Answer key · Unit 19

Stars and the 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

Starlight and temperature

A glowing body gives out a continuous spectrum that depends only on its temperature. Wien's law links the peak to the temperature, the Stefan–Boltzmann law the power, and together they read a star's temperature from its colour.

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The 6 steps and their answers

Step 1 of 6
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The colour of heat
Practice · Choose
Question
Orion has a red star, Betelgeuse, and a bluish-white one, Rigel. Which of them is hotter at its surface?

Correct answer
- Correct Rigel, the bluish one
Setup, choices and feedback
The Scene at this step
Temperature of the lump
950 K
Show the whole spectrum
Visible light
Choices
Shown in a random order.
- Correct Rigel, the bluish one
- Wrong Betelgeuse, the red one
  Misconception: Thinks red stars are the hottest, as red means hot in everyday life.
  Feedback from the catalogue
  In everyday life red marks hot and blue marks cold, but for a glowing body it is the other way round. A body starts to glow red at about 800 K, and the hotter it gets, the more the peak of its light moves to shorter wavelengths: orange, yellow, white, and in the end bluish. Red stars like Betelgeuse are the cool ones, about 3500 K, and blue stars like Rigel are hotter than 10,000 K.
- Wrong You cannot tell from the colour, because the colour depends only on what the star is made of
  Feedback
  The colour of a glowing body depends almost only on its temperature, not on the material: that is what makes it a thermometer.
Feedback for a right answer
Rigel, about 12,000 K, and Betelgeuse only about 3500 K. The Scene shows the same order: red, orange, yellow, and white as it heats up.

Worked solution
1. A glowing body gives out a continuous spectrum, and its peak moves to shorter wavelengths the hotter the body is (Wien’s law).
2. A body whose peak is in the red or the infrared looks red; with the peak in the green, all the colours are nearly equal and it looks white; with the peak in the ultraviolet, it looks bluish.
3. So the bluish star is the hotter: Rigel about 12,000 K, and Betelgeuse about 3500 K.
In everyday life red marks hot, but for a glowing body red is the coolest colour.
Step 2 of 6
Open this step
Wien's law
Practice · Calculate
Question
Temperature of the lump is locked at 1300 K, and it glows orange-red. At what wavelength is the peak of its radiation? $b = 2.90 \times 10^{-3}\ \mathrm{m \cdot K}$ ⁠.

Correct answer
2229 nm ± 2%
Setup, choices and feedback
The Scene at this step
Temperature of the lump
1300 Kfixed
Show the whole spectrum
0.5 (Visible light → Whole spectrum)
Accepted within
± 2%
Computed from the Scene at this step’s setting
Where the glow peaks, Wien's law b/T/nodes/furnace/outputs/peakWavelength
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 2822 nm ± 1.5%
  Misconception: Puts the temperature into Wien's law in degrees Celsius instead of kelvin.
  Feedback from the catalogue
  In Wien’s law, $\lambda_{\max} = \dfrac{b}{T}$ ⁠, and in the Stefan–Boltzmann law, $T$ ⁠ is an absolute temperature, in kelvin: $T_K = T_C + 273$ ⁠. At the temperatures of stars the difference is small, but for a lump at 1300 K putting in $1027$ ⁠ instead of $1300$ ⁠ changes the answer by more than $25\%$ ⁠.
- 700 nm ± 60 nm
  Misconception: Thinks a red-hot body gives out most of its light in the red, the colour we see.
  Feedback from the catalogue
  The colour we see is only the end of the curve that reaches into the visible range. A red-hot lump at about 1300 K gives out most of its energy in the infrared, peaking at about 2200 nm, far beyond the red. The eye sees only the tail of the curve, where it touches 700 nm. By Wien’s law the peak comes into the visible range only above about 4000 K.
Feedback for a right answer
About 2230 nm, in the infrared. What we see as red is only the tail of the curve that reaches into the visible range.

Hint
Wien’s law: $\lambda_{\max} = \dfrac{b}{T}$ ⁠, with $T$ ⁠ in kelvin.

Worked solution
1. The givens: $T = 1300\ \mathrm{K}$ ⁠, $b = 2.90 \times 10^{-3}\ \mathrm{m \cdot K}$ ⁠.
2. Wien’s law: $\lambda_{\max} = \dfrac{b}{T}$ ⁠.
3. Substituting: $\lambda_{\max} = \dfrac{2.90 \times 10^{-3}}{1300} \approx 2.23 \times 10^{-6}\ \mathrm{m} = 2230\ \mathrm{nm}$ ⁠.
That is deep in the infrared, three times the red end of visible light. Answering about 700 nm means assuming the peak is the colour you see; getting about 2820 nm means putting in degrees Celsius.
Step 3 of 6
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Hotter and shorter
Practice · Predict, then look
Question
The lump is red hot, at 950 K. It is heated to 1800 K, yellow hot. Which way will the peak of the curve move?

Correct answer
- Correct To a shorter wavelength
Setup, choices and feedback
The Scene at this step
Temperature of the lump
950 K
Show the whole spectrum
0.5 (Visible light → Whole spectrum)
Choices
- Correct To a shorter wavelength
- Wrong To a longer wavelength
  Misconception: Thinks red stars are the hottest, as red means hot in everyday life.
  Feedback from the catalogue
  In everyday life red marks hot and blue marks cold, but for a glowing body it is the other way round. A body starts to glow red at about 800 K, and the hotter it gets, the more the peak of its light moves to shorter wavelengths: orange, yellow, white, and in the end bluish. Red stars like Betelgeuse are the cool ones, about 3500 K, and blue stars like Rigel are hotter than 10,000 K.
- Wrong It will stay put, and only get taller
  Feedback
  The curve does get much taller, but it moves too: the peak follows the temperature.
Once the prediction is locked in, the Scene changes
Temperature of the lump
1800 K
The Scene then shows: 1610 nm · Where the glow peaks, Wien's law b/T/nodes/furnace/outputs/peakWavelength
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The peak moved from about 3050 to about 1610 nm: hotter, shorter wavelength.

Worked solution
1. By Wien’s law, $\lambda_{\max} = \dfrac{b}{T}$ ⁠: the peak goes inversely with the temperature.
2. At 950 K: $\lambda_{\max} \approx 3050\ \mathrm{nm}$ ⁠.
3. At 1800 K: $\lambda_{\max} \approx 1610\ \mathrm{nm}$ ⁠, almost half.
The temperature nearly doubled, and the peak moved to a wavelength nearly half as long. The curve also grew taller, and more of it reaches into the visible range, so the lump turns yellow and bright.
Step 4 of 6
Open this step
One micrometre
Practice · Reach the target
Question
Change Temperature of the lump until the peak of its radiation is at exactly 1000 nm, one micrometre, and press Check.

Goal
- 1000 nm ± 1%
  At the start of the step: 2229 nm
Setup, choices and feedback
The Scene at this step
Temperature of the lump
1300 K
Show the whole spectrum
0.5 (Visible light → Whole spectrum)
Read from the Scene at
Where the glow peaks, Wien's law b/T/nodes/furnace/outputs/peakWavelength
One setting that reaches it
Temperature of the lump
2870 K
Feedback for a right answer
About 2900 K: $\dfrac{2.90 \times 10^{-3}}{10^{-6}}$ ⁠. A bulb’s filament, and its peak is still in the infrared.

