Answer key · Unit 26

The extreme universe.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 5

Lesson 1 of 5 · About 25 minutes

Nebulae

Clouds of gas and dust between the stars: why some glow red, some blue and some are black patches, and how big a bubble a hot star keeps ionised.

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The 5 steps and their answers

Step 1 of 5
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A glowing cloud
Practice · Choose
Question
The star in the middle of the cloud is a Hot O star, blue-white, yet the gas round it glows red and pink. Where does the red light come from?

Correct answer
- Correct From hydrogen atoms the star has split: when an electron rejoins a proton it gives out hydrogen’s red line
Setup, choices and feedback
The Scene at this step
Star's temperature
39000 K
Gas density
100 atoms/cm³
Split its light
As we see it
Choices
Shown in a random order.
- Correct From hydrogen atoms the star has split: when an electron rejoins a proton it gives out hydrogen’s red line
- Wrong The cloud reflects the star’s light
  Misconception: Thinks an emission nebula glows red because it reflects its star’s light.
  Feedback from the catalogue
  The star in the middle of an emission nebula is blue-white, yet the cloud glows red and pink. That light is not reflected from the star: the gas gives it out itself. The star’s ultraviolet splits the hydrogen atoms, and when an electron rejoins a proton it steps down the energy levels and gives out particular lines. The strongest of them in visible light is hydrogen’s red line, at 656 nm.
- Wrong The gas is red-hot like embers, so it glows red
  Feedback
  The gas is hot, about 10,000 K, but it is far too thin to give out a continuous spectrum like embers. It gives out single lines.
- Wrong The dust in the cloud absorbs the blue and leaves only red
  Feedback
  Dust does redden stars behind it, but it cannot make the cloud glow. The glow comes from the gas itself.
Feedback for a right answer
Right: the gas glows like a neon sign, at hydrogen’s wavelengths.

Worked solution
1. The star is very hot, and most of its light is ultraviolet, with photons above $13.6$ ⁠ eV that split hydrogen atoms into a proton and an electron.
2. Now and then a proton captures an electron again, and it steps down the atom’s energy levels in a few jumps.
3. Each jump gives out a photon carrying the difference between the levels. The jump from the third level to the second gives hydrogen’s red line, at 656 nm.
Step 2 of 5
Open this step
Light that splits atoms
Practice · Calculate
Question
Splitting a hydrogen atom in its ground state takes a photon of at least $13.6$ ⁠ eV. What is the longest wavelength that can still do it, in nm? $hc = 1240\ \mathrm{eV \cdot nm}$ ⁠.

Correct answer
91.2 nm ± 2%
Setup, choices and feedback
The Scene at this step
Star's temperature
39000 K
Gas density
100 atoms/cm³
Split its light
Its atoms
Accepted within
± 2%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
91.2 nm, far ultraviolet: only very hot stars give out enough of it.

Hint
$\lambda = \dfrac{hc}{E}$ ⁠.

Worked solution
1. $\lambda = \dfrac{hc}{E} = \dfrac{1240}{13.6} \approx 91.2\ \mathrm{nm}$ ⁠.
That is far ultraviolet, well below the 400 nm of the violet end. A star like the Sun gives out almost none of it, so only very hot stars, of type O and early B, light up emission nebulae.
Step 3 of 5
Open this step
The bubble
Practice · Calculate
Question
At a Gas density of $100$ ⁠ atoms per cm³ the star holds an ionised bubble about $10.1$ ⁠ light-years in radius. The density is now locked at $800$ ⁠, eight times as much. Using $R \propto n^{-2/3}$ ⁠, what is the bubble’s radius, in light-years?

Correct answer
2.52 ly ± 3%
Setup, choices and feedback
The Scene at this step
Star's temperature
39000 Kfixed
Gas density
800 atoms/cm³fixed
Split its light
Its atoms
Accepted within
± 3%
Computed from the Scene at this step’s setting
Strömgren radius in metres: the bubble the star keeps ionised/nodes/cloud/outputs/stromgrenRadius
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1.26 ly ± 3%
  Misconception: Thinks the glowing bubble shrinks in proportion to the gas density: eight times denser, eight times smaller.
  Feedback from the catalogue
  The bubble grows until, each second, as many electrons and protons rejoin inside it as the star splits apart. The rate of rejoining in each cubic centimetre grows like $n^2$ ⁠, because an electron and a proton have to meet, and the bubble’s volume grows like $R^3$ ⁠. So $R^3 n^2$ ⁠ stays fixed, and $R \propto n^{-2/3}$ ⁠: $8$ ⁠ times denser makes the bubble $8^{2/3} = 4$ ⁠ times smaller, not $8$ ⁠.
Feedback for a right answer
About $2.5$ ⁠ light-years: eight times denser, a bubble four times smaller, but glowing brighter.

Hint
$8^{2/3} = \left(\sqrt[3]{8}\right)^2$ ⁠.

Worked solution
1. In the bubble, splittings per second equal rejoinings: $Q = \tfrac{4}{3}\pi R^3 n^2 \alpha$ ⁠, so $R \propto n^{-2/3}$ ⁠.
2. $8^{2/3} = 2^2 = 4$ ⁠.
3. $R = \dfrac{10.1}{4} \approx 2.5$ ⁠ light-years.
In dense gas each atom rejoins quickly, so the star can keep only a small bubble ionised. Answering about $1.3$ ⁠ divides by $8$ ⁠, as if the radius fell like the density.
Step 4 of 5
Open this step
Blue dust
Practice · Predict, then look
Question
What will happen to the cloud when Star’s temperature drops from the O star to a Blue-white B star, at about 14,000 K?

