the radius or overall structure of the Earth, then you would expect no significant change of g with time except in the very early history of the Earth. If you're interested in gravity anomalies in the past, the small scale variations in the field will have been very different then. Mikenorton (talk) 17:16, 17 March 2019 (UTC)[reply]

You say "except in the very early history of Earth". This is what I mean by "before the supposed collision with Theia/Orpheus". The collision that allegedly created the Moon. In the Mesozoic/Antediluvian Era, the Earth's surface gravity might have been different too (weaker) since dinosaurs, other animals and plants were so big. 212.186.7.232 (talk) 09:53, 18 March 2019 (UTC)[reply]

This post is a very useful look at evidence for changes in gravity with time - in summary there is no evidence that supports lower gravity during the Mesozoic and lost of evidence that supports gravity similar to the present. Mikenorton (talk) 10:52, 18 March 2019 (UTC)[reply]

The size of dinosaurs, etc., is no guide to changes in the Earth's gravity; for at least the suggested timing is backwards. The collision with Theia was very early in the evolution of the Earth, over 4 billion years ago; the Mesozoic Era was a (mere) 250-66 million years ago. George963 au (talk) 01:27, 4 November 2023 (UTC)[reply]

Yes i agree we need more knowledge on this topic 2601:300:4100:75F0:E8F3:8332:B5B1:EB5C (talk) 02:29, 23 March 2023 (UTC)[reply]

I have re-read the section related to gravitational anomalies by geography and believe it is backwards. I'm not sure if the mgals are also backwards. It shows red being positive, but red on the map seems to correspond with lower gravity areas.

The lowest gravity falls on the mountain range in Peru on the west coast of South America. That whole range is lit up in red.

The oceans are overwhelmingly blue and should be some of the highest gravity areas. Peter Bailey (talk) 06:16, 27 April 2019 (UTC)[reply]

I was going to start a topic on this as well. If you look at the animated globe, all the mountains are in red but the supporting text says red is higher gravity.

And adding to the confusion, your post ends with saying the oceans should be some of the highest gravity areas which other then appearing to be wrong contradicts the point. The WSmart (talk) 10:49, 18 March 2022 (UTC)[reply]

Maybe red means higher gravity at the geoid? —Tamfang (talk) 02:39, 29 March 2023 (UTC)[reply]

If we're talking about satellite gravity (like the GRACE results) then what affects the satellite is not the same as the value of gravity that would be measured at the Earth's surface directly below the satellite at the same time. The "Altitude" section (third paragraph) attempts to cover this apparent contradiction. In summary a satellite is affected by high elevation areas in the opposite sense to how a ground observer would be. Mikenorton (talk) 16:03, 29 March 2023 (UTC)[reply]

"The first correction to be applied to the model is the free air correction (FAC) that accounts for heights above sea level. Near the surface of the Earth (sea level), gravity decreases with height such that linear extrapolation would give zero gravity at a height of one half of the Earth's radius - (9.8 m·s−2 per 3,200 km.[19])"

The reference is not a reference, it says "The rate of decrease is calculated by differentiating g(r) with respect to r and evaluating at r=rEarth."

This was moved here on 27th May 2007 from the standard gravity page, which was moved their from the g-force page 6 March 2007.

We know the earth's gravity acts on Geostationary satellites and the moon, far beyond many times the Earth's radius (and this made someone smart I trust think science is not all it's meant to be). This clarification then needs to give context to where it applies and where it doesn't, or remove it if it's not something the Free Air Correction does actually predict.

I don't like that the reference is just an explanation! I suspect that the approximates used for calculations work very well within the 10km high differential of our crust, but are not meant to be applied over thousands of kilometres. So I want to back that up or find out more context. Any ideas? Greg (talk) 07:44, 4 October 2020 (UTC)[reply]

The free-air correction is intended to remove the effects of topography on observations above the reference ellipsoid, which equates with sea level. The resulting free-air gravity anomaly is just one step to reaching the ultimate aim of showing the gravity field as it would be measured if the whole earth was at sea level - the other main correction is the bouguer correction, which accounts for the fact that the material between the point of observation and sea level is rock, not air. The "reference" that you mentioned is actually a "note" to help explain, although I tend to agree that it's not that helpful. Mikenorton (talk) 15:19, 4 October 2020 (UTC)[reply]

https://academic.oup.com/gji/article/154/1/35/604237 http://geopixel.co.uk/g4g_lab1.html

So does the "such that" in the ongoing sentence help in any way in understanding? ("gravity decreases with height such that linear extrapolation would give zero gravity at a height of one half of the Earth's radius")

The intent of the sentence's clarification seems to be: "gravity decreases with height. The FAC does not work and is not intended for higher altitudes, a linear extrapolation would give zero gravity at a height of one half of the Earth's radius". Greg (talk) 03:38, 23 October 2020 (UTC)[reply]

Shouldn't the section "Estimating g from the law of universal gravitation" include a remark that the inertial mass (the term that arises in Newton's 2nd law of motion) is assumed equal to the gravitational mass (the term that arises in the inverse square law). This does strike me as an extremely serious omission Wikipedia ought not be guilty of. Соловей поет (talk) 15:30, 11 August 2021 (UTC)[reply]