Feedback for a wrong answer
Not yet. If the peak is at too long a wavelength, heat the lump; if too short, cool it.

Hint
Turn Wien’s law round: $T = \dfrac{b}{\lambda_{\max}}$ ⁠.

Worked solution
1. $T = \dfrac{b}{\lambda_{\max}} = \dfrac{2.90 \times 10^{-3}}{1.00 \times 10^{-6}} = 2900\ \mathrm{K}$ ⁠.
About the temperature of the filament in an old light bulb. Its peak is in the infrared too, which is why most of the electricity a filament bulb uses comes out as heat rather than light.
Step 5 of 6
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The fourth power
Practice · Predict, then look
Question
At 1300 K each square metre of the lump gives out about 162 kW. The temperature is doubled to 2600 K. How many times more power will it give out?

Correct answer
- Correct $16$ ⁠ times
Setup, choices and feedback
The Scene at this step
Temperature of the lump
1300 K
Show the whole spectrum
0.5 (Visible light → Whole spectrum)
Choices
- Wrong $2$ ⁠ times
  Misconception: Thinks the power a hot body radiates grows in proportion to its temperature.
  Feedback from the catalogue
  By the Stefan–Boltzmann law, the power each square metre gives out is $P = \sigma T^4$ ⁠: it grows as the fourth power of the temperature. Twice the temperature gives $2^4 = 16$ ⁠ times the power, not $2$ ⁠ times. That is why a slightly hotter star is far brighter.
- Wrong $4$ ⁠ times
  Misconception: Thinks the power a hot body radiates grows in proportion to its temperature.
  Feedback from the catalogue
  By the Stefan–Boltzmann law, the power each square metre gives out is $P = \sigma T^4$ ⁠: it grows as the fourth power of the temperature. Twice the temperature gives $2^4 = 16$ ⁠ times the power, not $2$ ⁠ times. That is why a slightly hotter star is far brighter.
- Correct $16$ ⁠ times
Once the prediction is locked in, the Scene changes
Temperature of the lump
2600 K
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$16$ ⁠ times: $\sigma T^4$ ⁠, and the reading jumped from about 0.16 to about 2.6 MW per square metre.

Worked solution
1. The Stefan–Boltzmann law: the power per square metre is $P = \sigma T^4$ ⁠, with $\sigma = 5.67 \times 10^{-8}\ \mathrm{W/(m^2 \cdot K^4)}$ ⁠.
2. Twice the temperature: $P$ ⁠ grows $2^4 = 16$ ⁠ times.
3. At 2600 K: $5.67 \times 10^{-8} \cdot 2600^4 \approx 2.6 \times 10^6\ \mathrm{W/m^2}$ ⁠, $16$ ⁠ times 162 kW.
The power reading beside the Scene jumped to match. A star twice as hot as the Sun gives out $16$ ⁠ times as much from each square metre.
Step 6 of 6
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The Sun's temperature
Practice · Calculate
Question
Show the whole spectrum has added the curve of sunlight, and its peak is at about 502 nm, in blue-green light. What is the temperature of the Sun’s surface?

Correct answer
5770 K ± 2%
Setup, choices and feedback
The Scene at this step
Temperature of the lump
3000 K
Show the whole spectrum
Whole spectrum
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
About 5770 K. Astronomers measure the temperatures of millions of stars this way: from their colour alone.

Hint
$T = \dfrac{b}{\lambda_{\max}}$ ⁠, with $\lambda$ ⁠ in metres.

Worked solution
1. $\lambda_{\max} = 502\ \mathrm{nm} = 5.02 \times 10^{-7}\ \mathrm{m}$ ⁠.
2. $T = \dfrac{b}{\lambda_{\max}} = \dfrac{2.90 \times 10^{-3}}{5.02 \times 10^{-7}} \approx 5770\ \mathrm{K}$ ⁠.
The Sun does not look green, because its peak is broad and all the colours arrive at almost the same strength: together they look white. This is how the temperature of any star is measured, without touching it.

Lesson 2 of 5 · About 25 minutes

Spectral lines

A glowing gas gives out only certain wavelengths: jumps between the atom's energy levels. Each element gives a barcode of its own, and that is how we know what stars are made of, even from the dark lines in their light.

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The 6 steps and their answers

Step 1 of 6
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A barcode
Practice · Choose
Question
A tube of hydrogen glows, and the prism spreads its light out. On the screen there are only a few coloured lines, not a continuous rainbow. Why?

Correct answer
- Correct A hydrogen atom has only certain energy levels, and each photon carries the difference between two of them
Setup, choices and feedback
The Scene at this step
Gas in the tube
Hydrogen
Electron energy
25 eV
Show the energy jumps
Spectrum
Choices
Shown in a random order.
- Correct A hydrogen atom has only certain energy levels, and each photon carries the difference between two of them
- Wrong The prism lets only a few colours through and absorbs the rest
  Misconception: Thinks the prism or the spectroscope picks out the bright lines, rather than the gas giving out only those colours.
  Feedback from the catalogue
  The prism only separates the colours: white light from a filament lamp makes a continuous rainbow through it. The lines come from the gas. Each atom can only be in certain energy levels, and a photon given out in a jump from one level to another carries exactly the difference between them, $E = hf = \dfrac{hc}{\lambda}$ ⁠. So each element gives only its own wavelengths.
- Wrong There are too few atoms in the tube to fill the whole rainbow
  Feedback
  Even with more gas, or a higher pressure, the lines stay in the same places. They are set by the structure of the atom, not by how much gas there is.
- Wrong The eye sees only the strong colours, and the rest of the rainbow is too faint
  Feedback
  Between the lines there is no light at all: even a sensitive camera finds nothing there.
Feedback for a right answer
Fixed energy levels, so fixed wavelengths. Each element gets a barcode of its own.

Worked solution
1. An electron in an atom can be only in certain energy levels, like the rungs of a ladder.
2. When an electron drops from one level to another, the atom gives out a photon with exactly the difference: $E = E_{\text{upper}} - E_{\text{lower}}$ ⁠.
3. A photon’s energy sets its wavelength: $\lambda = \dfrac{hc}{E}$ ⁠. A limited number of jumps gives a limited number of wavelengths, and each one is a line.
The prism only separates the colours from each other: light from a filament lamp passes through it as a continuous rainbow.
Step 2 of 6
Open this step
The energy of a photon
Practice · Calculate
Question
Hydrogen’s strong red line comes from the jump from level $n = 3$ ⁠, 12.09 eV above the ground state, to $n = 2$ ⁠, at 10.20 eV. What is its wavelength? $hc = 1240\ \mathrm{eV \cdot nm}$ ⁠.

Correct answer
656.5 nm ± 1%
Setup, choices and feedback
The Scene at this step
Gas in the tube
Hydrogenfixed
Electron energy
25 eV
Show the energy jumps
Energy jumps
Accepted within
± 1%
Computed from the Scene at this step’s setting
Vacuum wavelength of the brightest visible line showing (0 when none)/nodes/bench/outputs/brightest
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 102.6 nm ± 2%
  Misconception: Uses the energy of one level for the photon, instead of the difference between the two levels of the jump.
  Feedback from the catalogue
  The photon takes the energy the atom loses in the jump: the difference between the upper level and the lower one, $E = E_{\text{upper}} - E_{\text{lower}}$ ⁠. A level’s energy is measured from the ground state, so it matches the photon only for a jump all the way down to the ground state, and that is usually ultraviolet light.
Feedback for a right answer
$\dfrac{1240}{1.89} \approx 656\ \mathrm{nm}$ ⁠: the red line, the brightest in hydrogen’s barcode.