Correct answer
- Correct The red glow will almost vanish, and the dust will look bluish
Setup, choices and feedback
The Scene at this step
Star's temperature
39000 K
Gas density
100 atoms/cm³
Split its light
As we see it
Choices
- Correct The red glow will almost vanish, and the dust will look bluish
- Wrong The red bubble will just shrink a little
  Feedback
  The bubble shrinks tens of times: at 14,000 K the star gives out very few photons above $13.6$ ⁠ eV.
- Wrong The cloud will glow a stronger red
  Misconception: Thinks an emission nebula glows red because it reflects its star’s light.
  Feedback from the catalogue
  The star in the middle of an emission nebula is blue-white, yet the cloud glows red and pink. That light is not reflected from the star: the gas gives it out itself. The star’s ultraviolet splits the hydrogen atoms, and when an electron rejoins a proton it steps down the energy levels and gives out particular lines. The strongest of them in visible light is hydrogen’s red line, at 656 nm.
Once the prediction is locked in, the Scene changes
Star's temperature
14000 K
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A reflection nebula: dust scattering starlight, blue like the sky.

Worked solution
1. The number of ionising photons, above $13.6$ ⁠ eV, falls steeply with temperature: a B star makes less than a hundredth of an O star’s.
2. So the ionised bubble shrinks to a fraction of a light-year, and the red glow almost vanishes.
3. The cloud’s dust still scatters the star’s visible light, and blue scatters more than red, as in the daytime sky. That is a reflection nebula, like the one round the Pleiades.
Step 5 of 5
Open this step
A black cloud
Practice · Reach the target
Question
The star is like the Sun, and the cloud is so thin you can hardly see it. Turn it into a dark nebula, a black patch hiding the stars behind it, and press Check.

Goal
- 1 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Star's temperature
5800 K
Gas density
20 atoms/cm³
Split its light
As we see it
Read from the Scene at
0 thin, 1 dark, 2 reflection, 3 emission, 4 ionised/nodes/cloud/outputs/regime
One setting that reaches it
Star's temperature
3000 K
Gas density
50 atoms/cm³
Feedback for a right answer
A dark nebula: dense dust absorbing the light of the stars behind it.

Feedback for a wrong answer
Not yet. If the cloud is still see-through, it needs more dust; if it glows, the star is too hot.

Hint
What hides starlight: the star, or the dust in the cloud?

Worked solution
1. A cool star like the Sun does not ionise the gas or light up enough dust to be seen.
2. What is left is the dust itself, which absorbs the light of the stars behind it. The denser the cloud, the more it absorbs.
3. Raise Gas density to about $100$ ⁠ atoms per cm³ or more: the cloud becomes a black hole in the star field, and the stars at its edge turn red.
The Horsehead Nebula in Orion is a dark nebula like this, against a glowing emission nebula.

Lesson 2 of 5 · About 20 minutes

Neutron stars and pulsars

What is left of a heavy star that exploded: a ball the width of a city that spins hundreds of times a second and sweeps us like a lighthouse.

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The 6 steps and their answers

Step 1 of 6
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A lighthouse in space
Practice · Choose
Question
A radio telescope receives short, regular pulses from the pulsar. What makes them?

Correct answer
- Correct A steady radio beam that turns with the star and sweeps across the Earth each turn
Setup, choices and feedback
The Scene at this step
Spin rate
30 turns/s
Beam tilt
75 °
Show the pulses
Off
Choices
Shown in a random order.
- Correct A steady radio beam that turns with the star and sweeps across the Earth each turn
- Wrong The star switches on and off again and again
  Misconception: Thinks a pulsar’s star itself flashes on and off, rather than a steady beam sweeping past us as it turns.
  Feedback from the catalogue
  The neutron star does not flash. It sends out two steady beams of radio waves from its magnetic poles, and when its magnetic axis is tilted from its spin axis, the beams turn with it like a lighthouse’s lamp. A pulse arrives only when a beam sweeps across our line of sight. If the beams miss the Earth, the star keeps shining, but we receive no pulse at all.
- Wrong The star swells and shrinks like a beating heart
  Misconception: Thinks a pulsar’s star itself flashes on and off, rather than a steady beam sweeping past us as it turns.
  Feedback from the catalogue
  The neutron star does not flash. It sends out two steady beams of radio waves from its magnetic poles, and when its magnetic axis is tilted from its spin axis, the beams turn with it like a lighthouse’s lamp. A pulse arrives only when a beam sweeps across our line of sight. If the beams miss the Earth, the star keeps shining, but we receive no pulse at all.
- Wrong Signals from another civilisation
  Feedback
  Jocelyn Bell Burnell and her colleagues jokingly called the first pulsar LGM-1, “little green men”, but soon found more and more like it, all over the sky.
Feedback for a right answer
A cosmic lighthouse: the beam is steady, and the turning makes the pulses.

Worked solution
1. A neutron star sends out two radio beams from its magnetic poles.
2. The magnetic axis is tilted from the spin axis, so the beams turn with the star, like a lighthouse’s lamp.
3. Each turn, if a beam sweeps across our line of sight, a pulse arrives. The time between pulses is the time for one turn.
Step 2 of 6
Open this step
The Crab
Practice · Calculate
Question
Spin rate is locked at the Crab Nebula’s pulsar, $30$ ⁠ turns a second. How many milliseconds pass between two pulses?

Correct answer
33.33 ms ± 2%
Setup, choices and feedback
The Scene at this step
Spin rate
30 turns/sfixed
Beam tilt
75 °
Show the pulses
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
Time for one turn, 1 / spin rate/nodes/pulsar/outputs/period
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
33.3 ms: thirty pulses a second, from a star heavier than the Sun.

Hint
$T = \dfrac{1}{f}$ ⁠.

Worked solution
1. $T = \dfrac{1}{f} = \dfrac{1}{30} \approx 0.0333\ \mathrm{s}$ ⁠.
2. In ms: $33.3$ ⁠.
The Crab pulsar was born in a supernova that Chinese and Arab astronomers recorded in 1054, and it still turns $30$ ⁠ times a second.
Step 3 of 6
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Spinning up
Practice · Calculate
Question
Spin rate is locked at the Fastest known, $716$ ⁠ turns a second. A neutron star has a radius of about 12 km. How fast does its equator move, in km/s?

