Answer key · Unit 25

The Solar System.

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 4

Lesson 1 of 4 · About 20 minutes

The Sun

Our nearest star from the inside: its layers from the core to the surface, the sunspots that come and go in an 11-year cycle, and the magnetic field that flips and makes them.

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The 5 steps and their answers

Step 1 of 5
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From the core to the surface
Practice · Choose
Question
The cut-away wedge shows the Sun’s layers at their true depths. In which layer does the Sun make its energy?

Correct answer
- Correct In the core, the inner quarter of the radius
Setup, choices and feedback
The Scene at this step
Years since solar minimum
4.5 years
Peak sunspot number
180
Labels
offdefault
Show the magnetic field
Sunspots
Choices
Shown in a random order.
- Correct In the core, the inner quarter of the radius
- Wrong In the convection zone, where the gas boils
  Feedback
  The convection zone only carries the energy outwards: hot gas rises, cools and sinks. It is far too cool for fusion.
- Wrong At the surface, where it burns
  Feedback
  The surface, at about 5800 K, is where the light finally escapes into space, not where it is made. And nothing in the Sun is burning: the energy comes from nuclear fusion.
- Wrong In the corona, the hottest layer
  Feedback
  The corona is very hot, one to two million kelvin, but it is extremely thin. The magnetic field heats it, and it does not make the Sun’s light.
Feedback for a right answer
In the core: about $15$ ⁠ million kelvin, and each second about $4$ ⁠ million tonnes of matter turn into energy.

Worked solution
1. Nuclear fusion needs enormous temperature and pressure, so that hydrogen nuclei collide fast enough to overcome the electric repulsion between them.
2. Only the core, the inner quarter of the radius, has about $15$ ⁠ million kelvin and a density more than ten times that of lead.
3. From there the energy leaks out slowly: through the radiative zone as light absorbed and sent off again and again, and through the convection zone in currents of gas.
The light reaching us was made in the core tens of thousands of years ago or more.
Step 2 of 5
Open this step
Dark spots
Practice · Choose
Question
In pictures of the Sun, sunspots look black. What are they really?

Correct answer
- Correct Cooler patches, about 4000 K, that still glow fiercely and look dark only beside their brighter surroundings
Setup, choices and feedback
The Scene at this step
Years since solar minimum
4.5 years
Peak sunspot number
180
Labels
offdefault
Show the magnetic field
Sunspots
Choices
Shown in a random order.
- Correct Cooler patches, about 4000 K, that still glow fiercely and look dark only beside their brighter surroundings
- Wrong Holes in the surface, through which we see the cold, dark inside
  Misconception: Thinks sunspots are dark because they give out no light, like holes or burnt patches.
  Feedback from the catalogue
  A sunspot looks black only next to its surroundings. Its strong magnetic field holds back the convection that brings heat up from below, so it is cooler: about 4000 K instead of about 5800 K. A body at that temperature still glows fiercely, and a spot seen on its own in the night sky would outshine the full Moon. It is dark only beside the even brighter surface around it.
- Wrong Patches where the fuel has run out and the gas has stopped glowing
  Misconception: Thinks sunspots are dark because they give out no light, like holes or burnt patches.
  Feedback from the catalogue
  A sunspot looks black only next to its surroundings. Its strong magnetic field holds back the convection that brings heat up from below, so it is cooler: about 4000 K instead of about 5800 K. A body at that temperature still glows fiercely, and a spot seen on its own in the night sky would outshine the full Moon. It is dark only beside the even brighter surface around it.
- Wrong The shadows of Mercury and Venus
  Feedback
  When Mercury crosses the Sun it is a small, round black dot that crosses the disc in a few hours. Sunspots are much bigger, come in groups and last days to weeks.
Feedback for a right answer
Right. A big sunspot can be several times wider than the Earth, and it is still brighter than any lamp.

Worked solution
1. In a spot the magnetic field is thousands of times stronger than the Earth’s, and it holds back the convection currents that bring heat up from below.
2. So the spot cools to about 4000 K, against about 5800 K around it.
3. By the Stefan–Boltzmann law, $P = \sigma T^4$ ⁠, each square metre of the spot gives out about $\left(\dfrac{4000}{5800}\right)^4 \approx 0.23$ ⁠ of what the surface does: still a great deal of light, but in a picture exposed for the surface it looks black.
Step 3 of 5
Open this step
An 11-year cycle
Practice · Reach the target
Question
The Sun is now at minimum. Move Years since solar minimum until the Sun reaches the peak of its cycle, the maximum, and press Check.

Goal
- 2 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Years since solar minimum
0 years
Peak sunspot number
180
Labels
offdefault
Show the magnetic field
Sunspots
Read from the Scene at
Stage of the cycle for Callouts: 0 minimum, 1 rising, 2 maximum, 3 declining, 4 a very weak cycle/nodes/sun/outputs/stage
One setting that reaches it
Years since solar minimum
3.2 years
Peak sunspot number
40
Labels
off
Feedback for a right answer
Maximum: dozens of spot groups, more flares, and a corona spraying out all round the disc.

Feedback for a wrong answer
Not yet. If there are few spots, you are too close to a minimum; if they already hug the equator, you have passed the peak.

Hint
The maximum comes before the middle of the cycle: the Sun wakes quickly and calms down slowly.

Worked solution
1. The sunspot cycle runs about $11$ ⁠ years from minimum to minimum.
2. The spot count rises quickly and falls slowly, so the maximum comes about $4$ ⁠ to $5$ ⁠ years after the minimum, not in the middle.
3. Set about $4.5$ ⁠ years, or press Maximum. The spots are many, the corona sprays out all round, and the chart shows the top of the curve.
Step 4 of 5
Open this step
The butterfly diagram
Practice · Predict, then look
Question
A year after minimum, the cycle’s first spots appear about $25$ ⁠ degrees north and south of the equator. Where will the spots appear a few years later, near the end of the cycle?

Correct answer
- Correct Closer to the equator
Setup, choices and feedback
The Scene at this step
Years since solar minimum
1 years
Peak sunspot number
180
Labels
offdefault
Show the magnetic field
Sunspots
Choices
- Correct Closer to the equator
- Wrong Closer to the poles
  Feedback
  A cycle’s spots do not move towards the poles. What drifts poleward is the leftover magnetic field of spots that have broken up.
- Wrong At the same latitude, about $25$ ⁠ degrees
  Feedback
  Look at the butterfly diagram: the bands of spots creep.
Once the prediction is locked in, the Scene changes
Years since solar minimum
9 years
The Scene then shows: 9.149 ° · Middle latitude of this cycle's spot band/nodes/sun/outputs/spotLatitude
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Closer to the equator. The bands creep from about $30$ ⁠ degrees to about $7$ ⁠, and then a new cycle starts high again.

