Equatorial bulge: Difference between revisions

{{for|the feature on some of [[Saturn]]'s moons|equatorial ridge}}

{{main|Earth's ellipsoidradii}}

{{further|Earth ellipsoid|Figure of the Earth}}

The planet [[Earth]] has a rather slight equatorial bulge:; itits equatorial diameter is about {{cvt|43|km|mi}} widergreater atthan theits equatorpolar than pole-to-polediameter, with a difference whichof is close toabout {{fract|1/300|298}} of the equatorial diameter. If the Earth were scaled down to a globe with an equatorial diameter of {{convert|1 meter at the equator|m|ft}}, that difference would be only {{cvt|3 millimeters|mm|in}}. While too small to notice visually, that difference is still more than twice the largest deviations of the actual surface from the ellipsoid, including the [[list of mountain peaks by prominence|tallest mountains]] and deepest [[oceanic trenchestrench]]es.

The[[Earth's rotation of the earth]] also affects the [[sea level]], the imaginary surface thatused isas useda [[reference frame]] from which to measure [[altitude]]s from. This surface coincides with the mean water surface level in oceans, and is extrapolated over land by taking into account the local [[gravity of Earth|gravitational potential]] and the centrifugal force.

The difference of the [[radius|radii]] is thus about {{cvt|21 |km|mi}}. An observer standing at sea level on either [[geographical pole|pole]], therefore, is {{cvt|21 |km|mi}} closer to Earth's center than if standing at sea level on the Equator. As a result, the highest point on Earth, measured from the center and outwards, is the peak of Mount [[Chimborazo]] in [[Ecuador]] rather than [[Mount Everest]]. But since the ocean also bulges, like Earth and [[atmosphere of Earth|its atmosphere]], Chimborazo is not as high above sea level as Everest is. Similarly the lowest point on Earth, measured from the center and outwards, is the [[Litke Deep]] in the [[Arctic Ocean]] rather than [[Challenger Deep]] in the [[Pacific Ocean]]. But since the ocean also flattens, like Earth and its atmosphere, Litke Deep is not as low below sea level as Challenger Deep is.

More precisely, theEarth's surface of the [[Earth]] is usually approximated by an ideal [[oblate ellipsoid]], for the purposes of defining precisely the [[latitude]] and [[longitude]] grid for [[cartography]], as well as the "center of the Earth". In the [[World Geodetic System|WGS-84]] standard [[Earth ellipsoid]], widely used for map-making and the [[GPS]] system, the[[Earth radius of the |Earth's radius]] is assumed to be {{cvt|6378.137|km|mi|comma=gaps}} atto the equatorEquator and {{cvt|6356.7523142|km|mi|comma=gaps}} center-to- either pole;, meaning a difference of {{cvt|21.3846858|km|mi|comma=gaps}} inbetween the radii andor {{cvt|42.7693716|km|mi|comma=gaps}} inbetween the diameters, and a relative [[flattening]] of 1/298.257223563. The sea level[[ocean surface]] is much closer to this standard ellipsoid than the surface[[Earth's ofcrust|solid thesurface]] solidof Earth is.

[[Gravity]] tends to contract a celestial body into a [[sphere]], the shape for which all the mass is as close to the center of gravity as possible. [[Rotation]] causes a distortion from this spherical shape; a common measure of the distortion is the [[flattening]] (sometimes called ellipticity or oblateness), which can depend on a variety of factors including the size, [[angular velocity]], [[density]], and [[Elasticity (physics)|elasticity]].

As long as the proto-planet is still too oblate to be in equilibrium, the release of [[gravitational energy|gravitational potential energy]] on contraction keeps driving the increase in rotational kinetic energy. As the contraction proceeds, the rotation rate keeps going up, hence the required force for further contraction keeps going up. There is a point where the increase of rotational kinetic energy on further contraction would be larger than the release of gravitational potential energy. The contraction process can only proceed up to that point, so it halts there.

[[ImageFile:Forces oblate spheroid2.gif|frame|right|The forces at play in the case of a planet with an equatorial bulge due to rotation.<br/>

Red arrow: gravity <br/>

Green arrow,: the [[normal force]] <br/>

Blue arrow: the resultant force <br/>

The resultant force provides required centripetal force. Without this centripetal force frictionless objects would slide towards the equator. <br/>

In calculations, when a coordinate system is used that is co-rotating with the Earth, the vector of the notional [[centrifugal force (rotating reference frame)|centrifugal force]] points outward, and is just as large as the vector representing the centripetal force.]]

Because of a planet's rotation around its own axis, the [[gravitational acceleration]] is less at the [[equator]] than at the [[geographical pole|poles]]. In the 17th century, following the invention of the [[pendulum clock]], French scientists found that clocks sent to [[French Guiana]], on the northern coast of [[South America]], ran slower than their exact counterparts in Paris. Measurements of the acceleration due to gravity at the equator must also take into account the planet's rotation. Any object that is stationary with respect to the surface of the Earth is actually following a circular trajectory, circumnavigating the Earth's axis. Pulling an object into such a circular trajectory requires a force. The acceleration that is required to circumnavigate the Earth's axis along the equator at one revolution per [[sidereal day]] is 0.0339&nbsp;m/s²². Providing this acceleration decreases the effective gravitational acceleration. At the equatorEquator, the effective gravitational acceleration is 9.7805&nbsp;m/s². This means that the true gravitational acceleration at the equatorEquator must be 9.8144&nbsp;m/s² (9.7805&nbsp;+&nbsp;0.0339&nbsp;=&nbsp;9.8144).

At the poles, the gravitational acceleration is 9.8322&nbsp;m/s². The difference of 0.0178&nbsp;m/s² between the gravitational acceleration at the poles and the true gravitational acceleration at the equatorEquator is because objects located on the equatorEquator are about {{cvt|21 kilometers|km|mi}} further away from the [[center of mass]] of the Earth than at the poles, which corresponds to a smaller gravitational acceleration.

In summary, there are two contributions to the fact that the effective gravitational acceleration is less strong at the equator than at the poles. About 70 percent% of the difference is contributed by the fact that objects circumnavigate the Earth's axis, and about 30 percent% is due to the non-spherical shape of the Earth.

The diagram illustrates that on all latitudes the effective gravitational acceleration is decreased by the requirement of providing a centripetal force; the decreasing effect is strongest on the equatorEquator.

}}</ref><ref>{{cite journal

}}</ref><ref>{{cite journal

}}</ref> They depend on the orientation of the Earth's symmetry axis in the inertial space, and, in the general case, affect ''all'' the Keplerian [[orbital elements]] with the exception of the [[semimajor axis]]. If the reference ''z'' axis of the coordinate system adopted is aligned along the Earth's symmetry axis, then only the [[longitude of the ascending node]] Ω, the [[argument of pericenter]] ω and the [[mean anomaly]] ''M'' undergo secular precessions.<ref>{{cite journal

*<math>a_e = a\, (1 + \tfrac{f}{3})</math> and <math>a_p = a\,(1 - \tfrac{2f}{3})</math> are respectively the equatorial and polar radius,{{Dubious|date=April 2023}}