Correct answer
53990 km/s ± 2%
Setup, choices and feedback
The Scene at this step
Spin rate
716 turns/sfixed
Beam tilt
75 °
Show the pulses
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
Speed of the equator of a 12 km neutron star/nodes/pulsar/outputs/equatorSpeed
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 54,000 km/s, $18\%$ ⁠ of the speed of light.

Hint
Each turn the equator travels the circumference, $2\pi r$ ⁠.

Worked solution
1. The circumference: $2\pi r = 2\pi \cdot 12 \approx 75.4\ \mathrm{km}$ ⁠.
2. $v = 2\pi r f = 75.4 \cdot 716 \approx 54{,}000\ \mathrm{km/s}$ ⁠.
3. Compared with the speed of light, $\dfrac{54{,}000}{300{,}000} \approx 0.18$ ⁠: almost a fifth of light speed.
An ordinary star would be torn apart by such a spin. Only a neutron star’s enormous gravity holds it together.
Step 4 of 6
Open this step
Tilt decides what we see
Practice · Predict, then look
Question
What will happen to the pulse train on the plot when Beam tilt changes to just $20^\circ$ ⁠ between the magnetic axis and the spin axis?

Correct answer
- Correct The pulses will vanish: the beams no longer sweep across the Earth
Setup, choices and feedback
The Scene at this step
Spin rate
30 turns/s
Beam tilt
75 °
Show the pulses
On
Choices
- Correct The pulses will vanish: the beams no longer sweep across the Earth
- Wrong The star will go on pulsing as before
  Misconception: Thinks a pulsar’s star itself flashes on and off, rather than a steady beam sweeping past us as it turns.
  Feedback from the catalogue
  The neutron star does not flash. It sends out two steady beams of radio waves from its magnetic poles, and when its magnetic axis is tilted from its spin axis, the beams turn with it like a lighthouse’s lamp. A pulse arrives only when a beam sweeps across our line of sight. If the beams miss the Earth, the star keeps shining, but we receive no pulse at all.
- Wrong The pulses will come faster
  Feedback
  The pulse rate is set by the spin rate, which has not changed. The tilt only decides whether the beam hits us.
Once the prediction is locked in, the Scene changes
Beam tilt
20 °
The Scene then shows: 0 · Beams that sweep across the line of sight each turn: 0, 1 or 2/nodes/pulsar/outputs/pulsesPerTurn
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The line goes flat: the beam still turns, but misses us.

Worked solution
1. Our line of sight is $80$ ⁠ degrees from the spin axis, and each beam spreads about $15$ ⁠ degrees either side.
2. A beam hits us only if the magnetic axis passes within $15$ ⁠ degrees of the line of sight: $|\alpha - 80^\circ| < 15^\circ$ ⁠.
3. At $20$ ⁠ degrees the beam turns in a small circle round the pole, $60$ ⁠ degrees from us, and never hits.
The star shines exactly as before; we are simply outside the beam. That is why the Galaxy holds many more pulsars than we see.
Step 5 of 6
Open this step
Two beams
Practice · Reach the target
Question
Some pulsars send a second, fainter pulse each turn, half a turn after the first. Change Beam tilt until both beams reach the Earth every turn, and press Check.

Goal
- 2 ± 0.1
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Spin rate
30 turns/s
Beam tilt
75 °
Show the pulses
On
Read from the Scene at
Beams that sweep across the line of sight each turn: 0, 1 or 2/nodes/pulsar/outputs/pulsesPerTurn
One setting that reaches it
Spin rate
0.5 turns/s
Beam tilt
86 °
Feedback for a right answer
Two pulses every turn: the extra one comes from the other pole.

Feedback for a wrong answer
Not yet. The second beam still misses: tilt the magnetic axis further.

Hint
The second beam leaves the other pole, pointing the opposite way. When do both pass near the equator?

Worked solution
1. The north beam hits us when $|\alpha - 80^\circ| < 15^\circ$ ⁠, that is $65^\circ < \alpha < 95^\circ$ ⁠.
2. The south beam points $180^\circ - \alpha$ ⁠ from the axis and hits when $|180^\circ - \alpha - 80^\circ| < 15^\circ$ ⁠, that is $\alpha > 85^\circ$ ⁠.
3. Both together: $\alpha$ ⁠ above $85$ ⁠ degrees, nearly $90^\circ$ ⁠.
Then the magnetic axis lies almost in the plane of the equator, and both beams sweep the same band of sky, half a turn apart.
Step 6 of 6
Open this step
Counting pulses
Practice · Reach the target
Question
Beam tilt is locked at $90^\circ$ ⁠, so both beams reach the Earth every turn. Change the Spin rate until the telescope receives 100 pulses a second, and press Check.

Goal
- 100 ± 0.6
  At the start of the step: 60
Setup, choices and feedback
The Scene at this step
Spin rate
30 turns/s
Beam tilt
90 °fixed
Show the pulses
On
Read from the Scene at
Pulses received each second/nodes/pulsar/outputs/pulseRate
One setting that reaches it
Spin rate
49.75 turns/s
Feedback for a right answer
50 turns a second: two beams, two pulses each turn.

Hint
How many pulses arrive each turn?

Worked solution
1. Each turn both beams sweep past us: two pulses a turn.
2. To get 100 pulses a second, the star has to turn $\dfrac{100}{2} = 50$ ⁠ times a second.
An astronomer counting pulses has to know whether one beam or two is reaching us. Otherwise they would think the star spins twice as fast as it really does.

Lesson 3 of 5 · About 30 minutes

Black holes

Gravity not even light escapes: how big the horizon is, how it grows with mass, and why a black hole’s shadow is bigger than its horizon.