How long is gravity range 2409:4042:785:F569:0:0:29D0:10A1 (talk) 15:57, 21 May 2022 (UTC)[reply]

• The acceleration experienced due to Earth's gravity is inversely proportional to the square of the distance from the center of the Earth. The Earth's gravitational field does not abruptly end anywhere. Rather, it gradually fades into nothingness. The "range" is infinite, although if one is far enough from Earth then Earth's gravity will be so small as to be unnoticeable. Crossover1370 (talk | contribs) 17:17, 25 May 2022 (UTC)[reply]

This article was the subject of a Wiki Education Foundation-supported course assignment, between 12 February 2024 and 14 June 2024. Further details are available on the course page. Student editor(s): Maaatttthhheeewww (article contribs). Peer reviewers: WikiIsaacPedia.

— Assignment last updated by Ahlluhn (talk) 00:58, 31 May 2024 (UTC)[reply]

Bahhshhdhdhdhjdbw jdjdhd sgvs s gss. Sv 152.57.221.183 (talk) 12:40, 9 June 2024 (UTC)[reply]

An approximate value for gravity at a distance r from the center of the Earth can be obtained by assuming that the Earth's density is spherically symmetric.

Don't assume the Earth's density profile is symmetric, even for simplification purposes; it muddies the water terribly and we miss a number of clues.

As the PREM chart for density shows through analysis of P&S waves, matter is denser toward the core and is greater at the poles than at equatorial, Peruvian mountaintops.

It should be a great clue to the error in thinking here that we can measure a higher acceleration at the poles where there is less physical matter underfoot than at the top of an equatorial mountain where there's considerably more mass underfoot.

This speaks volumes to the lack of understanding involving gravity wells that pervades modern academia and leads to regurgitated disinformation as found in this article.

The gravity depends only on the mass inside the sphere of radius r

It's a Newtonian notion of mass attracting mass that has you arriving at an acceleration of zero at the core. That's incorrect.

We know that matter curves spacetime. A gravity well like the Earth is a four-dimensional hole that three-dimensional material falls into and then gets stratified by density. The fourth dimension - time - is what is getting stretched.

An object in motion tangent to the Earth will have its path curved by the warpage of spacetime. However, an object sitting motionless on the surface has a spacetime line that runs from space straight through the top of the object, out of the bottom, making a direct line to the core. The object's straightest line of travel through spacetime is now a straight line radial to the core, which happens to be blocked here at the surface, 3963 miles from the center. Your spacetime line sitting at your desk reading this is pinched in from outer space to our core and back out again. You can't travel there because denser matter is between you and it, so your body is stuck, constantly feeling the deceleration of your fall like an elevator continually coming to a stop at a floor.

When you theorize on a subject where time is the agent of change specifically, you cannot use a formula that does not consider the main variable. This is precisely the case with 'gravity.' Newton didn't know about time dilation, but Einstein did. You cannot develop a theory of what acceleration looks like at the center of any gravity well without first asking the simple question, "What time is it there relative to the surface?"

from phys.org "A trio of researchers in Denmark has calculated the relative ages of the surface of the Earth versus its core and has found that the core is 2.5 years younger than the crust. [it's likely considerably younger than even this] During one of his famous lectures at Caltech in the 1960's, Richard Feynman remarked that due to time dilation, the Earth's core is actually younger than its crust. General relativity suggests that really big objects, like planets and stars, actually warp the fabric of spacetime, which results in a gravitational pull capable of slowing down time. Thus, an object closer to Earth's center would feel a stronger pull—a clock set near the core would run slower than one placed at the surface, which means that the material that makes up the core is actually younger than the material that makes up the crust. In this new effort, the research trio ran the math to discover the actual number involved. They found that over the course of our planet's 4.5-billion-year history, the pull of gravity causes the core to be approximately 2.5 years younger than the crust—ignoring geological processes, of course." -phys.org

Time cannot be slower at the core and simultaneously be at zero acceleration. That's not how relativity works.

As a thought experiment, consider the Earth, as it is with its stratified layers - a dense core with progressively less dense layers on top until you get to the crust and out into the atmosphere. Now take the moon and shrink it down to the size of a softball. Retain the mass of the moon, but now it’s close to a neutron star in density. Hold this ultra-dense object directly over the surface of the Earth and let it fall. The deepest portion of Earth’s core and the center of the softball-sized moon will quickly displace the less dense materials between them and merge, with the little moon traveling the most distance and the core moving slightly for the merging. There will be some oscillation, but the dropped "softball moon" will quickly occupy the center of the center, driving the gravity well ever deeper with its added mass. And due to its ultra-density, it will reside at the point of greatest acceleration - the center.

So, for essentially the same volume, Earth’s surface acceleration is now over 10 m/s² due to being in a deeper gravity well because the mass of the Earth has increased an entire moon’s worth without gaining any significant volume. An acceleration tapering to zero at the core is a physics recipe for a hollow earth rather than the densest matter in the gravity well.

Thanks, Joe