12.P — Worked Problems: Astrophysics and Cosmology

Fourteen problems across Part 12. Every solution shows the arithmetic. Attempt each before opening it.

Problem 1 — The Sun's remaining lifetime

The Sun converts 0.71 % of the mass involved in fusion to energy, and about 10 % of its hydrogen is available. Find its total main-sequence lifetime and how much remains.

Solution

Available hydrogen:

m = 0.10\times0.74\times1.989\times10^{30} = 1.472\times10^{29}\ \text{kg}

Energy available:

E = 0.0071\,mc^2 = 0.0071(1.472\times10^{29})(9\times10^{16}) = 9.40\times10^{43}\ \text{J}

Lifetime at current luminosity:

t = \frac{9.40\times10^{43}}{3.85\times10^{26}} = 2.44\times10^{17}\ \text{s}

= \frac{2.44\times10^{17}}{3.156\times10^{7}} = 7.7\times10^{9}\ \text{years}

Remaining: 7.7-4.6 = 3.1 billion years.

What to notice. The standard figure is about 5 billion years remaining, and the difference is that the Sun's luminosity is not constant. It has brightened by about 30 % since formation and will continue to brighten, so the average output over the remaining life exceeds the current value — but the core also becomes more efficient as helium accumulates and it contracts.

The number that actually matters for Earth is much sooner. In about 1 billion years the Sun will be roughly 10 % brighter, which is enough to push Earth past a runaway greenhouse threshold (Chapter 11.2). Earth's habitability ends long before the Sun does.

Problem 2 — Central temperature from hydrostatic equilibrium

Estimate the central temperature of a star with 10 solar masses and 4 solar radii, assuming a mean particle mass of 0.6m_p.

Solution

Central pressure estimate:

P_c \approx \frac{GM\bar{\rho}}{R}

\bar{\rho} = \frac{M}{\frac{4}{3}\pi R^3} = \frac{1.989\times10^{31}}{\frac{4}{3}\pi(2.784\times10^{9})^3} = \frac{1.989\times10^{31}}{9.037\times10^{28}} = 220\ \text{kg/m}^3

P_c = \frac{(6.674\times10^{-11})(1.989\times10^{31})(220)}{2.784\times10^{9}} = \frac{2.921\times10^{23}}{2.784\times10^{9}} = 1.05\times10^{14}\ \text{Pa}

Temperature, taking the central density as roughly 50 times the average (as in the Sun):

T = \frac{P\mu m_p}{\rho k_B} = \frac{(1.05\times10^{14})(0.6)(1.673\times10^{-27})}{(1.1\times10^{4})(1.381\times10^{-23})}

= \frac{1.054\times10^{-13}}{1.519\times10^{-19}} = 6.9\times10^{5}\ \text{K}

That is too low, because the crude P_c estimate underestimates by about a hundredfold (Chapter 12.1). Scaling the pressure up accordingly:

T \approx 3\times10^{7}\ \text{K}

What to notice. Around 3\times10^{7} K, which is above the 1.8\times10^{7} K crossover for the CNO cycle (Chapter 12.1). A 10-solar-mass star burns hydrogen predominantly by CNO, not by the pp chain.

And the T^{20} dependence means the energy generation is concentrated in a tiny central region, which drives convection in the core — the opposite of the Sun, which has a radiative core and a convective envelope.

Problem 3 — Mass–luminosity and lifetime

A star has 15 solar masses. Estimate its luminosity, main-sequence lifetime, and surface temperature if its radius is 5 solar radii.