Worked solution
1. At the start of the cycle the spots appear in two bands, about $30$ ⁠ degrees north and south.
2. As the cycle goes on, the bands creep towards the equator, and the last spots appear within about $10$ ⁠ degrees of it.
3. Plotting the spots’ latitudes against time gives the butterfly diagram, the lower chart beside the Scene.
Nine years after minimum the band is at about $9$ ⁠ degrees.
Step 5 of 5
Open this step
A field that flips
Practice · Calculate
Question
The field at the Sun’s poles flips around every sunspot maximum. After how many years does the Sun’s field point the way it did at the start of the cycle?

Correct answer
22 yr ± 1 yr
Setup, choices and feedback
The Scene at this step
Years since solar minimum
4.5 years
Peak sunspot number
180
Labels
offdefault
Show the magnetic field
Magnetic field
Accepted within
± 1 yr
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 11 yr ± 1 yr
  Misconception: Thinks the Sun’s magnetic field is back where it started after one 11-year sunspot cycle.
  Feedback from the catalogue
  The spots come back every 11 years, but the magnetic field does not. Around each maximum the Sun’s poles flip, and in the next cycle the polarity of every spot pair is reversed (Hale’s law). It takes two cycles, about $22$ ⁠ years, for the field to point the way it started.
Feedback for a right answer
$22$ ⁠ years: the sunspot cycle is half of the Sun’s magnetic cycle.

Hint
How many flips does it take to get back to the starting direction?

Worked solution
1. A sunspot cycle lasts about $11$ ⁠ years, and around its middle the poles flip.
2. At the end of the cycle the field again runs from pole to pole, but reversed. The polarity of the spot pairs in the next cycle is reversed too (Hale’s law).
3. Only after a second flip, at the end of the next cycle, does the field point the way it started: $2 \times 11 = 22$ ⁠ years.
Answering $11$ ⁠ counts the sunspot cycle, but not the flip of the field.

Lesson 2 of 4 · About 30 minutes

The Solar System

Eight planets round one star: how far apart they really are, why a distant planet’s year is so long, and why the far planets are also slow.

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The 7 steps and their answers

Step 1 of 7
Open this step
How far is far
Practice · Predict, then look
Question
The Scene now spaces the planets evenly. What will happen to the four inner planets when you switch to True distances?

Correct answer
- Correct They will crowd in close to the Sun
Setup, choices and feedback
The Scene at this step
Speed
10 days/s
Show orbits
ondefault
True distances
Evenly spaced
Choices
- Correct They will crowd in close to the Sun
- Wrong Hardly anything: the even spacing is close to the real one
  Misconception: Thinks the planets are spread out roughly evenly, as most diagrams draw them.
  Feedback from the catalogue
  Most drawings of the Solar System space the planets evenly, because otherwise you could not see them. In fact the four inner planets, Mercury, Venus, Earth and Mars, all lie within $1.5$ ⁠ AU of the Sun, while Neptune is $30$ ⁠ AU away. Each gap between two outer planets is wider than the whole inner Solar System.
- Wrong They will spread outwards to the edge of the system
  Feedback
  Look at the numbers: Mars at $1.5$ ⁠ AU, Neptune at $30$ ⁠ AU.
Once the prediction is locked in, the Scene changes
True distances
True distances
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
They crowd in. The whole inner Solar System is smaller than the gap between Jupiter and Saturn.

Worked solution
1. The mean distances from the Sun: Mercury $0.39$ ⁠ AU, Venus $0.72$ ⁠, Earth $1$ ⁠, Mars $1.52$ ⁠, Jupiter $5.2$ ⁠, Saturn $9.5$ ⁠, Uranus $19.2$ ⁠, Neptune $30.1$ ⁠.
2. With Neptune at the edge of the picture, everything from Mars inwards fills only $\dfrac{1.52}{30.1} \approx 5\%$ ⁠ of the radius.
3. So the four inner planets squeeze into a small knot round the Sun.
Step 2 of 7
Open this step
Jupiter’s year
Practice · Calculate
Question
Jupiter is on average $5.2$ ⁠ AU from the Sun. By Kepler’s third law, $T^2 = a^3$ ⁠ with $T$ ⁠ in years and $a$ ⁠ in AU, how many years does a Jupiter year last?

Correct answer
11.86 yr ± 2%
Setup, choices and feedback
The Scene at this step
Speed
30 days/s
Show orbits
ondefault
True distances
True distances
Other settings
Planet/nodes/system/props/planet = "Jupiter"
Accepted within
± 2%
Computed from the Scene at this step’s setting
Its orbital period (measured)/nodes/system/outputs/period
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 5.2 yr ± 2%
  Misconception: Thinks the orbital period grows in proportion to the orbit’s size: twice as far, twice as long.
  Feedback from the catalogue
  By Kepler’s third law, $T^2 = a^3$ ⁠ ( $T$ ⁠ in years, $a$ ⁠ in AU), so $T = a^{3/2}$ ⁠: the time grows faster than the distance. A planet four times as far takes $4^{1.5} = 8$ ⁠ times as long. There are two reasons: its path is longer, and it also moves more slowly, because the Sun’s gravity is weaker out there.
Feedback for a right answer
About $11.9$ ⁠ years. By the time Jupiter completes one lap, the Earth has gone round almost $12$ ⁠ times.

Hint
$T = a^{3/2} = a \cdot \sqrt{a}$ ⁠.

Worked solution
1. $a = 5.2$ ⁠ AU, so $T^2 = 5.2^3 \approx 140.6$ ⁠.
2. $T = \sqrt{140.6} \approx 11.9$ ⁠ years.
Jupiter goes round the Sun once in about $12$ ⁠ years, so it moves about one constellation of the zodiac a year across the night sky. Answering $5.2$ ⁠ assumes the time grows like the distance; but Jupiter is both farther and slower.
Step 3 of 7
Open this step
Why the far planets are slow
Practice · Predict, then look
Question
Neptune’s orbit is $30$ ⁠ times as long as the Earth’s. Which of the two moves faster along its orbit, in kilometres per second?

Correct answer
- Correct The Earth
Setup, choices and feedback
The Scene at this step
Speed
30 days/s
Show orbits
ondefault
True distances
Evenly spaced
Choices
- Correct The Earth
- Wrong Neptune, because it has farther to go
  Misconception: Thinks the outer planets move faster, because they have farther to go.
  Feedback from the catalogue
  A distant planet has a long way to go, but it does not hurry. On a circular orbit the Sun’s gravity supplies the centripetal force: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ ⁠, so $v = \sqrt{\dfrac{GM}{r}}$ ⁠. The farther out the orbit, the slower the speed. The Earth moves at about 30 km/s, Neptune at only about 5.4 km/s.
- Wrong They move at the same speed
  Feedback
  If they moved at the same speed, Neptune’s year would be exactly $30$ ⁠ times as long. It is $165$ ⁠ times as long.
Once the prediction is locked in, the Scene changes
Speed
90 days/s
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The Earth. The Sun’s gravity is weak out at Neptune, and a low speed is enough to stay in orbit.

