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Luminosity of the Sun

  1. Sep 1, 2010 #1
    I want to make an order-of-magnitude estimate of the Sun's present-day luminosity.

    For this, I want to use just the following two pieces of information in a simple idealised model of (the mechanism of energy production in a star):

    1) the mass of a helium nucleus is about 1 percent less than the mass of four hydrogen nuclei.

    2) the Sun’s hydrogen-burning lifetime is believed to be 10 Gyr.


    This is my attempt at the problem:

    The Sun is passing through its mid-life at the moment.
    So, I am assuming that half of the hydrogen nuclei have to converted to helium.

    Therefore, (0.5)(mass of all H nuclie) + (0.5)(mass of all He nuclie) = (present) Mass of the Sun

    "Mass of a helium nuclues = (0.99)(4*Mass of a hydrogen nucleus)" implies that the mass of all helium nuclei in the Sun is 0.8*1030 kg.

    So, the intial mass of the Sun = (2*Present mass of the hydrogen nuclie) = 2.481030 kg.

    So, total number of reactions (that will have occurred by the time the Sun dies) =
    initial mass of Sun / 4 = 0.6*1030.

    Energy release per reaction = (delta-m)(c2) = (4*(1-0.99))(c2)
    = 6.0*10-12 J.

    So, (average) luminosity = Total energy released / energy release per reaction = 3.3*1026 W.

    I would be grateful if anyone could point out any flaws in the argument.
    Also, I am not sure whether I have arrived at a reasonably good answer with incorrect methods. :confused:
     
  2. jcsd
  3. Sep 1, 2010 #2
    I think it would be easier and more accurate to calculate it based on the known solar energy that arrives Earth, somewhere around 1.2kW per square meter. We know the Sun-Earth distance so it should be pretty easy to calculate the energy at the Sun (i.e 'shrink the sphere').
     
  4. Sep 1, 2010 #3

    phyzguy

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    The solar luminosity is measured to be 3.8x10^26 W, so your estimate isn't bad. I think the argument is reasonable for a rough estimate. Of course, much more detailed models exist that can tell you how much H has burned to He as a function of radius. For example:

    http://www.pha.jhu.edu/~france/PAPERS/solmodel.pdf [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Sep 1, 2010 #4
    My method gave me a roughly approximate of 3.374×10^26 W, using the surface area of a sphere of radius = 1AU, and 1.2kW/m2. Very similar to yours but still not the 3.8×10^26 W.
     
    Last edited: Sep 1, 2010
  6. Sep 2, 2010 #5

    Chronos

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    A small amount of fusion, other than the hydrogen-helium cycle, is also possible. It may be enough to make up the difference.
     
  7. Sep 2, 2010 #6
    The Solar Constant is defined as ~1368 W/sq.m, thus why the luminosity you got is incorrect. That's an arbitary average, whereas the actual Sun oscillates by a percent either side over several decades.

    Over the course of the Sun's Main Sequence the luminosity rises gradually from 0.7 of its current value to 1.8 times as it leaves the Main Sequence. Then, for about 2 billion years, it'll rise to about 2-3 times present before starting its climb up the Red Giant Peak, when it'll hit ~2700 times its present value. Contrary to various TV Sci-fi shows the transition from Main Sequence to Red Giant won't happen in a brief timespan, but takes ~2 billion years, with the most dramatic rise in the last ~0.1-0.01 billion. Once it has climbed the RG Peak, then the Core will explode (the Helium Flash) but the Sun won't. Virtually all the energy released goes into heaving the Core of the Sun onto the Helium Main Sequence, which'll last ~0.1 billion years at a steady ~50 times present luminosity, before a brief period in which the Sun alternates between ballooning and looking like a fat White Dwarf. That end period is known as the Asymptotic Giant Branch and lasts mere millions of years.
     
  8. Sep 2, 2010 #7
    Thanks for that value. Using it (1368W/sqr.m), on a sphere of radius = 1 AU, I get 3.847×10^26 W which is way more accurate.
     
  9. Sep 3, 2010 #8
    Firstly, the mass difference is ~0.7%

    Second, the amount burnt during the Main Sequence is only what's available in the Core, about 16% of the total in the Sun.

    Finally, the initial ratio of H/He4, by mass, was about 3:1 (i.e. 75% H, and 25% He.)

    Thus 0.007*0.16*0.75 of the Sun's mass is converted into pure energy, radiated from its surface over the ~10 billion years of the Main Sequence. During that time the luminosity gradually rises from 70% of the present levels to over 180%. The average output is roughly ~125%. Since the mass-fraction is so tiny the Sun's mass is roughly a constant 2E+30 kg over that period. Thus roughly ~1.7E+27 kg of mass-energy has been liberated, which is ~1.5E+44 J total. Total time ~3.2E+17 seconds, thus 100/125 of the average puts the present day luminosity at ~3.75E+26 W.


    The problem is the initial mass ratio is roughly 3:1 as I mentioned.

    Ah. How did you get that? Half of 1% is just 1/200, thus the Sun's initial mass is, at most, just 2E+30 kg/(0.995) ~2E+30 kg, with two digit accuracy applied.

    Total what reactions? Individual protons mass just 1.67E-27 kg, thus 2E+30 kg of them is 1.2E+57 protons, and reacting half is 6E+56 reactions. Four protons don't get together all at once to make 1 helium. Instead two combine to make deuterium, then deuterium reacts with deuterium, the helium-3 or tritium products of those reactions then react with deuterium and make helium-4. Very rarely two deuteriums bang together and make a helium-4 directly, with release of a gamma-ray, but never four protons at once.

    Oh well you didn't do too bad. Lost some digits along the way I think. Easy to do.
     
  10. Sep 3, 2010 #9
    One thing that you should do is to try to see how much your numbers vary if you change your assumptions. One thing that I think you'll find the number that you get out doesn't change too much if you change the assumptions for the amount of helium/hydrogen/lifetime of the sun. What then is happening is that because your estimates are all within an order of magnitude of the right answers, and some are overestimates, and some are underestimates, you get something pretty close to the right answer.
     
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