Solution

Luminosity:

\frac{L}{L_\odot} = \left(\frac{M}{M_\odot}\right)^{3.5} = 15^{3.5}

\ln 15 = 2.708, \quad 3.5\times2.708 = 9.478, \quad e^{9.478} = 1.31\times10^{4}

L = 1.31\times10^{4}L_\odot = 5.04\times10^{30}\ \text{W}

Lifetime:

t \propto \frac{M}{L} = \frac{15}{1.31\times10^{4}} = 1.145\times10^{-3}

t = 1.145\times10^{-3}\times10^{10} = 1.15\times10^{7}\ \text{years}

Surface temperature, from Stefan–Boltzmann:

L = 4\pi R^2\sigma T^4 \quad\Longrightarrow\quad T = \left(\frac{L}{4\pi R^2\sigma}\right)^{1/4}

R = 5\times6.96\times10^{8} = 3.48\times10^{9}\ \text{m}

4\pi R^2 = 4\pi(1.211\times10^{19}) = 1.522\times10^{20}\ \text{m}^2

T^4 = \frac{5.04\times10^{30}}{(1.522\times10^{20})(5.67\times10^{-8})} = \frac{5.04\times10^{30}}{8.630\times10^{12}} = 5.84\times10^{17}

T = (5.84\times10^{17})^{1/4}

\sqrt{5.84\times10^{17}} = 7.64\times10^{8}, \qquad \sqrt{7.64\times10^{8}} = 2.76\times10^{4}

T = 27{,}600\ \text{K}

What to notice. A B-type star (Chapter 11.9), blue-white, 13,000 times the Sun's output, lasting 11.5 million years.

Compare with the Sun's 10 billion. Fifteen times the mass gives a thousandth of the lifetime.

And its Wien peak (Chapter 3.7) is at 2.898\times10^{-3}/27600 = 105 nm — far ultraviolet. Most of its output is invisible, and it strongly ionises the surrounding gas, which is why massive young stars sit inside glowing emission nebulae.

Problem 4 — The Chandrasekhar limit for a different composition

The Chandrasekhar mass scales as \mu_e^{-2}, where \mu_e is the number of nucleons per electron. Compute it for a helium white dwarf and for an iron core.

Solution

For carbon–oxygen, \mu_e = 2 and M_{\text{Ch}} = 1.44M_\odot.

Helium-4: 4 nucleons, 2 electrons, so \mu_e = 2. Same limit, 1.44 M_\odot.

Iron-56: 56 nucleons, 26 electrons:

\mu_e = \frac{56}{26} = 2.154

M_{\text{Ch}} = 1.44\times\left(\frac{2}{2.154}\right)^2 = 1.44\times(0.9285)^2 = 1.44\times0.862 = 1.24\,M_\odot

What to notice. An iron core collapses at a lower mass than a carbon one, because iron has proportionally fewer electrons per nucleon and therefore less degeneracy pressure per unit mass.

This is why massive stars collapse when their iron core reaches about 1.2–1.4 M_\odot (Chapter 12.2), and it is also why the exact value depends on the core's composition and temperature.

For pure hydrogen, \mu_e = 1 and the limit would be 4\times1.44 = 5.8M_\odot — but a hydrogen white dwarf cannot exist, because hydrogen at those densities would fuse.

Problem 5 — Neutron star surface gravity and escape velocity

A neutron star has 1.4 solar masses and a radius of 11 km. Find the surface gravity, escape velocity, and gravitational redshift.

Solution

M = 1.4\times1.989\times10^{30} = 2.785\times10^{30}\ \text{kg}

Surface gravity:

g = \frac{GM}{R^2} = \frac{(6.674\times10^{-11})(2.785\times10^{30})}{(1.1\times10^{4})^2} = \frac{1.859\times10^{20}}{1.21\times10^{8}} = 1.54\times10^{12}\ \text{m/s}^2

Which is 1.57\times10^{11} times Earth's gravity.

Escape velocity:

v_{\text{esc}} = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2(1.859\times10^{20})}{1.1\times10^{4}}} = \sqrt{3.380\times10^{16}} = 1.84\times10^{8}\ \text{m/s}

= 0.61c

Gravitational redshift (Chapter 6.9):

\frac{\lambda_\infty}{\lambda_0} = \frac{1}{\sqrt{1-r_s/R}}

r_s = 2.95\times1.4 = 4.13\ \text{km}

\frac{r_s}{R} = \frac{4.13}{11} = 0.3755

\frac{\lambda_\infty}{\lambda_0} = \frac{1}{\sqrt{0.6245}} = \frac{1}{0.7903} = 1.265

A 26.5 % redshift.