Hint
The photon carries the difference between the two levels, and $\lambda = \dfrac{hc}{E}$ ⁠.

Worked solution
1. The photon’s energy: $E = 12.09 - 10.20 = 1.89\ \mathrm{eV}$ ⁠.
2. The wavelength: $\lambda = \dfrac{hc}{E} = \dfrac{1240}{1.89} \approx 656\ \mathrm{nm}$ ⁠.
This is hydrogen’s red line, $\mathrm{H}\alpha$ ⁠, which colours the nebulae where stars are born red. Getting about 103 nm means using the upper level’s energy alone: that is the photon of a jump all the way down to the ground state, in the ultraviolet.
Step 3 of 6
Open this step
Enough energy
Practice · Reach the target
Question
The electrons in the tube now bring 8 eV to a collision, not enough to excite a single atom. Raise Electron energy until exactly one visible line appears, and press Check.

Goal
- 1 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Gas in the tube
Hydrogenfixed
Electron energy
8 eV
Show the energy jumps
0.3 (Spectrum → Energy jumps)
Read from the Scene at
Visible lines in the barcode/nodes/bench/outputs/visibleLines
One setting that reaches it
Electron energy
12.1 eV
Feedback for a right answer
From 12.09 eV: enough to lift an atom to level $n = 3$ ⁠, from where it drops to $n = 2$ ⁠ and gives out the red line.

Feedback for a wrong answer
Not yet. If there is no visible line, raise the energy; if there is more than one, lower it a little.

Hint
The first visible line comes from level $n = 3$ ⁠. How much energy does it take to lift an atom there?

Worked solution
1. For an atom to give out a photon in the jump $3 \to 2$ ⁠, an electron first has to lift it from the ground state to level $n = 3$ ⁠: 12.09 eV.
2. From 10.2 eV atoms are already excited to level $n = 2$ ⁠, but the jump back from there gives out ultraviolet light, which the eye cannot see.
3. Between 12.09 and 12.75 eV, level $n = 4$ ⁠, only the red line appears.
This is the experiment of Franck and Hertz, from 1914: energy levels show up as sharp thresholds in the electrons’ energy.
Step 4 of 6
Open this step
Fingerprints
Practice · Choose
Question
The gas in the tube is hidden. Its barcode: a pair of close yellow lines at 589 nm, far brighter than all the rest, and a few faint lines. What gas is in the tube?

Correct answer
- Correct Sodium
Setup, choices and feedback
The Scene at this step
Gas in the tube
Sodiumhidden
Electron energy
25 eV
Show the energy jumps
Spectrum
Choices
Shown in a random order.
- Correct Sodium
- Wrong Hydrogen
  Feedback
  Hydrogen’s strong line is red, at 656 nm, and it has no yellow line at all.
- Wrong Helium
  Feedback
  Helium has a yellow line, at 588 nm, but it is single, with a row of blue, green and red lines beside it.
- Wrong Mercury
  Feedback
  Mercury’s strong lines are blue and green, at 436 and 546 nm.
Feedback for a right answer
Sodium: the yellow pair at 589 nm. The same line shows as a dark line in sunlight, which is how we know there is sodium in the Sun.

Worked solution
1. Each element gives out only the wavelengths of its own jumps, so the barcode is a fingerprint.
2. Sodium’s bright line is a close pair at 589.0 and 589.6 nm: the jump from the first excited level, 2.1 eV above the ground state.
3. None of the other elements here gives a line like that.
This is the orange-yellow light of old street lamps, and also the dark line $D$ ⁠ that Fraunhofer found in sunlight.
Step 5 of 6
Open this step
An element from the Sun
Practice · Calculate
Question
In 1868 a yellow line was found in sunlight at 587.6 nm that no known element gave. That is how helium was discovered. What is the energy of a photon in that line? $hc = 1240\ \mathrm{eV \cdot nm}$ ⁠.

Correct answer
2.11 eV ± 1%
Setup, choices and feedback
The Scene at this step
Gas in the tube
Heliumfixed
Electron energy
25 eV
Show the energy jumps
Energy jumps
Accepted within
± 1%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 23.07 eV ± 1%
  Misconception: Uses the energy of one level for the photon, instead of the difference between the two levels of the jump.
  Feedback from the catalogue
  The photon takes the energy the atom loses in the jump: the difference between the upper level and the lower one, $E = E_{\text{upper}} - E_{\text{lower}}$ ⁠. A level’s energy is measured from the ground state, so it matches the photon only for a jump all the way down to the ground state, and that is usually ultraviolet light.
Feedback for a right answer
2.11 eV: the difference between two of helium’s levels. The element is named after the Sun, helios in Greek, and was found on Earth only in 1895.

Hint
$E = \dfrac{hc}{\lambda}$ ⁠.

Worked solution
1. $E = \dfrac{hc}{\lambda} = \dfrac{1240}{587.6} \approx 2.11\ \mathrm{eV}$ ⁠.
2. On the ladder of Show the energy jumps you can see the jump: from a level at 23.07 eV to a level at 20.96 eV, a difference of 2.11 eV.
Answering 23.07 eV means reading the height of the upper level; the photon carries only what the atom loses in the jump.
Step 6 of 6
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Dark lines
Practice · Choose
Question
When sunlight is spread out in a good spectroscope, its rainbow is full of thin dark lines. What are they?

Correct answer
- Correct Light absorbed by cooler gas in the Sun’s outer layers, at exactly the wavelengths it would give out
Setup, choices and feedback
The Scene at this step
Gas in the tube
Hydrogen
Electron energy
25 eV
Show the energy jumps
Spectrum
Choices
Shown in a random order.
- Correct Light absorbed by cooler gas in the Sun’s outer layers, at exactly the wavelengths it would give out
- Wrong The wavelengths of elements the Sun does not contain
  Misconception: Thinks dark lines in a star's spectrum mean those elements are missing from it.
  Feedback from the catalogue
  The other way round: a dark line shows the element is there. Fairly cool gas in a star’s outer layers absorbs from the light passing through it exactly the wavelengths it would give out, because those are the jumps its atoms can make. That is how a new element, helium, was found in the Sun’s spectrum in 1868, twenty-seven years before it was found on Earth.
- Wrong Flaws in the glass of the prism
  Misconception: Thinks the prism or the spectroscope picks out the bright lines, rather than the gas giving out only those colours.
  Feedback from the catalogue
  The prism only separates the colours: white light from a filament lamp makes a continuous rainbow through it. The lines come from the gas. Each atom can only be in certain energy levels, and a photon given out in a jump from one level to another carries exactly the difference between them, $E = hf = \dfrac{hc}{\lambda}$ ⁠. So each element gives only its own wavelengths.
- Wrong The shadows of sunspots
  Feedback
  Sunspots are cooler patches on the Sun’s surface. They dim its light in every colour, not at particular wavelengths.
Feedback for a right answer
Absorption: the gas absorbs exactly its own lines. Each dark line is an element that is there, not one that is missing.

