Answer key · Unit 21

Atmosphere and weather.

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

Layers of the atmosphere

Rise through the air from the ground to 200 km: it cools, warms, cools again and turns hot. Measure how fast it cools, find where half the air lies, and see why each layer is warm or cold.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Not colder all the way up
Practice · Predict, then look
Question
At 11 km the probe reads −56.5 °C, and below, in the troposphere, the air cooled steadily with height. Raise the Altitude to 30 km. What will the temperature be there?

Correct answer
- Correct About the same, and even a little warmer
Setup, choices and feedback
The Scene at this step
Altitude
11 km
Latitude
45 °
Show the temperature
Air
Choices
- Wrong Much colder, about −180 °C, because the air keeps cooling at the same rate
  Misconception: Thinks the air keeps getting colder with height all the way to space.
  Feedback from the catalogue
  The air cools with height only in the troposphere, up to about 11 km. Above it the temperature stops falling and then rises again, because ozone absorbs ultraviolet light and turns it into heat. The air is warm where sunlight is absorbed: at the ground, in the ozone layer and in the thin gas at the top of the atmosphere.
- Correct About the same, and even a little warmer
- Wrong Warm, above 0 °C, because the air is nearer the Sun
  Misconception: Thinks air high up is warmer (or should be) because it is nearer the Sun.
  Feedback from the catalogue
  A few kilometres are nothing next to the 150 million kilometres to the Sun: sunlight is almost exactly as strong at every height. What sets the temperature is where the light is absorbed. Air is almost transparent to visible light, so the troposphere is heated from below, by the ground. In the stratosphere ozone absorbs ultraviolet, which is why the air there warms with height.
Once the prediction is locked in, the Scene changes
Altitude
30 km
The Scene then shows: -46.64 °C · Air temperature at the probe/nodes/air/outputs/temperature
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About −47 °C: above the tropopause the cooling stops, and in the stratosphere the air warms with height.

Worked solution
1. In the troposphere the air cools about 6.5 °C per kilometre, and at that rate it would reach about −180 °C at 30 km.
2. But above the tropopause, at 11 km, the cooling stops: up to about 20 km the temperature holds steady.
3. Above 20 km it rises again, and at 30 km it is about −47 °C, a little warmer than at 11 km.
This is the stratosphere: a layer that warms with height, because its ozone absorbs ultraviolet light.
Step 2 of 6
Open this step
How fast the air cools
Practice · Calculate
Question
The latitude is locked at 45°. Read the temperature at the ground, at an Altitude of 0 km, and the temperature at 5 km. By how many degrees does the air cool for each kilometre of climb in the troposphere?

Correct answer
6.5 °C/km ± 3%
Setup, choices and feedback
The Scene at this step
Altitude
5 km
Latitude
45 °fixed
Show the temperature
Air
Accepted within
± 3%
Computed from the Scene at this step’s setting
How fast the air cools with height at the probe (negative where it warms)/nodes/air/outputs/lapseRate
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 9.8 °C/km ± 2%
  Misconception: Mixes up how fast a rising thermal cools with how fast the surrounding air cools with height.
  Feedback from the catalogue
  There are two different rates here. A rising bubble of air, a thermal, always cools at 9.8 °C/km, because it expands. The air around it is a given profile, which in the Scene cools at 6.5 °C/km higher up, and near the ground depends on how much the ground has warmed it. The thermal keeps rising as long as it is warmer than its surroundings.
Feedback for a right answer
6.5 °C per kilometre: that is why mountain tops are cold, even though they are nearer the Sun.

Hint
Divide the difference in temperature by the difference in height.

Worked solution
1. At the ground: 15 °C. At 5 km: −17.5 °C.
2. The difference: $15 - (-17.5) = 32.5$ ⁠ degrees.
3. Per kilometre: $\dfrac{32.5}{5} = 6.5$ ⁠, that is 6.5 °C/km.
This is the average of the standard atmosphere. A rising bubble of air cools faster, at 9.8 °C/km, but that is the bubble’s rate, not that of the air around it.
Step 3 of 6
Open this step
Where the air is
Practice · Reach the target
Question
The pressure at any height is the weight of all the air above it, so it tells how much of the atmosphere lies below the probe. Raise the Altitude until exactly half of the atmosphere’s mass is below the probe, and press Check.