What to notice. Escape velocity is 61 % of light speed, so the surface is deep in the relativistic regime and Newtonian calculations are only approximate.

The redshift is directly measurable and is one of the tools for constraining the neutron star equation of state — measuring M and the redshift gives R, which discriminates between models of matter above nuclear density.

And r_s/R = 0.38 means the star is only 2.7 times its own Schwarzschild radius. Neutron stars sit close to the boundary of becoming black holes, which is exactly why their maximum mass is so uncertain (Chapter 6.8).

Problem 6 — Hawking temperature and evaporation

Find the Hawking temperature and evaporation lifetime of a black hole with the mass of Mount Everest, 10^{15} kg.

Solution

T_H = \frac{\hbar c^3}{8\pi GMk_B}

= \frac{(1.055\times10^{-34})(2.7\times10^{25})}{8\pi(6.674\times10^{-11})(10^{15})(1.381\times10^{-23})}

Numerator: 2.849\times10^{-9}

Denominator: 8\pi\times6.674\times10^{-11} = 1.677\times10^{-9}; times 10^{15} = 1.677\times10^{6}; times 1.381\times10^{-23} = 2.316\times10^{-17}

T_H = \frac{2.849\times10^{-9}}{2.316\times10^{-17}} = 1.23\times10^{8}\ \text{K}

123 million kelvin — hotter than the Sun's core.

Lifetime:

t \approx 2\times10^{67}\left(\frac{M}{M_\odot}\right)^3\ \text{years} = 2\times10^{67}\left(\frac{10^{15}}{1.989\times10^{30}}\right)^3

= 2\times10^{67}\times(5.028\times10^{-16})^3 = 2\times10^{67}\times1.271\times10^{-46}

= 2.54\times10^{21}\ \text{years}

Its Schwarzschild radius:

r_s = \frac{2GM}{c^2} = \frac{2(6.674\times10^{-11})(10^{15})}{9\times10^{16}} = 1.48\times10^{-12}\ \text{m}

What to notice. A mountain compressed to a thousandth the size of an atom, at 123 million kelvin, radiating for 10^{21} years.

But it does not evaporate now. It absorbs the cosmic microwave background at 2.725 K — negligible compared with its own emission, so this one is net-evaporating, unlike a stellar black hole.

The mass that evaporates in exactly the age of the universe is about 10^{12} kg. Primordial black holes of that mass would be exploding today, and searches for the resulting gamma-ray bursts have found nothing — constraining their abundance to under 10^{-8} of the dark matter.

Problem 7 — Flat rotation curve mass

A galaxy has a flat rotation curve at 220 km/s out to 30 kpc. Find the enclosed mass, and compare with the visible mass of 6\times10^{10}M_\odot.

Solution

M = \frac{v^2r}{G}

r = 30\times3.086\times10^{19} = 9.258\times10^{20}\ \text{m}

v^2 = (2.2\times10^{5})^2 = 4.84\times10^{10}

M = \frac{(4.84\times10^{10})(9.258\times10^{20})}{6.674\times10^{-11}} = \frac{4.481\times10^{31}}{6.674\times10^{-11}} = 6.71\times10^{41}\ \text{kg}

= 3.38\times10^{11}M_\odot

\frac{M_{\text{total}}}{M_{\text{visible}}} = \frac{3.38\times10^{11}}{6\times10^{10}} = 5.6

What to notice. A factor of 5.6, which is the standard result for spiral galaxies at this radius and grows further out.

Now check what MOND predicts (Chapter 12.7):

v = (GMa_0)^{1/4}

With M = 6\times10^{10}M_\odot = 1.19\times10^{41} kg and a_0 = 1.2\times10^{-10}:

v^4 = (6.674\times10^{-11})(1.19\times10^{41})(1.2\times10^{-10}) = 9.53\times10^{20}

v = (9.53\times10^{20})^{1/4}

\sqrt{9.53\times10^{20}} = 3.09\times10^{10}, \qquad \sqrt{3.09\times10^{10}} = 1.76\times10^{5}

v = 176\ \text{km/s}

Against the observed 220. Within 20 % using only the visible mass and one universal constant, with no free parameters.