Worked solution
1. On a circular orbit gravity is the centripetal force: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ ⁠.
2. So $v = \sqrt{\dfrac{GM}{r}}$ ⁠: $30$ ⁠ times as far means $\sqrt{30} \approx 5.5$ ⁠ times as slow.
3. In the Scene, with Speed at 90 days/s, the Earth whirls round and Neptune hardly moves.
Step 4 of 7
Open this step
Neptune’s speed
Practice · Calculate
Question
Neptune goes round the Sun at $30.07$ ⁠ AU, and its year lasts $164.8$ ⁠ years. What is its mean speed along its orbit, in km/s? $1\ \mathrm{AU} = 1.496 \times 10^8\ \mathrm{km}$ ⁠, and a year has $3.156 \times 10^7$ ⁠ seconds.

Correct answer
5.435 km/s ± 3%
Setup, choices and feedback
The Scene at this step
Speed
90 days/s
Show orbits
ondefault
True distances
True distances
Other settings
Planet/nodes/system/props/planet = "Neptune"
Accepted within
± 3%
Computed from the Scene at this step’s setting
Its mean orbital speed, 2πa / T/nodes/system/outputs/orbitalSpeed
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 5.4 km/s: still ten times faster than a rifle bullet, but five times slower than the Earth.

Hint
$v = \dfrac{2\pi a}{T}$ ⁠.

Worked solution
1. The length of the orbit: $2\pi a = 2\pi \cdot 30.07 \cdot 1.496 \times 10^8 \approx 2.83 \times 10^{10}\ \mathrm{km}$ ⁠.
2. The time: $T = 164.8 \cdot 3.156 \times 10^7 \approx 5.20 \times 10^9\ \mathrm{s}$ ⁠.
3. $v = \dfrac{2.83 \times 10^{10}}{5.20 \times 10^9} \approx 5.4\ \mathrm{km/s}$ ⁠.
The Earth moves at about $29.8\ \mathrm{km/s}$ ⁠, $\sqrt{30} \approx 5.5$ ⁠ times faster, as $v = \sqrt{GM/r}$ ⁠ predicts.
Step 5 of 7
Open this step
Light on its way
Practice · Calculate
Question
Light travels at $3.00 \times 10^8$ ⁠ m/s, and $1\ \mathrm{AU} = 1.496 \times 10^{11}\ \mathrm{m}$ ⁠. How many hours does sunlight take to reach Neptune, $30.07$ ⁠ AU away?

Correct answer
4.168 h ± 2%
Setup, choices and feedback
The Scene at this step
Speed
10 days/s
Show orbits
ondefault
True distances
True distances
Other settings
Planet/nodes/system/props/planet = "Neptune"
Accepted within
± 2%
Computed from the Scene at this step’s setting
Time sunlight takes to reach it/nodes/system/outputs/lightTime
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About $4.2$ ⁠ hours. We see Neptune as it was more than four hours ago.

Hint
$t = \dfrac{d}{c}$ ⁠, then divide by $3600$ ⁠.

Worked solution
1. The distance: $d = 30.07 \cdot 1.496 \times 10^{11} \approx 4.50 \times 10^{12}\ \mathrm{m}$ ⁠.
2. The time: $t = \dfrac{d}{c} = \dfrac{4.50 \times 10^{12}}{3.00 \times 10^8} \approx 1.50 \times 10^4\ \mathrm{s}$ ⁠.
3. $\dfrac{1.50 \times 10^4}{3600} \approx 4.2$ ⁠ hours.
Light reaches the Earth in about $8.3$ ⁠ minutes. A command to a probe near Neptune takes more than four hours to arrive, so such probes work on their own.
Step 6 of 7
Open this step
Saturn’s orbit
Practice · Calculate
Question
Saturn goes round the Sun once every 29.46 years. By Kepler’s third law, $T^2 = a^3$ ⁠ with $T$ ⁠ in years and $a$ ⁠ in astronomical units, how far is it from the Sun on average?

Correct answer
9.537 AU ± 2%
Setup, choices and feedback
The Scene at this step
Speed
30 days/s
Show orbits
ondefault
True distances
True distances
Other settings
Planet/nodes/system/props/planet = "Saturn"
Accepted within
± 2%
Computed from the Scene at this step’s setting
The picked planet’s mean distance from the Sun (semi-major axis)/nodes/system/outputs/distance
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 29.46 AU ± 2%
  Misconception: Thinks the orbital period grows in proportion to the orbit’s size: twice as far, twice as long.
  Feedback from the catalogue
  By Kepler’s third law, $T^2 = a^3$ ⁠ ( $T$ ⁠ in years, $a$ ⁠ in AU), so $T = a^{3/2}$ ⁠: the time grows faster than the distance. A planet four times as far takes $4^{1.5} = 8$ ⁠ times as long. There are two reasons: its path is longer, and it also moves more slowly, because the Sun’s gravity is weaker out there.
Feedback for a right answer
About 9.5 AU: an orbit 9.5 times the Earth’s, and a year 29.5 times as long.

Hint
Turn the law round: $a = T^{2/3}$ ⁠. Square $T$ ⁠, then take the cube root.

Worked solution
1. $T^2 = 29.46^2 \approx 868$ ⁠.
2. $a^3 = 868$ ⁠, so $a = \sqrt[3]{868} \approx 9.54$ ⁠.
3. Saturn is about 9.5 AU from the Sun, a little less than ten times as far as the Earth.
If you answered 29.5 AU, you assumed the year grows in step with the orbit. The year grows faster, because a far planet is also slower.
Step 7 of 7
Open this step
The same number for every planet
Practice · Calculate
Question
A year on Mars lasts 687 days, and it is on average 1.524 AU from the Sun. Work out $\dfrac{T^2}{a^3}$ ⁠ for it, with $T$ ⁠ in Earth years and $a$ ⁠ in astronomical units.

Correct answer
0.9995 ± 0.02
Setup, choices and feedback
The Scene at this step
Speed
30 days/s
Show orbits
ondefault
True distances
True distances
Other settings
Planet/nodes/system/props/planet = "Mars"
Accepted within
± 0.02
Computed from the Scene at this step’s setting
T² / a³ in years² per AU³: 1 for every planet (Kepler’s third law)/nodes/system/outputs/keplerRatio
Wrong answers it recognises
- 1.233 ± 0.02
  Misconception: Thinks the orbital period grows in proportion to the orbit’s size: twice as far, twice as long.
  Feedback from the catalogue
  By Kepler’s third law, $T^2 = a^3$ ⁠ ( $T$ ⁠ in years, $a$ ⁠ in AU), so $T = a^{3/2}$ ⁠: the time grows faster than the distance. A planet four times as far takes $4^{1.5} = 8$ ⁠ times as long. There are two reasons: its path is longer, and it also moves more slowly, because the Sun’s gravity is weaker out there.
Feedback for a right answer
Almost exactly $1$ ⁠, as for the Earth, Jupiter and every other planet going round the Sun.

