- #1

Jamison Lahman

- 143

- 35

## 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