Goal
- 0.5 ± 0.015
  At the start of the step: 1.11 × 10⁻¹⁶
Setup, choices and feedback
The Scene at this step
Altitude
0 km
Latitude
45 °fixed
Show the temperature
Air
Read from the Scene at
0..1 of the atmosphere’s mass below the probe/nodes/air/outputs/airBelow
One setting that reaches it
Altitude
5.5 km
Feedback for a right answer
About 5.5 km: half of all the air in the world lies lower than the top of Everest.

Feedback for a wrong answer
Not yet. If more than half the air is below the probe, lower it; if less, raise it.

Hint
Half the mass below means half the pressure at the ground: about 507 hPa.

Worked solution
1. At the ground the pressure is 101,325 Pa: the weight of all the air.
2. Half the air below the probe means half is left above it, a pressure of about 50,700 Pa.
3. That happens at about 5.5 km, lower than the top of Everest.
The air is not spread evenly up to the edge of space at 100 km: it is squeezed down under its own weight, and the pressure halves about every 5.5 km.
Step 4 of 6
Open this step
High over the tropics, low near the poles
Practice · Predict, then look
Question
Over the tropics the airliner, at 10.5 km, flies deep in the troposphere. Choose Polar under Latitude. Which layer will the plane fly in there?

Correct answer
- Correct In the stratosphere
Setup, choices and feedback
The Scene at this step
Altitude
10.5 kmfixed
Latitude
5 °
Show the temperature
Air
Choices
- Correct In the stratosphere
- Wrong Still in the troposphere, because the tropopause is at the same height everywhere
  Misconception: Thinks the tropopause, the top of the troposphere, is at the same height everywhere.
  Feedback from the catalogue
  The tropopause is high over the tropics, about 16–17 km, and low near the poles, about 8–9 km. Over the hot tropical ground air rises higher and keeps cooling for longer, so the troposphere there is deep, and its tropopause is the coldest, about −80 °C.
- Wrong In the mesosphere
  Feedback
  The mesosphere begins only at about 50 km, at every latitude.
Once the prediction is locked in, the Scene changes
Latitude
75 °
The Scene then shows: 8.5 km · Height of the tropopause/nodes/air/outputs/tropopause
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
In the stratosphere: near the poles the tropopause drops to about 8.5 km.

Worked solution
1. Over the tropics the tropopause is at about 16.5 km, so 10.5 km is deep in the troposphere.
2. Near the poles the ground is cold, air does not rise high, and the tropopause drops to about 8.5 km.
3. The plane, at 10.5 km, is above it: in the stratosphere.
That is why long flights over the poles often cruise in the calm, dry air of the stratosphere.
Step 5 of 6
Open this step
A layer that warms
Practice · Choose
Question
The Altitude is 40 km, and Show the temperature is at Temperature, showing where sunlight is absorbed. Above 20 km the air warms with height, to about 0 °C at 50 km. Why?

Correct answer
- Correct Ozone absorbs ultraviolet light from the Sun, and its energy stays in the air as heat
Setup, choices and feedback
The Scene at this step
Altitude
40 km
Latitude
45 °
Show the temperature
Temperature
Choices
Shown in a random order.
- Correct Ozone absorbs ultraviolet light from the Sun, and its energy stays in the air as heat
- Wrong The air there is nearer the Sun
  Misconception: Thinks air high up is warmer (or should be) because it is nearer the Sun.
  Feedback from the catalogue
  A few kilometres are nothing next to the 150 million kilometres to the Sun: sunlight is almost exactly as strong at every height. What sets the temperature is where the light is absorbed. Air is almost transparent to visible light, so the troposphere is heated from below, by the ground. In the stratosphere ozone absorbs ultraviolet, which is why the air there warms with height.
- Wrong Warm air from the ground rises and collects up there
  Misconception: Thinks warm air from the ground rises and collects at the top, so the upper air is warm.
  Feedback from the catalogue
  Warm air does rise, but as it rises the pressure on it falls, so it expands and cools. It does not arrive high up warm. The stratosphere’s warmth does not come from below: it is made there, as ozone absorbs ultraviolet light from the Sun.
Feedback for a right answer
Ozone: the air is warm where sunlight is absorbed.