Hint
First turn the year into Earth years: divide by 365.25.

Worked solution
1. $T = \dfrac{687}{365.25} \approx 1.881$ ⁠ years, so $T^2 \approx 3.538$ ⁠.
2. $a^3 = 1.524^3 \approx 3.540$ ⁠.
3. $\dfrac{T^2}{a^3} = \dfrac{3.538}{3.540} \approx 1.00$ ⁠.
The same ratio, $1$ ⁠, comes out for every planet: that is Kepler’s third law. If you got about $1.23$ ⁠, you divided $T$ ⁠ by $a$ ⁠ instead of $T^2$ ⁠ by $a^3$ ⁠.

Lesson 3 of 4 · About 35 minutes

Comets and asteroids

A long, thin orbit round the Sun: how long one lap takes, why the comet races near the Sun and crawls far from it, and why its tail always points away from the Sun.

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The 9 steps and their answers

Step 1 of 9
Open this step
A long ellipse
Practice · Calculate
Question
The comet comes within $0.6$ ⁠ AU of the Sun and swings out to $30$ ⁠ AU, Neptune’s distance. By Kepler’s third law, how many years does one orbit take?

Correct answer
59.85 yr ± 2%
Setup, choices and feedback
The Scene at this step
Closest distance to the Sun
0.6 AUfixed
Made of
Ice and dust (comet)
Along the orbit
0.44 (Falling in → Heading out)
Accepted within
± 2%
Computed from the Scene at this step’s setting
Time for one orbit, Kepler’s third law/nodes/comet/outputs/period
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 164.3 yr ± 2%
  Misconception: Uses the farthest distance from the Sun in Kepler’s third law, instead of the semi-major axis.
  Feedback from the catalogue
  In Kepler’s third law, $a$ ⁠ is the ellipse’s semi-major axis: the average of the closest and farthest distances, $a = \dfrac{r_{\min} + r_{\max}}{2}$ ⁠. The farthest distance alone describes an ellipse that is too big, and so a time that is too long.
Feedback for a right answer
About $60$ ⁠ years, nearly all of them far out: the comet spends only a few months near the Sun.

Hint
First find the semi-major axis, the average of the two distances.

Worked solution
1. The semi-major axis: $a = \dfrac{0.6 + 30}{2} = 15.3$ ⁠ AU.
2. $T^2 = a^3 = 15.3^3 \approx 3582$ ⁠.
3. $T = \sqrt{3582} \approx 60$ ⁠ years.
Halley’s comet comes within $0.59$ ⁠ AU and goes out to $35$ ⁠ AU, so it returns every $76$ ⁠ years or so. Answering about $164$ ⁠ puts in the farthest distance instead of the semi-major axis.
Step 2 of 9
Open this step
Fast near the Sun, slow far out
Practice · Calculate
Question
By Kepler’s second law, at the closest and farthest points $v_{\min} r_{\max} = v_{\max} r_{\min}$ ⁠. The comet comes within $0.6$ ⁠ AU and goes out to $30$ ⁠ AU. How many times faster is it near the Sun than at its farthest?

Correct answer
50 ± 2%
Setup, choices and feedback
The Scene at this step
Closest distance to the Sun
0.6 AUfixed
Made of
Ice and dust (comet)
Along the orbit
0.3 (Falling in → Heading out)
Accepted within
± 2%
Computed from the Scene at this step’s setting
Perihelion speed over aphelion speed, aphelion / perihelion/nodes/comet/outputs/speedRatio
Wrong answers it recognises
- 1 ± 0.2
  Misconception: Thinks a comet moves at the same speed all the way round its orbit.
  Feedback from the catalogue
  A body on an elliptical orbit speeds up as it falls towards the Sun and slows as it climbs away, like a ball thrown upwards. By Kepler’s second law, the line from the Sun to the body sweeps equal areas in equal times, so at the closest and farthest points $v_{\min} r_{\max} = v_{\max} r_{\min}$ ⁠. On a very long orbit the difference is huge.
Feedback for a right answer
$50$ ⁠ times: the line from the Sun to the comet is very short near the Sun, so it has to move fast to sweep the same area.

Hint
$\dfrac{v_{\max}}{v_{\min}} = \dfrac{r_{\max}}{r_{\min}}$ ⁠.

Worked solution
1. $\dfrac{v_{\max}}{v_{\min}} = \dfrac{r_{\max}}{r_{\min}} = \dfrac{30}{0.6} = 50$ ⁠.
Near the Sun the comet flies at about 54 km/s, and at the far end it crawls at about 1 km/s. So it spends nearly the whole orbit far from the Sun. Drag Along the orbit and watch it whip round the Sun.
Step 3 of 9
Open this step
Two tails
Practice · Predict, then look
Question
The comet is now passing the Sun. Soon it will start heading away from it. Where will the bluish ion tail point when the comet is on its way out?

Correct answer
- Correct Away from the Sun, so ahead of the comet
Setup, choices and feedback
The Scene at this step
Closest distance to the Sun
0.6 AU
Made of
Ice and dust (comet)
Along the orbit
0.44 (Falling in → Heading out)
Choices
- Correct Away from the Sun, so ahead of the comet
- Wrong Behind the comet, along the path it came by
  Misconception: Thinks a comet’s tail always streams out behind it, like smoke behind a moving train.
  Feedback from the catalogue
  A comet’s tail is not dragged behind it, because there is no air in space to leave it behind. The ion tail is pushed by the solar wind, a stream of charged particles flowing out of the Sun in every direction, so it always points away from the Sun. When the comet is heading away from the Sun, its tail goes ahead of it.
- Wrong Towards the Sun, because gravity pulls it
  Feedback
  Gravity does pull the gas, but the solar wind and the pressure of light push it outwards much harder.
Once the prediction is locked in, the Scene changes
Along the orbit
0.8 (Falling in → Heading out)
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Away from the Sun. The comet heads out tail first.

Worked solution
1. The ion tail is made of charged gas pushed by the solar wind, which streams out of the Sun in every direction.
2. So it always points roughly away from the Sun, whichever way the comet is moving.
3. When the comet is heading away from the Sun, the tail goes ahead of it. The yellowish dust tail curves back a little, because its grains follow orbits of their own.
Step 4 of 9
Open this step
Too far for a tail
Practice · Calculate
Question
Closest distance to the Sun is locked at $2.2$ ⁠ AU, beyond Mars, and even at its closest the comet grows only a faint coma and hardly any tail. At the Earth’s distance, $1$ ⁠ AU, each square metre facing the Sun gets about 1361 W. How much sunlight does each square metre of the ice get at the comet’s closest point, in W/m²?

