WS 10.1, Problem 1: 'Til the Fusion Ends

We've talked about the birth of stars in molecular clouds. We also briefly discussed the main
sequence, on which stars are in hydrostatic equilibrium owing to energy generated by nuclear fusion in their cores. Now let's investigate what happens when a star like the Sun can no longer support itself via nuclear fusion.

a) At what rate is the Sun generating energy?

At first, we were confused by this question.  We knew it couldn't possibly be as simple as we were making it.  Then we realized we were being typical Harvard students and overthinking things, and that luminosity is given in ergs per second.  So, the sun is generating energy at a rate of

$L_\odot=4\times 10^{33}$ ergs/s

b) If fusion converts matter into energy with a 0.7% e fficiency, and if the Sun has 10% of its mass available for fusion (in the core only), how long does it take to use up its fuel supply? What is the general relationship between the mass of a star and its main-sequence lifetime?

The first step was to find the mass of the sun available for fusion.  

$(.1)(M_\odot)=(1\times 10^-1)(2\times 10^{33}g)= 2\times 10^{32}g$

The second step was to find the energy produces by fusion in the core using the equation for energy that I always knew, but never really knew when to use.  

$E=M_{fus} c^2=(2\times 10^{32}g)(3\times 10^{10}cm/s)$

But, the question states that the sun's fusion is only 0.7% efficient, so we have to multiply the energy by the efficiency.  

$E=M_{fus} c^2=(7\times 10^{-3})(2\times 10^{32}g)(3\times 10^{10}cm/s)=1.4\times 10^{51} ergs$

Finally, using dimensional analysis, we were able to find the final equation for the time.  

$t=\frac{E}{L_\odot}=4\times 10^{17}s\approx 1\times 10^{10} yrs$

d) The core will collapse until there is a force available to hold it up. One such force is supplied by degeneracy pressure. The pressure inside of a white dwarf star is provided by the motion of electrons. The electrons are in a tough situation: they can't occupy the exact same state, but there's not much room for them to coexist easily inside of a dense white dwarf. As a result, they must always be in motion to avoid other electrons (roughly speaking). This e ffect becomes important when the inter-particle spacing is of order the de Broglie wavelength $\lambda$, which is related to the momentum via 

$\lambda=frac{h}{v}$  

For a stellar core of a given temperature, which particles reach this critical density fi rst: electrons or protons?

The foundation of our solution to this problem was the equation for momentum:

$p=mv$

We had to compare the momentums of the electrons and photons in the star.  

$p_e=m_e v_e$

$p_p=m_p v_p=1800m_e v_p$ 

Because protons are just so much more massive than electrons, the momemtum of a proton is a lot bigger than the momentum of an electron.  Therefore, protons have a shorter de Broglie wavelength, so the protons reach their critical density first.   

e) If a typical white dwarf has roughly half the mass of the Sun and the radius of the Earth, what is the typical density of a white dwarf in grams per cubic centimeter? What is the volume of white dwarf material that weighs as much as a car?

This was pretty simple. 

$\rho_{WD}=\frac{m}{v}=1.5\times 10^6 g/cm^3$

We then figured that the average car is about 1 metric ton, so about $1\times 10^6$g.  

$v=\frac{m_{car}}{\rho_{WD}}=0.67$ cm^3