Worked solution
1. Sunlight passes through most of the air without being absorbed. It is absorbed in three places: at the ground, in the ozone layer and at the top of the atmosphere.
2. In the stratosphere ozone absorbs most of the ultraviolet, and every photon absorbed warms the air around it.
3. The top of the layer is warmest: there is little air there to heat, and the light is absorbed there first.
Being 40 km nearer the Sun changes nothing, and air rising from the ground cools on the way, it does not warm.
Step 6 of 6
Open this step
Hot, but it feels cold
Practice · Choose
Question
At an Altitude of 200 km, in the thermosphere, the probe reads about 580 °C. What would an astronaut’s hand feel there, in the shade?

Correct answer
- Correct Cold: there are so few molecules that they hardly hand over any heat, and the hand loses more by radiating
Setup, choices and feedback
The Scene at this step
Altitude
200 km
Latitude
45 °
Show the temperature
Temperature
Choices
Shown in a random order.
- Correct Cold: there are so few molecules that they hardly hand over any heat, and the hand loses more by radiating
- Wrong A burn: 580 °C is hotter than an oven
  Misconception: Thinks air at hundreds of degrees would feel hot even when it is very thin.
  Feedback from the catalogue
  Temperature measures how fast the molecules move, not how much heat they can hand over. In the thermosphere the molecules are very fast, but there are so few of them that very few hit a hand. The hand loses far more heat by radiating than it gains from them, so it cools.
- Wrong Nothing, because the temperature there is not real
  Feedback
  The temperature is real: the molecules really do move very fast. There are just very few of them.
Feedback for a right answer
Cold: the temperature is high, but there is hardly any heat to hand over.

Worked solution
1. Temperature measures how fast the molecules move. In the thermosphere they absorb the Sun’s harshest light, and few of them share it, so they are very fast.
2. But at 200 km the pressure is over a billion times lower than at the ground, and very few molecules hit the hand.
3. The hand radiates heat to space far faster than it gains it from them, so it cools.
Like sparks from a sparkler: they are very hot, but so small that they do not burn your hand.

Lesson 2 of 5 · About 25 minutes

Storms and weather

One afternoon ends with a few white puffs, another in a thunderstorm. See how the ground's heat, the moisture and the wind decide what grows: thermals, the cloud base, the latent heat that feeds a storm, and the wind shear that sets how long it lives.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Thermals
Practice · Predict, then look
Question
On a cool day, 20 °C, the thermals stop a few hundred metres above the ground. Slide Ground temperature up to 33 °C. What will happen to the thermals?

Correct answer
- Correct They will rise much higher
Setup, choices and feedback
The Scene at this step
Ground temperature
293.1 K
Dew point
291.1 K
Wind shear
14 m/s
Build the storm
Towering
Choices
- Correct They will rise much higher
- Wrong They will stop at the same height, because the air aloft has not changed
  Feedback
  The air aloft is the same, but the bubble leaving the ground is warmer, so it stays warmer than its surroundings for longer.
- Wrong They will stop lower, because hot air cools faster
  Misconception: Mixes up how fast a rising thermal cools with how fast the surrounding air cools with height.
  Feedback from the catalogue
  There are two different rates here. A rising bubble of air, a thermal, always cools at 9.8 °C/km, because it expands. The air around it is a given profile, which in the Scene cools at 6.5 °C/km higher up, and near the ground depends on how much the ground has warmed it. The thermal keeps rising as long as it is warmer than its surroundings.
Once the prediction is locked in, the Scene changes
Ground temperature
306.1 K
The Scene then shows: 13.25 km · Height where the rising air stops: the thermals or the cloud top/nodes/storm/outputs/top
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Much higher: on a hot day the air near the ground stays warmer than its surroundings for a long way up.