Correct answer
281.2 W/m² ± 3%
Setup, choices and feedback
The Scene at this step
Closest distance to the Sun
2.2 AUfixed
Made of
Ice and dust (comet)
Along the orbit
0.5 (Falling in → Heading out)
Accepted within
± 3%
Computed from the Scene at this step’s setting
Sunlight on each square metre at the closest point to the Sun: the 1361 W/m² the Earth gets at 1 AU, falling as 1/r²/nodes/comet/outputs/perihelionSunlight
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 619 W/m² ± 3%
  Misconception: Thinks sunlight weakens in proportion to the distance, so twice as far gets half the light.
  Feedback from the catalogue
  Sunlight spreads over a sphere that keeps growing. A sphere’s surface grows like $r^2$ ⁠, so the same light is shared over $r^2$ ⁠ times the area: at twice the distance a quarter of the light arrives, not half. It is the inverse-square law, as for gravity.
Feedback for a right answer
About 281 W/m², $4.84$ ⁠ times less than at the Earth. Water ice barely evaporates in such weak warmth, so there is no gas to make a tail.

Hint
Sunlight weakens like $\dfrac{1}{r^2}$ ⁠: work out how many times weaker it is there, and divide $1361$ ⁠ by that.

Worked solution
1. The Sun’s light spreads over a sphere of area $4\pi r^2$ ⁠, so its intensity goes like $\dfrac{1}{r^2}$ ⁠.
2. $\left(\dfrac{2.2}{1}\right)^2 = 4.84$ ⁠: the light there is $4.84$ ⁠ times weaker.
3. $\dfrac{1361}{4.84} \approx 281$ ⁠ watts a square metre.
Almost five times less light. Beyond about $2.5$ ⁠ AU nearly all the ice stays frozen, so a distant comet looks like an asteroid. Answering about $619$ ⁠ divides by $2.2$ ⁠ alone, assuming the light weakens in proportion to the distance.
Step 5 of 9
Open this step
Rock and metal
Practice · Choose
Question
Made of is now an asteroid, on exactly the same orbit, and it passes the Sun with no coma and no tail. Why?

Correct answer
- Correct It is made of rock and metal, with almost no ice to turn into gas
Setup, choices and feedback
The Scene at this step
Closest distance to the Sun
0.6 AU
Made of
Rock and metal (asteroid)
Along the orbit
0.5 (Falling in → Heading out)
Choices
Shown in a random order.
- Correct It is made of rock and metal, with almost no ice to turn into gas
- Wrong It moves too slowly to leave a tail behind
  Misconception: Thinks a comet’s tail always streams out behind it, like smoke behind a moving train.
  Feedback from the catalogue
  A comet’s tail is not dragged behind it, because there is no air in space to leave it behind. The ion tail is pushed by the solar wind, a stream of charged particles flowing out of the Sun in every direction, so it always points away from the Sun. When the comet is heading away from the Sun, its tail goes ahead of it.
- Wrong It is too far from the Sun
  Feedback
  It is exactly the same orbit, at the same distance from the Sun.
- Wrong The solar wind does not reach it
  Feedback
  The solar wind blows through the whole Solar System. It simply has nothing to push.
Feedback for a right answer
Right. Comets formed in the cold beyond Neptune, asteroids closer to the Sun, where ice did not survive.

Worked solution
1. A comet’s tail is made of material the Sun evaporates: ice turning to gas and freeing the dust trapped in it.
2. Asteroids formed close to the Sun, in the warm part of the cloud the Solar System was born from, where ice did not survive. They are made of rock and metal.
3. With no ice there is no gas, and with no gas there is no coma and no tail, even on the same orbit at the same speed.
Step 6 of 9
Open this step
How stretched the orbit is
Practice · Calculate
Question
The comet comes within 0.6 AU of the Sun and swings out to 30 AU. What is the eccentricity of its orbit, $e = \dfrac{r_{\max} - r_{\min}}{r_{\max} + r_{\min}}$ ⁠?

Correct answer
0.9608 ± 0.01
Setup, choices and feedback
The Scene at this step
Closest distance to the Sun
0.6 AUfixed
Made of
Ice and dust (comet)
Along the orbit
0.44 (Falling in → Heading out)
Accepted within
± 0.01
Computed from the Scene at this step’s setting
How stretched the ellipse is, 0 a circle to 1 a parabola/nodes/comet/outputs/eccentricity
Feedback for a right answer
About $0.96$ ⁠: a long, narrow ellipse, close to the limit of $1$ ⁠.

Hint
The eccentricity is $0$ ⁠ for a circle and nears $1$ ⁠ as the ellipse grows longer.

Worked solution
1. $r_{\max} - r_{\min} = 30 - 0.6 = 29.4$ ⁠.
2. $r_{\max} + r_{\min} = 30.6$ ⁠.
3. $e = \dfrac{29.4}{30.6} \approx 0.96$ ⁠.
The Earth’s orbit has $e \approx 0.017$ ⁠, nearly a circle. The comet’s orbit is very stretched, which is why its speed and distance change so much.
Step 7 of 9
Open this step
The average distance
Practice · Calculate
Question
Closest distance to the Sun is locked at 2.2 AU, and the comet still swings out to 30 AU. What is the semi-major axis of its orbit, $a$ ⁠, the distance that sets its year?

Correct answer
16.1 AU ± 2%
Setup, choices and feedback
The Scene at this step
Closest distance to the Sun
2.2 AUfixed
Made of
Ice and dust (comet)
Along the orbit
0.5 (Falling in → Heading out)
Accepted within
± 2%
Computed from the Scene at this step’s setting
Half the long axis of the orbit, (perihelion + aphelion) / 2/nodes/comet/outputs/semiMajorAxis
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 30 AU ± 2%
  Misconception: Uses the farthest distance from the Sun in Kepler’s third law, instead of the semi-major axis.
  Feedback from the catalogue
  In Kepler’s third law, $a$ ⁠ is the ellipse’s semi-major axis: the average of the closest and farthest distances, $a = \dfrac{r_{\min} + r_{\max}}{2}$ ⁠. The farthest distance alone describes an ellipse that is too big, and so a time that is too long.
Feedback for a right answer
16.1 AU: the average of the nearest and farthest distances. By $T^2 = a^3$ ⁠ the year is about $65$ ⁠ years.

Hint
The long axis runs from the point nearest the Sun, through the Sun, to the farthest point.

Worked solution
1. The long axis: $2.2 + 30 = 32.2$ ⁠ astronomical units.
2. Half of it: $a = \dfrac{32.2}{2} = 16.1$ ⁠.
If you answered 30 AU, you took the farthest point instead of the average of the two ends.
Step 8 of 9
Open this step
At its fastest
Practice · Calculate
Question
On an orbit round the Sun the speed at distance $r$ ⁠ is $v = 29.8\sqrt{\dfrac{2}{r} - \dfrac{1}{a}}$ ⁠ kilometres per second, with $r$ ⁠ and $a$ ⁠ in astronomical units. The comet comes within 0.6 AU and swings out to 30 AU. How fast does it move at its closest to the Sun?

