Lecture 20 Solutions: How Stars Die

Practice Problem Solutions

Solutions to the Lecture 20 practice problems.

Note

Student note: These are model solutions written to show setup, units, and reasoning clearly. Your own work can be shorter as long as the physics, units, and logic are sound.

Use these after you try the problems yourself. The point is to compare your reasoning to a worked example, not to skip the thinking step.

Core Concepts

Problem 1: Planetary Nebula Timescale

Given: Radius \(r = 0.5\) light-years; expansion speed \(v = 15\) km/s.

Find: Approximate age \(t\).

Equation: \(t = r/v\)

Steps:

Convert the radius to meters:

\[ r = 0.5 \times 9.46 \times 10^{15}\,\mathrm{m} = 4.73 \times 10^{15}\,\mathrm{m} \]

Convert the speed:

\[ v = 15\,\mathrm{km/s} = 1.5 \times 10^4\,\mathrm{m/s} \]

Now divide:

\[ t = \frac{4.73 \times 10^{15}}{1.5 \times 10^4} \approx 3.15 \times 10^{11}\,\mathrm{s} \]

Convert seconds to years:

\[ t \approx \frac{3.15 \times 10^{11}}{3.15 \times 10^7} \approx 1.0 \times 10^4\,\mathrm{yr} \]

Unit check: meters divided by meters per second gives seconds.

Sanity check: The result is about \(10^4\) years, which matches the stated typical planetary-nebula lifetime.

Answer: The nebula is about 10,000 years old.

Problem 2: White Dwarf Density

Given: \(M_\odot = 2 \times 10^{30}\,\mathrm{kg}\); \(R_\oplus = 6.4 \times 10^6\,\mathrm{m}\).

Find: Average density \(\rho = M/V\).

Equation:

\[ V = \frac{4}{3}\pi R^3, \qquad \rho = \frac{M}{V} \]

Steps:

First find the volume:

\[ V = \frac{4}{3}\pi (6.4 \times 10^6)^3 \approx 1.10 \times 10^{21}\,\mathrm{m^3} \]

Then find the density:

\[ \rho = \frac{2 \times 10^{30}}{1.10 \times 10^{21}} \approx 1.8 \times 10^9\,\mathrm{kg/m^3} \]

Compare to water:

\[ \frac{1.8 \times 10^9}{10^3} = 1.8 \times 10^6 \]

Sanity check: A white dwarf should be millions of times denser than ordinary matter, so this is the right scale.

Answer: The average density is about \(1.8 \times 10^9\,\mathrm{kg/m^3}\), or roughly 1.8 million times the density of water.

Problem 3: Chandrasekhar Limit

Given: White dwarf mass \(= 1.5\,M_\odot\).

Find: Can electron degeneracy pressure support it?

Reasoning:

The Chandrasekhar limit is about \(1.4\,M_\odot\). A white dwarf above this mass is unstable: electron degeneracy pressure can no longer balance gravity.

What happens next depends on the physical setting:

• If the white dwarf is carbon-oxygen rich and pushed upward by accretion, carbon ignites explosively and the star can become a Type Ia supernova.
• More generally, the star cannot remain a white dwarf. Collapse must continue once the degeneracy support fails.

Answer: A \(1.5\,M_\odot\) white dwarf is not stable. Electron degeneracy pressure cannot support it, so it must either collapse further or explode, depending on its composition and environment.

Working with Core Collapse

Problem 4: The Iron Catastrophe

Key idea: Iron sits near the peak of the binding-energy-per-nucleon curve.

That means:

• Fusing light elements up to iron releases energy.
• Fusing iron into heavier nuclei requires energy.

So an iron core is the end of the line for ordinary stellar fusion. Once the core becomes iron rich, fusion no longer supplies the outward pressure needed to support the star.

Answer: Iron is special because it has one of the highest binding energies per nucleon. An iron core cannot release energy by fusing; further fusion is endothermic, so the core loses support and collapses.

Problem 5: Collapse Timescale

Given: Initial radius \(R_i \approx 6400\) km; final radius \(R_f \approx 20\) km; time \(\Delta t = 0.3\) s.

Find: Average infall speed.

