Lecture 21 Solutions: Neutron Stars and Black Holes

Practice Problem Solutions

Solutions to the Lecture 21 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 Problems

Problem 1: Neutron Star Density

Given: Mass \(M = 2.8 \times 10^{30}\,\mathrm{kg}\); radius \(R = 10\,\mathrm{km} = 1.0 \times 10^4\,\mathrm{m}\).

Find: Density, comparison to nuclear density, and teaspoon mass.

Equation:

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

Steps:

Find the volume:

\[ V = \frac{4}{3}\pi (1.0 \times 10^4)^3 \approx 4.19 \times 10^{12}\,\mathrm{m^3} \]

Then the density:

\[ \rho = \frac{2.8 \times 10^{30}}{4.19 \times 10^{12}} \approx 6.7 \times 10^{17}\,\mathrm{kg/m^3} \]

Compare to nuclear density:

\[ \frac{6.7 \times 10^{17}}{2.3 \times 10^{17}} \approx 2.9 \]

So the mean density is about three times nuclear density.

For a teaspoon of volume \(5 \times 10^{-6}\,\mathrm{m^3}\):

\[ m = \rho V = (6.7 \times 10^{17})(5 \times 10^{-6}) \approx 3.4 \times 10^{12}\,\mathrm{kg} \]

Answer: The density is about \(6.7 \times 10^{17}\,\mathrm{kg/m^3}\). That is about 3 times nuclear density, and a teaspoon would contain about \(3 \times 10^{12}\,\mathrm{kg}\) of matter.

Problem 2: Schwarzschild Radius

Given: \(R_s \approx 2.95\,\mathrm{km}\,(M/M_\odot)\)

Find: The Schwarzschild radius for a stellar black hole, Sgr A*, and Earth

Use the shortcut

\[ R_s \approx 2.95\,\mathrm{km}\times \left(\frac{M}{M_\odot}\right) \]

(a) 20-solar-mass black hole

\[ R_s \approx 2.95 \times 20 \approx 59\,\mathrm{km} \]

(b) Sgr A* with \(4.1 \times 10^6\,M_\odot\)

\[ R_s \approx 2.95 \times 4.1 \times 10^6 \approx 1.21 \times 10^7\,\mathrm{km} \]

(c) Earth

Earth has mass \(M_\oplus \approx 5.97 \times 10^{24}\,\mathrm{kg}\), or

\[ \frac{M_\oplus}{M_\odot} \approx 3.0 \times 10^{-6} \]

So

\[ R_s \approx 2.95 \times 3.0 \times 10^{-6}\,\mathrm{km} \approx 8.9 \times 10^{-6}\,\mathrm{km} = 8.9\,\mathrm{mm} \]

Unit check: The mass ratio is dimensionless, so the output keeps the kilometer units built into the shortcut.

Sanity check: Schwarzschild radius scales linearly with mass, so a supermassive black hole should have a much larger event horizon than a stellar black hole, while Earth’s would be tiny.

Answer:
- \(20\,M_\odot\): about 59 km
- Sgr A*: about \(1.2 \times 10^7\) km
- Earth: about 9 mm

Problem 3: Pulsar Rotation

Given: Period \(P = 0.033\,\mathrm{s}\); radius \(R = 10\,\mathrm{km} = 1.0 \times 10^4\,\mathrm{m}\).

(a) Rotations per second

\[ f = \frac{1}{P} = \frac{1}{0.033} \approx 30.3\,\mathrm{s^{-1}} \]

(b) Equatorial speed

\[ v = \frac{2\pi R}{P} = \frac{2\pi (1.0 \times 10^4)}{0.033} \approx 1.9 \times 10^6\,\mathrm{m/s} \]

That is about \(1900\,\mathrm{km/s}\).

(c) Fraction of the speed of light

\[ \frac{v}{c} = \frac{1.9 \times 10^6}{3.0 \times 10^8} \approx 6.3 \times 10^{-3} \]

Answer: The Crab Pulsar rotates about 30 times per second, its equator moves at about \(1900\,\mathrm{km/s}\), and that is about 0.6% of the speed of light.

Problem 4: Orbital Mechanics at the Galactic Center

Use Kepler’s third law in solar units:

\[ M = \frac{a^3}{P^2} \]

with \(a\) in AU, \(P\) in years, and \(M\) in solar masses.

Given: \(a = 1000\,\mathrm{AU}\), \(P = 16\,\mathrm{yr}\).

\[ M = \frac{1000^3}{16^2} = \frac{10^9}{256} \approx 3.9 \times 10^6\,M_\odot \]

Interpretation: That is millions of solar masses confined to a tiny region near the Galactic Center. No normal star cluster can pack that much mass into such a small volume and remain dark and stable.

Answer: The central mass is about \(4 \times 10^6\,M_\odot\), which is strong evidence that Sgr A* is a supermassive black hole.

Problem 5: Accretion Disk Power

Given: Accretion rate \(\dot{M} = 10^{-9}\,M_\odot/\mathrm{yr}\) and efficiency \(\eta = 0.1\).

Convert the mass rate to SI:

\[ \dot{M} = \frac{10^{-9}(2 \times 10^{30}\,\mathrm{kg})}{3.15 \times 10^7\,\mathrm{s}} \approx 6.35 \times 10^{13}\,\mathrm{kg/s} \]

Use

\[ P = \eta \dot{M} c^2 \]

so

\[ P = 0.1 \times 6.35 \times 10^{13} \times (3.0 \times 10^8)^2 \approx 5.7 \times 10^{29}\,\mathrm{W} \]

Compare to the Sun:

\[ \frac{5.7 \times 10^{29}}{3.8 \times 10^{26}} \approx 1.5 \times 10^3 \]

Answer: The accretion disk radiates about \(5.7 \times 10^{29}\,\mathrm{W}\), or roughly 1500 times the Sun’s luminosity.