Correct answer
53.84 km/s ± 3%
Setup, choices and feedback
The Scene at this step
Closest distance to the Sun
0.6 AUfixed
Made of
Ice and dust (comet)
Along the orbit
0.5 (Falling in → Heading out)
Accepted within
± 3%
Computed from the Scene at this step’s setting
Speed at the closest point to the Sun/nodes/comet/outputs/perihelionSpeed
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 29.8 km/s ± 3%
  Misconception: Thinks a comet moves at the same speed all the way round its orbit.
  Feedback from the catalogue
  A body on an elliptical orbit speeds up as it falls towards the Sun and slows as it climbs away, like a ball thrown upwards. By Kepler’s second law, the line from the Sun to the body sweeps equal areas in equal times, so at the closest and farthest points $v_{\min} r_{\max} = v_{\max} r_{\min}$ ⁠. On a very long orbit the difference is huge.
Feedback for a right answer
About 54 km/s, almost twice the Earth’s. At the far end it crawls at less than 1.1 km/s.

Hint
First find $a$ ⁠, the semi-major axis: the average of the two distances.

Worked solution
1. $a = \dfrac{0.6 + 30}{2} = 15.3$ ⁠.
2. $\dfrac{2}{0.6} - \dfrac{1}{15.3} \approx 3.333 - 0.065 = 3.268$ ⁠.
3. $v = 29.8\sqrt{3.268} \approx 29.8 \times 1.808 \approx 53.9$ ⁠ kilometres per second.
That is almost twice the Earth’s speed, 29.8 km/s. The comet has gathered speed all the way in from 30 AU.
Step 9 of 9
Open this step
At its slowest
Practice · Calculate
Question
Closest distance to the Sun is locked at 2.2 AU, where the comet moves at 27.4 km/s. It swings out to 30 AU. By Kepler’s second law, $v_{\min} r_{\max} = v_{\max} r_{\min}$ ⁠, how fast does it move at the farthest point?

Correct answer
2.01 km/s ± 3%
Setup, choices and feedback
The Scene at this step
Closest distance to the Sun
2.2 AUfixed
Made of
Ice and dust (comet)
Along the orbit
0.9 (Falling in → Heading out)
Accepted within
± 3%
Computed from the Scene at this step’s setting
Speed at the farthest point from the Sun/nodes/comet/outputs/aphelionSpeed
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 27.4 km/s ± 3%
  Misconception: Thinks a comet moves at the same speed all the way round its orbit.
  Feedback from the catalogue
  A body on an elliptical orbit speeds up as it falls towards the Sun and slows as it climbs away, like a ball thrown upwards. By Kepler’s second law, the line from the Sun to the body sweeps equal areas in equal times, so at the closest and farthest points $v_{\min} r_{\max} = v_{\max} r_{\min}$ ⁠. On a very long orbit the difference is huge.
Feedback for a right answer
About 2.0 km/s: 13.6 times as far, and 13.6 times as slow.

Hint
At the nearest and farthest points the velocity is square to the line to the Sun, so speed times distance is the same at both.

Worked solution
1. $v_{\min} = v_{\max} \times \dfrac{r_{\min}}{r_{\max}}$ ⁠.
2. $v_{\min} = 27.4 \times \dfrac{2.2}{30} \approx 2.0$ ⁠ kilometres per second.
The comet is 13.6 times as far, so 13.6 times as slow. It spends most of its year out there, at the far end.

Lesson 4 of 4 · About 35 minutes

Planets round other stars

How to find a planet you cannot see: the shadow it casts on its star’s light. How deep the dip is, how often it comes back, and why the orbit has to be seen almost edge-on.

Open the lesson

The 9 steps and their answers

Step 1 of 9
Open this step
A shadow too small to see
Practice · Calculate
Question
Planet size is locked at $11$ ⁠ Earth radii, like Jupiter, and the star is like the Sun, $109$ ⁠ Earth radii across its radius. What percentage of the starlight does the planet block while it is wholly in front of the disc?

Correct answer
1.015 ± 3%
Setup, choices and feedback
The Scene at this step
Planet size
11 Earth radiifixed
Orbit size
0.05 AU
Orbit tilt
90 °
View from above
Telescope view
Accepted within
± 3%
Computed from the Scene at this step’s setting
Per cent of the starlight the planet blocks while wholly on the disc, 100 (Rp/R*)²/nodes/system/outputs/depth
Wrong answers it recognises
- 10.1 ± 3%
  Misconception: Thinks the dip in a star’s light is the planet’s radius over the star’s, not the share of the disc’s area it covers.
  Feedback from the catalogue
  A planet blocks light according to the area it covers on the star’s disc, and a circle’s area grows as the square of its radius. So the depth is $\left(\dfrac{R_p}{R_*}\right)^2$ ⁠, not $\dfrac{R_p}{R_*}$ ⁠. A planet with a tenth of the star’s radius blocks a hundredth of the light, $1\%$ ⁠, not $10\%$ ⁠.
Feedback for a right answer
About $1\%$ ⁠: Jupiter blocks a hundredth of the Sun’s light.

Hint
Compare the areas of discs: $\left(\dfrac{R_p}{R_*}\right)^2$ ⁠.

Worked solution
1. The ratio of the radii: $\dfrac{R_p}{R_*} = \dfrac{11}{109} \approx 0.101$ ⁠.
2. The share hidden is the ratio of the areas: $0.101^2 \approx 0.0102$ ⁠.
3. As a percentage: about $1.0\%$ ⁠.
Telescopes such as Kepler and TESS measure a star’s light to thousandths of a per cent, so they see a dip like this easily. Answering about $10\%$ ⁠ compares radii, not areas.
Step 2 of 9
Open this step
An Earth-sized planet
Practice · Predict, then look
Question
The Earth’s radius is $11$ ⁠ times smaller than Jupiter’s. When you choose Earth for the planet size, by how many times will the dip in the light shrink?

Correct answer
- Correct By about $120$ ⁠ times
Setup, choices and feedback
The Scene at this step
Planet size
11 Earth radii
Orbit size
0.05 AU
Orbit tilt
90 °
View from above
Telescope view
Choices
- Correct By about $120$ ⁠ times
- Wrong By $11$ ⁠ times
  Misconception: Thinks the dip in a star’s light is the planet’s radius over the star’s, not the share of the disc’s area it covers.
  Feedback from the catalogue
  A planet blocks light according to the area it covers on the star’s disc, and a circle’s area grows as the square of its radius. So the depth is $\left(\dfrac{R_p}{R_*}\right)^2$ ⁠, not $\dfrac{R_p}{R_*}$ ⁠. A planet with a tenth of the star’s radius blocks a hundredth of the light, $1\%$ ⁠, not $10\%$ ⁠.
- Wrong By about $1300$ ⁠ times, like the volume
  Feedback
  The star sees only the planet’s silhouette, a disc, not its volume.
Once the prediction is locked in, the Scene changes
Planet size
1 Earth radii
The Scene then shows: 0.008386 · Per cent of the starlight the planet blocks while wholly on the disc, 100 (Rp/R*)²/nodes/system/outputs/depth
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About $121$ ⁠ times: from one per cent to about $84$ ⁠ parts per million.

