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The power radiated by the sun is 3.90 1026 W. The earth orbits the sun in a nearly circular orbit of radius 1.50 1011 m. The earth's axis of rotation is tilted by 27° relative to the plane of the orbit (see the drawing), so sunlight does not strike the equator perpendicularly. What power strikes a 0.575-m2 patch of flat land at the equator at point Q?

Answer in W

  • Physics -

    Find the intensity at the Earth surface.

    Power= .575/(4PI*(1.50E11)^2 ) * 3.9E26 watts.

    Now, the area is not perpendicular, so the perpendicular area is power above multiplied by cos27

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