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Heating the surface of Earth

  1. Oct 25, 2006 #1

    tony873004

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    To what extent is Earth's surface heated by heat from the interior? There were differing opinions in my Astrobiology class today. I would guess that solar radiation is by far the dominant source, perhaps 100:1 vs interior heat. Does anyone else have a better guess, or perhaps a source or equation to figure this out?
     
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  3. Oct 25, 2006 #2

    Bystander

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    Crustal heat flows are measured in mW/m2, average runs somewhere between 1-10.
     
  4. Oct 26, 2006 #3
    yeh compare that to 1370w/m2 from the sun is a big difference.
     
  5. Oct 26, 2006 #4

    LURCH

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    You have an Astrobiology course?! I'm a bit jealous.
     
  6. Oct 26, 2006 #5

    tony873004

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    It's a very cool class, taught by Debra Fisher, one of the key astronomers in the detection of exosolar planets. And we get lots of great guest lecturers too. Nick Platts, Frank Drake, Geoff Marcy, and others.

    Here's my attempt to throw some numbers at the temp problem.

    Earth’s average temperature: 287K.

    Therefore, Earth radiates:

    [tex]
    \begin{array}{l}
    E = \sigma T^4 \\
    E = \left( {5.7 \times 10^{ - 8} W/\left( {m^2 } \right)} \right) \times \left( {287} \right)^{} = 387\,W/m^2 \\
    \end{array}
    [/tex]

    On average, the Earth intercepts its 2-D cross section of the Sun’s flux:

    Earth's cross section:
    [tex]
    \pi r^2 = \pi \left( {6378000m} \right)^2 = 1.28 \times 10^{14} m^2
    [/tex]

    Solar flux at Earth's distance:
    [tex]
    \frac{{3.8 \times 10^{26} W}}{{4\pi \left( {149580000000m} \right)^2 }} = 1352\,W/m^2
    [/tex]

    Earth receives:
    [tex]
    1.28 \times 10^{14} m^2 \, \times 1352\,W/m^2 = 1.7 \times 10^{17} W
    [/tex]

    Earth's surface area:
    [tex]
    4\pi r^2 = 4\pi \left( {6378000m} \right)^2 = 5.1 \times 10^{14} m^2
    [/tex]

    So each square meter of Earth's surface receives an average of
    [tex]
    \frac{{1.7 \times 10^{17} W}}{{5.1 \times 10^{14} m^2 }} = 338\,W/m^2
    [/tex]

    Earth’s albedo is 0.367, so therefore each square meter absorbs:
    [tex]
    \left( {338\,W/m^2 } \right) \times \left( {1 - 0.367} \right) = 214\,W/m^2
    [/tex]

    So Earth is absorbing 214 watts per square meter, but radiating 387 watts per square meter?
    My logic is probably flawed. The greehouse effect probably accounts for the difference, not internal heat escaping. Any thoughts?
     
  7. Oct 26, 2006 #6

    SpaceTiger

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  8. Oct 26, 2006 #7

    Bystander

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    Not too wise to "marry" the number given for albedo; use of 1 for emissiivity when calculating black body radiation for what is a very "tattletale gray" body like a planet accounts for the big part of your discrepancy; the heat transfer problem hidden in the "greenhouse" analogy is not well defined, modelled, or measured.
     
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