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The 8 steps and their answers

Step 1 of 8
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Gravity that wins
Practice · Choose
Question
Imagine the Sun suddenly turned into a black hole of exactly the same mass. What would happen to the Earth’s orbit?

Correct answer
- Correct Nothing: the Earth would go round on the same orbit, in the dark
Setup, choices and feedback
The Scene at this step
Mass
10 Suns
Viewing angle
78 °default
Straighten the light
Bent light
Choices
Shown in a random order.
- Correct Nothing: the Earth would go round on the same orbit, in the dark
- Wrong The Earth would be sucked in
  Misconception: Thinks a black hole sucks in everything round it, pulling harder than a star of the same mass at the same distance.
  Feedback from the catalogue
  A black hole is not a vacuum cleaner. At a given distance its gravity is $F = \dfrac{GMm}{r^2}$ ⁠, exactly as for any other body of the same mass. What is special is that all the mass is squeezed into so small a ball that you can get very close to it, and there gravity is enormous. Far away, planets would go on circling it on the same orbits.
- Wrong The Earth would fly off, because a black hole does not pull on distant things
  Feedback
  A black hole pulls like any mass. Far from it, its gravity is the same as the Sun’s.
Feedback for a right answer
Exactly the same orbit, but cold and dark.

Worked solution
1. The gravity on the Earth is $F = \dfrac{GMm}{r^2}$ ⁠, and it depends only on the mass $M$ ⁠ and the distance $r$ ⁠.
2. Neither the mass nor the distance has changed, so neither has the force, nor the orbit.
3. The only change is that the Sun would be squeezed into a ball about 3 km in radius. Only very close to such a ball is gravity enormous.
Step 2 of 8
Open this step
The horizon
Practice · Calculate
Question
Mass is locked at $10$ ⁠ solar masses. What is the Schwarzschild radius, $R_s = \dfrac{2GM}{c^2}$ ⁠, in kilometres? $G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$ ⁠, $M_\odot = 1.99 \times 10^{30}\ \mathrm{kg}$ ⁠, $c = 3.00 \times 10^8\ \mathrm{m/s}$ ⁠.

Correct answer
29.53 km ± 2%
Setup, choices and feedback
The Scene at this step
Mass
10 Sunsfixed
Viewing angle
78 °default
Straighten the light
Bent light
Accepted within
± 2%
Computed from the Scene at this step’s setting
Radius of the event horizon, Rs = 2GM/c²/nodes/hole/outputs/schwarzschildRadius
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 29.5 km: that is the horizon, the boundary from which not even light returns.

Hint
$M = 10 \cdot 1.99 \times 10^{30}\ \mathrm{kg}$ ⁠.

Worked solution
1. $M = 1.99 \times 10^{31}\ \mathrm{kg}$ ⁠.
2. $R_s = \dfrac{2 \cdot 6.67 \times 10^{-11} \cdot 1.99 \times 10^{31}}{(3.00 \times 10^8)^2} = \dfrac{2.65 \times 10^{21}}{9.00 \times 10^{16}} \approx 2.95 \times 10^4\ \mathrm{m}$ ⁠.
3. About $29.5$ ⁠ kilometres.
Ten Suns in a ball the width of a city. For the Sun itself, the Schwarzschild radius is only about $3$ ⁠ kilometres.
Step 3 of 8
Open this step
Heavier, bigger
Practice · Predict, then look
Question
What will happen to the horizon’s radius when Mass doubles from $10$ ⁠ to $20$ ⁠ solar masses?

Correct answer
- Correct It will double
Setup, choices and feedback
The Scene at this step
Mass
10 Suns
Viewing angle
78 °default
Straighten the light
Straight light
Choices
- Correct It will double
- Wrong It will grow only $1.26$ ⁠ times, like a ball whose mass doubled
  Misconception: Thinks a black hole’s size grows like the cube root of its mass, as a ball of fixed density does.
  Feedback from the catalogue
  A ball of ordinary matter, at a fixed density, grows like the cube root of its mass. The Schwarzschild radius, $R_s = \dfrac{2GM}{c^2}$ ⁠, is proportional to the mass itself: twice the mass, a horizon of twice the radius. That makes a heavier black hole less dense: the mean density inside the horizon falls like $\dfrac{1}{M^2}$ ⁠.
- Wrong It will not change: all black holes are the same size
  Feedback
  There are black holes of a few Suns, and at the centres of galaxies there are holes of millions and billions of Suns, as big as the Solar System.
Once the prediction is locked in, the Scene changes
Mass
20 Suns
The Scene then shows: 59.07 km · Radius of the event horizon, Rs = 2GM/c²/nodes/hole/outputs/schwarzschildRadius
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It doubles: every ring in the Scene grows in proportion to the mass.

Worked solution
1. $R_s = \dfrac{2GM}{c^2}$ ⁠ is proportional to $M$ ⁠.
2. Twice the mass: twice $R_s$ ⁠, from about $29.5$ ⁠ to about $59$ ⁠ kilometres.
The horizon’s volume grows $8$ ⁠ times and the mass only $2$ ⁠, so a heavier black hole is less dense. The black hole at the centre of our Galaxy, about $4$ ⁠ million Suns, has a mean density lower than water’s.
Step 4 of 8
Open this step
The glowing disc
Practice · Calculate
Question
Straighten the light is at Straight light, and the labelled rings show. The disc ends at the innermost stable circular orbit, at $3 R_s$ ⁠. For $10$ ⁠ solar masses, $R_s \approx 29.5$ ⁠ km. How many kilometres from the hole’s centre is the disc’s inner edge?

Correct answer
88.6 km ± 2%
Setup, choices and feedback
The Scene at this step
Mass
10 Sunsfixed
Viewing angle
78 °default
Straighten the light
Straight light
Accepted within
± 2%
Computed from the Scene at this step’s setting
Inner edge of the disc, the innermost stable circular orbit, 3 Rs/nodes/hole/outputs/innerEdge
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 88.6 km, three times the horizon’s radius.

