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How WGS 84 defines Earth

Besides being a map/chart datum WGS 84 (World Geodetic System of 1984) also defines the shape and size of the ellipsoid of revolution (an oblate spheroid) that is considered to be the best mathematical model of Earth:

 Flattening = f = Semi-major axis = equatorial radius = a = 1/298.257223563  (≈ 3.35 ‰) 6 378 137.0 m From these two numbers it is possible to calculate: Semi-minor axis = polar distance = b = (1−f)a = 6 356 752.3142 m Difference between equatorial radius and polar distance = a−b = 21 384.6858 m Axis ratio = b/a = 1−f = 0.996 647 189 335 First eccentricity squared = e2 = 1−(b/a)2 = (2−f)f = First eccentricity = e = 0.006 694 379 990 14 0.081 819 190 842 621 Arithmetic mean radius of Earth = (2a+b)/3 = (1−f/3)a = 6 371 008.7714 m Surface area of Earth = A = 2πa2+π(b2/e)ln[(1+e)/(1−e)] = Radius of sphere of equal surface area = ½√(A/π) = 510 065 621.724 km2 6 371 007.1809 m (= 2π[a2+(b2/e)arctanh(e)]) Volume of Earth = V = 4πa2b/3 = Radius of sphere of equal volume = (�V/π)¹⁄³ = (a2b)¹⁄³ = 1 083 207 319 801 km3 6 371 000.7900 m (= geometric mean radius) Maximum circumference of Earth = circumference of Earth at the equator = circumference of parallel of latitude at 0° latitude = 2πa = Radius of sphere of equal circumference = a 40 075.017 km(see above) Minimum circumference of Earth = circumference of Earth through the poles = 4 × (distance from the equator to a pole) = 4 × 10001.966 km = Radius of sphere of equal circumference = 40 007 863 m/2π = 40 007.863 km 6 367 449.1458 m See the main table. Difference between maximum and minimum circumference = 67.154 km Radius of curvature at the poles = a/(1−e2)¹⁄² =Radius of curvature in a meridian plane at the equator = a(1−e2) = 6 399 593.6258 m 6 335 439.3273 m (= a2/b)(= b2/a) Difference between maximum and minimum radius of curvature = 64 154.2985 m Latitude where latitudal widths are equal to equator widths = (e.g. width of one minute of latitude equals width of one minute along the equator). 54.14432°54° 46.858′54° 46′ 51.5″ (1 min. of lat.    = 1855.325 m) (1 min. of long. = 1072.371 m) If Earth had been a perfect sphere, this would had happened at all latitudes. Latitude halfway between the equator and one of the poles = 45.78096°45° 08.659′45° 08′ 39.5″ (1 min. of lat.    = 1852.243 m) (1 min. of long. = 1310.811 m) If Earth had been a perfect sphere, this would had happened at exactly 45° of lati­tude (½ × 90° = 45°). Latitude where latitudal widths and longitudal widths are equal = (e.g. width of one minute of latitude equals width of one minute of longitude). 06.58980°06° 35.388′06° 35′ 23.3″ (1 min. of lat.    = 1843.148 m) (1 min. of long. = 1843.148 m) If Earth had been a perfect sphere, this would had happened at the equator.

 Links: PDF files: NIMA Technical Report TR8350.2, Department of Defense World Geodetic System 1984, Its Definition and Relationships with Local Geodetic Systems (especially chapter 3). DMA Technical Report TR8350.2-A, Supplement To Department of Defense World Geodetic System 1984 Technical Report Methods, Techniques, and Data Used in WGS 84 Development (especially chapter 3).

Calculated by Sigurd Humerfelt (25th January 2000, last updated 26th October 2010) (133).