# 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 km^{2}, or 0.0286 of the earth's surface area A_E = 5.1E8 km^{2} 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.