This is why MOND is not dismissed despite its failures at cluster and cosmological scales. It fits galaxy rotation curves remarkably well, and any dark matter model must explain why the halo properties correlate so tightly with the visible matter.

Problem 8 — Critical density and the number of atoms

Compute the critical density for H_0 = 70 km/s/Mpc, the mean baryon density, and the number of atoms per cubic metre.

Solution

H_0 = \frac{70\times10^{3}}{3.086\times10^{22}} = 2.268\times10^{-18}\ \text{s}^{-1}

\rho_c = \frac{3H_0^2}{8\pi G} = \frac{3(5.144\times10^{-36})}{8\pi(6.674\times10^{-11})} = \frac{1.543\times10^{-35}}{1.677\times10^{-9}}

\rho_c = 9.20\times10^{-27}\ \text{kg/m}^3

Baryon density, with \Omega_b = 0.049:

\rho_b = 0.049\times9.20\times10^{-27} = 4.51\times10^{-28}\ \text{kg/m}^3

Atoms per cubic metre, taking the mean mass as 1.22m_p for a 75:25 hydrogen–helium mix:

n = \frac{4.51\times10^{-28}}{1.22\times1.673\times10^{-27}} = \frac{4.51\times10^{-28}}{2.041\times10^{-27}} = 0.221\ \text{m}^{-3}

What to notice. About one atom per five cubic metres, averaged over the whole universe including galaxies.

The best laboratory vacuum reaches about 10^{-12} Pa, which at room temperature is 2.4\times10^{8} molecules per cubic metre — a billion times denser than the cosmic average.

Intergalactic space is emptier than anything achievable on Earth by nine orders of magnitude.

And the total baryonic mass in the observable universe:

M = \rho_b\times\frac{4}{3}\pi(4.4\times10^{26})^3 = (4.51\times10^{-28})(3.57\times10^{80}) = 1.6\times10^{53}\ \text{kg}

About 10^{80} atoms, which is the origin of that frequently quoted number.

Problem 9 — Redshift and lookback

A quasar has z = 3.0. Find the scale factor when its light was emitted, the CMB temperature at that time, and the wavelength at which its Lyman-alpha emission is observed.

Solution

Scale factor:

a = \frac{1}{1+z} = \frac{1}{4} = 0.25

The universe was a quarter of its present size.

CMB temperature. Since T \propto 1/a:

T = 2.725\times(1+z) = 2.725\times4 = 10.9\ \text{K}

Lyman-alpha, emitted at 121.6 nm:

\lambda_{\text{obs}} = 121.6\times4 = 486.4\ \text{nm}

Blue-green visible light, from an ultraviolet transition.

What to notice. This shift is what makes high-redshift quasars observable at all. Lyman-alpha at 121.6 nm is blocked by the Earth's atmosphere (Chapter 11.9), so it is unobservable from the ground for nearby objects — and at z = 3 it has been shifted into the visible band.

And the CMB at 10.9 K means the universe was noticeably warmer. At z = 50 it was at 139 K; at z = 137 it was at 373 K — the boiling point of water.

There was an epoch, roughly z = 100 to 137, when the entire universe was between the freezing and boiling points of water. Liquid water could have existed anywhere with sufficient pressure, about 10 to 17 million years after the Big Bang. Whether anything could have used it is doubtful — there were no heavy elements yet — but it is a real feature of the thermal history.

Problem 10 — Primordial helium

Verify the predicted helium mass fraction from the neutron-to-proton ratio at nucleosynthesis.

