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Core Mass-Luminosity Relationship in RGB Stars

  1. Oct 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Low-mass stars like the Sun obey the core mass-luminosity relationship as they burn H in a shell and climb the RBG (Red Giant Branch). What is the energy released per unit mass when fusing hydrogen into helium?

    2. Relevant equations
    The core mass-luminosity relationship:
    $$
    L=2.3 \times 10^{5}L_{\odot}\left(\frac{M_{c}}{M_{\odot}}\right)^{6}
    $$
    My professor also gave the timescale for a 1##M_{\odot}## star on the RGB as ##5 \times 10^{8}s##. He also gave the mass of the core for a 1##M_{\odot}## star at the tip of the RGB as ##M_{c}=.45M_{\odot}## though I am doubtful this is useful.
    3. The attempt at a solution
    Since Luminosity is ##\frac{Energy}{Time}## and ##M_{\odot}## is a constant, the equation can be written as
    $$
    \frac{E}{M_{c}^{6}} = \frac{2.3 \times 10^{5}L_{\odot}}{M_{\odot}^{6}}t=
    \frac{2.3 \times 10^{5}(3.839 \times 10^{26}W)(5 \times 10^{8}s)}{(1.9891 \times 10^{30}kg)^{6}}=7.13 \times 10^{-142}J/kg^{6}
    $$
    As you can see, the number I got was extremely small and in terms of ##J/kg^{6}## not ##J/kg## as is desired.

    Additionally, using the ##M_{c}=.45M_{\odot}##,
    $$
    E=2.3 \times 10^{5}L_{\odot}\left(\frac{M_{c}}{M_{\odot}}\right)^{6}t=
    2.3 \times 10^{5}(3.839 \times 10^{26}W)(.45)^{6}(5 \times 10^{8}s)
    $$$$
    =3.67 \times 10^{38} J
    $$
    Then divided by the mass of the core yields:
    $$
    \frac{E}{M_{c}} = \frac{E}{.45M_{\odot}} = \frac{3.67 \times 10^{38}}{.45(1.9891 \times 10^{30})} = 1.843 \times 10^{8} J/kg
    $$
    This seems somewhat reasonable, but the next four parts of the question depend on this answer and I'm not sure if I am allowed to use the mass of core for a star at the tip of the RGB for the entirety of the RGB. Thanks in advanced for any help
     
  2. jcsd
  3. Oct 24, 2016 #2

    kuruman

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    Science Advisor
    Homework Helper
    Gold Member

    Just by looking at this question (as a physicist), I think of the proton-proton cycle that starts with 6 protons as in here
    https://en.wikipedia.org/wiki/Proton–proton_chain_reaction
    Your number is in on the very low side. For example, the dissociation energy for water is about 500 kJ/mol. To dissociate 1 kg of water (56 mol) you will need 56(mole/kg)*500,000(J/mole) =3×107 J/kg. That's only one order of magnitude below your answer. Surely the Sun can do better than that.
     
  4. Nov 1, 2016 #3
    In deed you were correct. Turns out the problem was a simple E=mc2 problem. About .7% of the mass of hydrogen is released during fusion, so the Energy/mass=.007c2 o0)
     
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