The Earth's shape, like that of all major planets, approximates a sphere. A true sphere has a unique radius, but on Earth the distance from the mean sea level at each point on the surface to the center (the "radius of the Earth at a point on the surface") varies slightly from place to place. With few exceptions, this variation ranges from 6,356.750 km to 6,378.135 km (≈3,949.901 — 3,963.189 mi), which are the "polar radius" and the "equatorial radius", respectively. Strictly speaking only a sphere has a true radius, but it is usual to speak of the "radius of the Earth" to refer to various fixed distances and to various mean radii, discussed below. The numerical differences between the different "radii" vary by well less than one percent. For all planets, including Earth, the systematic sources of the distortion from spherical are rotation, variation of mass density within the planet, and tidal forces. [The center of the Earth is somewhat model dependent. Exceptions to the cited range occur near the South Pole and along the equator. Also, differences due to variation of mass density within the planet and tidal forces require data for the entire surface of the Earth and are not included here. For details see Figure of the Earth, Geoid, and Earth tide.]

"Note: Earth radius is sometimes used as a unit of distance, especially in astronomy and geology. It is usually denoted by $R_oplus$.

Introduction

Rotation of a planet causes it to approximate an "oblate ellipsoid/spheroid" with a bulge at the equator and flattening at the North and South Poles, so that the "equatorial radius" $a$ is larger than the "polar radius" $b$ by approximately $a q$ where the "oblateness constant" $q$ is:::$q=frac\left\{a^3 omega^2\right\}\left\{GM\right\}, ,!$where $omega$ is the angular frequency, $G$ is the gravitational constant, and $M$ is the mass of the planet. This follows from the International Astronomical Union definition rule (2): a planet assumes a shape due to hydrostatic equilibrium where gravity and centrifugal forces are nearly balanced. [http://www.iau2006.org/mirror/www.iau.org/iau0603/index.html IAU 2006 General Assembly: Result of the IAU Resolution votes] ] For the Earth $q^\left\{-1\right\}approx 289$, which is close to the measured inverse flattening $f^\left\{-1\right\}approx 298.257$.Additionally, the bulge at the equator shows slow variations. The bulge had been declining, but since 1998 the bulge has increased, possibly due to redistribution of ocean mass via currents. [ [http://www.gsfc.nasa.gov/topstory/20020801gravityfield.html Satellites Reveal A Mystery Of Large Change In Earth's Gravity Field] , Aug. 1, 2002, Goddard Space Flight Center. ] The variation in density and crustal thickness causes gravity to vary on the surface, so that the mean sea level will differ from the ellipsoid. This difference is the "geoid height", positive above or outside the ellipsoid, negative below or inside. The geoid height variation is under 110 m on Earth. The geoid height can have abrupt changes due to earthquakes (such as the Sumatra-Andaman earthquake) or reduction in ice masses (such as Greenland). [ [http://www.spaceref.com/news/viewpr.html?pid=18567 NASA's Grace Finds Greenland Melting Faster, 'Sees' Sumatra Quake] , December 20, 2005, Goddard Space Flight Center. ]

The tides from the gravity of the Moon and Sun cause the surface of the Earth to rise and fall by tenths of meters at a point over a nearly 12 hr period.

Therefore, the values defined below are based on a "general purpose" model, refined as globally precisely as possible within 5 m of reference ellipsoid height, and to within 100 m of mean sea level (neglecting geoid height).

Additionally, the radius can be estimated from the curvature of the Earth at a point. Like a torus the curvature at a point will be largest (tightest) in one direction (North-South on Earth) and smallest (flattest) perpendicularly (East-West). The corresponding radius of curvature depends on location and direction of measurement from that point. A consequence is that a distance to the true horizon at the equator is slightly shorter in the north/south direction than in the east-west direction.

In summary, local variations in terrain prevent the definition of a single absolutely "precise" radius. One can only adopt an idealized model. Since the estimate by Eratosthenes, many models have been created. Historically these models were based on regional topography, giving the best reference ellipsoid for the area under survey. As satellite remote sensing and especially the Global Positioning System rose in importance, true global models were developed which, while not as accurate for regional work, best approximate the earth as a whole.

The following radii are fixed, and do not include a variable location dependence.

Equatorial radius: $a$

The Earth's equatorial radius, or semi-major axis, is the distance from its center to the equator and equals 6,378.137 km (≈3,963.191 mi; ≈3,443.918 nmi). At coord|0|S|121.83|E|, the geoid height rises to 63.42 m above the reference ellipsoid (), giving a total radius of 6,378.200 km.The equatorial radius is often used to compare Earth with other planets.

Polar radius: $b$

The Earth's polar radius, or semi-minor axis, is the distance from its center to the North and South Poles, and equals 6,356.7523 km (≈3,949.903 mi; ≈3,432.372 nmi). The geoid height (WGS-84) at the North Pole is 13.6 m above the reference ellipsoid, and at the South Pole 29.5 m below the reference, giving the more exact 6,356.766 km and 6,356.723 km, respectively.

Radius at a given geodetic latitude

The Earth's radius at geodetic latitude, $phi,!$, is:

:$R=R\left(phi\right)=sqrt\left\{frac\left\{\left(a^2cos\left(phi\right)\right)^2+\left(b^2sin\left(phi\right)\right)^2\right\}\left\{\left(acos\left(phi\right)\right)^2+\left(bsin\left(phi\right)\right)^2;,!$

These are based on a oblate ellipsoid.

