# Power radiated by the Sun at different locations [closed]

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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## closed as off-topic by Rob Jeffries, Kyle Kanos, ACuriousMind, John Rennie, NeuneckAug 7 at 10:10

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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