Hint
$r = 3 R_s$ ⁠.

Worked solution
1. $r = 3 R_s = 3 \cdot 29.5 \approx 88.6\ \mathrm{km}$ ⁠.
Inside that distance there is no stable orbit: gas that gets there plunges into the hole within moments. Before that it circles at about half the speed of light and heats to millions of degrees, which is why discs round black holes shine in X-rays.
Step 5 of 8
Open this step
Bent light
Practice · Choose
Question
With light bending as it really does, the black shadow in the middle looks $2.6$ ⁠ times bigger than the horizon itself. Why?

Correct answer
- Correct Light rays passing close to the hole bend inwards and are swallowed, so the shadow is wider than the horizon
Setup, choices and feedback
The Scene at this step
Mass
10 Suns
Viewing angle
78 °default
Straighten the light
Bent light
Choices
Shown in a random order.
- Correct Light rays passing close to the hole bend inwards and are swallowed, so the shadow is wider than the horizon
- Wrong A cloud of black dust wraps the hole
  Feedback
  There is no dust round the hole in the Scene. The shadow comes from the geometry of light alone.
- Wrong The hole is sucking in the disc and so growing
  Misconception: Thinks a black hole sucks in everything round it, pulling harder than a star of the same mass at the same distance.
  Feedback from the catalogue
  A black hole is not a vacuum cleaner. At a given distance its gravity is $F = \dfrac{GMm}{r^2}$ ⁠, exactly as for any other body of the same mass. What is special is that all the mass is squeezed into so small a ball that you can get very close to it, and there gravity is enormous. Far away, planets would go on circling it on the same orbits.
Feedback for a right answer
Right: gravity magnifies the shadow like a lens.

Worked solution
1. Near the hole gravity bends light. At $1.5 R_s$ ⁠, the photon sphere, light can even circle the hole.
2. A ray aimed from the viewer that passes within $\tfrac{\sqrt{27}}{2} R_s \approx 2.6 R_s$ ⁠ of the centre bends down to the horizon and is swallowed.
3. So no light reaches us from any of those directions, and the shadow is $2.6$ ⁠ times as wide as the horizon. That is what the first picture of a black hole showed, in 2019.
Step 6 of 8
Open this step
The photon sphere
Practice · Calculate
Question
Mass is locked at 20 solar masses, and its Schwarzschild radius is about 59 km. At $1.5 R_s$ ⁠ from the centre gravity is so strong that light can circle the hole: the photon sphere. What is its radius, in kilometres?

Correct answer
88.6 km ± 3%
Setup, choices and feedback
The Scene at this step
Mass
20 Sunsfixed
Viewing angle
78 °default
Straighten the light
Straight light
Accepted within
± 3%
Computed from the Scene at this step’s setting
Radius where light can circle the hole, 1.5 Rs/nodes/hole/outputs/photonSphere
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 89 km. Light circling there is balanced on a knife-edge: a small nudge sends it in or out.

Hint
Multiply the Schwarzschild radius by $1.5$ ⁠.

Worked solution
1. $R_s \approx 59.1$ ⁠ kilometres for 20 solar masses.
2. $1.5 \times 59.1 \approx 88.6$ ⁠ kilometres.
Every distance round a black hole grows in step with its mass: twice as heavy, a photon sphere twice as big.
Step 7 of 8
Open this step
The shadow
Practice · Calculate
Question
Mass is locked at 10 solar masses, $R_s \approx 29.5$ ⁠ kilometres. Because the light bends, the radius of the black shadow you see is $\dfrac{\sqrt{27}}{2} R_s$ ⁠. What is it, in kilometres?

Correct answer
76.73 km ± 3%
Setup, choices and feedback
The Scene at this step
Mass
10 Sunsfixed
Viewing angle
78 °default
Straighten the light
Bent light
Accepted within
± 3%
Computed from the Scene at this step’s setting
Apparent radius of the dark shadow, √27/2 Rs/nodes/hole/outputs/shadowRadius
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 29.5 km ± 3%
  Misconception: Thinks the black shadow seen round a black hole is the event horizon itself, the same size.
  Feedback from the catalogue
  Gravity bends the light passing near the hole. Light coming close to the photon sphere, $1.5 R_s$ ⁠, is swallowed, so from every direction within $\dfrac{\sqrt{27}}{2} R_s \approx 2.6 R_s$ ⁠ of the centre no light reaches us. The shadow is 2.6 times the horizon.
Feedback for a right answer
About 77 km: 2.6 times the horizon itself.

Hint
$\sqrt{27} \approx 5.20$ ⁠.

Worked solution
1. $\dfrac{\sqrt{27}}{2} \approx \dfrac{5.196}{2} \approx 2.60$ ⁠.
2. $2.60 \times 29.5 \approx 76.7$ ⁠ kilometres.
The shadow is 2.6 times the horizon, because even light passing a little outside the photon sphere falls in. That is how the shadow of the black hole in the galaxy M87 was photographed.
Step 8 of 8
Open this step
How dense
Practice · Predict, then look
Question
Divide the hole’s mass by the volume of a ball of radius $R_s$ ⁠, and you get the mean density inside the horizon. For 10 solar masses it is about 1.8 × 10¹⁷ kg/m³, like an atom’s nucleus. Predict: what will happen to it when Mass doubles to 20 solar masses?

Correct answer
- Correct It will fall to a quarter
Setup, choices and feedback
The Scene at this step
Mass
10 Suns
Viewing angle
78 °default
Straighten the light
Straight light
Choices
- Wrong It will double, because there is twice the matter
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Wrong It will not change, because the mass and the horizon grow together
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Correct It will fall to a quarter
- Wrong It will fall to a half
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Mass
20 Suns
The Scene then shows: 46070000000000000 · Mean density inside the horizon, kg/m³: the mass over a ball of radius Rs/nodes/hole/outputs/density
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A quarter: the mass grows 2 times, and the volume inside the horizon 8 times.