Worked solution
1. A thermal rises as long as it is warmer than the air around it, and cools at 9.8 °C per kilometre.
2. On a cool day the ground is no warmer than the air above it, and the bubble is soon as cold as its surroundings.
3. On a hot day the bubble leaves much warmer, and stays warmer than its surroundings to a great height. Here it reaches its dew point, and a cloud forms that climbs to the tropopause.
This temperature gradient, how fast the air cools with height, is the first ingredient of a storm.
Step 2 of 6
Open this step
The surrounding air's gradient
Practice · Calculate
Question
A hot day, 33 °C at the ground. The thermals have stirred the low air. The chart beside the Scene shows the temperature of the surrounding air with height. By how many degrees per kilometre does that air cool over the lowest 3 km?

Correct answer
8.683 °C/km ± 5%
Setup, choices and feedback
The Scene at this step
Ground temperature
306.1 Kfixed
Dew point
291.1 K
Wind shear
14 m/s
Build the storm
Towering
Accepted within
± 5%
Computed from the Scene at this step’s setting
The same lapse over the lowest 3 km, in kelvin per metre/nodes/storm/outputs/lapseRate
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 9.8 °C/km ± 2%
  Misconception: Mixes up how fast a rising thermal cools with how fast the surrounding air cools with height.
  Feedback from the catalogue
  There are two different rates here. A rising bubble of air, a thermal, always cools at 9.8 °C/km, because it expands. The air around it is a given profile, which in the Scene cools at 6.5 °C/km higher up, and near the ground depends on how much the ground has warmed it. The thermal keeps rising as long as it is warmer than its surroundings.
- 6.5 °C/km ± 2%
  Misconception: Mixes up how fast a rising thermal cools with how fast the surrounding air cools with height.
  Feedback from the catalogue
  There are two different rates here. A rising bubble of air, a thermal, always cools at 9.8 °C/km, because it expands. The air around it is a given profile, which in the Scene cools at 6.5 °C/km higher up, and near the ground depends on how much the ground has warmed it. The thermal keeps rising as long as it is warmer than its surroundings.
Feedback for a right answer
About 8.7 °C/km: the hot ground has made the low air steep, and thermals rise through it easily.

Hint
Read the temperature at the ground and at 3 km on the chart, and divide by three.

Worked solution
1. At the ground: 33 °C. At 3 km the air is about 7 °C.
2. The difference: about $26$ ⁠ degrees over $3$ ⁠ kilometres.
3. Per kilometre: $\dfrac{26}{3} \approx 8.7$ ⁠, that is about 8.7 °C/km.
That lies between the Scene’s air aloft, at 6.5 °C/km, and a rising thermal, at 9.8 °C/km. The steeper the gradient, the more readily air rises.
Step 3 of 6
Open this step
The cloud base
Practice · Reach the target
Question
The Ground temperature is locked at 31 °C. A thermal closes on its dew point by about 8 °C per kilometre, so the cloud base is about 125 m above the ground for every degree between the temperature and the dew point. Change the Dew point so that the cloud base is 1 km up, and press Check.

Goal
- 1 km ± 0.08 km
  At the start of the step: 1.723 km
Setup, choices and feedback
The Scene at this step
Ground temperature
304.1 Kfixed
Dew point
291.1 K
Wind shear
14 m/s
Build the storm
Cumulus
Read from the Scene at
Height where the thermals reach their dew point; NaN when they stop below it/nodes/storm/outputs/cloudBase
One setting that reaches it
Dew point
296.6 K
Wind shear
0 m/s
Feedback for a right answer
A cloud base at 1 km: humid air condenses low down.

Feedback for a wrong answer
Not yet. If the base is too high, raise the dew point; if too low, lower it.

Hint
$\dfrac{1000}{125} = 8$ ⁠: a dew point about 8 °C below the temperature.

Worked solution
1. The cloud base is roughly $125\ \mathrm{m}$ ⁠ times the gap between the temperature and the dew point.
2. For 1000 m the gap must be $\dfrac{1000}{125} = 8$ ⁠ degrees.
3. A dew point of about $31 - 8 = 23\ ^\circ\mathrm{C}$ ⁠. In the Scene the base reaches 1 km at a dew point of about 23–24 °C.
Humid air reaches its dew point low down, which is why clouds start close to the ground on a muggy day.
Step 4 of 6
Open this step
Without moisture
Practice · Predict, then look
Question
A hot day, 33 °C, and a dew point of 18 °C: a thunderstorm. Slide Dew point down to 5 °C, without changing the heat. What will happen?

