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Finding the surface flux of the sun

  1. Sep 17, 2007 #1

    The flux from the Sun above the Earth's atmosphere is about 1370 watts/m^2. This quantity is called the solar constant S and equals pi*f(sun). Use the angular radius of the sun as seen from the Eart to find pi*F , the surface flux of the sun.

    2. Relevant equations

    possible equations that may be relevant: Flux=sigma*T^4, T being the temeperature dependence and sigma=5.669*10^-8/m^2 *K^4; Luminosity = flux*area, a

    3. The attempt at a solution

    I have to find the angular radius of the sun in order to find the surface flux of the sun. To find the angular radius of the sun, I would probably have to used the fact that theta=lambda/Diameter of the sun; theta=lambda/2*radius of the sun. L=4*pi*R^2 *sigma*T^4. There are still 3 unknown variables: lambda, and the Temperature. Perhaps I should approach a solution to this problem in a different fashion: I know the radius of the sun is 6.96*10^5 km and luminosity of the sun is 3.90*10^26 W and the flux from the sun over the earth's atmosphere is 1370 watts/m^2. L/A=flux => (3.90*10^26 W)/(4pi(6.96*10^8 m)^2) = 64067276.69 W/m^2. But that calculation gives me the total flux of the sun, it doesn't give me the surface flux of the sun. how do you find the surface flux of the sun?
  2. jcsd
  3. Sep 18, 2007 #2
    I think you can just use the area of spheres, and the sun as a point source to calculate this.
  4. Sep 18, 2007 #3


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    I don't think you want to treat the Sun as a point source if your aim is to find its surface flux (the angular radius is not that tiny).
  5. Sep 18, 2007 #4


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    I believe you're supposed to proceed from the angular radius, *without* knowing the radius of the Sun or the distance to it from here. As BlackWyvern says, you want to use the surface areas of spheres. You want to consider that one square meter versus the entire sphere at Earth's distance, then consider how that relates to the area of the Sun's surface. Using the definition of angular radius (in the small-angle approximation), you'll notice that it can be identified in your proportional relations.

    You shouldn't need to know anything about blackbody radiation or the Sun's surface temperature either.

    I was puzzled by one other thing you said: what definition are they using in your course for "surface flux"? I've been checking around and that term generally means power or luminosity per surface area ( W/[m^2] ).
    Last edited: Sep 18, 2007
  6. Sep 18, 2007 #5
    I just realized that wouldn't work because of the distance differences.

    But you could do pretty much the same thing, just know that since it's a radiation, it's intensity will depreciate proportional to 1/d^2. Use this and the surface area of spheres (I think it's A = 4 (pi) r^2 ) to derive it.
  7. Sep 18, 2007 #6


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    Actually, these two statements are equivalent. This is because L = F x A, which is the key to the solution. Since intensity is power/area, it has a simple relation to flux; in fact, depending on which specialists' definition you're using, that can *be* the same thing. (That's why I asked what definition was intended in the problem.)
  8. Sep 18, 2007 #7


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    I found one source that might help with this. In Foukal's _Solar Astrophysics_ (pp.40-41), what we will find is the net flux through the Sun's surface. If we call *that* f, then f = pi*F, where F is known as the "astrophysical flux".
  9. Sep 18, 2007 #8
    I don't think the definition of angular radius is stated clearly in my textbook , but is it theta=lambda/2*radius. how do you find theta and lambda in order to find radius of the sun. Also should I assume that the luminosity of the sun is 3.90*10^26 Watts

    well according to my book, theta = lambda/d , d being the size of the apeture. theta= 5e-7 rad. so the two unkwowns are lambda and d. I don't understand why d and lambda are relevant in helping me find the radius of the sun.

    also, would r just be the distance from one point on earth to another point on the sun, which is just one astronomical unit ?

    if so, then the surface area sun would be: SA=4*pi*r^2 = 4*pi*(1.496e11 m)^2 = 2.812e23 m^2

    Since the flux of the earth is given in the problem, I can now find the luminosity of the sun, which is just L = SA *F = (2.81e21m^2)(1370 W/m^2) = 3.84e24 watts.

    Now how would I find the surface flux of the Sun , now that I calculated its luminosity? Could I also used the fact that sin(theta)= R/AU. How would I find theta ?

    I don't understand why my professor wants me to go through numerous calculations for finding the radius of the sun when it is given in the book. I mean whats the point of have the radius of the sun in the appendix of my textbook if I'm going to have to going to have rederived it through tedious calculations.
    Last edited: Sep 18, 2007
  10. Sep 18, 2007 #9


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    I think they're giving you an apparent radius, based on the wavelength of observation. The theta = lambda/d is an approximate value for the angular size of the Airy disk for a point source.

    In any case, that isn't what we need here. For the physical angular size of the Sun, we'd just want the angle subtended by the Sun's radius at the distance of Earth (1 AU, as you've said). Use the small-angle approximation, since this angle is much less than 0.1 radian.

    I think you have the radius of Earth's orbit in there...

    So you have the surface area of a sphere of 1 AU radius, but something's off because that value for the luminosity is a factor of 100 low...

    You can drop the sine because the small-angle approximation will be valid (it gives entirely adequate precision).

    You don't need to *find* theta, actually, as I'll explain below. Go back to the equation you gave, L = F x A. Consider that the same power leaving the Sun's surface also passes through that sphere with 1 AU radius. So you can equate F x A for each sphere and solve for F for the Sun's surface; theta will appear in your expression.

    Are you in an astrophysics or experimental physics course? The point of the exercise is that you are calculating the nex flux of the Sun from two *directly measurable* quantities, the solar radiation flux at the "top" of Earth's atmosphere and the angular diameter of the Sun. Quantities calculated only from directly measurable ones are more reliable than those which are computed from inferred quantities, such as the solar radius or the astronomical unit (there is a whole historical literature you can look at about how these and similar quantities were found and how we got to the currently accepted values).

    When you get past the introductory courses (1000-level in some systems) in the physical sciences, you start getting more into *how* various quantities and equations are derived, rather than just being told, "Take our word for this and use it!"

    All those nicely-tabulated quantities listed in your textbook or handbooks of reference data are the result of uncounted person-years of effort and debate. It is always a good idea to know where the numbers you use came from and how reliable (precise, or even just accurate) they really are...

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