Talk:Equatorial bulge

Proposal: merge the physics of the oblate article
I propose that the physics content of the oblate article is merged into the equatorial bulge article. --Cleonis | Talk 11:47, 17 March 2006 (UTC)

Ocean and Atmosphere

 * The equatorial bulge of Atmosphere is much more important than the ocean and litosphere bulges. This is why the cumulonimbus top level may reach from the ground 20 km high along the ICT and much less at middle and high latitudes. That also means that the equatorial bulge in oceans is more important than in the solid continental Earth level. --Fev 01:42, 7 April 2007 (UTC)

Can you elaborate on what you wrote here? The solid Earth mantle is ductile. Over periods of millions of years, gravity remodels local features. For example, when there are ice ages, the weight of ice mass on for example the north of scandinavian countrys causes that part of the Earth do be pressed down. Since the end of the last ice age, the scandiniavian countries are rising again.

Over periods of millions of years, this kind of movement can be understood in terms of buoyancy. The earth mantle tends to develop in the direction of hydrostatic equilibrium.

In that sense it is to be expected that the oceans have the same equatorial bulge as the solid continental Earth. In don't know about the atmosphere. The atmosphere is compressible, and the density of air mass is much more dependent on the temperature than is the case for the oceans or the solid Earth. --Cleonis | Talk 06:15, 7 April 2007 (UTC)

A simple explanation

 * Arthur N. Strahler (Physical Geography, 1960) explains well the differences between the geoid and the spheroid of reference (The Earth as a geoid, p. 9). He says: "If we could crisscross the continents with canals or tunnels at sea level, permitting the ocean water to seek its level in the hearth of the continent, the geoid could be stablished". He also states that "Under the deep ocean basins, where a large mass of rock is replaced by less dense water, the force of gravity at sea level is greater, depressing the surface of geoid beneath the surface of the reference spheroid". Even Strahler do not explain why is this so, it is very easy to understand that the force of gravity in every point of the Earth's surface depends on the distance to the center of our planet: the higher we go, the less force of gravity we have. That is why it has been stablished that the Chimborazo volcano in Ecuador is higher than the Everest if we use the sea level at Bombay as reference to both mountains.
 * The considerable difference between sea level at the equator and the middle latitudes was established many years ago in the French Guiana: a precision pendule clock from Paris (city placed at a short altitude above sea level) was brought to Cayenne, almost at the same altitude above sea level, but the clock started going slow and the difference went at the same pace until some arrangements need to be made to correct the real geoid altitude. By the same token, a Foucault pendule will have a faster oscillation at the poles than at the equator.
 * Within the Atmosphere, temperature is not the only factor in modifying the density of air mass. Think about this: a place in Algeria has the highest temperature record in the whole world. However the air density there is much higher than along the equatorial circle, where temperatures never go as high. The enormous altitude of many cumulonimbus at the equator (20 km) is not due to the ITC (Inter Tropical Convergence Zone, a concept Strahler didn't know, he talked about the Equatorial Trough) but to the strong centrifugal force at the equator. As you said, air is compressible, but also expands a lot as it happens along the equator. And this expansion upward is the reason explaining Trade winds both in the N. and S. hemispheres that are originated by a kind of "suction".
 * One of the best proofs of what I am saying is that neither ocean currents nor Trade Winds cross the Ecuator. Trade winds, after arriving to the ITCZ, go up (20 km, more or less) and go back to the Tropics, at great altitude, without clouds and slowly descending. And the equatorial currents along oceans bring water to a big altitude above the spheroid of reference and this fact avoids currents to cross equator.
 * Another excellent proof of the higher altitude of oceans at the equatorial belt is that tidal movements are much higher at the middle latitutes (Nova Scotia, France) than in the Equatorial zone, where sea level is already higher and the atraction by the moon need to win overimpossing to this higher level, an unthinkable idea. This is the reason why deltas (and not estuaries) occur at lower latitudes (Niger, Orinoco, Amazonas and so forth).
 * Unfortunately, my English is not as good as I wish. In the WP in Spanish, where I have the same registered name, I explain much better thes ideas. --Fev 02:05, 10 April 2007 (UTC)