Correct answer
- Correct The sky will stay clear, with no cloud
Setup, choices and feedback
The Scene at this step
Ground temperature
306.1 K
Dew point
291.1 K
Wind shear
14 m/s
Build the storm
Mature
Choices
- Correct The sky will stay clear, with no cloud
- Wrong Still a storm, because the heat is what drives it
  Misconception: Thinks a hot enough day is enough for a thunderstorm, whatever the moisture.
  Feedback from the catalogue
  Heat alone makes thermals, not cloud. To make cloud a thermal has to cool to its dew point, and in dry air that is very far up. And with no condensation there is no fuel for a storm either: the heat released as water vapour condenses. That is why a hot desert can stay clear while a hot, humid day builds a storm.
- Wrong A bigger storm, because dry air is lighter
  Feedback
  Humid air is in fact slightly lighter than dry air. And without vapour there is no condensation, and without condensation no cloud.
Once the prediction is locked in, the Scene changes
Dew point
278.1 K
The Scene then shows: 3.07 km · Height where the rising air stops: the thermals or the cloud top/nodes/storm/outputs/top
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Clear: the thermals stop before they reach their dew point.

Worked solution
1. With a dew point of 5 °C the gap from 33 °C is $28$ ⁠ degrees, so the cloud base would have to be at about $28 \times 125 = 3500$ ⁠ metres.
2. The thermals stop at about 3.1 km, before they get there.
3. With no condensation there is no cloud, and none of the heat that condensing releases to feed a storm.
That is how a hot desert can stay clear and bumpy, without a drop of rain.
Step 5 of 6
Open this step
The fuel
Practice · Choose
Question
Step Build the storm from Cumulus to Towering. Above the cloud base the amber beads brighten: the rising air is warmer and warmer than its surroundings. What warms it?

Correct answer
- Correct The heat released as water vapour condenses into droplets
Setup, choices and feedback
The Scene at this step
Ground temperature
304.1 K
Dew point
291.1 K
Wind shear
14 m/s
Build the storm
Cumulus
Choices
Shown in a random order.
- Correct The heat released as water vapour condenses into droplets
- Wrong Sunlight warming the white cloud
  Misconception: Thinks sunlight warming the cloud is what keeps a storm cloud rising.
  Feedback from the catalogue
  Sunlight mostly heats the ground, and that is what starts the thermals. Inside the cloud the fuel is latent heat: water vapour took in about 2.5 MJ per kilogram when it evaporated, and when it condenses in the cloud that heat goes back into the air. So cloudy air cools more slowly as it rises and stays warmer than its surroundings all the way to the tropopause.
- Wrong Friction between the droplets and the air
  Feedback
  That friction is tiny, and in any case it slows the updraft, it does not push it.
Feedback for a right answer
Latent heat: the energy the Sun put into evaporation comes back in the cloud.

Worked solution
1. Evaporating a kilogram of water takes about 2.5 MJ, which came from the Sun at the sea or a wet field.
2. When the vapour condenses in the cloud, all of that heat goes back into the air.
3. So cloudy air cools only 4–7 °C per kilometre instead of 9.8, and stays warmer than its surroundings up to the tropopause.
This is latent heat, a thunderstorm’s fuel. Forecasters measure it as CAPE.
Step 6 of 6
Open this step
Wind shear
Practice · Choose
Question
Build the storm is at An hour on. Slide Wind shear between 0, 14 and 25 m/s. Which storm is still going an hour later?

