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Homework Help: Stefan's Law Applied to the Sun

  1. Sep 24, 2011 #1
    I am trying to understand an example in my Modern Physics textbook (Example 3.1, page 5 in thishttp://phy240.ahepl.org/Chp3-QT-of-Light-Serway.pdf" [Broken] or pg 69 using the book numbering)

    I don't understand why the average earth-sun distance is being used in the conservation of energy equation instead of the radius of the earth. Isn't the idea that whatever total power is emitted from the sun must equal the total power received at the earth? [I think e_total can be power received, too, right? It just depends on context?] So, to make sure the power at each end of the journey is equal, we multiply the power per unit area (the values we have) by the surface area of each body. But in that case we would want 4pi*(Earth Radius) not 4pi*(Earth-Sun Distance)

    I am also a bit worried that no energy is "lost" on the way to the Earth. The book doesn't really talk about that. How do we *know* that energy is conserved in this way?

    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 24, 2011 #2
    To make my second question a bit more precise, how can the total power emitted by the sun equal the total power received at the Earth, since presumably at least half of the power radiated by the Sun goes off in a direction opposite the Earth?
  4. Sep 24, 2011 #3

    D H

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    Absolutely not! The Earth intercepts a tiny, tiny fraction of the Sun's output. The Earth's cross section to solar radiation is approximately the area of a circle with the radius of the Earth. The Sun's radiation output is pretty much the uniform with respect to direction. By the time the radiation gets to 1 AU it is spread over the surface of a sphere 2 AU in diameter. The fraction of the Sun's output that is received by the Earth is

    [tex]f = \frac{2\pi r_e^2}{4\pi R_e^2}[/tex]

    where [itex]r_e[/itex] is the radius of the Earth, 6378 km, and [itex]R_e[/itex] is the radius of the Earth's orbit, 149,598,000 kilometers. The value of this f is about 9×10-10. Tiny.

    The total power emitted by the Sun is equal to the total power crossing this 2 AU diameter sphere.
  5. Sep 24, 2011 #4
    Okay, I think I understand this now. Thank you!
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