Hint
The radius grows like the mass. How much does a ball’s volume grow when its radius doubles?

Worked solution
1. $R_s$ ⁠ grows 2 times, so the volume grows $2^3 = 8$ ⁠ times.
2. The mass grows only 2 times.
3. The density: $\dfrac{2}{8} = \dfrac{1}{4}$ ⁠ of before, about 4.6 × 10¹⁶ kg/m³.
The density goes like $\dfrac{1}{M^2}$ ⁠. For a black hole of a hundred million solar masses it is close to the density of water: what makes it a black hole is how much matter there is, not how tightly packed it is.

Lesson 4 of 5 · About 25 minutes

Gravitational waves

Two black holes spiral together and merge, sending ripples through space itself: how much mass turns into waves, why a heavy pair “sounds” low, and how tiny the change the detector measures is.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Ripples in space
Practice · Calculate
Question
Heavier hole is $36$ ⁠ solar masses, and Lighter hole $80\%$ ⁠ of it, $28.8$ ⁠ solar masses. In the merger about $4.7\%$ ⁠ of the total mass leaves as gravitational waves. How many solar masses has the hole that is left?

Correct answer
61.74 M☉ ± 1%
Setup, choices and feedback
The Scene at this step
Heavier hole
36 Suns
Lighter hole
80 % of heavier
Follow to a detector
Ripples
Accepted within
± 1%
Computed from the Scene at this step’s setting
Mass of the hole left after the merger/nodes/pair/outputs/finalMass
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 64.8 M☉ ± 0.5%
  Misconception: Thinks the black hole left by a merger has the two holes’ masses added together.
  Feedback from the catalogue
  Gravitational waves carry energy, and by $E = mc^2$ ⁠ energy that leaves is mass that goes. When black holes merge, a few per cent of their mass leaves as waves, so the final hole is lighter than the two together. In the first detection about three Suns’ worth left.
Feedback for a right answer
About $61.7$ ⁠: three Suns’ worth turned into waves in space itself.

Hint
First add the two masses, then take off $4.7\%$ ⁠.

Worked solution
1. The total mass: $36 + 28.8 = 64.8$ ⁠ solar masses.
2. Leaving as waves: $0.047 \cdot 64.8 \approx 3.1$ ⁠ solar masses.
3. Left: $64.8 - 3.1 \approx 61.7$ ⁠ solar masses.
On 14 September 2015 the LIGO detectors caught just such a merger: holes of $36$ ⁠ and $29$ ⁠ solar masses, and a final hole of $62$ ⁠. Answering $64.8$ ⁠ forgets that the waves carry mass.
Step 2 of 5
Open this step
Mass into waves
Practice · Calculate
Question
In the first detection’s merger about $3.06$ ⁠ solar masses left as gravitational waves. How much energy is that, in joules? $M_\odot = 1.99 \times 10^{30}\ \mathrm{kg}$ ⁠, $c = 3.00 \times 10^8\ \mathrm{m/s}$ ⁠.

Correct answer
5.471 × 10⁴⁷ J ± 3%
Setup, choices and feedback
The Scene at this step
Heavier hole
36 Suns
Lighter hole
80 % of heavier
Follow to a detector
Ripples
Accepted within
± 3%
Computed from the Scene at this step’s setting
Energy carried off by the waves, E = mc²/nodes/pair/outputs/radiatedEnergy
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About $5.5 \times 10^{47}$ ⁠ joules, all as a shiver in space and none as light.

Hint
$E = mc^2$ ⁠.

Worked solution
1. $m = 3.06 \cdot 1.99 \times 10^{30} \approx 6.09 \times 10^{30}\ \mathrm{kg}$ ⁠.
2. $E = mc^2 = 6.09 \times 10^{30} \cdot 9.00 \times 10^{16} \approx 5.5 \times 10^{47}\ \mathrm{J}$ ⁠.
The Sun will give out about $10^{44}$ ⁠ joules over its whole ten-billion-year life, thousands of times less. In the split second of the merger, the pair sent out more power than all the stars in the visible universe together.
Step 3 of 5
Open this step
Pitch
Practice · Predict, then look
Question
The detector “hears” the merger as a chirp that rises in pitch and ends on the final hole’s ringing. What will happen to that frequency when Heavier hole grows from $36$ ⁠ to $60$ ⁠ solar masses?

Correct answer
- Correct It will fall
Setup, choices and feedback
The Scene at this step
Heavier hole
36 Suns
Lighter hole
80 % of heavier
Follow to a detector
At a detector
Choices
- Correct It will fall
- Wrong It will rise
  Misconception: Thinks a heavier pair of black holes sends out waves of higher frequency.
  Feedback from the catalogue
  In a heavier pair every length and every time grows with the mass: the horizons are bigger, and the holes merge while farther apart. A bigger orbit takes longer, so the frequency is lower, $f \propto \dfrac{1}{M}$ ⁠. As with instruments: a big drum sounds lower than a small one.
- Wrong It will not change
  Feedback
  The frequency depends on the mass: every time in the pair grows with it.
Once the prediction is locked in, the Scene changes
Heavier hole
60 Suns
The Scene then shows: 165.4 Hz · Frequency of the waves as the final hole rings down/nodes/pair/outputs/ringFrequency
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It falls: a heavy pair “sounds” lower, like a big drum.

