Ellipsoid surface area

Posted on by John

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 \phi = \frac{c}{a}

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 numpy import pi, sin, cos, arccos
from scipy.special import ellipkinc, ellipeinc

# values in meters based on GRS 80
# https://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.

Update 2: This post gives a simple but accurate approximation for the area of an ellipsoid.

2 thoughts on “Ellipsoid surface area”

  1. eric

    How did you decide to use the average of the equatorial radius and the polar radius as the radius of the sphere used for comparison? I would have thought the question “How much difference does the earth’s equatorial bulge make in its surface area?” would only have meaning when related to two possible ‘states’ the earth could have. Therefore, I think a more natural basis of comparison would be between two shapes that have the same volume (I’m fine with ignoring possible density variations caused by rotation + gravity, so volume would be equivalent to mass). So, given a fixed value for Earth’s mass (volume), what would it’s radius be if it were a sphere? It seems to me that that radius should be the one used for comparison. I think this value for spherical radius should be the cube root of the product of the ellipsoidal radii.

  2. Nikhil

    Eric, yes, the identity would be r = ∛abc, which happens to be larger than (a+c)/2. Using this model, the elliptical surface area is only 0.000201% larger than the spherical surface area!

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