 * Can you please specify the following: when you mention 'the higher altitude of oceans at the equatorial belt'; higher relative to what?
 * The Geoid is at the equator 21 kilometers further away from the geometrical center than at the poles.
 * The very definition of the Geoid is water level. On very small scale, the levelness of, say, the foundations for a house can be checked by taking a transparent flexible tube, putting one end at one corner, and the other end at another corner, and then you compare. Of course that kind of procedure is totally impractible for surveying the Geoid over distances of thousands of kilometers, but the procedures that surveyors use are designed to find the shape of the Geoid as defined that way.
 * In that sense it is inherently impossible for the sea-level to deviate from the Geoid, sea-level is the Geoid: at the equator 21 kilometers further away from the Earth's geometrical center than at the poles.


 * It is the solid Earth that deviates locally from the Geoid; ocean floors are klometers below sea-level, some mountain ranges are kilometers above sea-level.


 * Hence my question: higher altitude of oceans relative to what?


 * About tides. We have that the sea mass around Antarctica is the only body of water that runs around the earth along a full circle. We can regard the tide running around the full circle of the antarctic seas as two waves with a period of half a day. The ring of the antarctic sea is the only body of ocean where the tide wave can build up, where it can be reinforced cycle after cycle. From the antarctic sea, waves branch of into the other ocean basins. For example the branche of wave that travels into the atlantic ocean takes about two days to reach the latitude of Europe. Effectively, that means that the tide around England the tide of two days ago.
 * In all, the tidal motions in the atlantic ocean basin are pretty complicated: there is some locally induced sloshing, and a lot of action coming in from the southern ice sea. --Cleonis | Talk 20:34, 10 April 2007 (UTC)

Answering Cleonis

 * First question: I already answered your question in the second paragraph of "A simple explanation" (please read it again). You sould remember that spheroid of reference is a regular shape and the actual geoid is not. Thus, the equatorial bulge of oceans is greater than the solid Earth's bulge at the equatorial circle, the same way it happens with equatorial bulge of atmosphere. These differences are due, not to the Earth's shape but to its rotation movement and the centrifugal force it generates over both oceans and atmosphere.


 * As you said, surveyors use procedures to measure where and how much geoid’s shape differs from spheroid’s. Later, we have to explain these differences. However, you said that geoid’s shape has the same equatorial bulge than the solid Earth has. This is not true, since ocean waters (like it happens with atmosphere) are fluid and continents are not. Continent’s sea level is at the equator 21 km further away from the center of the Earth than at North and South poles. These differences affect gravity, being greater at the poles and lesser at the equator. With atmosphere happens the same thing. Therefore, both atmosphere and oceans have an equatorial bulge because of less gravity at very low latitudes. Besides this, centrifugal force brings an additional reason to the equatorial bulge in fluids.
 * The Geoid's article states that "the geoid varies by only about ±100 m about the reference ellipsoid of revolution" as everybodu can see. It is very clear to see that the positive difference of geoid's level over the spheroid of reference is, obviously, at the equatorial circle, where both, low gravity and high centrifugal force are at the maximum.
 * I never said that sea level deviates from the geoid. It is the spheroid of reference the immaginary figure with a 21 km equatorial bulge. If geoid's shape is about 100 m. over (or under) this immaginary figure and geoid's level is also sea level, this sea level will be, at the equator, above the spheroid of reference because of the equatorial bulge in oceans. Spheroid of reference is a regular figure and geoid's (sea level) is not. That means the sea level at the equator should be 21 km away from the center of the Earth, than sea level at the poles. Is it so?. No, because sea level "climbs" up the continents along equator because equatorial bulge due to the Earth's rotation movement. --Fev 01:59, 15 April 2007 (UTC)
 * I think tidal motions in the Atlantic Ocean Basin are not as complicated as Cleonis thinks, since they are easily predictable for every coastal place. And Atlantic oceanic currents follow a very simple 2 circuit pattern: the Northern Atlantic circuit includes the Gulf Stream, the Canaries Current, the North Equatorial Current, which deviates NW by the coastal shape of northern Brazil and the Guianas and later turns around the Caribbean Sea to become again the starting point of the Gulf Stream current. Part of the warm Gulf Stream Current enters in the Arctic Ocean, which is almost a closed ocean and after moving around it, goes out at the Bering Strait (Oya Shivo current, cold; and Labrador Current, also cold). And Southern Atlantic circuit is almost the same, being this similarity a reason to explain all ocean currents are a response to the rotation movement of the Earth. If it is not the same is because continents shape: the Antarctic drift current westward around the Antarctica is only a circular current and in the northern hemisphere there are two circuits (North Atlantic and North Pacific). Without Europe and North America, Kuro Shivo and Gulf Stream currents would be also one circular current.
 * Even you do not want to talk about equatorial bulge of Atmosphere, you should read articles about Troposphere (the meteorological layer of Atmosphere) and Tropopause (transition layer between Troposphere and Stratosphere). You'll see tropopause is much higher at the equator because of higher centrifugal force there due to the Earth's rotation movement. Understanding this fact will help to understand ocean currents, since both, water and air, are fluids. --Fev 01:59, 15 April 2007 (UTC)
 * My apologies for interupting, but I believe I just read that the Equatorial bulge on Earth is 42.72 Kilometres, not 21. (Stat-ist-ikk (talk) 13:04, 15 April 2010 (UTC))

