#format jsmath
= 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.