Blog #30 - Death of a Star

Death of a Star 

1. Stellar death. We’ve now covered the main portions of a star’s life: it’s“birth” from the collapse of molecular clouds as well as the majority of it’s“life” on the Main Sequence in hydrostatic equilibrium. Now we’re going to take a look at what happens when a star like the Sun“dies” after it can no longer support itself via nuclear fusion.

(a) Life. The sun’s luminosity is L = 1033 erg/s. If fusion converts matter into energy with a 0.7%
efficiency and the Sun has 10% of its mass (the mass at the center/core) available for fusion, how long does it take to use up its fuel supply?

We start with the relationship between mass, luminosity and energy from fusion \[L \approx M^4 \approx \frac{E}{t} \] From this we can use the mass-energy equivalence to convert the energy relation to the mass of the core \[E = m*c^2 \rightarrow L \approx \frac{mc^2}{t} \] From here we can use the 0.7% conversion relation to find the time until all of the fusion energy is used. \[t = (0.007) \frac{.1 M_{\odot} c^2}{L_{\odot}} = 1.26*10^{18} seconds \approx 40 billion years\]

(b) Death. We previously derived that the time it takes for a cloud of gas to collapse under its own weight with no opposing force of pressure (i.e. the free-fall time tf f) is: \[t_{ff} = \sqrt{\frac{3 \pi}{32 G \rho}} \]

How long does it take the core of the Sun (again containing 10% of its mass) to collapse under its own weight once it runs out of fuel and there is no longer any opposing pressure?

First we solve for the density of the core \[\rho_c = \frac{M_{\odot}}{\frac{4}{3} \pi R_{\odot}^3} = \frac{3 M_{\odot}}{4 \pi R_{\odot}^3} \] Plugging this density into the equation for collapse time we get \[t_{ff} = \sqrt{\frac{3 \pi}{32 G \rho_c}} = \sqrt{\frac{\pi^2 R_{\odot}^3}{8 G M_{\odot}}} \] Solving for this value we get the time of collapse to be \[t_{ff} = 10^4 \space seconds \]This result is pretty shocking but thankfully we have fusion energy to keep our Sun alive!