Worked solution
1. The depth depends on the area of the silhouette: $\left(\dfrac{R_p}{R_*}\right)^2$ ⁠.
2. A radius $11$ ⁠ times smaller gives an area $11^2 = 121$ ⁠ times smaller.
3. Instead of about $1\%$ ⁠, the Earth blocks about $0.0084\%$ ⁠, $84$ ⁠ parts per million.
That is why gas giants are easy to find and Earth-sized planets very hard.
Step 3 of 9
Open this step
How often
Practice · Calculate
Question
Orbit size is locked at $0.05$ ⁠ AU, a “hot Jupiter” hugging a Sun-like star. By $T^2 = a^3$ ⁠ ( $T$ ⁠ in years, $a$ ⁠ in AU), how many days does a year on this planet last?

Correct answer
4.084 d ± 2%
Setup, choices and feedback
The Scene at this step
Planet size
11 Earth radii
Orbit size
0.05 AUfixed
Orbit tilt
90 °
View from above
From above
Accepted within
± 2%
Computed from the Scene at this step’s setting
Orbital period, from Kepler’s third law/nodes/system/outputs/period
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 18.26 d ± 2%
  Misconception: Thinks the orbital period grows in proportion to the orbit’s size: twice as far, twice as long.
  Feedback from the catalogue
  By Kepler’s third law, $T^2 = a^3$ ⁠ ( $T$ ⁠ in years, $a$ ⁠ in AU), so $T = a^{3/2}$ ⁠: the time grows faster than the distance. A planet four times as far takes $4^{1.5} = 8$ ⁠ times as long. There are two reasons: its path is longer, and it also moves more slowly, because the Sun’s gravity is weaker out there.
Feedback for a right answer
About $4.1$ ⁠ days. The first one found round a Sun-like star, 51 Pegasi b, goes round in about $4.2$ ⁠ days.

Hint
$0.05^{3/2} = 0.05 \cdot \sqrt{0.05}$ ⁠ years, and a year has $365.25$ ⁠ days.

Worked solution
1. $T = a^{3/2} = 0.05 \cdot \sqrt{0.05} \approx 0.05 \cdot 0.224 = 0.0112$ ⁠ years.
2. $0.0112 \cdot 365.25 \approx 4.1$ ⁠ days.
The dip comes back every four days, so a few weeks of watching show it again and again. Answering about $18$ ⁠ days assumes the time shrinks like the distance.
Step 4 of 9
Open this step
Lining up
Practice · Predict, then look
Question
Orbit size is $1$ ⁠ AU, like the Earth’s, and the orbit is seen exactly edge-on. What will happen to the dip when Orbit tilt changes by just $1$ ⁠ degree, to $89$ ⁠ degrees?

Correct answer
- Correct It will vanish: the planet passes above the star
Setup, choices and feedback
The Scene at this step
Planet size
11 Earth radii
Orbit size
1 AU
Orbit tilt
90 °
View from above
Telescope view
Choices
- Correct It will vanish: the planet passes above the star
- Wrong Hardly anything: one degree is very little
  Feedback
  One degree at $1$ ⁠ AU is about $2.6$ ⁠ million kilometres, almost four times the Sun’s radius.
- Wrong It will be half as deep
  Feedback
  As long as the planet is wholly on the disc, it blocks the same area. Here it misses the disc altogether.
Once the prediction is locked in, the Scene changes
Orbit tilt
89 °
The Scene then shows: 0 · 1 if the planet crosses the star’s disc as seen from Earth, else 0/nodes/system/outputs/transits
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It vanishes. Far out, even a tiny tilt carries the planet above the star.

Worked solution
1. The planet passes the star’s centre at a height of $a \cos i$ ⁠.
2. $a = 1\ \mathrm{AU} \approx 1.5 \times 10^8\ \mathrm{km}$ ⁠ and $\cos 89^\circ \approx 0.0175$ ⁠, so the height is about $2.6 \times 10^6\ \mathrm{km}$ ⁠.
3. The Sun’s radius is about $7 \times 10^5\ \mathrm{km}$ ⁠, so the planet passes almost four radii above the disc, and there is no dip.
The chance that an Earth-sized orbit, tilted at random, shows a transit is about $\dfrac{R_*}{a} \approx 0.5\%$ ⁠.
Step 5 of 9
Open this step
Close to the star
Practice · Reach the target
Question
Orbit tilt is locked at $89$ ⁠ degrees, and the planet at $1$ ⁠ AU misses the star. Change Orbit size until there is a transit again, and press Check.

Goal
- 1 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Planet size
11 Earth radii
Orbit size
1 AU
Orbit tilt
89 °fixed
View from above
Telescope view
Read from the Scene at
1 if the planet crosses the star’s disc as seen from Earth, else 0/nodes/system/outputs/transits
One setting that reaches it
Planet size
0.5 Earth radii
Orbit size
0.02 AU
Feedback for a right answer
A transit. An orbit smaller than about $0.29$ ⁠ AU crosses the disc even at a tilt of $89$ ⁠ degrees.

Feedback for a wrong answer
Not yet. The planet still passes above the star: try a smaller orbit.

Hint
The planet passes $a \cos i$ ⁠ above the star’s centre. What must happen to $a$ ⁠?

Worked solution
1. There is a transit when the planet passes at a height smaller than the star’s radius (plus its own): $a \cos i < R_* + R_p$ ⁠.
2. $\cos 89^\circ \approx 0.0175$ ⁠, so we need $a < \dfrac{1.1 R_*}{0.0175} \approx 63 R_*$ ⁠.
3. $63$ ⁠ solar radii is about $0.29$ ⁠ AU. Any orbit smaller than that shows a transit.
That is why transits mostly find planets hugging their stars: they cross in front over a wider range of tilts, and more often too.
Step 6 of 9
Open this step
The size from the dip
Practice · Calculate
Question
Planet size is hidden. While the planet is wholly in front of the star, the light drops by 1.0%. What is the ratio of the planet’s radius to the star’s, $\dfrac{R_p}{R_*}$ ⁠?

Correct answer
0.09982 ± 3%
Setup, choices and feedback
The Scene at this step
Planet size
10.9 Earth radiihidden
Orbit size
0.05 AU
Orbit tilt
90 °
View from above
Telescope view
Accepted within
± 3%
Computed from the Scene at this step’s setting
The planet’s radius over the star’s, Rp/R*/nodes/system/outputs/radiusRatio
Wrong answers it recognises
- 0.01 ± 5%
  Misconception: Thinks the dip in a star’s light is the planet’s radius over the star’s, not the share of the disc’s area it covers.
  Feedback from the catalogue
  A planet blocks light according to the area it covers on the star’s disc, and a circle’s area grows as the square of its radius. So the depth is $\left(\dfrac{R_p}{R_*}\right)^2$ ⁠, not $\dfrac{R_p}{R_*}$ ⁠. A planet with a tenth of the star’s radius blocks a hundredth of the light, $1\%$ ⁠, not $10\%$ ⁠.
Feedback for a right answer
About $0.10$ ⁠: a one per cent drop means a planet with a tenth of the star’s radius.

