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| 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 with the northern edge at latitude $ \phi $ = 1.331 radians = 76.26°S . The cap area is $ A_C = 2\pi R^2(1-\sin(\phi)) $, so we can compute the latitude of the cap from $ \phi = \sin^{-1} ( 1 - 2 A_C / A_E ) $ . The portion of the sun-facing disk occupied by this cap is $ 0.5 - ( \phi + 0.5 \sin{ 2\phi } / \pi $ | 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. |
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.