Worked solution
1. The Sun’s hot, dense surface gives out a continuous spectrum, like the lump in the furnace.
2. The light passes through the Sun’s atmosphere, a thinner and cooler gas. An atom can absorb only a photon that matches exactly one of its jumps, and those are the same wavelengths it would give out.
3. So less light arrives at those wavelengths, and there dark lines appear.
Where the dark lines are says which elements are in the Sun, and how deep they are says how much. That is how the Sun was found to be made mostly of hydrogen and helium.

Lesson 3 of 5 · About 25 minutes

The life of a star

A star shines by nuclear fusion, and its mass decides everything: how bright it is, how long it lives, and how it ends, as a white dwarf, a neutron star or a black hole.

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The 6 steps and their answers

Step 1 of 6
Open this step
A ball of gas
Practice · Choose
Question
The Sun has shone for about $4.6$ ⁠ billion years, at almost the same power. Where does its energy come from?

Correct answer
- Correct From hydrogen nuclei fusing into helium in its core
Setup, choices and feedback
The Scene at this step
Starting mass
1 Suns
Age the star
0.3 (Birth → End)
Choices
Shown in a random order.
- Correct From hydrogen nuclei fusing into helium in its core
- Wrong From burning gas, like a giant flame
  Misconception: Thinks the Sun shines because something in it is burning, as a fire does.
  Feedback from the catalogue
  Burning is a chemical reaction, and it gives out a few electronvolts per molecule. If the Sun were made of burning coal, it would have run out after a few thousand years. The Sun has shone for $4.6$ ⁠ billion years thanks to nuclear fusion: in its core, at about $15$ ⁠ million kelvin, hydrogen nuclei join into helium, and each fusion gives out millions of electronvolts, by $E = mc^2$ ⁠.
- Wrong From the heat left over from its formation, slowly cooling
  Feedback
  Shrinking and its first heat would have lasted only a few tens of millions of years. That was Kelvin’s calculation, and geologists knew the Earth was older.
- Wrong From splitting uranium, as in a nuclear reactor
  Feedback
  There is almost no uranium in the Sun. It is made mostly of hydrogen, and the energy comes from joining light nuclei, not splitting heavy ones.
Feedback for a right answer
Nuclear fusion: every second the Sun turns about $600$ ⁠ million tonnes of hydrogen into helium, and loses about $4$ ⁠ million tonnes of mass as light.

Worked solution
1. In the Sun’s core the temperature is about $15$ ⁠ million kelvin and the density is huge, so hydrogen nuclei collide fast enough to overcome their electric repulsion and fuse.
2. Four hydrogen nuclei become one helium nucleus, whose mass is about $0.7\%$ ⁠ less than theirs together.
3. The missing mass comes out as energy, $E = mc^2$ ⁠: about 27 MeV for each helium nucleus, millions of times more than a chemical reaction.
While there is hydrogen in its core, the star is on the main sequence, steady and at almost constant power.
Step 2 of 6
Open this step
Mass and power
Practice · Calculate
Question
Starting mass is locked at $10$ ⁠ solar masses. On the main sequence, a star’s power grows roughly like $L \propto M^{3.5}$ ⁠. How many times brighter than the Sun does it shine?

Correct answer
3162 ± 3%
Setup, choices and feedback
The Scene at this step
Starting mass
10 Sunsfixed
Age the star
0.3 (Birth → End)
Accepted within
± 3%
Computed from the Scene at this step’s setting
Power on the main sequence, in Suns: M^3.5/nodes/star/outputs/luminosity
Wrong answers it recognises
- 10 ± 1
  Misconception: Thinks a star's luminosity grows in proportion to its mass, so twice the mass shines twice as bright.
  Feedback from the catalogue
  In a heavier star gravity squeezes the core harder, so it is hotter, and fusion runs far faster. On the main sequence the power grows roughly like $L \propto M^{3.5}$ ⁠: a star $10$ ⁠ times heavier shines about $3000$ ⁠ times as bright. If the power were in proportion to the mass, every star would live just as long.
Feedback for a right answer
About $3160$ ⁠: $10$ ⁠ times the mass, more than three thousand times the power.

Hint
$10^{3.5} = 10^3 \cdot \sqrt{10}$ ⁠.

Worked solution
1. $\dfrac{L}{L_\odot} = \left(\dfrac{M}{M_\odot}\right)^{3.5} = 10^{3.5}$ ⁠.
2. $10^{3.5} = 10^3 \cdot 10^{0.5} = 1000 \cdot 3.16 \approx 3160$ ⁠.
In a heavy star gravity squeezes the core harder, it is hotter, and the rate of fusion climbs steeply. That is why heavy stars are so bright, and blue.
Step 3 of 6
Open this step
Heavy and brief
Practice · Predict, then look
Question
The Sun will spend about ten billion years on the main sequence. A star of $10$ ⁠ solar masses has $10$ ⁠ times the fuel. How long will it live compared with the Sun?

Correct answer
- Correct Far less
Setup, choices and feedback
The Scene at this step
Starting mass
1 Suns
Age the star
0.3 (Birth → End)
Choices
- Correct Far less
- Wrong $10$ ⁠ times longer, because it has $10$ ⁠ times the fuel
  Misconception: Thinks a heavier star lives longer because it has more fuel.
  Feedback from the catalogue
  A heavy star has more fuel, but it burns it far faster. Its lifetime is the fuel divided by the rate of burning, $t \propto \dfrac{M}{M^{3.5}} = M^{-2.5}$ ⁠. The Sun will live about ten billion years, a star $10$ ⁠ times heavier only about $30$ ⁠ million, and a light red dwarf trillions of years.
- Wrong Just as long, because the fuel and the rate of burning both grow $10$ ⁠ times
  Misconception: Thinks a star's luminosity grows in proportion to its mass, so twice the mass shines twice as bright.
  Feedback from the catalogue
  In a heavier star gravity squeezes the core harder, so it is hotter, and fusion runs far faster. On the main sequence the power grows roughly like $L \propto M^{3.5}$ ⁠: a star $10$ ⁠ times heavier shines about $3000$ ⁠ times as bright. If the power were in proportion to the mass, every star would live just as long.
Once the prediction is locked in, the Scene changes
Starting mass
10 Suns
The Scene then shows: 0.03162 · Time on the main sequence, billions of years: 10 × M^−2.5 (the fuel M burnt at a rate M^3.5)/nodes/star/outputs/lifetime
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About $32$ ⁠ million years, less than one per cent of the Sun’s life: $10$ ⁠ times the fuel, but burnt $3000$ ⁠ times as fast.

Worked solution
1. The lifetime is the fuel divided by the rate of burning: $t \propto \dfrac{M}{L}$ ⁠.
2. $L \propto M^{3.5}$ ⁠, so $t \propto \dfrac{M}{M^{3.5}} = M^{-2.5}$ ⁠.
3. For $10$ ⁠ solar masses: $t = 10 \cdot 10^{-2.5} \approx 0.032$ ⁠ billion years, about $32$ ⁠ million years.
The number the Scene shows is the lifetime in billions of years. That is why the bright blue stars we see are all young: the old ones blew up long ago.
Step 4 of 6
Open this step
How long
Practice · Calculate
Question
Starting mass is locked at $2$ ⁠ solar masses. Using $t \approx 10 \cdot M^{-2.5}$ ⁠ billion years, with $M$ ⁠ in solar masses, how many billion years will the star spend on the main sequence?