Correct answer
- Correct The one with strong shear: the updraft leans away from its rain and turns
Setup, choices and feedback
The Scene at this step
Ground temperature
304.1 K
Dew point
291.1 K
Wind shear
0 m/s
Build the storm
An hour on
Choices
Shown in a random order.
- Correct The one with strong shear: the updraft leans away from its rain and turns
- Wrong The one with no shear, because no wind tears it apart
  Misconception: Thinks wind shear always weakens a thunderstorm by blowing it apart.
  Feedback from the catalogue
  In a thunderstorm over land, wind shear actually makes it last longer. With no shear the rain falls back down through the updraft, and the cold downdraft chokes it within the hour. Moderate shear moves the gust front so that it lifts new cells, and strong shear tilts the updraft away from its own rain and sets it turning: a supercell that lives for hours.
- Wrong All the same, because they all have the same heat and moisture
  Feedback
  They all have the same fuel, but the shear decides whether the rain falls into the updraft and chokes it.
Feedback for a right answer
Strong shear: a supercell, whose rain does not choke its updraft.

Worked solution
1. With no shear the updraft stands straight. Its rain falls back down through it, and the cold downdraft chokes it: a single cell, which rains itself out within the hour.
2. With moderate shear, 14 m/s, the gust front lifts warm air at its edge, and a new cell grows as the old one dies.
3. With strong shear, 25 m/s, the updraft leans, its rain falls beside it, and it starts to turn: a supercell that lives for hours.
So wind shear, which sounds like something that would tear storms apart, is in fact what lets them live.

Lesson 3 of 5 · About 25 minutes

The Coriolis effect

Throw a ball straight across a turning carousel and it misses. On the Earth the ground turns about its upright at a rate set by the latitude, which bends winds and spins storms: right in the north, left in the south, and not at all at the equator.

Open the lesson

The 8 steps and their answers

Step 1 of 8
Open this step
A throw that misses
Practice · Choose
Question
From the sky, Earth, London. The ball is thrown from the middle of the carousel to the catcher at the rim. What is its path, seen from above?

Correct answer
- Correct A straight line: the ground, and the catcher with it, turn beneath it
Setup, choices and feedback
The Scene at this step
Rotation rate
1 turns/day
Latitude
51.5 ° N
Ride along
From the sky
Choices
Shown in a random order.
- Correct A straight line: the ground, and the catcher with it, turn beneath it
- Wrong An arc to the right, because the Coriolis force pushes it sideways
  Misconception: Thinks a real sideways force pushes the moving ball or air off course.
  Feedback from the catalogue
  From the sky the ball is seen flying in a straight line: nothing pushes it sideways. What moves is the ground, turning beneath it. Only someone who measures everything relative to the turning ground sees a curving path, and has to add a “Coriolis force” to make their equations come out. It is a fictitious force: no body exerts it.
- Wrong A straight line that lands right in the catcher’s hands
  Feedback
  The ball flies straight, but during the flight the catcher has been carried on with the ground.
Feedback for a right answer
A straight line: all that curves is the description from the turning ground.

Worked solution
1. Once the ball has left the hand nothing pushes it sideways, so from the sky it is seen flying in a straight line.
2. During the flight the carousel, and the catcher on it, turn.
3. So the ball lands where the catcher was, to their right.
Only someone turning with the ground sees the ball curve. That is the Coriolis effect: a result of describing motion from a turning platform, not a force anything exerts.
Step 2 of 8
Open this step
Right in the north, left in the south
Practice · Predict, then look
Question
Riding along: in London the ball curves and lands to the right of the catcher. Choose Cape Town, at 34° south, under Latitude. Where will the ball land?

Correct answer
- Correct To the left of the catcher
Setup, choices and feedback
The Scene at this step
Rotation rate
1 turns/day
Latitude
51.5 ° N
Ride along
Riding along
Choices
- Correct To the left of the catcher
- Wrong To the right of the catcher, as in London
  Feedback
  In the southern hemisphere the ground turns clockwise, the other way from the north, so the bend is reversed too.
- Wrong Right in the hands, because there is no effect in the south
  Feedback
  The effect vanishes only at the equator. In Cape Town it is almost as strong as in London.
Once the prediction is locked in, the Scene changes
Latitude
-33.9 ° N
The Scene then shows: 25.1 ° · How far round from the catcher a throw lands: the angle the ground turns during it/nodes/spin/outputs/deflection
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
To the left: in the south the ground turns clockwise.

