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Astronomy question

  1. Oct 4, 2006 #1
    I'm taking Astr. 321 right now. We were supposed to read the first 3 chapters. They go through the different planets (some detail, but nothing too complicated), the sun, asteroids, etc. The lectures went over this in more detail and went through things like which elements were created first, how the rest were created, etc. Basic stuff.

    Then I get the homework:

    Question 1: If your hand is 7.5cm away from a 100 watt light bulb, it gets the same energy as it would get from the sun on a sunny day. Determine the wattage of the sun. Easy ratio (plus some squares).

    Next problem: How much mass is it fusing per second (easy, E=mc^2).

    3rd: " Estimate how long the Sun can generate this power knowing that mass of 4 protons that effectively fuse to form a He nucleus in the core of the sun is 1.008 times the mass of a helium nucleus. The missing mas is converted to energy. Nuclear fusion is somewhat inefficient in the Sun because it only occurs in the core- so assume that only 10% of the Sun's mass is ever hot enough to burn H into He and 90% of the Sun's hydrogen is untouched. Assume the Sun starts off as pure hydrogen (not quite true) and that the Sun's luminosity is constant during most of its life, its "main sequence lifetime" when it burns H into He. (answer in years) "

    While typing this I came up with an idea: do I take the mass of the Sun from the book (it's 1.99e^30 I think) and then only take Mass - (1/1.008) as the mass released as energy, and at 10% of that speed of fusion?

    If not, then I am stumped. I've been working on this since like 2 days ago with a friend.

    EDIT: My prof. is Donald Brownlee, btw. ;)
     
  2. jcsd
  3. Oct 4, 2006 #2

    Astronuc

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    Staff: Mentor

    Assume the wattage of the sun (power) is derived from fusion.

    One can calculate the energy/ reaction.

    Dividing the power by energy/reaction yields the reaction rate, and knowing the mass consumed per reaction.

    Dividing the mass of H in the sun, by mass consumption rate would allow one to estimate the lifetime of the sun.

    With respect to the ratio, 1.008, that is the ratio of mass of 4 protons/He-4 nucleus, or (mass 4p)/(mass He-4) = 1.008.

    If the sun's energy is generated by 10% of the mass, most of which is hydrogen, what happens when less than 10% of the mass is hydrogen?
     
  4. Oct 4, 2006 #3
    I have the wattage from question 1. Am I supposed to find out how much per reaction is "created" by assuming a proton is 1g/mol, finding out how many mols per second are fused (calculating from mass/second), then how many protons/second, then divide by 4 to get reactions/second, then do wattage/(reactions/second) to get energy/reactions?

    Ok, that makes sense. Since you're only left with reaction/second.

    So I do have to use the mass of the sun from the book, I can't calculate it here?

    I don't understand when I need to use this. So far I haven't had to... at least if I explained it correctly above.

    EDIT: I think I figured it out. Since I figured out the mass transformed into energy, I use 1.008 (ratio) to find the TOTAL mass used (before any transformation) per reaction, and then do the rest of the steps once I have the total mass, right?

    Errr... red giant?
     
    Last edited: Oct 4, 2006
  5. Oct 4, 2006 #4
    Ok, nevermind, I screwed up. What I was supposed to do was take "mass converted" per second and use it as a ratio with 1.008 to find "mass total used" per second, THEN simply divide the Sun's mass by that, right? Because even though the mass isn't converted, it's turned into He and therefore useless (for now :p).

    I still don't get the 10% thing, though. Do I only need to divide 90% of the Sun's mass by my calculated "total mass used" per second? I understand that if the Sun can't fuse any more H into He, it gets mad and grows big, ending the "main sequence lifetime", but the wording of the problem confuses me. It almost makes it seem that I only use 10% of the total mass... but that doesn't make sense, either.
     
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