Correct answer
1.768 ± 3%
Setup, choices and feedback
The Scene at this step
Starting mass
2 Sunsfixed
Age the star
0.3 (Birth → End)
Accepted within
± 3%
Computed from the Scene at this step’s setting
Time on the main sequence, billions of years: 10 × M^−2.5 (the fuel M burnt at a rate M^3.5)/nodes/star/outputs/lifetime
Wrong answers it recognises
- 20 ± 3%
  Misconception: Thinks a heavier star lives longer because it has more fuel.
  Feedback from the catalogue
  A heavy star has more fuel, but it burns it far faster. Its lifetime is the fuel divided by the rate of burning, $t \propto \dfrac{M}{M^{3.5}} = M^{-2.5}$ ⁠. The Sun will live about ten billion years, a star $10$ ⁠ times heavier only about $30$ ⁠ million, and a light red dwarf trillions of years.
- 10 ± 3%
  Misconception: Thinks a star's luminosity grows in proportion to its mass, so twice the mass shines twice as bright.
  Feedback from the catalogue
  In a heavier star gravity squeezes the core harder, so it is hotter, and fusion runs far faster. On the main sequence the power grows roughly like $L \propto M^{3.5}$ ⁠: a star $10$ ⁠ times heavier shines about $3000$ ⁠ times as bright. If the power were in proportion to the mass, every star would live just as long.
Feedback for a right answer
About $1.8$ ⁠ billion years. Stars like that do not last long enough to grow life round them as on Earth.

Hint
$2^{2.5} = 2^2 \cdot \sqrt 2$ ⁠.

Worked solution
1. $t = 10 \cdot 2^{-2.5}$ ⁠ billion years.
2. $2^{2.5} = 4 \cdot 1.414 \approx 5.66$ ⁠.
3. $t = \dfrac{10}{5.66} \approx 1.8$ ⁠ billion years.
A star like Sirius, of about two solar masses, will live less than a fifth as long as the Sun. Answering $20$ ⁠ means thinking twice the fuel gives twice the life; answering $10$ ⁠ means assuming the power grows like the mass.
Step 5 of 6
Open this step
The end of the Sun
Practice · Choose
Question
Age the star has reached the end of the life of a star of one solar mass. What will be left of the Sun in about eight billion years?

Correct answer
- Correct A white dwarf: a hot core the size of the Earth, after a red giant and a planetary nebula
Setup, choices and feedback
The Scene at this step
Starting mass
1 Suns
Age the star
End
Choices
Shown in a random order.
- Correct A white dwarf: a hot core the size of the Earth, after a red giant and a planetary nebula
- Wrong A black hole, after a supernova
  Misconception: Thinks the Sun will end in a supernova or as a black hole.
  Feedback from the catalogue
  The Sun is too light. Only a star that starts with more than about $8$ ⁠ solar masses ends up with an iron core that collapses in a supernova, and only above about $20$ ⁠ solar masses does the collapse leave a black hole. The Sun will swell into a red giant, puff its outer layers off as a planetary nebula, and leave a white dwarf, a core the size of the Earth that slowly cools.
- Wrong A neutron star, after a supernova
  Misconception: Thinks every star ends its life in an explosion.
  Feedback from the catalogue
  Most stars do not explode. Stars like the Sun shed their outer layers gently, and red dwarfs, most of the stars in the Galaxy, have never yet come to the end of their lives: they will go on burning for trillions of years. Only stars that start with more than about $8$ ⁠ solar masses end in a supernova.
- Wrong Nothing: it will use itself up until nothing is left
  Feedback
  Only a small part of the mass turns into energy. About half of the Sun will remain as a dense core.
Feedback for a right answer
A white dwarf. Supernovae, neutron stars and black holes are kept for much heavier stars.

Worked solution
1. When the hydrogen in the core runs out, the core will shrink and heat up, and the outer layers will swell and cool: a red giant, roughly as wide as the Earth’s orbit.
2. The outer layers will drift slowly away and glow round the core: a planetary nebula.
3. The core, half the Sun’s mass in the size of the Earth, will remain as a white dwarf that slowly cools.
To collapse in a supernova, a star has to start with more than about $8$ ⁠ solar masses; to leave a black hole, more than about $20$ ⁠.
Step 6 of 6
Open this step
Heavy endings
Practice · Reach the target
Question
Change Starting mass until the star ends its life in a supernova and leaves a neutron star, and press Check.

Goal
- 2 ± 0.1
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Starting mass
1 Suns
Age the star
End
Read from the Scene at
What it leaves: 0 nothing yet (a red dwarf outlasts the universe so far), 1 a white dwarf, 2 a neutron star, 3 a black hole/nodes/star/outputs/remnant
One setting that reaches it
Starting mass
8 Suns
Feedback for a right answer
Between $8$ ⁠ and $20$ ⁠ solar masses: a supernova and a neutron star, a teaspoon of which weighs a billion tonnes.

Feedback for a wrong answer
Not yet. Below $8$ ⁠ solar masses a white dwarf is left; above $20$ ⁠, a black hole.

Hint
Heavy enough to collapse, and light enough not to become a black hole.

Worked solution
1. Below about $8$ ⁠ solar masses the star sheds its outer layers gently and leaves a white dwarf.
2. Above $8$ ⁠ solar masses fusion carries on all the way to iron, from which no more energy can be had. The core collapses in a fraction of a second, and the star blows up in a supernova.
3. Between $8$ ⁠ and $20$ ⁠ solar masses the collapsing core stops as a ball of neutrons about $20$ ⁠ kilometres across; above $20$ ⁠ it carries on collapsing into a black hole.
Any mass between $8$ ⁠ and $20$ ⁠ solar masses is an answer.

Lesson 4 of 5 · About 30 minutes

Measuring distances in space

The distance ladder: parallax for nearby stars, parsecs and light years, and the inverse-square law with standard candles, Cepheids and supernovae, for distant galaxies.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Parallax
Practice · Predict, then look
Question
A nearby star seems to move against the background of distant stars between January and July, as the Earth goes to the other side of the Sun. The star is moved from $890$ ⁠ to $1780$ ⁠ light years away, twice as far. What will happen to its parallax?

Correct answer
- Correct It will get $2$ ⁠ times smaller
Setup, choices and feedback
The Scene at this step
Star distance
890 light years
Cepheid period
30 daysdefault
Galaxy distance
60 million light yearsdefault
Climb the ladder
Parallax
Choices
- Correct It will get $2$ ⁠ times smaller
- Wrong It will get $4$ ⁠ times smaller
  Misconception: Thinks a star's parallax shrinks with the square of its distance, like its brightness.
  Feedback from the catalogue
  Parallax is an angle, and it shrinks like $\dfrac{1}{d}$ ⁠: the baseline, the Earth’s orbit, is fixed, and from twice as far it is seen at half the angle. Only brightness falls like $\dfrac{1}{d^2}$ ⁠, because light spreads over an area, not a length.
- Wrong It will not change, because the Earth’s orbit has not changed
  Feedback
  The orbit is the same orbit, but from twice as far it is seen at half the angle.
Once the prediction is locked in, the Scene changes
Star distance
1780 light years
The Scene then shows: 0.001832 · The nearby star’s parallax in arcseconds: one over its distance in parsecs/nodes/ladder/outputs/parallaxArcsec
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$2$ ⁠ times: $p \propto \dfrac{1}{d}$ ⁠. The number beside the Scene is the parallax in arcseconds.

