Core Mass-Luminosity Relationship in RGB Stars

[Jamison Lahman]

Homework Statement

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?

Homework 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.

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

 

[kuruman]

What is the energy released per unit mass when fusing hydrogen into helium?

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.

 

[Jamison Lahman]

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.

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