Challenge Problems

Problem 1: Tidal Forces and Spaghettification

Use

\[ a_{\rm tidal} \approx \frac{2GM\Delta r}{r^3} \]

with \(\Delta r = 1.7\,\mathrm{m}\).

(a) At the event horizon of a \(10\,M_\odot\) black hole

The horizon radius is

\[ R_s \approx 29.5\,\mathrm{km} = 2.95 \times 10^4\,\mathrm{m} \]

Then

\[ a_{\rm tidal} \approx \frac{2G(10M_\odot)(1.7)}{R_s^3} \approx 1.8 \times 10^8\,\mathrm{m/s^2} \]

That is enormously larger than Earth’s gravity.

(b) Distance where the tidal acceleration equals \(9.8\,\mathrm{m/s^2}\)

Solve

\[ 9.8 = \frac{2GM\Delta r}{r^3} \]

so

\[ r = \left(\frac{2GM\Delta r}{9.8}\right)^{1/3} \approx 7.7 \times 10^6\,\mathrm{m} \]

That is about 7700 km from the black hole’s center.

(c) Sgr A*

For \(M = 4.1 \times 10^6\,M_\odot\),

\[ r \approx \left(\frac{2GM\Delta r}{9.8}\right)^{1/3} \approx 5.8 \times 10^8\,\mathrm{m} \]

But Sgr A* has

\[ R_s \approx 1.2 \times 10^{10}\,\mathrm{m} \]

So the radius where tidal acceleration reaches \(g\) is well inside the event horizon.

Answer: A stellar-mass black hole tears you apart well outside the horizon, while a supermassive black hole has much gentler tidal forces at the horizon, so you could cross it before feeling extreme stretching.

Problem 2: Neutron Star Mass Limit

Use

\[ M_{\max} \approx 0.7 \frac{c^2 R_{\rm ns}}{G} \]

with \(R_{\rm ns} = 10\,\mathrm{km} = 1.0 \times 10^4\,\mathrm{m}\).

\[ M_{\max} \approx 0.7 \times \frac{(3.0 \times 10^8)^2(1.0 \times 10^4)}{6.67 \times 10^{-11}} \]

\[ M_{\max} \approx 9.4 \times 10^{30}\,\mathrm{kg} \]

Convert to solar masses:

\[ \frac{9.4 \times 10^{30}}{2.0 \times 10^{30}} \approx 4.7\,M_\odot \]

Interpretation: This simple model gives a maximum mass around \(4.7\,M_\odot\), which is higher than the observed TOV limit of roughly \(2\) to \(3\,M_\odot\). That mismatch tells you the real physics of neutron matter and strong gravity is more restrictive than this toy estimate.

Problem 3: Gravitational Wave Energy

Given: Energy released \(= 3\,M_\odot c^2\) over about \(0.2\) s.

Compute the energy:

\[ E = 3(2.0 \times 10^{30})(3.0 \times 10^8)^2 \approx 5.4 \times 10^{47}\,\mathrm{J} \]

Average power:

\[ P_{\rm avg} = \frac{E}{0.2} \approx 2.7 \times 10^{48}\,\mathrm{W} \]

Compare to the Sun:

\[ \frac{2.7 \times 10^{48}}{3.8 \times 10^{26}} \approx 7 \times 10^{21} \]

So the average power was about \(7 \times 10^{21}\) Suns.

Because the inspiral accelerates rapidly near merger, the peak power must be higher than the average by at least an order-unity factor. An order-of-magnitude estimate for the peak is therefore around

\[ 10^{49}\,\mathrm{W} \]

Answer: The average gravitational-wave power was about \(2.7 \times 10^{48}\,\mathrm{W}\), and the peak was plausibly of order \(10^{49}\,\mathrm{W}\).

Problem 4: Multi-Messenger Astronomy

Restatement: We want to interpret what the gravitational-wave chirp and delayed electromagnetic emission reveal about the merger.

Key elements: - what the rising frequency means - what the time delay means - what we infer about ejecta and the merger aftermath

The gravitational-wave signal sweeping from low to high frequency is the classic chirp of an inspiraling compact binary:

• rising frequency means the orbit is shrinking
• rising amplitude means the objects are accelerating and radiating more strongly

The delay between the gravitational-wave signal and the electromagnetic signal tells us the light did not come from the inspiral itself. Instead, it came from ejecta, shocks, and radioactive material produced after merger.

So the combined signal lets us infer:

• the orbital inspiral dynamics from the gravitational waves
• the production of ejecta and heavy elements from the delayed light
• the geometry and expansion of the outflow from the time delay and spectral evolution

Observable -> model -> inference: - Observable: gravitational-wave chirp followed by delayed light across the spectrum - Model: inspiral and merger of two neutron stars, plus radioactive ejecta and afterglow emission - Inference: the event produced both compact-object dynamics and matter-rich outflows, including the conditions needed for heavy-element production

Answer: The frequency evolution shows a shrinking, accelerating binary orbit. The delayed electromagnetic emission shows that the merger launched ejecta and powered a kilonova/afterglow, making GW170817 a textbook case of multi-messenger astronomy.