Worked solution
1. The parallax $p$ ⁠ is half the angle through which the star seems to move, that is, the angle at which the radius of the Earth’s orbit, 1 AU, is seen from the star.
2. For small angles $p \approx \dfrac{1\ \mathrm{AU}}{d}$ ⁠: inversely with the distance.
3. Twice the distance: $p$ ⁠ gets $2$ ⁠ times smaller, from about $0.0037$ ⁠ to about $0.0018$ ⁠ arcseconds.
The star’s brightness gets $4$ ⁠ times smaller at twice the distance, but parallax is an angle, not an area.
Step 2 of 6
Open this step
Parsecs
Practice · Calculate
Question
Star distance is locked at $890$ ⁠ light years, like Delta Cephei. One parsec is $3.26$ ⁠ light years. What is the star’s parallax, in arcseconds?

Correct answer
0.003665 ± 3%
Setup, choices and feedback
The Scene at this step
Star distance
890 light yearsfixed
Cepheid period
30 daysdefault
Galaxy distance
60 million light yearsdefault
Climb the ladder
Parallax
Accepted within
± 3%
Computed from the Scene at this step’s setting
The nearby star’s parallax in arcseconds: one over its distance in parsecs/nodes/ladder/outputs/parallaxArcsec
Wrong answers it recognises
- 0.001124 ± 3%
  Misconception: Uses p = 1/d with the distance in light years instead of parsecs.
  Feedback from the catalogue
  The formula $p = \dfrac{1}{d}$ ⁠, with $p$ ⁠ in arcseconds, holds only with $d$ ⁠ in parsecs: that is the definition of a parsec, the distance at which the parallax is one arcsecond. One parsec is $3.26$ ⁠ light years, so a distance in light years must first be divided by $3.26$ ⁠, or written $p = \dfrac{3.26}{d_{\text{ly}}}$ ⁠.
Feedback for a right answer
About $0.0037$ ⁠ arcseconds: $\dfrac{3.26}{890}$ ⁠. A tiny angle, but Gaia measures it.

Hint
First convert to parsecs, then $p = \dfrac{1}{d}$ ⁠.

Worked solution
1. The distance in parsecs: $d = \dfrac{890}{3.26} \approx 273\ \mathrm{pc}$ ⁠.
2. The parallax: $p = \dfrac{1}{d} = \dfrac{1}{273} \approx 0.0037$ ⁠ arcseconds.
That is about the angle a small coin makes from $1000$ ⁠ kilometres away. The Gaia satellite measures angles like that to about half a per cent. Getting about $0.0011$ ⁠ means putting light years straight into the parsec formula.
Step 3 of 6
Open this step
The nearest star
Practice · Calculate
Question
Proxima Centauri, the nearest star to the Sun, has the largest parallax ever measured: $0.768$ ⁠ arcseconds. How far away is it?

Correct answer
4.25 ly ± 2%
Setup, choices and feedback
The Scene at this step
Star distance
4.4 light years
Cepheid period
30 daysdefault
Galaxy distance
60 million light yearsdefault
Climb the ladder
Parallax
Accepted within
± 2%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1.302 ly ± 3%
  Misconception: Uses p = 1/d with the distance in light years instead of parsecs.
  Feedback from the catalogue
  The formula $p = \dfrac{1}{d}$ ⁠, with $p$ ⁠ in arcseconds, holds only with $d$ ⁠ in parsecs: that is the definition of a parsec, the distance at which the parallax is one arcsecond. One parsec is $3.26$ ⁠ light years, so a distance in light years must first be divided by $3.26$ ⁠, or written $p = \dfrac{3.26}{d_{\text{ly}}}$ ⁠.
Feedback for a right answer
About $4.25$ ⁠ light years: $\dfrac{3.26}{0.768}$ ⁠. Light leaving it today will reach us in more than four years.

Hint
$d = \dfrac{1}{p}$ ⁠ in parsecs, and one parsec is $3.26$ ⁠ light years.

Worked solution
1. $d = \dfrac{1}{p} = \dfrac{1}{0.768} \approx 1.30\ \mathrm{pc}$ ⁠.
2. In light years: $1.30 \cdot 3.26 \approx 4.25$ ⁠ light years.
Every other star is farther away, so every other star has a parallax of less than one arcsecond. Answering $1.30$ ⁠ means stopping at parsecs.
Step 4 of 6
Open this step
Standard candles
Practice · Choose
Question
Beyond a few thousand light years the parallax is too small to measure. Why do astronomers then look for pulsating stars of the Cepheid kind in particular?

Correct answer
- Correct The period of the pulse reveals how much light the Cepheid really gives out, so its brightness in the sky reveals its distance
Setup, choices and feedback
The Scene at this step
Star distance
890 light yearsdefault
Cepheid period
30 days
Galaxy distance
60 million light years
Climb the ladder
Cepheids
Choices
Shown in a random order.
- Correct The period of the pulse reveals how much light the Cepheid really gives out, so its brightness in the sky reveals its distance
- Wrong Because they are the brightest stars in the sky, and so the nearest
  Misconception: Thinks a star that looks brighter must be nearer.
  Feedback from the catalogue
  How bright a star looks depends on two things: how much light it really gives out, its luminosity, and how far away it is. A powerful, distant star can look brighter than a weak, nearby one. To measure a distance from brightness you need a standard candle: a body whose true luminosity is known, like a Cepheid from its period.
- Wrong Because every Cepheid gives out exactly the same amount of light
  Feedback
  Different Cepheids give out very different amounts of light. What they share is the link between their period and their luminosity.
- Wrong Because their pulsing light reaches us faster
  Feedback
  All light travels at the same speed. The pulse matters because of what it reveals about the star.
Feedback for a right answer
The period gives the true luminosity, the apparent brightness gives the distance. That is how Hubble found in 1924 that Andromeda is another galaxy.

Worked solution
1. How bright a star looks depends on how much light it gives out, $L$ ⁠, and on its distance: $b = \dfrac{L}{4\pi d^2}$ ⁠.
2. In 1912 Henrietta Swan Leavitt found that Cepheids with a longer period give out more light: Leavitt’s law. Measure the period, and you know $L$ ⁠.
3. Measure the brightness $b$ ⁠, and work out $d = \sqrt{\dfrac{L}{4\pi b}}$ ⁠.
A body whose light output is known is called a standard candle. Leavitt’s law is calibrated on nearby Cepheids, whose distances are measured by parallax: that is the next rung of the ladder.
Step 5 of 6
Open this step
The inverse-square law
Practice · Calculate
Question
Type Ia supernovae all reach the same luminosity at their peak. The calibrating galaxy is $30$ ⁠ million light years away, and the target’s Galaxy distance is locked at $60$ ⁠ million. How many times fainter will the target’s supernova look?

Correct answer
4 ± 0.05
Setup, choices and feedback
The Scene at this step
Star distance
890 light yearsdefault
Cepheid period
30 daysdefault
Galaxy distance
60 million light yearsfixed
Climb the ladder
Supernovae
Accepted within
± 0.05
Computed from the Scene at this step’s setting
How many times fainter the target supernova looks than the calibrator’s/nodes/ladder/outputs/faintness
Wrong answers it recognises
- 2 ± 0.05
  Misconception: Thinks a source twice as far looks half as bright.
  Feedback from the catalogue
  Light spreads out over a growing sphere, whose area grows like $4\pi d^2$ ⁠. At twice the distance the same light is spread over $4$ ⁠ times the area, so the brightness is $4$ ⁠ times smaller: $b = \dfrac{L}{4\pi d^2}$ ⁠. At $10$ ⁠ times the distance, it is $100$ ⁠ times smaller.
Feedback for a right answer
$4$ ⁠ times: twice the distance, the same light over $4$ ⁠ times the area.

