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Another Problem-Radiation

  1. Dec 10, 2005 #1
    2. The rate at which radiant energy reaches the surface of the earth from the sun is 1.40 kW/m^2. The distance from the earth to the sun is 1.49 x 10^11 m, and the radius of the sun is 6.95 x 10^8 m.

    a. What is the rate of radiation of energy per unit area from the sun's surface?

    b. If the sun radiates as an ideal blackbody, what is the temperature of its surface?

    o_O, I know I don't have any actual WORK for this problem, but that's because I have absolutely NO CLUE where to start. I have the equation for the rate at which radiant energy reaches the surface, but I've already been given a value for that. If anyone who is good at this kind of thermodynamics, any help is appreciated. Thanks.
  2. jcsd
  3. Dec 10, 2005 #2


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    Staff: Mentor

    Let the total power of the Sun = P.

    The Sun's radius is rs. The area at the Sun's surface is S = 4[itex]\pi[/itex]rs2.

    Then the power flux at the sun's surface is P/S.

    Now way out at some long distance R, the same total power passes through an area of 4[itex]\pi[/itex]R2. That distance could be any distance, e.g. the orbit of Venus, Earth, Mars, etc.

    On part b, think Stefan-Boltzmann.
  4. Dec 10, 2005 #3
    "Let the total power of the Sun = P. "

    When you say that, do you mean the rate at which radiant energy reaches the surface of the earth is P?

    "Then the power flux at the sun's surface is P/S."

    By Power flux are you saying that the Rate of Radiation of Energy Per Unit Area from the sun's furface is P/S?

    Sorry for all the questions, but I'm just trying to be clear. I'm not sure what a Power Flux is thats all.
  5. Dec 10, 2005 #4


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    Staff: Mentor

    When I say the total power, I mean the rate at which the energy is emitted (radiated) from the sun.

    Power (W) is the rate of energy (J), i.e. 1 W = 1 J/s.

    Flux is power divided by area W/m2.

    The total power of the sun, P, is radiated in all directions, and let one assume that it is equally distributed in all directions.

    The flux at the sun's surface is simply P/S.

    Further out that same power, P, is distributed over a sphere of larger area.

  6. Dec 10, 2005 #5
    So lets see if I get this.

    Setting the Power Flux as R

    R = P/S, where R = 1.40 kw/m[tex]^2[/tex]

    and S = 4[itex]\pi[/itex]r^2

    Therefore, rearranging the equation R = P/S, you want to solve for P, the total power P = RS

    Am I correct so far for part A?
  7. Dec 10, 2005 #6


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    Staff: Mentor

    Yes, remembering that 1.4 kW/m2 applies to the flux at the earth's surface or rather, at that distance from the sun.

    One can also think of Total Power, P = Flux1* Area1 = Flux2* Area2
  8. Dec 10, 2005 #7
    So P = RS = R(4[itex]\pi[/itex]r[tex]^2[/tex])

    But part A says per unit area, which means the demoninator of my units should be an m[tex]^2[/tex] but the equation P = RS comes out to have units of kW/m because the m on the numerator and the m[tex]^2[/tex] cancel out to be 1/m.

    What did I do wrong?
  9. Dec 10, 2005 #8


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    Homework Helper

    As you said in post #5, R = 1.40 [kw/m^2] , an intensity.
    So R times Area [m^2] is Power [Watts].

    Power is Energy per time interval.
    Energy is conserved, and in space none is absorbed,
    so the same Energy per time interval crosses any spherical surface.

    Intensity is Power [Watts] per surface Area that's pierced by the flow.

    the surrface Area of a sphere is 4 pi r^2 .
  10. Dec 10, 2005 #9
    Alright. I got 8.5 x 10^18 kw/m[tex]^2[/tex] for my answer to part a.

    And Astronuc said something about Stefan-Boltzmann as a hint to part b. I looked up him in my textbook and he is known for the Stefan-Boltzmann constant which = 5.6699 x 10^-8 W/m[tex]^2[/tex]K[tex]^4[/tex]

    It doesn't really say anything much of black bodies except that its an ideal radiator, with an emissivity of unity, which is also an ideal absorber, absorbing all of the radiation that strikes it.

    How can I make such a qualitative description into an equality used to solve for the temperature of the Sun?
  11. Dec 10, 2005 #10


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    Homework Helper

    We are about 200 times as far from the Sun's center as the Sun's surface is.
    So the Energy has spread out about 40000 times as much by Earth's orbit.
    This does NOT imply 8.5E18 [kW/m^2] at the Sun's surface!

    Astronuc meant the "Stephan-Boltzmann Radiation Law", which uses that constant (not the person Stephan nor the person Boltzmann).
    You can figure out the radiated Intensity as a function of Temperature
    (in Kelvin!) from the UNITS of the constant ... emissivity = 1 for the Sun.
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