Worked solution
1. In the northern hemisphere the ground turns about its upright anticlockwise, seen from above.
2. In the southern hemisphere $\sin\varphi$ ⁠ is negative: the ground turns clockwise.
3. So the catcher is carried the other way, and the ball lands to their left, about 25° from the aim.
That is why paths bend right and storms spin anticlockwise in the north, and left and clockwise in the south.
Step 3 of 8
Open this step
The sine of the latitude
Practice · Calculate
Question
The Latitude is locked at 30° north, and the Rotation rate at Earth, one turn a day. The label that prints the bend has been hidden. Each throw lasts $3$ ⁠ hours. By how many degrees does the ground turn under the ball during the throw, that is, how far round from the catcher does it land?

Correct answer
22.5 ° ± 0.5 °
Setup, choices and feedback
The Scene at this step
Rotation rate
1 turns/dayfixed
Latitude
30 ° Nfixed
Ride along
Riding along
Other settings
Labels/nodes/spin/props/labels = false
Accepted within
± 0.5 °
Computed from the Scene at this step’s setting
How far round from the catcher a throw lands: the angle the ground turns during it/nodes/spin/outputs/deflection
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 45 ° ± 0.5 °
  Misconception: Thinks the ground turns about its own upright once a day everywhere on Earth, as it does at a pole.
  Feedback from the catalogue
  Only at a pole does the ground turn about its own upright a full turn a day. At the equator it does not turn about it at all, and in between it turns at the Earth’s rate times $\sin\varphi$ ⁠, the sine of the latitude. At 30°, $\sin 30^\circ = 0.5$ ⁠: half a turn a day.
Feedback for a right answer
22.5°: half the bend at a pole.

Hint
The ground turns $\sin\varphi$ ⁠ turns a day. $3$ ⁠ hours is an eighth of a day.

Worked solution
1. The ground at 30° turns about its upright at the Earth’s rate times $\sin 30^\circ = 0.5$ ⁠: half a turn a day, $180^\circ$ ⁠.
2. $3$ ⁠ hours is $\dfrac{3}{24} = \dfrac{1}{8}$ ⁠ of a day.
3. $180^\circ \times \dfrac{1}{8} = 22.5^\circ$ ⁠.
An answer of $45^\circ$ ⁠ takes a full turn a day, as at a pole. Only there does the ground turn that fast.
Step 4 of 8
Open this step
Finding a latitude
Practice · Reach the target
Question
The Rotation rate is locked at Earth. Find a latitude in the northern hemisphere where a 3-hour throw lands 30° to the right of the catcher, and press Check.

Goal
- 30 ° ± 0.5 °
  At the start of the step: 0 °
- 1 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Rotation rate
1 turns/dayfixed
Latitude
0 ° N
Ride along
Riding along
Read from the Scene at
How far round from the catcher a throw lands: the angle the ground turns during it/nodes/spin/outputs/deflection
Read from the Scene at
Where the throw lands: 1 right of the catcher, −1 left, 0 in the catcher’s hands/nodes/spin/outputs/side
One setting that reaches it
Latitude
41 ° N
Feedback for a right answer
About 41.8° north: $\sin\varphi = \dfrac{2}{3}$ ⁠.

Feedback for a wrong answer
Not yet. If the ball lands too far from the catcher, move towards the equator; if too close, move further north.

Hint
At a pole the bend is 45°. You need $\sin\varphi = \dfrac{30}{45}$ ⁠.

Worked solution
1. At a pole a 3-hour throw lands 45° from the catcher, an eighth of a turn.
2. At latitude $\varphi$ ⁠ the bend is $45^\circ \sin\varphi$ ⁠.
3. $\sin\varphi = \dfrac{30}{45} = 0.67$ ⁠, so $\varphi \approx 41.8^\circ$ ⁠ north, about the latitude of Rome.
In the south, at 41.8° south, the ball would land the same distance off, but to the left.
Step 5 of 8
Open this step
Foucault's pendulum
Practice · Calculate
Question
A Foucault pendulum swings in a plane fixed in space, and the floor turns beneath it. The Latitude is locked at 31.8° north, Jerusalem. How many hours does the floor take to turn once under the pendulum?