Worked solution
1. In a pair of black holes every length is proportional to the total mass $M$ ⁠, and so is every time.
2. So the frequency of the waves is proportional to $\dfrac{1}{M}$ ⁠.
3. The total mass grows from $64.8$ ⁠ to $108$ ⁠ solar masses, $1.67$ ⁠ times, and the frequency falls $1.67$ ⁠ times: from about 276 Hz to about 165 Hz.
Step 4 of 5
Open this step
A balanced pair
Practice · Reach the target
Question
Heavier hole is locked at $36$ ⁠ solar masses, and the lighter one is only $12\%$ ⁠ of it: the pair radiates about one per cent of its mass. Change Lighter hole until the waves carry off at least $4.8\%$ ⁠ of the total mass, and press Check.

Goal
- 0.0482 ± 0.0002
  At the start of the step: 0.01011
Setup, choices and feedback
The Scene at this step
Heavier hole
36 Sunsfixed
Lighter hole
12 % of heavier
Follow to a detector
Ripples
Read from the Scene at
Share of the total mass radiated, 0..1/nodes/pair/outputs/radiatedShare
One setting that reaches it
Lighter hole
88 % of heavier
Feedback for a right answer
A nearly equal pair radiates the largest share of its mass.

Feedback for a wrong answer
Not yet. The pair is still too lopsided: bring the masses closer.

Hint
Which pair swings harder: a small hole round a giant, or two similar holes?

Worked solution
1. A small hole beside a giant hardly moves it, so the pair barely shakes space.
2. The closer the masses, the wider the two holes swing round each other, and the stronger the waves.
3. Raise Lighter hole above about $88\%$ ⁠ of the heavier: the share radiated reaches about $4.8\%$ ⁠, and the most, at $100\%$ ⁠, is $4.84\%$ ⁠.
Step 5 of 5
Open this step
Catching a wave
Practice · Choose
Question
Follow to a detector is at At a detector. When the first detection’s wave passed the Earth, by how much did one 4 km arm of a LIGO detector change? Use the chart.

Correct answer
- Correct By a few thousandths of a proton’s width, under one per cent of it
Setup, choices and feedback
The Scene at this step
Heavier hole
36 Suns
Lighter hole
80 % of heavier
Follow to a detector
At a detector
Choices
Shown in a random order.
- Correct By a few thousandths of a proton’s width, under one per cent of it
- Wrong By about the width of a hair
  Feedback
  A hair is about $10^{13}$ ⁠ times wider than the real change.
- Wrong By about the width of an atom
  Feedback
  An atom is a hundred thousand times wider than a proton, so tens of millions of times wider than the change.
- Wrong Not at all: gravitational waves do not change distances
  Feedback
  A gravitational wave is exactly a stretching and squeezing of distances, by turns, in two directions at right angles.
Feedback for a right answer
About two hundred and seventy times less than a proton: one of the most delicate measurements people have ever made.

Worked solution
1. The wave’s peak strength, $h$ ⁠, is about $3 \times 10^{-21}$ ⁠: every distance changes by that fraction of itself.
2. In a 4 km arm the change is about $\tfrac{1}{2} h L \approx 6 \times 10^{-18}\ \mathrm{m}$ ⁠.
3. A proton is about $1.7 \times 10^{-15}\ \mathrm{m}$ ⁠ across, about $270$ ⁠ times more.
To measure such a change, LIGO bounces laser light hundreds of times along each arm and compares the two arms at right angles.

Lesson 5 of 5 · About 25 minutes

Gravitational lensing

A galaxy bending the light of a distant one behind it into a ring, arcs or a cross: why it happens, how the ring’s size depends on mass, and how it weighs matter no one can see.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
A ring
Practice · Choose
Question
The distant galaxy lies almost exactly behind the lens galaxy, and we see it as a complete ring round it. Why a ring?

Correct answer
- Correct Its light bends round the lens equally on every side, so it reaches us from every direction round it
Setup, choices and feedback
The Scene at this step
Lens mass
1 trillion Suns
Offset from the line
0.08 arcsec
Flattening of the lens
0.1
Show where it really is
What we see
Choices
Shown in a random order.
- Correct Its light bends round the lens equally on every side, so it reaches us from every direction round it
- Wrong The distant galaxy really is ring-shaped
  Feedback
  Drag Show where it really is to Where it really is: it is an ordinary small spiral.
- Wrong The light passes through the lens’s gas and is refracted, as in glass
  Misconception: Thinks a gravitational lens bends light by refraction, as light passing through the galaxy’s gas, like glass.
  Feedback from the catalogue
  A gravitational lens does not need the light to pass through matter. The galaxy’s mass curves the space round it, and the light simply takes the straightest path through curved space. That is why dark matter, which neither absorbs nor gives out light, bends light too, and why every colour bends by the same angle, with no rainbow as from a prism.
Feedback for a right answer
An Einstein ring: perfect symmetry round the line of sight.

Worked solution
1. Mass curves the space round it, and light passing close by bends towards it.
2. When the source, the lens and we lie in one line, no side is special: light passing above, below and to either side bends by the same angle and reaches us.
3. So we see the source in every direction round the lens, at the same distance from it: an Einstein ring.
Step 2 of 5
Open this step
The ring’s size
Practice · Calculate
Question
A lens of a trillion solar masses makes a ring about $2.0$ ⁠ arcseconds in radius. Lens mass is now locked at $4$ ⁠ trillion. The ring’s radius is proportional to $\sqrt{M}$ ⁠. What is the radius now, in arcseconds?

Correct answer
4.036 arcsec ± 3%
Setup, choices and feedback
The Scene at this step
Lens mass
4 trillion Sunsfixed
Offset from the line
0.08 arcsec
Flattening of the lens
0.1
Show where it really is
What we see
Accepted within
± 3%
Computed from the Scene at this step’s setting
Einstein radius/nodes/lens/outputs/einsteinRadius
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 8.07 arcsec ± 3%
  Misconception: Thinks an Einstein ring’s radius grows in proportion to the lens mass.
  Feedback from the catalogue
  The angle through which light bends near a mass is proportional to $\dfrac{M}{b}$ ⁠, where $b$ ⁠ is how far the ray passes from the mass. The light making the ring bends just enough to reach us, and from that the ring’s radius is $\theta_E \propto \sqrt{M}$ ⁠. A lens four times as heavy gives a ring twice as wide.
Feedback for a right answer
About $4.0$ ⁠ arcseconds: four times the mass, twice the ring.