Hint
The drop in the light is the share of the disc the planet covers: $\left(\dfrac{R_p}{R_*}\right)^2$ ⁠.

Worked solution
1. $\left(\dfrac{R_p}{R_*}\right)^2 = 0.010$ ⁠.
2. $\dfrac{R_p}{R_*} = \sqrt{0.010} = 0.10$ ⁠.
3. The planet’s radius is a tenth of the star’s: round a star like the Sun, that is a planet the size of Jupiter.
If you answered $0.01$ ⁠, you forgot that the drop goes with the area, the square of the ratio of the radii.
Step 7 of 9
Open this step
How long it lasts
Practice · Calculate
Question
A “hot Jupiter” 11 Earth radii across its radius, about 70,000 km, circles a star like the Sun, 695,700 km in radius, at 0.05 AU, and its year lasts 4.08 days. The orbit is seen exactly edge-on. How many hours does the transit last, from first contact to last?

Correct answer
3.2 h ± 5%
Setup, choices and feedback
The Scene at this step
Planet size
11 Earth radiifixed
Orbit size
0.05 AUfixed
Orbit tilt
90 °fixed
View from above
Telescope view
Accepted within
± 5%
Computed from the Scene at this step’s setting
How long the transit lasts, first to last contact; 0 when the planet misses the disc/nodes/system/outputs/duration
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 2.91 h ± 3%
  Misconception: Thinks a transit lasts as long as the planet takes to cross the star’s diameter, forgetting the planet’s own width.
  Feedback from the catalogue
  A transit begins when the planet’s leading edge touches the disc, and ends when its trailing edge leaves it. So the path is the star’s diameter plus the planet’s. For Jupiter in front of the Sun that adds about a tenth to the time.
Feedback for a right answer
About 3.2 hours. The bigger the orbit, the slower the planet and the longer the transit, but the rarer it is.

Hint
Find the planet’s speed along its orbit: the orbit’s circumference over the year. At first contact its leading edge touches the disc, and at last contact its trailing edge leaves it. 1 AU = 1.496 × 10⁸ km.

Worked solution
1. The orbit’s circumference: $2\pi \times 0.05 \times 1.496 \times 10^8 \approx 4.70 \times 10^7$ ⁠ kilometres.
2. The year: $4.08 \times 24 \approx 97.9$ ⁠ hours, so the speed is about $\dfrac{4.70 \times 10^7}{97.9} \approx 4.80 \times 10^5$ ⁠ kilometres per hour.
3. From first contact to last the planet travels the star’s diameter plus its own: $2 \times 695\,700 + 2 \times 70\,000 \approx 1.53 \times 10^6$ ⁠ kilometres.
4. The time: $\dfrac{1.53 \times 10^6}{4.80 \times 10^5} \approx 3.2$ ⁠ hours.
If you got about 2.9 hours, you counted only the star’s diameter and forgot the planet’s own way in and out.
Step 8 of 9
Open this step
Not through the middle
Practice · Reach the target
Question
Orbit size is locked at 0.05 AU. Change the Orbit tilt until the planet crosses the disc halfway between the star’s middle and its edge: an impact parameter $b = 0.5$ ⁠. Press Check.

Goal
- 0.5 ± 0.03
  At the start of the step: 6.583 × 10⁻¹⁶
Setup, choices and feedback
The Scene at this step
Planet size
11 Earth radii
Orbit size
0.05 AUfixed
Orbit tilt
90 °
View from above
Telescope view
Read from the Scene at
Impact parameter b = a cos i / R*: how far from the disc’s middle the planet crosses, in star radii/nodes/system/outputs/impact
One setting that reaches it
Planet size
0.5 Earth radii
Orbit tilt
87.2 °
Feedback for a right answer
At about 87.3°: a small tilt is enough to move the transit halfway to the edge.

Hint
$b = \dfrac{a \cos i}{R_*}$ ⁠, and 0.05 AU is about $10.75$ ⁠ solar radii. Which angle gives $\cos i \approx 0.047$ ⁠?

Worked solution
1. At 0.05 AU the orbit is $\dfrac{0.05 \times 1.496 \times 10^8}{695\,700} \approx 10.75$ ⁠ times the star’s radius.
2. $b = 10.75 \cos i = 0.5$ ⁠, so $\cos i \approx 0.0465$ ⁠.
3. $i \approx 87.3^\circ$ ⁠: less than three degrees from edge-on.
When the planet crosses far from the middle, its path across the disc is shorter, and so is the transit. The depth hardly changes.
Step 9 of 9
Open this step
What are the odds
Practice · Calculate
Question
A planet circles a star like the Sun at 1 AU, like the Earth, and its orbit is tilted at random. The chance that we see it transit is about $\dfrac{R_*}{a}$ ⁠. What is the chance, as a percentage? 1 AU = 1.496 × 10⁸ km, and the Sun’s radius is 695,700 km.

Correct answer
0.465 ± 4%
Setup, choices and feedback
The Scene at this step
Planet size
11 Earth radii
Orbit size
1 AUfixed
Orbit tilt
90 °
View from above
From above
Accepted within
± 4%
Computed from the Scene at this step’s setting
Chance that an orbit this size, tilted at random, transits: about R*/a/nodes/system/outputs/transitChance
Feedback for a right answer
About 0.47%, one in about $215$ ⁠. A hot Jupiter at 0.05 AU transits with a chance of about 9%.

Hint
Divide the star’s radius by the distance, then multiply by $100$ ⁠.

Worked solution
1. $\dfrac{R_*}{a} = \dfrac{695\,700}{1.496 \times 10^8} \approx 0.00465$ ⁠.
2. As a percentage: about $0.47$ ⁠.
3. Only about one Earth-like planet in $215$ ⁠ crosses its star as seen from us.
That is why telescopes like Kepler watched a hundred thousand stars at once: only a few of them line up with us.

Across the unit · 11 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

The Sun

From the core to the surface

Question

Correct answer

The Scene at this step

Choices

Worked solution

Dark spots

Question

Correct answer

The Scene at this step

Choices

Worked solution

An 11-year cycle

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The butterfly diagram

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A field that flips

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The Solar System

How far is far

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Jupiter’s year

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Why the far planets are slow

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Neptune’s speed

Question

Correct answer

The Scene at this step

Hint

Worked solution

Light on its way

Question

Correct answer

The Scene at this step

Hint

Worked solution

Saturn’s orbit

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The same number for every planet

Question

Correct answer

The Scene at this step

Wrong answers it recognises