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I am wondering can someone help to solve second part which extends first part;

The power radiated by the Sun is ${3.9*10^{26}}_{watt}$. The earth orbits the sun in a nearly circular orbit of radius $1.5 *10^{11}m$. The earth’s axis of rotation is tilted by $27^o$ relative to the plane of the orbit , so sunlight does not strike the equator perpendicularly. What power strikes a $0.75$ $m^2$patch of flat land at the equator?

$P_0 = I*4*π*r^2$

$P = I*A*cos(\alpha)$

then

$P=\frac{P_0 * A *cos(\alpha)}{ 4*π *r^2}$

then

$P=\frac{(3.9*10^{26})(0.75)(cos(27))}{(4*π*(1.5*10^{11})^2)}=920_{watt}$

~ How much power strikes a patch of the same area located at latitude (at the same time of year) of 43° S (like Christchurch, New Zealand) or 43° N?

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How do you find the second part of the question to be different from the first? –  kleingordon Apr 18 '13 at 2:14
    
I assume you are talking about noon at a solstice, yes? At noon on an equinox, the sun's radiation does strike a point on the equator perpendicularly. –  WhatRoughBeast Aug 7 at 1:19

1 Answer 1

There was a similar question asked at How do you calculate power at the focal point of a mirror?

The answer isn't based just on simple trigonometry because the atmosphere has a significant impact on the power reaching the Earth's surface. You want to look into the "AM1" data which is a measurement of solar energy after passing through one atmosphere worth of air. Keep in mind that the amount of atmosphere (the total thickness of air) light must pass through depends on the angle the light approaches the Earth so it varies by latitude, elevation, time of day, and day of the year.

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I'm guessing this is a homework question, in which students are expected to neglect atmospheric effects –  kleingordon Apr 18 '13 at 2:12

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