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Finding the lifetime of a main sequence star.

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Given the following data, calculate the main sequence lifetime of the Sun (in years), assuming that all the initial mass is hydrogen and all of it is converted into helium.
    Mass of the Sun = M = 2x1030kg
    Luminosity of the Sun = L = 4x1032W
    Energy released in the proton-proton reaction (4H [itex]\rightarrow He[/itex]) is Efusion = 4x10-12J
    Comparing the mass of the final helium-4 atom with the masses of the four protons reveals that 0.7% of the mass of the original protons has been lost.


    2. Relevant equations

    L = [itex]\frac{E}{t}[/itex]

    where L is the stars Luminosity, E is the total energy supplied by hydrogen fusion and t is the stars main-sequence lifetime.

    The main-sequence lifetime of a star is also given by:

    t = [itex]\frac{fMc^2}{L}[/itex]

    where f is fraction of of the stars mass converted into energy and c being the speed of light.


    3. The attempt at a solution
    - First attempt...

    Using the data given, I can just punch it into the equation above, which gave me:

    t = [0.007*2x1030*(3x108)2] ÷ 4x1026

    This gives me an answer of 3.15x1018 seconds.

    Converting this into years yields...

    3.15x1018 x [itex]\frac{1}{60*60*24*365}[/itex]

    which gives 99.9x109 years.

    This means the main sequence lifetime of the Sun is about 100 billion years, which can't be correct because in almost every textbook I look in, it states its main sequence lifetime should be 10 billion years. I know that I am by a factor of 10 out somewhere, but I cant seem to find where I have gone wrong.


    - Second attempt...
    The amount of fusion reactions that take place in the Sun in order to provide the necessary Luminosity is...

    L ÷ Efusion = (4*1026) ÷ (4x10-12) = 1038 s-1

    Using the above, I can find the total mass of hydrogen converted into helium per unit time (using Wikipedia to find the mass of hydrogen nuclei)...

    R = rate at which hydrogen is converted = 4(1.67x10^-27)(10^38) ≈ 670x10^9 kg s-1

    I multiplied by 4 in the above because 4 hydrogen nuclei are required for 1 fusion reaction.

    The lifetime of the Sun can now be found by dividing the mass of the Sun by the rate at which the hydrogen is converted to helium, so...

    [itex]\frac{M}{R}[/itex] = 2x1030 ÷ 670x109 ≈ 3x1018s.

    Converting this into years yields...

    3.1x1018 x [itex]\frac{1}{60*60*24*365}[/itex]

    which gives 99.9x109 years.


    This is the exact same answer that I got in the first attempt! Either what I have done is correct or I have gone wrong in the two attempts somewhere, but I know that I must be going wrong somewhere because the answer is out by a factor of 10 and its quite frustrating now.

    Thanks in advance for any replies and sorry for the bad layout of the 'attempt at a solution' section.
     
  2. jcsd
  3. Nov 20, 2011 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF constantinou1!

    There is nothing wrong with your arithmetic, but rather with your astrophysics. You are assuming that the main sequence lifetime will end when ALL of the mass of the star is converted from hydrogen to helium. However, this is not necessary. In practice, it is sufficient for the mass in the CORE of the star (the only place that is hot enough for fusion to be occurring in the first place) to be entirely converted from hydrogen to helium. At this point, core fusion ceases, and hence the star goes off the main sequence. So, your value for "M" here would have to be the core mass, rather than the total stellar mass. I'm not sure how much of the mass of the sun is in its core, but presumably the fraction is ~1/10.

    EDIT: It looks like your second solution method has the same problem.
     
  4. Nov 20, 2011 #3
    Ahh ok, I understand now, thanks alot cepheid for making that clear. Taking that into consideration, the answer that I am coming to now seems to make alot more sense. :smile:
     
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