Hint
$b = \dfrac{L}{4\pi d^2}$ ⁠: the same $L$ ⁠, a different distance.

Worked solution
1. Both supernovae have the same luminosity $L$ ⁠.
2. The apparent brightness is $b = \dfrac{L}{4\pi d^2}$ ⁠, so $\dfrac{b_{\text{calibrator}}}{b_{\text{target}}} = \left(\dfrac{d_{\text{target}}}{d_{\text{calibrator}}}\right)^2$ ⁠.
3. $\left(\dfrac{60}{30}\right)^2 = 2^2 = 4$ ⁠.
Answering $2$ ⁠ means forgetting that the light spreads over an area, which grows as the square of the distance.
Step 6 of 6
Open this step
A hundred times fainter
Practice · Reach the target
Question
A supernova in a distant galaxy looks $100$ ⁠ times fainter than the one in the calibrating galaxy, $30$ ⁠ million light years away. Change Galaxy distance until its supernova looks $100$ ⁠ times fainter, and press Check.

Goal
- 100 ± 2%
  At the start of the step: 4
Setup, choices and feedback
The Scene at this step
Star distance
890 light yearsdefaultfixed
Cepheid period
30 daysdefaultfixed
Galaxy distance
60 million light years
Climb the ladder
Supernovae
Read from the Scene at
How many times fainter the target supernova looks than the calibrator’s/nodes/ladder/outputs/faintness
One setting that reaches it
Galaxy distance
297 million light years
Feedback for a right answer
$300$ ⁠ million light years: $10$ ⁠ times as far, $100$ ⁠ times as faint.

Feedback for a wrong answer
Not yet. If the supernova is still too bright, move the galaxy farther away; if too faint, bring it closer.

Hint
$100$ ⁠ times fainter means $\sqrt{100}$ ⁠ times farther.

Worked solution
1. $\left(\dfrac{d}{d_{\text{calibrator}}}\right)^2 = 100$ ⁠, so $\dfrac{d}{d_{\text{calibrator}}} = 10$ ⁠.
2. $d = 10 \cdot 30 = 300$ ⁠ million light years.
That far, Cepheids can no longer be picked out, but a supernova as bright as billions of suns can be seen from billions of light years. In 1998 this is how the universe was found to be speeding up its expansion.

Lesson 5 of 5 · About 30 minutes

The expanding universe

Galaxies are moving apart, the distant ones faster: Hubble's law. The universe has no centre, space itself stretches and stretches light with it, and the Hubble time says roughly when it all began.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Galaxies move apart
Practice · Predict, then look
Question
From our galaxy, the Milky Way, all the other galaxies seem to be moving away, the distant ones faster. We switch to watching from another galaxy, $7$ ⁠ billion light years from us. What will we see from there?

Correct answer
- Correct From there too every galaxy moves away, the distant ones faster
Setup, choices and feedback
The Scene at this step
Time since the Big Bang
13.8 billion years
Watch from
Milky Way
Show speeds and light
0.4 (Galaxies → Speeds and light)
Choices
- Correct From there too every galaxy moves away, the distant ones faster
- Wrong The galaxies on our side move towards it, and the rest move away
  Misconception: Thinks the galaxies all moving away from us means the Milky Way is at the centre of the expansion.
  Feedback from the catalogue
  From every galaxy you see the same thing: all the others moving away, faster the farther they are. That is what an even stretching of space looks like, like spots on rising dough: from every spot, all the others move away. The universe has no centre that everything started from, because the Big Bang happened everywhere at once.
- Wrong They all move the same way, away from the Milky Way
  Misconception: Thinks the galaxies all moving away from us means the Milky Way is at the centre of the expansion.
  Feedback from the catalogue
  From every galaxy you see the same thing: all the others moving away, faster the farther they are. That is what an even stretching of space looks like, like spots on rising dough: from every spot, all the others move away. The universe has no centre that everything started from, because the Big Bang happened everywhere at once.
Once the prediction is locked in, the Scene changes
Watch from
Another galaxy
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The same picture from everywhere: the universe has no centre that everything moves away from.

Worked solution
1. In an even stretching, every distance grows in the same ratio: if all distances grow by $1\%$ ⁠, a galaxy twice as far moves away twice as fast.
2. That holds from any point of view: the distances from the other galaxy grow in exactly the same ratio.
3. So from every galaxy you see the same picture: all of them moving away, at a speed in proportion to the distance.
It is like raisins in a rising cake: every raisin sees all the others moving away from it. There is no centre.
Step 2 of 6
Open this step
Hubble's law
Practice · Calculate
Question
Today, by the Scene, $H_0 \approx 67.7$ ⁠ kilometres per second per megaparsec. A galaxy is $100$ ⁠ megaparsecs from us, about $326$ ⁠ million light years. How fast is it moving away? Give the answer in kilometres per second.

Correct answer
6769 ± 2%
Setup, choices and feedback
The Scene at this step
Time since the Big Bang
13.8 billion yearsfixed
Watch from
Milky Way
Show speeds and light
0.4 (Galaxies → Speeds and light)
Accepted within
± 2%
Computed from the Scene at this step’s setting
Expansion rate H(t), km/s per megaparsec/nodes/space/outputs/hubble
Feedback for a right answer
About 6770 km/s: $67.7 \cdot 100$ ⁠. A galaxy twice as far moves away twice as fast.

Hint
Hubble’s law: $v = H_0 d$ ⁠.

Worked solution
1. Hubble’s law: $v = H_0 d$ ⁠.
2. $v = 67.7\ \dfrac{\mathrm{km/s}}{\mathrm{Mpc}} \cdot 100\ \mathrm{Mpc} = 6770\ \mathrm{km/s}$ ⁠.
About $2\%$ ⁠ of the speed of light. In a galaxy like that the hydrogen lines are shifted to the red by about $2\%$ ⁠: that is how the speed is measured, and then divided by the distance to find $H_0$ ⁠.
Step 3 of 6
Open this step
Faster than light
Practice · Choose
Question
By Hubble’s law, a galaxy more than about $14$ ⁠ billion light years away is moving away from us faster than light. Does that contradict relativity?

Correct answer
- Correct No: the galaxy is not flying through space, the space between us is stretching
Setup, choices and feedback
The Scene at this step
Time since the Big Bang
13.8 billion years
Watch from
Milky Way
Show speeds and light
0.4 (Galaxies → Speeds and light)
Choices
Shown in a random order.
- Correct No: the galaxy is not flying through space, the space between us is stretching
- Wrong Yes, so Hubble’s law cannot hold at such distances
  Misconception: Thinks galaxies receding faster than light break relativity, as if they were flying through space.
  Feedback from the catalogue
  Relativity forbids moving through space faster than light, relative to things near you. Galaxies hardly move relative to the space around them: the space between them stretches. The speed in Hubble’s law is the rate at which the distance grows, and it can exceed the speed of light without anything overtaking a ray of light on the way.
- Wrong No, because as a galaxy nears the speed of light it slows down
  Misconception: Thinks galaxies receding faster than light break relativity, as if they were flying through space.
  Feedback from the catalogue
  Relativity forbids moving through space faster than light, relative to things near you. Galaxies hardly move relative to the space around them: the space between them stretches. The speed in Hubble’s law is the rate at which the distance grows, and it can exceed the speed of light without anything overtaking a ray of light on the way.
- Wrong No, because there are no galaxies that far away
  Feedback
  There are: the light of such galaxies is seen all the time, with redshifts above $1.5$ ⁠.
Feedback for a right answer
Space stretches, and nothing overtakes light on the way. Einstein’s limit holds speeds locally, not the rate at which huge distances grow.

