How much difference does the earth’s equatorial bulge make in its surface area?
To first approximation, the earth is a sphere. The next step in sophistication is to model the earth as an ellipsoid.
The surface area of an ellipsoid with semi-axes a ≥ b ≥ c is
The functions E and F are incomplete elliptic integrals
implemented in SciPy as
ellipkinc. Note that the SciPy functions take m as their second argument rather its square root k.
For the earth, a = b and so m = 1.
The following Python code computes the ratio of earth’s surface area as an ellipsoid to its area as a sphere.
from scipy import pi, sin, cos, arccos from scipy.special import ellipkinc, ellipeinc # values in meters based on GRS 80 # http://en.wikipedia.org/wiki/GRS_80 equatorial_radius = 6378137 polar_radius = 6356752.314140347 a = b = equatorial_radius c = polar_radius phi = arccos(c/a) # in general, m = (a**2 * (b**2 - c**2)) / (b**2 * (a**2 - c**2)) m = 1 temp = ellipeinc(phi, m)*sin(phi)**2 + ellipkinc(phi, m)*cos(phi)**2 ellipsoid_area = 2*pi*(c**2 + a*b*temp/sin(phi)) # sphere with radius equal to average of polar and equatorial r = 0.5*(a+c) sphere_area = 4*pi*r**2 print(ellipsoid_area/sphere_area)
This shows that the ellipsoid model leads to 0.112% more surface area relative to a sphere.
Source: See equation 19.33.2 here.
Update: It was suggested in the comments that it would be better to compare the ellipsoid area to that of a sphere of the same volume. So instead of using the average of the polar and equatorial radii, one would take the geometric mean of the polar radius and two copies of the equatorial radius. Using that radius, the ellipsoid has 0.0002% more area than the sphere.
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