I'll end this explanation in about two days. Thank you. --Fev 20:35, 13 April 2007 (UTC). So far, so good --Fev 01:59, 15 April 2007 (UTC)


 * Over time, the solid Earth develops in the direction of hydrostatic equilibrium.
 * An important factor in local deviation from hydrostatic equilibrium of solid Earth is plate tectonics. The plate that the continent of India rests on presses against Asia, pushing up the Himalaya. Importantly, in reshaping the Himalaya region, the solid Earth isn't crushed, it is remodeled, the mountains are folds. The remolding of solid Earth in the Himalayan region is such that for every kilometer of rise there has also been a kilometer of sinking. The sheer weight of the mountains causes the region to sink.
 * In that sense it is inevitable that the solid Earth, being ductile, has overall the same equatorial bulge as the oceans.
 * To see what is meant with the assertion that the solid Earth is ductile, see the pitch drop experiment
 * "The first Professor of Physics at the University of Queensland, Professor Thomas Parnell, began an experiment in 1927 to illustrate that everyday materials can exhibit quite surprising properties. The experiment demonstrates the fluidity and high viscosity of pitch, a derivative of tar once used for waterproofing boats.  At room temperature pitch feels solid - even brittle - and can easily be shattered with a blow from a hammer (see the video clip below).  It's quite amazing then, to see that pitch at room temperature is actually fluid!"


 * "In 1927 Professor Parnell heated a sample of pitch and poured it into a glass funnel with a sealed stem. Three years were allowed for the pitch to settle, and in 1930 the sealed stem was cut. From that date on the pitch has slowly dripped out of the funnel - so slowly that now, 77 years later, the ninth drop is only just forming."


 * Over the course of millions of years, the solid Earth settles to an equilibrium that is the same as the equilibrium of an everyday fluid. It is crucial to not overlook the fluidity of the solid Earth. --Cleonis | Talk 06:55, 14 April 2007 (UTC)

This is not fair
Cleonis: I will thank you if you stop writing in my answer and bringing up some themes they don't have anything to do with what I am trying to say here. If you have something to say, I will like to see it in a paragraph with some title like I did before. Thank you. --Fev 15:36, 14 April 2007 (UTC)

How the Geoid differs from the reference spheroid
I copy and paste from above: "That also means that the equatorial bulge in oceans is more important than in the solid continental Earth level. --Fev 01:42, 7 April 2007 (UTC)"

First a thought experiment: a rotating planet with the property that its composition is perfecly homogenous. Such a rotating planet would assume a perfect spheroid shape.

The Geoid, however, is bumpy. Locally, the difference between the Geoid and the reference spheroid can be as much as 100 meter.

To my knowledge, the location on earth where this difference of 100 is reached is at a location in the Indian ocean. There is a part in the basin of the Indian ocean where the solid Earth underneath the water happens to consist of unusually dense material. This dense material exerts a stronger gravitational force than the usual density crust material, and right above that area with stronger gravitation the sea level differs from the reference spheroid more than on any other place on earth. All deviations of the Geoid from the reference ellipsoid are due to such local differences in gravitational pull.