Hint
By how much does $\sqrt{M}$ ⁠ grow when $M$ ⁠ grows four times?

Worked solution
1. $\theta_E \propto \sqrt{M}$ ⁠, and the mass grows $4$ ⁠ times.
2. $\sqrt{4} = 2$ ⁠, so the radius doubles: $2 \cdot 2.02 \approx 4.0$ ⁠ arcseconds.
An arcsecond is $\tfrac{1}{3600}$ ⁠ of a degree, about the size of a coin seen from four kilometres away. Answering about $8$ ⁠ assumes the radius grows like the mass.
Step 3 of 5
Open this step
Arcs
Practice · Predict, then look
Question
What will happen to the ring when the distant galaxy moves a little to one side, to 0.4 arcsec?

Correct answer
- Correct It will break into two arcs, on opposite sides of the lens
Setup, choices and feedback
The Scene at this step
Lens mass
1 trillion Suns
Offset from the line
0.08 arcsec
Flattening of the lens
0.1
Show where it really is
What we see
Choices
- Correct It will break into two arcs, on opposite sides of the lens
- Wrong It will move sideways and stay whole
  Feedback
  A whole ring needs full symmetry round the line of sight. When the source moves, light reaches us only along two special directions.
- Wrong It will vanish, leaving only the lens
  Feedback
  At a small offset there is still strong bending. The galaxy behind leaves the lens only when it is far off the line, and even then we see it once.
Once the prediction is locked in, the Scene changes
Offset from the line
0.4 arcsec
The Scene then shows: 2 · Separate images of the source (a ring counts as one)/nodes/lens/outputs/images
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Two arcs, the brighter on the side the galaxy moved to.

Worked solution
1. When the source is not exactly behind the lens, the light has two paths: one passing the near side of the lens and one the far side.
2. Each path makes an image, so we see two arcs on opposite sides of the lens.
3. The arc on the side the source moved to is farther from the middle and brighter; the other is close to the lens and faint.
Step 4 of 5
Open this step
A cross
Practice · Reach the target
Question
A flattened lens can split the galaxy behind it into four images, an “Einstein cross”. Change the Scene until you see four separate images, and press Check.

Goal
- 4 ± 0.1
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Lens mass
1 trillion Suns
Offset from the line
0.08 arcsec
Flattening of the lens
0.1
Show where it really is
What we see
Read from the Scene at
Separate images of the source (a ring counts as one)/nodes/lens/outputs/images
One setting that reaches it
Lens mass
2.85 trillion Suns
Offset from the line
0 arcsec
Flattening of the lens
0.2
Feedback for a right answer
An Einstein cross: four images of one galaxy, round the lens.

Feedback for a wrong answer
Not yet. With two or three images, the galaxy is too far off the line or the lens is too round.

Hint
Keep the galaxy nearly in line with the lens, and change the lens’s shape.

Worked solution
1. A round lens makes a ring or two images: it has no special direction.
2. An oval lens bends light by different amounts along its long and short axes, and the ring breaks into four.
3. Keep Offset from the line near 0 and raise Flattening of the lens to $0.4$ ⁠ or more.
Step 5 of 5
Open this step
Weighing what we cannot see
Practice · Choose
Question
From the size of the ring you can work out the lens’s mass. Doing it for real galaxies gives a mass several times that of all their stars and gas. What follows?

Correct answer
- Correct Most of a galaxy’s mass is matter that gives no light, dark matter, and it bends light like any mass
Setup, choices and feedback
The Scene at this step
Lens mass
1 trillion Suns
Offset from the line
0.08 arcsec
Flattening of the lens
0.1
Show where it really is
0.5 (What we see → Where it really is)
Choices
Shown in a random order.
- Correct Most of a galaxy’s mass is matter that gives no light, dark matter, and it bends light like any mass
- Wrong The galaxy’s gas refracts the light more than was thought
  Misconception: Thinks a gravitational lens bends light by refraction, as light passing through the galaxy’s gas, like glass.
  Feedback from the catalogue
  A gravitational lens does not need the light to pass through matter. The galaxy’s mass curves the space round it, and the light simply takes the straightest path through curved space. That is why dark matter, which neither absorbs nor gives out light, bends light too, and why every colour bends by the same angle, with no rainbow as from a prism.
- Wrong Relativity is wrong, and light bends more than it predicts
  Feedback
  The bending relativity predicts has been measured precisely near the Sun, where the mass is known. Galaxies’ rotation curves show the same missing mass too.
Feedback for a right answer
Dark matter: gravitational lenses are one of the main ways to weigh it.

Worked solution
1. The ring’s radius depends only on the lens’s total mass and the distances, not on what it is made of.
2. The mass that comes out is several times what the stars and gas account for.
3. The accepted explanation: the galaxy sits in a halo of dark matter, which neither gives out nor absorbs light, but has gravity. That is how gravitational lenses map matter no one can see.

Across the unit · 10 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Nebulae

A glowing cloud

Question

Correct answer

The Scene at this step

Choices

Worked solution

Light that splits atoms

Question

Correct answer

The Scene at this step

Hint

Worked solution

The bubble

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Blue dust

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A black cloud

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Neutron stars and pulsars

A lighthouse in space

Question

Correct answer

The Scene at this step

Choices

Worked solution

The Crab

Question

Correct answer

The Scene at this step

Hint

Worked solution

Spinning up

Question

Correct answer

The Scene at this step

Hint

Worked solution

Tilt decides what we see

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Two beams

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Counting pulses

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Black holes

Gravity that wins

Question

Correct answer

The Scene at this step

Choices