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| Antarctica is 14.6E6 km^2^, or 0.0286 of the 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 ( $ 2\pi R^2(1-sin(\phi)) $ ). The portion of the sun-facing disk occupied by this cap is $ R^2 ( \pi ( \pi - 2\phi ) - 0.5 \sin{ 2\phi } ) $ | Antarctica is 14.6E6 km^2^, or 0.0286 of the 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 ( $ 2\pi R^2(1-sin(\phi)) $ ). The portion of the sun-facing disk occupied by this cap is $ 0.5 - ( \phi + 0.5 \sin{ 2\phi } / 2 \pi $ |
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 5.1E8 km2 of the earth. We can approximate this as a circular cap with the northern edge at latitude \phi = 1.331 radians = 76.26°S ( 2\pi R^2(1-sin(\phi)) ). The portion of the sun-facing disk occupied by this cap is 0.5 - ( \phi + 0.5 \sin{ 2\phi } / 2 \pi