Equation:

\[ v_{\rm avg} = \frac{\Delta r}{\Delta t} \]

Steps:

Compute the distance fallen:

\[ \Delta r = 6400 - 20 = 6380\,\mathrm{km} \]

Now divide by the time:

\[ v_{\rm avg} = \frac{6380}{0.3} \approx 2.13 \times 10^4\,\mathrm{km/s} \]

Compare to the speed of light:

\[ \frac{2.13 \times 10^4}{3.0 \times 10^5} \approx 0.071 \]

Sanity check: A core collapse should be a substantial fraction of \(c\), but still less than \(c\).

Answer: The average infall speed is about \(2.1 \times 10^4\,\mathrm{km/s}\), or about 7% of the speed of light.

Problem 6: Neutrino Physics

Neutrinos interact weakly with matter, but in a collapsing stellar core two things become enormous at once:

• the number of neutrinos produced
• the density of material behind the stalled shock

Even with a tiny interaction cross section, a small fraction of an enormous neutrino flux can deposit enough energy into the surrounding matter to reheat it. That heating helps revive the shock and drive the explosion.

Answer: Neutrino heating works because the core is so dense and the neutrino flux is so huge that even a weak interaction can transfer a meaningful amount of energy.

Nucleosynthesis and Element Origins

Problem 7: Where Did You Come From?

One valid example is iron.

• Stellar environment: Iron is assembled late in the life of massive stars and is also produced in supernova nucleosynthesis.
• Dispersal mechanism: A core-collapse supernova ejects that material into the interstellar medium.
• Entry into the solar nebula: The iron mixed into the gas and dust that later collapsed to form the Sun and planets about 4.6 billion years ago.

Observable -> model -> inference: We measure iron lines in stellar and nebular spectra, use nucleosynthesis models, and infer that the iron in rocky planets came from earlier generations of dying stars.

Problem 8: Supernova Energy Budget

Given: Total energy \(E_{\rm tot} = 10^{46}\,\mathrm{J}\); kinetic fraction \(= 1\%\).

Find: Energy in ejecta motion.

Steps:

\[ E_{\rm kin} = 0.01 \times 10^{46} = 10^{44}\,\mathrm{J} \]

Compare to the Sun’s lifetime output:

\[ E_\odot \approx L_\odot \times t_\odot \approx (3.8 \times 10^{26})(10^{10}\,\mathrm{yr}) \]

Convert \(10^{10}\) yr to seconds:

\[ 10^{10}\,\mathrm{yr} \approx 3.15 \times 10^{17}\,\mathrm{s} \]

So

\[ E_\odot \approx 3.8 \times 10^{26} \times 3.15 \times 10^{17} \approx 1.2 \times 10^{44}\,\mathrm{J} \]

Answer: The ejecta carry about \(10^{44}\,\mathrm{J}\) of kinetic energy, which is of order one solar lifetime of radiated energy.

Unit check: A percentage of a quantity in joules is still measured in joules.

Sanity check: Since only 1% of the total supernova energy goes into ejecta motion, the answer should be two orders of magnitude below \(10^{46}\,\mathrm{J}\), which it is.

Comparing Supernovae

Problem 9: Type Ia vs. Type II

Observable differences you would expect:

• Light curve: Type Ia events usually show a smoother rise and decline; Type II events often show a plateau or a broader, more diverse behavior because of their hydrogen envelopes.
• Spectra: Type II supernovae show hydrogen lines. Type Ia supernovae do not show hydrogen and instead show strong features from intermediate-mass and iron-peak elements.
• Ejecta velocity: Both kinds are fast, but the measured line widths and how they evolve differ because Type Ia ejecta come from a compact white dwarf while Type II ejecta expand through an extended hydrogen envelope.
• Progenitor context: Type II comes from a massive star core collapse; Type Ia comes from a white dwarf in a binary.

Why these differences exist: The explosions start from very different structures. Type II includes a hydrogen-rich stellar envelope. Type Ia destroys a degenerate white dwarf with no hydrogen envelope.