Solution

Freeze-out ratio at k_BT = 0.86 MeV, with \Delta mc^2 = 1.293 MeV:

\frac{n}{p} = e^{-1.293/0.86} = e^{-1.5035} = 0.2224

Neutron decay over about 180 s, with \tau = 879 s:

\frac{n}{p} = 0.2224\times e^{-180/879} = 0.2224\times e^{-0.2048} = 0.2224\times0.8148 = 0.1812

Take 1000 protons, so there are 181 neutrons.

All neutrons go into helium-4, each nucleus using 2 neutrons and 2 protons:

N_{\text{He}} = \frac{181}{2} = 90.5

Protons consumed: 2\times90.5 = 181

Protons remaining: 1000-181 = 819

Mass fraction:

Y = \frac{90.5\times4}{90.5\times4+819\times1} = \frac{362}{362+819} = \frac{362}{1181} = 0.307

That is higher than the standard 0.25, because using k_BT = 0.86 MeV overstates the freeze-out ratio slightly. With the standard freeze-out value of n/p = 1/7 after decay:

Y = \frac{2}{1+n/p\ \text{inverse}} = \frac{2(1/7)}{1+1/7} = \frac{2}{8} = 0.25

What to notice. The result is very sensitive to two numbers: the neutron–proton mass difference and the neutron lifetime.

And it is sensitive to the number of neutrino species. More species means more relativistic energy density, faster expansion, earlier freeze-out, a higher n/p ratio, and more helium.

Nucleosynthesis constrained the number of neutrino families to three before LEP measured it (Chapter 8.6). Two completely different experiments — a cosmological abundance measurement and a particle collider — agreeing on a count of fundamental particle families.

Problem 11 — Age of the universe from H_0

Compute the Hubble time for H_0 = 70 and for H_0 = 73 km/s/Mpc, and comment on the Hubble tension.

Solution

For H_0 = 70:

\frac{1}{H_0} = \frac{1}{2.268\times10^{-18}} = 4.409\times10^{17}\ \text{s} = 13.97\ \text{Gyr}

For H_0 = 73:

H_0 = \frac{73\times10^{3}}{3.086\times10^{22}} = 2.366\times10^{-18}\ \text{s}^{-1}

\frac{1}{H_0} = 4.227\times10^{17}\ \text{s} = 13.39\ \text{Gyr}

Difference: 0.58 Gyr, about 4 %.

What to notice. The Hubble time is not the age; the actual age depends on the expansion history. With \Omega_m = 0.315 and \Omega_\Lambda = 0.685, the true age is 0.956 times the Hubble time:

t_0 = 0.956\times13.97 = 13.36\ \text{Gyr} \quad\text{for } H_0 = 70

t_0 = 0.956\times13.39 = 12.80\ \text{Gyr} \quad\text{for } H_0 = 73

The Planck value of 13.797 Gyr comes from the full parameter fit with H_0 = 67.4.

And here is why the tension matters beyond the number itself. The oldest globular clusters are dated at 12.5\pm1 Gyr from stellar evolution models. At H_0 = 73, the universe is 12.8 Gyr old, which leaves under a gigayear between the Big Bang and the oldest stars.

That is uncomfortably tight and it is an independent reason to take the tension seriously rather than assuming the local measurement is simply wrong.

Problem 12 — Escape from a galaxy cluster

The Coma Cluster has a velocity dispersion of 1000 km/s and a radius of 3 Mpc. Estimate its mass by the virial theorem, and compare with the visible mass of 3\times10^{13}M_\odot.

Solution

M \approx \frac{5\sigma^2R}{G}

R = 3\times3.086\times10^{22} = 9.258\times10^{22}\ \text{m}

\sigma^2 = (10^{6})^2 = 10^{12}

M = \frac{5(10^{12})(9.258\times10^{22})}{6.674\times10^{-11}} = \frac{4.629\times10^{35}}{6.674\times10^{-11}} = 6.94\times10^{45}\ \text{kg}

= 3.49\times10^{15}M_\odot

\frac{M_{\text{total}}}{M_{\text{visible}}} = \frac{3.49\times10^{15}}{3\times10^{13}} = 116

What to notice. A factor of over 100, which is larger than the accepted value of about 10.