The average sealevel at the equator (averaged all around the ring of the equator) is the same as the height of the reference spheroid at the equator. In other words: at the equator, the average height of the Geoid with respect to the geometrical center is 6,378.137 km. The Geoid is bumpy all over the earth, hence it is also bumpy around the ring of the equator, but there is no structural deviation.

It is also interesting to follow this causal chain: Because of its rotation, the Earth has an equatorial bulge. Because of the equatorial bulge, objects located on the equator are further away from the Earth's center of mass than objects located on the poles. On the poles, the gravitational acceleration is 9.8322 m/s² On the equator, the true gravitational acceleration is 9.8144 m/s². There is less true gravitational acceleration at the equator, because of the greater distance to the Earth's center of mass. What is measured by gravimetric instruments at the equator is an effective gravitational force. The force that is required to circumnavigate the Earth's axis along the equator at one revolution per sidereal day is 0.0339 m/s². The effective gravitational force (the force that gravimetric instruments measure at the equator: 9.7805² m/s.), is the true gravitational force minus the required centripetal force.

Fev, your earlier posts gave me insufficient clues to ascertain whether you dispute any of the above statements. If you disagree with any of the above statemens, please indicate so. --Cleonis | Talk 16:50, 14 April 2007 (UTC)

Let me finish

 * First, I am going to finish what I was writing above, and later, if I do not have more important things to do,. I'll try to answer these new questions you are bringing up —The preceding unsigned comment was added by Fev (talk • contribs) 00:31, 15 April 2007 (UTC).

a mathematical challenge....?
If I want to fint the actual difference between two mountains in Norway, that being Galdhøpigen and Gaustadtoppen. I am given to understand that I put 59 point 85 on the calculator, then press the cosinus button. Minus is next, and then it is 61 point 62, then cosinus again. Then =, but what do I multiply with? Is it enough with 21360 metres, or does the ocean also bulge at the eq.-line? --82.134.28.194 (talk) 13:14, 14 January 2011 (UTC)


 * What do you mean by "the actual difference"? — Yes, the ocean also bulges; the reference ellipsoid is (approximately) mean sea level. —Tamfang (talk) 20:51, 15 July 2012 (UTC)

Alternative explanation
The alternative explanation is drawn down from observations of rotating stars where the mass of the star is roughly the same but a faster rotation generates a greater spherical deviation between equatorial and polar diameters -http://www.jpl.nasa.gov/releases/2001/release_2001_150.html .The uneven rotational gradient between equatorial and polar coordinates is a trait of differential rotation and generates an uneven spherical shape ,the faster the rotation,the more uneven the rotational gradient and subsequently the greater spherical deviation.It makes no sense to reference the Earth's spherical deviation from the dead center of the Earth when the mechanism borrowed from fluid dynamics does a better job of explaining the deviation notwithstanding that this mechanism fits neatly with a crustal evolution/motion.In some ways 'equatorial bulge' is a misnomer as the mechanism is spread out across the rotating globe hence the planet's spherical deviation is a better description.This explanation was given in 2005 and long before discussions of rotation entered plate tectonics but fundamentally differential rotations kills two birds with one stone so that it brings two planetary features close together using a common mechanism Gkell1 (talk) 14:23, 3 July 2011 (UTC)

which average
To get the Earth's mean radius, these two radii must be averaged. To get the Earth's mean radius, we must start with a definition of 'mean radius'. Is it the radius of a sphere of equivalent volume? (That, at least, is easy to compute.) Is it the average, taken over the surface area of the geoid, of distance to the core? If the latter, which distance: the obvious straight line, or the component parallel with local gravity? Likely there are other candidate definitions. —Tamfang (talk) 20:54, 15 July 2012 (UTC)

Calculating oblateness
I have an extra piece of information that I never learned but am pretty sure is true. Is it the case that when ever a planet is rotating, it's oblate but the oblateness changes the gravitational field of the planet and that in turn makes it even more oblate, so when ever a planet is rotating, it is more oblate than it would be if it had the gravitational field that it would have had if it was a sphere. Blackbombchu (talk) 02:51, 17 June 2013 (UTC)

