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Mars and Jupiter's atmosphere math question

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  1. Nov 12, 2013 #1
    1. The problem statement, all variables and given/known data

    What should be the temperature of a spherical asteroid located between Mars and Jupiter, twice as far from the Sun as Earth? The asteroid has no atmosphere and its albedo is .15.


    2. Relevant equations



    3. The attempt at a solution

    The asteroid is twice as far from the Sun as Earth, so it receives ¼ the solar energy. Thus
    T = [341 W/m2 (1 – 0.15)/(4 5.67e-8 W/(m2 K4))]1/4 = 189 K


    I know this is right because my professor did this as a solution but I am confused as how this was reached.
     
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  3. Nov 12, 2013 #2

    cepheid

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    Hellow courtneywetts,

    Which part of the solution are you confused about?

    As you stated, since the asteroid is twice the distance of Earth from the sun, it receives 1/4 of the power per unit area. This is because of the inverse-square law for the intensity of light. So that's where the 341 W/m2 comes from (we receive 4 times that amount of solar energy at the surface of the Earth, roughly).

    The (1 - 0.15) is because of the albedo. An albedo of 0.15 means that 15% of the incoming light is reflected back into space. This means that 85% of it (1 - 0.15) is absorbed, heating up the asteroid. Hence, we multiply the incoming solar radiation by this factor, to get the actual amount absorbed.

    Next, comes the only tricky part, conceptually. What you have to realize is that if the asteroid is in thermal equilibrium, then the energy coming in in the form of EM radiation must be equal to the energy going out. In other words, the asteroid re-radiates all of the solar energy that it receives back into space. If this were not true, and the input were not the same as the output, the asteroid would heat up or cool down until this equilibrium was reached.

    So, for the final part of the problem, you assume that the asteroid is itself a blackbody emitter, and apply the Stefan-Boltzmann law to it:$$P/A = \sigma T^4$$where you know what P/A is, it's just 0.85*(341 W/m2) (since it has to re-emit all of the solar power that it absorbs).

    Your professor solved this equation for T.
     
  4. Nov 12, 2013 #3

    haruspex

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    One small point... this analysis pretends the asteroid is uniformly irradiated. In practice it will almost surely have poles, day and night zones, and may even be locked, one side always in night. Because of the fourth power in the Stefan-Boltzmann law, this uneven irradiation means the actual average temperature will be a little less.
     
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