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Intensity of solar radiation

  1. Mar 24, 2017 #1
    1. The problem statement, all variables and given/known data
    What is the intensity of solar radiation in a place where is the full moon ?

    I know luminosity of the Sun and Earth's distance from the Sun RE and Moon´s distance from the Earth RM. Trajectory of Earth and Moon are circular.

    2. Relevant equations
    distance of the site from the sun is d = RE + RM

    3. The attempt at a solution
    Radiation intensity decreases with the square of the distance.
    Intensity of radiation is amount of radiant power per unit area
     
  2. jcsd
  3. Mar 24, 2017 #2

    berkeman

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    Staff: Mentor

    Sorry, your problem statement didn't translate very well. Is there a diagram associated with this question? Are they asking what is the intensity of sunlight at the combined distance Re + Rm? As if you were on the Moon and the Earth was nearly between the Moon and the Sun?
     
  4. Mar 24, 2017 #3
    sorry for my bad English :-(
    I think - full Moon - position is Sun - Earth - Moon
    in full Moon - Earth is between the Sun and Moon .........and distance between the Sun and the Moon is the sum of Earth's distance from the Sun RE and Moon´s distance from the Earth RM. At this distance I have to calculate solar radiation intensity.
     
  5. Mar 24, 2017 #4

    berkeman

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    Staff: Mentor

    Okay, makes sense. :smile:

    So it sounds like you have all the information you need to do the calculation. You said you have the radiant intensity at the Earth's surface, right? And you know the intensity decreases with the square of the distance. So can you show us how you would calculate the lower intensity when you add in the extra distance to the Moon?
     
  6. Mar 24, 2017 #5
    I know only luminosity of Sun......3,827.10 26 W.....but maybe I do not to calculate from this information.......
    but I think, I can find the radiant intensity at the Earth's surface in some tables...... IE = 1360 W/m2

    IE/IM = RE2 / ( RE + RM )2
     
  7. Mar 24, 2017 #6

    berkeman

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    That looks about right -- I always remember that it's about 1kW/m^2 at the Earth.
    That doesn't look quite right to me. Remember that the intensity is proportional to the inverse of the square of the distance. Intensity gets lower the farther away you get... :smile:
     
  8. Mar 24, 2017 #7
    conversely... :-)
    IM/IE = RE2 / ( RE + RM )2
     
  9. Mar 24, 2017 #8

    berkeman

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    Yeah, that looks better. What answer do you get? Can you show your work?
     
  10. Mar 24, 2017 #9
    RE = 150.106 km
    RM = 384.103 km
    IE = 1360 W/m2

    IM = 1353 W/m2 ?
     
  11. Mar 24, 2017 #10

    berkeman

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    Looks good to me. Do you have a way to check the answer? Is the value of the intensity at the Earth a standard value that you got from your class or textbook?
     
  12. Mar 24, 2017 #11
    value on Earth I found my physical tables..solar irradiance
     
  13. Mar 24, 2017 #12
    thank you very much for your help :-)
     
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