Polar Ice Cap

Question: How much of the Earth's sunlight hits land under the icecaps?

First approximation: Circular Antarctica, no axial tilt

Antarctica is 14.6E6 km2, or 0.0286 of the earth's surface area A_E = 5.1E8 km2 of the earth. We can approximate this as a circular cap. The cap area is A_C = 2\pi R^2(1-\sin(\phi)) . W can compute the latitude of northern edge of the cap from \phi = \sin^{-1} ( 1 - 2 A_C / A_E ) , which is 70.52°S or 1.23077 radians. The portion of the sun-facing disk occupied by this cap is 0.5 - ( \phi + 0.5 \sin( 2 \phi ) ) / \pi or 0.00815 . That means the average illumination of the cap is 28.5% (0.00815/0.0286) of the average illumination of the rest of the planet. Since the cap is mostly white from ice and snow, much of that reduced illumination bounces back out into space.

PolarIceCap (last edited 2012-06-16 01:52:42 by KeithLofstrom)