Ellipsoid surface area

The Endeavour 2014-07-06

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

$A = 2\pi \left( c^2 + \frac{ab}{\sin\phi} \left( E(\phi, k) \sin^2\phi + F(\phi, k) \cos^2 \phi\right)\right)$

where

$\cos$

and

$m = k^2 = \frac{a^2(b^2 - c^2)}{b^2(a^2 - c^2)}$

The functions E and F are incomplete elliptic integrals

$F(\phi, k) = \int_0^\phi \frac{d\theta}{\sqrt{1 - k^2 \sin^2\theta}}$

and

$E(\phi, k) = \int_0^\phi \sqrt{1 - k^2 \sin^2\theta}\,d\theta$

implemented in SciPy as ellipeinc and 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, arccosfrom scipy.special import ellipkinc, ellipeinc# values in meters based on GRS 80# http://en.wikipedia.org/wiki/GRS_80equatorial_radius = 6378137polar_radius = 6356752.314140347a = b = equatorial_radiusc = polar_radiusphi = 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)**2ellipsoid_area = 2*pi*(c**2 + a*b*temp/sin(phi))# sphere with radius equal to average of polar and equatorialr = 0.5*(a+c)sphere_area = 4*pi*r**2print(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.