Eratosthenes used two points, one exactly north of the other. The points are separated by distance $D$, and the vertical directions at the two points are known to differ by angle of $heta$, in radians.A formula based on Eratosthenes method is:::$R= frac\left\{D\right\}\left\{ heta\right\};,!$which gives an estimate of radius based on the north-south curvature of the Earth.

Meridional

:In particular the Earth's "radius of curvature in the (north-south) meridian" at $phi,!$ is::::$M=M\left(phi\right)=frac\left\{\left(ab\right)^2\right\}\left\{\left(\left(acos\left(phi\right)\right)^2+\left(bsin\left(phi\right)\right)^2\right)^\left\{3/2;,!$

Normal

:If one point had appeared due east of the other, one finds the approximate curvature in east-west direction. East-west directions can be misleading. Point "B" which appears due East from "A" will be closer to the equator than "A". Thus the curvature found this way is smaller than the curvature of a circle of constant latitude, except at the equator.West can exchanged for east in this discussion.] :This "radius of curvature in the prime vertical", which is perpendicular, or "normal", to "M" at geodetic latitude $phi,!$ is: ["N" is defined as the radius of curvature in the plane which is normal to both the surface of the ellipsoid at, and the meridian passing through, the specific point of interest.] :::$N=N\left(phi\right)=frac\left\{a^2\right\}\left\{sqrt\left\{\left(acos\left(phi\right)\right)^2+\left(bsin\left(phi\right)\right)^2;,!$Note that "N=R" at the equator:

The Earth's mean radius of curvature (averaging over all directions) at latitude $phi,!$ is::::$R_a=sqrt\left\{MN\right\}=frac\left\{a^2b\right\}\left\{\left(acos\left(phi\right)\right)^2+\left(bsin\left(phi\right)\right)^2\right\};,!$

The Earth's radius of curvature along a course at geodetic bearing (measured clockwise from north) $alpha,!$, at $phi,!$ is: [A related application of "M" and "N": if two nearby points have the difference in latitude of $dphi,!$ and longitude of $dlambda,!$ (in radians) with mean latitude $phi,!$, then the distance "D" between them is:::$Dapproxsqrt\left\{\left(Mdphi\right)^2+\left(Ncosphi dlambda\right)^2\right\}.,!$The quantities inside the parentheses are approximately $Dcosalpha$ and $Dsinalpha$, respectively. Thus $dphi$ and $dlambda$ be estimated from "D", "M", and "N".] :::$R_c=frac$}_{1{frac{cos(alpha)^2}{M}+frac{sin(alpha)^2}{N.,!

The Earth's equatorial radius of curvature in the meridian is::::$frac\left\{b^2\right\}\left\{a\right\},!$= 6335.437 km

The Earth's polar radius of curvature is::::$frac\left\{a^2\right\}\left\{b\right\},!$= 6399.592 km

Quadratic mean radius: $Q_r$

:::$Q_r=sqrt\left\{frac\left\{3a^2+b^2\right\}\left\{4;,!$

It is this radius that would be used to approximate the ellipsoid's average great ellipse (i.e., this is the equivalent spherical "great-circle" radius of the ellipsoid).
For Earth, $Q_r$ equals 6,372.7976 km (≈3,959.873 mi; ≈3,441.035 nmi).

Authalic mean radius: $A_r$

Earth's authalic ("equal area") mean radius is 6,371.0072 km (≈3,958.760 mi; ≈3,440.069 nmi). This number is derived by square rooting the average (latitudinally cosine corrected) geometric mean of the meridional and transverse equatorial, or "normal" (i.e., perpendicular), arcradii of all surface points on the spheroid, which can be reduced to a closed-form solution::::$A_r=sqrt\left\{frac\left\{a^2+frac\left\{ab^2\right\}\left\{sqrt\left\{a^2-b^2ln\left\{left\left(frac\left\{a+sqrt\left\{a^2-b^2b ight\right)\left\{2=sqrt\left\{frac\left\{A\right\}\left\{4pi;,!$where $A$ is the authalic surface area of Earth. This would be the radius of a hypothetical perfect sphere which has the same, geometric mean oriented surface area as the spheroid.

Volumetric radius: $V_r$

Another, less utilized, sphericalization is that of the volumetric radius, which is the radius of a sphere of equal volume::::$V_r=sqrt \left[3\right] \left\{a^2b\right\};,!$For Earth, the volumetric radius equals 6,371.0008 km (≈3,958.760 mi; ≈3,440.069 nmi).

Another radius mean is the meridional mean, which equals the radius used in finding the perimeter of an ellipse. It can also be found by just finding the average value of "M"::::$M_r=frac\left\{2\right\}\left\{pi\right\}int_\left\{0\right\}^\left\{90^circ\right\}!M\left(phi\right),dphi;approxleft \left[frac\left\{a^\left\{1.5\right\}+b^\left\{1.5\left\{2\right\} ight\right] ^\left\{1/1.5\right\};,!$

For Earth, this works out to 6367.4491 km (≈3,956.545 mi; ≈3,438.147 nmi).

ee also

*Figure of the Earth
*Earth tide

Notes and references

Wikimedia Foundation. 2010.

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