Centripetal Force
I know this is an old argument, and I suppose I am taking a politically incorrect view. But I have a great deal of difficulty accepting that rotation causes centripetal force. I have similar difficulty accepting that centripetal force could cause a bulge. (well I suppose the centripetal force of my belt causes my tummy to bulge). Seriously though, rotation tends to push objects away because of their linear momentum. That constitutes a centrifugal force, which results in the equatorial bulge. Gravity pulls everything in, thereby imposing a centripetal force. Ultimately equilibrium must be maintained within the planet, which is achieved when the mass of the bulge itself generates enough gravitational (centripetal) force to overcome the centrifugal force of rotation. no? Doc.Ian (talk) 21:04, 24 September 2016 (UTC)

Bad explanation
I removed the following "mathematical explanation" of the bulge:


 * The standard formula for this [centrifugal] force is the relationship $$F_c = Mv^2 / R$$. However, velocity at the surface is equal to the product of radius and rotational velocity, and therefore the force is directly proportional to radius. Viewing the globe as a series of rotating discs, the radius R toward the poles decreases and thus a smaller force is produced for the same rotational velocity (approaching zero at the pole). Moving towards the Equator, v^2 increases much faster than R, thus producing the greatest force at the Equator.
 * In addition, because Earth's dense metallic core is included in the cross-sectional disc at the Equator, it contributes more to the mass of the disc. Similarly, there is a bulge in the water envelope of the oceans surrounding Earth; this bulge is created by the greater centrifugal force at the Equator and is independent from tides. Sea level at the Equator is 21.36 km higher than that at either pole, in terms of distance from the center of the planet.

There are many things wrong with it. "Radius" is ambiguous; it should be "distance from the axis". One cannot analyze the disks separately, because every disk contributes gravity to every point on the surface with different force and direction, so that integration disk-by-disk is very difficult. A homogeneous sphere has the same external gravitational field as the same mass concentrated at its center, but that is not true for a disk. Moreover, each disk has different (and non-homogeneous) density. --Jorge Stolfi (talk) 02:07, 1 April 2019 (UTC)

Planet versus Celestial Body?
The summary states: "An equatorial bulge is a difference between the equatorial and polar diameters of a planet due to the centrifugal force exerted by the rotation about the body's axis."

Later it's stated that: "Generally any celestial body that is rotating (and that is sufficiently massive to draw itself into spherical or near spherical shape) will have an equatorial bulge matching its rotation rate."

I saw a number of papers that used the term "equatorial bulge" for celestial bodies such as the Sun, Titan (not believed to be an equatorial ridge), and the moon.

I propose changing the summary to say: "An equatorial bulge is a difference between the equatorial and polar diameters of a celestial body due to the centrifugal force exerted by rotation about the body's axis."

(I suck at Wiki markup, forgive me) JackW2 (talk) 12:11, 27 March 2021 (UTC)


 * I am not sure if this applies to any bodies other than planets. Asteroids, comets, etc. are too small to be gravitationally rounded and equatorial bulge cannot be defined for them, while stars as far as I know do not rotate like a rigid body (for example solar rotation differs between equator and poles by as much as 50%, while many giant stars seem to have an irregular shape conditioned more by local differences in convection than rotation) so this is again not applicable. And, rounded moons are for all purposes planets, planetary scientists generally include them when talking about planets, and not referring to them as such is a pervasive popular science thing (see https://arxiv.org/abs/2110.15285 for historical background of such usage, refer also to planetary-mass moon) 93.103.223.236 (talk) 09:22, 10 April 2023 (UTC)

Expression for equatorial and polar radius
The formulas given for equatorial and polar radius inline are $$a_e = a\, (1 + \tfrac{f}{3})$$ and $$a_p = a\,(1 - \tfrac{2f}{3})$$; however multiplying those together to get the volume does not match; $$a_e^2 a_p \neq a^3$$ (the terms involving $$f$$ only cancel out to linear order). I worked back from the mathematical definition of flattening $$f = \tfrac{a_e - a_p}{a_e}$$ and this way one can obtain different expressions $$a_p = a\,(1-f)^\frac{2}{3}$$ and $$a_e = a\,(1-f)^{-\frac{1}{3}}$$ which can indeed be linearized to the given expressions for small $$f$$. However I was unable to find any suitable literature to cite here so I'm unaware if this linear approximation is just how it's conventionally done for simplicity so I refrained from directly editing this in article. Additional input would be appreciated. 93.103.223.236 (talk) 09:11, 10 April 2023 (UTC)