Worked solution
1. Relativity says no body moves through space faster than light, relative to the things beside it.
2. A galaxy is nearly at rest relative to the space round it. What grows is space itself, everywhere between us and it.
3. The speed in Hubble’s law, $v = H_0 d$ ⁠, is the rate at which the distance grows. When the distance is large enough, $\dfrac{c}{H_0} \approx 14.4$ ⁠ billion light years, that rate passes $c$ ⁠.
No ray of light is overtaken, so there is no contradiction.
Step 4 of 6
Open this step
The Hubble time
Practice · Calculate
Question
If the galaxies had always moved apart at the same speeds, how long ago would they all have been in one place? That time, $\dfrac{1}{H_0}$ ⁠, is called the Hubble time. Work it out in billions of years, with $H_0 = 67.7$ ⁠ kilometres per second per megaparsec. One megaparsec is $3.09 \times 10^{19}$ ⁠ kilometres, and a year is $3.16 \times 10^7$ ⁠ seconds.

Correct answer
14.44 ± 2%
Setup, choices and feedback
The Scene at this step
Time since the Big Bang
13.8 billion yearsfixed
Watch from
Milky Way
Show speeds and light
Galaxies
Accepted within
± 2%
Computed from the Scene at this step’s setting
The Hubble time 1/H(t), billions of years: how long the expansion would have taken at its present rate/nodes/space/outputs/hubbleTime
Feedback for a right answer
About $14.4$ ⁠ billion years, close to the age of the universe. Hubble’s law alone already says roughly when it all began.

Hint
A galaxy at distance $d$ ⁠ moving at $v = H_0 d$ ⁠ covers the distance in a time $\dfrac{d}{v} = \dfrac{1}{H_0}$ ⁠. Convert $H_0$ ⁠ to units of one per second.

Worked solution
1. A galaxy at distance $d$ ⁠ moves at $v = H_0 d$ ⁠, so it covered the distance in a time $t = \dfrac{d}{v} = \dfrac{1}{H_0}$ ⁠, whatever $d$ ⁠ is.
2. $H_0 = \dfrac{67.7\ \mathrm{km/s}}{3.09 \times 10^{19}\ \mathrm{km}} \approx 2.19 \times 10^{-18}\ \mathrm{s^{-1}}$ ⁠.
3. $\dfrac{1}{H_0} \approx 4.56 \times 10^{17}\ \mathrm{s}$ ⁠, and divided by $3.16 \times 10^7$ ⁠ seconds in a year: about $14.4$ ⁠ billion years.
That is very close to the age of the universe, $13.8$ ⁠ billion years. The difference: the expansion slowed down at first and sped up later.
Step 5 of 6
Open this step
Stretched light
Practice · Calculate
Question
Show speeds and light follows light that left the marked galaxy at a wavelength of 420 nm, violet. On its way to us space stretched, and its redshift is $z \approx 0.54$ ⁠. At what wavelength does it arrive?

Correct answer
647 nm ± 1%
Setup, choices and feedback
The Scene at this step
Time since the Big Bang
13.8 billion yearsfixed
Watch from
Milky Way
Show speeds and light
Speeds and light
Accepted within
± 1%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 647 nm: violet that arrives red, because space grew $1.54$ ⁠ times while the light was on its way.

Hint
$\lambda_{\text{received}} = (1 + z)\, \lambda_{\text{emitted}}$ ⁠.

Worked solution
1. Redshift is defined as $z = \dfrac{\lambda_{\text{received}} - \lambda_{\text{emitted}}}{\lambda_{\text{emitted}}}$ ⁠, so $\lambda_{\text{received}} = (1 + z)\,\lambda_{\text{emitted}}$ ⁠.
2. $\lambda_{\text{received}} = 1.54 \cdot 420 \approx 647\ \mathrm{nm}$ ⁠: red light.
In an expanding universe $1 + z$ ⁠ is exactly the factor by which the universe has grown since the light left: the universe grew $1.54$ ⁠ times while the light was on its way, and the wavelength stretched with it.
Step 6 of 6
Open this step
Speeding up
Practice · Predict, then look
Question
Since about $7.6$ ⁠ billion years after the Big Bang the expansion has been speeding up: every galaxy moves away from us faster and faster. What will happen to $H$ ⁠, the ratio of speed to distance, $6$ ⁠ billion years from now?

Correct answer
- Correct It will get smaller
Setup, choices and feedback
The Scene at this step
Time since the Big Bang
13.8 billion years
Watch from
Milky Way
Show speeds and light
0.4 (Galaxies → Speeds and light)
Choices
- Correct It will get smaller
- Wrong It will get bigger, because the expansion is speeding up
  Misconception: Thinks that because the expansion is speeding up, the Hubble constant grows with time.
  Feedback from the catalogue
  $H$ ⁠ is the relative rate of expansion: by what percentage every distance grows each year. In an accelerating expansion the distances grow faster and faster, but the distances themselves grow faster still, so the ratio $H = \dfrac{v}{d}$ ⁠ keeps falling. It only approaches a limit, about 56 km/s per megaparsec, set by dark energy.
- Wrong It will stay $67.7$ ⁠, because it is a constant of nature
  Feedback
  Hubble’s “constant” is constant only across space, the same for every galaxy at one time. Over time it changes.
Once the prediction is locked in, the Scene changes
Time since the Big Bang
20 billion years
The Scene then shows: 59.92 · Expansion rate H(t), km/s per megaparsec/nodes/space/outputs/hubble
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$H$ ⁠ falls to about $60$ ⁠. The acceleration is that each galaxy’s $v$ ⁠ grows, not that $H$ ⁠ grows.

Worked solution
1. $H = \dfrac{v}{d}$ ⁠: a galaxy’s speed divided by its distance.
2. In an accelerating expansion $v$ ⁠ grows, but $d$ ⁠ grows too, and faster in proportion. So the ratio gets smaller.
3. In the Scene, at $20$ ⁠ billion years: about $60$ ⁠ kilometres per second per megaparsec, against $67.7$ ⁠ today.
In the far future $H$ ⁠ approaches about $56$ ⁠, the value set by dark energy, and from then on distances grow exponentially.

Across the unit · 19 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

Starlight and temperature

The colour of heat

Question

Correct answer

The Scene at this step

Choices

Worked solution

Wien's law

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Hotter and shorter

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

One micrometre

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The fourth power

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The Sun's temperature

Question

Correct answer

The Scene at this step

Hint

Worked solution

Spectral lines

A barcode

Question

Correct answer

The Scene at this step

Choices

Worked solution

The energy of a photon

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Enough energy

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Fingerprints

Question

Correct answer

The Scene at this step

Choices

Worked solution

An element from the Sun

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Dark lines

Question

Correct answer

The Scene at this step

Choices