Observable -> model -> inference: - Observable: hydrogen lines versus no hydrogen, plateau-like versus smoother light curves, and different line-width velocity evolution - Model: hydrogen-rich core-collapse star versus hydrogen-poor thermonuclear white dwarf - Inference: the two supernovae came from fundamentally different progenitors and explosion mechanisms

Answer: Type II supernovae are identified by hydrogen-rich spectra, envelope-influenced light curves, and velocity signatures shaped by a massive star’s extended outer layers. Type Ia supernovae lack hydrogen and show more uniform thermonuclear behavior because they come from exploding white dwarfs.

Problem 10: Standard Candles

Type Ia supernovae are comparatively uniform because they are triggered when a white dwarf approaches a fairly narrow mass scale near the Chandrasekhar limit. That tends to produce similar thermonuclear conditions and similar peak luminosities.

Type II supernovae are harder to standardize because their brightness depends on more variable ingredients:

• progenitor mass
• envelope size
• prior mass loss
• exact explosion energy

Answer: Type Ia events are standardizable because they begin from similar white-dwarf conditions; Type II events are more diverse because their progenitors are more diverse.

Challenge Problems

Problem 11: Reading a Supernova Spectrum

Strong hydrogen emission is the key clue here. Type Ia supernovae are hydrogen poor, while Type II supernovae come from massive stars that still retain hydrogen in their outer layers.

The absence of strong iron or nickel absorption does not override the hydrogen evidence. The hydrogen tells you the progenitor had a substantial stellar envelope.

Answer: This spectrum is more consistent with a Type II supernova. The hydrogen emission implies a massive hydrogen-rich progenitor star rather than a white dwarf.

Problem 12: The Fate of the Sun

A concise timeline is:

1. Now: Main-sequence Sun, about \(1\,L_\odot\), \(5800\,\mathrm{K}\), \(1\,R_\odot\).
2. About 5 billion years from now: The Sun leaves the main sequence and becomes a red giant. Its luminosity rises by orders of magnitude, its temperature drops to roughly a few thousand kelvin, and its radius grows to tens or more than a hundred solar radii.
3. Late giant phases: The Sun enters the asymptotic giant branch, loses mass, and becomes even more distended and unstable.
4. Planetary-nebula phase: The outer envelope is expelled. This glowing shell lasts only about \(10^4\) years.
5. White dwarf phase: The exposed core remains, roughly Earth-sized and initially very hot. Over billions and then trillions of years it cools and fades.
6. Black dwarf: In principle the final state is a black dwarf, but the universe is not yet old enough for any to exist.

Answer: The Sun does not become a neutron star or black hole. It becomes a red giant, then a planetary nebula plus white dwarf, and only on enormously long timescales would it cool toward a black dwarf.

Problem 13: Galactic Chemical Enrichment

Given: Total stars \(= 2 \times 10^{11}\); Type II progenitors \(= 1/300\) of that population; iron yield per supernova \(= 0.1\,M_\odot\).

Steps:

Number of Type II progenitors:

\[ N_{\rm SN} = \frac{2 \times 10^{11}}{300} \approx 6.7 \times 10^8 \]

Total iron yield:

\[ M_{\rm Fe} = 6.7 \times 10^8 \times 0.1\,M_\odot = 6.7 \times 10^7\,M_\odot \]

Convert to kilograms:

\[ M_{\rm Fe} \approx 6.7 \times 10^7 \times 2 \times 10^{30} \approx 1.3 \times 10^{38}\,\mathrm{kg} \]

Compare this to the Sun’s mass:

\[ \frac{M_{\rm Fe}}{M_\odot} = 6.7 \times 10^7 \]

So this toy model gives an iron yield equal to about 67 million Suns worth of mass.

Interpretation: That is a huge amount of iron, but this is only a back-of-the-envelope estimate for one enrichment channel. It should not be treated as the Milky Way’s complete iron inventory because Type Ia supernovae also make iron, some iron remains locked in stars and remnants, and the assumed progenitor fraction is only approximate.

Answer: This channel contributes about \(6.7 \times 10^7\,M_\odot\) of iron, or \(1.3 \times 10^{38}\,\mathrm{kg}\). That is about 67 million times the Sun’s mass, but it is only a toy-model estimate for one enrichment pathway, not the Galaxy’s full iron budget.