The discrepancy comes from the visible mass estimate. The stars in cluster galaxies are only about 10 % of the cluster's baryons — most of the ordinary matter is hot X-ray-emitting gas between the galaxies, which was unknown to Zwicky in 1933 and is why his original estimate was a factor of 400 rather than 10.

Including the gas, the baryonic mass is about 3\times10^{14}M_\odot, giving a dark-to-visible ratio of about 10 — consistent with cosmological measurements.

This is a good illustration of how a discrepancy can be real and its magnitude wrong. Zwicky was right that most of the mass is invisible, and wrong by a factor of 40 about how much, and the correction came from a component nobody had imagined in 1933.

Problem 13 — Wormhole exotic matter

Estimate the negative energy needed for a traversable wormhole with a throat radius of 10 metres, and compare with the Casimir energy achievable in a 1 m³ apparatus with 100 nm plate spacing.

Solution

Required:

E \sim -\frac{c^4r_0}{G} = -\frac{(3\times10^{8})^4(10)}{6.674\times10^{-11}} = -\frac{8.1\times10^{34}}{6.674\times10^{-11}}

E \sim -1.2\times10^{45}\ \text{J}

Equivalent mass: -1.35\times10^{28} kg — about seven Jupiters.

Casimir energy available. The energy density between plates at spacing d:

u = -\frac{\pi^2\hbar c}{720d^3}

At d = 10^{-7} m:

u = -\frac{(9.8696)(1.055\times10^{-34})(3\times10^{8})}{720(10^{-21})} = -\frac{3.124\times10^{-25}}{7.2\times10^{-19}} = -4.34\times10^{-7}\ \text{J/m}^3

In 1 m³ (packed with plates at that spacing):

E \approx -4.34\times10^{-7}\ \text{J}

\frac{1.2\times10^{45}}{4.34\times10^{-7}} = 2.8\times10^{51}

What to notice. Fifty-one orders of magnitude short, and that is with an idealised apparatus filling a cubic metre entirely with Casimir plates.

To make up the difference you would need 2.8\times10^{51} cubic metres of apparatus, which is a sphere about 90 light years across.

And the quantum inequalities (Chapter 12.9) make it worse still by limiting how long negative energy can persist.

The point of the calculation is not that it is hard. It is that the shortfall is not a factor of a thousand or a million — it is 10^{51}, which is a different category of obstacle entirely.

Problem 14 — When the last star dies

The lowest-mass stars are 0.08M_\odot. Using t \propto M/L and L \propto M^{3.5}, estimate their lifetime. Then find the fraction of cosmic history that has elapsed.

Solution

\frac{t}{t_\odot} = \left(\frac{M}{M_\odot}\right)^{-2.5} = (0.08)^{-2.5}

\ln(0.08) = -2.526, \quad -2.5\times(-2.526) = 6.315, \quad e^{6.315} = 553

t = 553\times10^{10} = 5.5\times10^{12}\ \text{years}

But this underestimates, because red dwarfs are fully convective and burn essentially all their hydrogen rather than just the core's 10 %.

t = 10\times5.5\times10^{12} = 5.5\times10^{13}\ \text{years}

About 10^{14} years, matching the standard figure.

Fraction elapsed:

\frac{1.38\times10^{10}}{5.5\times10^{13}} = 2.5\times10^{-4}

What to notice. We are 0.025 % of the way through the era in which stars shine, and the stelliferous era is itself a vanishing fraction of the universe's total history — which runs to at least 10^{100} years (Chapter 12.8).

On a logarithmic timeline, the present is extraordinarily early.

And if intelligent observers are more likely to arise later — when more heavy elements have accumulated and more red dwarfs have had time to develop planets — then finding ourselves this early is itself unusual. That observation is sometimes used as an argument that habitability declines, or that something ends the era of observers early.

It is speculative reasoning about a sample of one, and it is worth knowing that the argument exists and how weak its foundations are.