Correct answer
45.42 h ± 2%
Setup, choices and feedback
The Scene at this step
Rotation rate
1 turns/dayfixed
Latitude
31.8 ° Nfixed
Ride along
From the sky
Accepted within
± 2%
Computed from the Scene at this step’s setting
Time for the ground to turn once under a Foucault pendulum; NaN where it does not turn/nodes/spin/outputs/pendulumDay
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 24 h ± 2%
  Misconception: Thinks the ground turns about its own upright once a day everywhere on Earth, as it does at a pole.
  Feedback from the catalogue
  Only at a pole does the ground turn about its own upright a full turn a day. At the equator it does not turn about it at all, and in between it turns at the Earth’s rate times $\sin\varphi$ ⁠, the sine of the latitude. At 30°, $\sin 30^\circ = 0.5$ ⁠: half a turn a day.
Feedback for a right answer
About $45.5$ ⁠ hours: almost two days.

Hint
The floor turns $\sin\varphi$ ⁠ turns a day. How many days for one turn?

Worked solution
1. At 31.8°: $\sin 31.8^\circ \approx 0.527$ ⁠.
2. The floor turns $0.527$ ⁠ of a turn a day, so a full turn takes $\dfrac{1}{0.527} \approx 1.90$ ⁠ days.
3. $1.90 \times 24 \approx 45.5$ ⁠ hours.
Only at a pole is it one day, and at the equator the pendulum does not turn at all. In Paris, where Foucault showed it in 1851, it is about $31.8$ ⁠ hours.
Step 6 of 8
Open this step
Spinning storms
Practice · Choose
Question
Step Ride along to Winds and storms. Hurricanes almost never form within about 5° of the equator, even though the sea there is very warm. Why?

Correct answer
- Correct Near the equator the ground hardly turns about its upright, so nothing sets the air drawn in spinning
Setup, choices and feedback
The Scene at this step
Rotation rate
1 turns/day
Latitude
20 ° N
Ride along
Winds and storms
Choices
Shown in a random order.
- Correct Near the equator the ground hardly turns about its upright, so nothing sets the air drawn in spinning
- Wrong The Coriolis effect is too strong there, because the ground moves fastest, and it tears the storm apart
  Misconception: Thinks the Coriolis effect is strongest at the equator, where the ground moves fastest.
  Feedback from the catalogue
  At the equator the ground does move fastest round the Earth’s axis, about 1670 km/h, but it does not turn about its own upright. The Coriolis effect comes from that turning, so it grows with $\sin\varphi$ ⁠: zero at the equator and largest at the poles. That is why hurricanes almost never form within about 5° of the equator.
- Wrong The air over the equator is too cold
  Feedback
  The air and sea near the equator are among the warmest in the world. There is heat to spare; what is missing is spin.
Feedback for a right answer
No spin: at the equator $\sin\varphi = 0$ ⁠.

Worked solution
1. A storm is a patch of low pressure that draws air in from all sides.
2. The Coriolis effect bends each stream of air sideways, so the streams miss the centre and start to swing round it.
3. The effect grows with $\sin\varphi$ ⁠: at the equator it is zero, and 5° from it $\sin 5^\circ \approx 0.09$ ⁠, too weak.
At the equator the ground moves fast round the Earth’s axis, but it does not turn about its own upright, and only that turning spins up winds.
Step 7 of 8
Open this step
The sink myth
Practice · Choose
Question
It is said that water drains from a sink one way in the north and the other way in the south, because of the Coriolis effect. What is true?

Correct answer
- Correct Not in a sink: the water leaves within seconds, and the ground hardly turns in that time
Setup, choices and feedback
The Scene at this step
Rotation rate
1 turns/day
Latitude
51.5 ° N
Ride along
From the sky
Choices
Shown in a random order.
- Correct Not in a sink: the water leaves within seconds, and the ground hardly turns in that time
- Wrong It is true, just as for hurricanes
  Misconception: Thinks sinks and toilets drain one way in the north and the other way in the south because of the Coriolis effect.
  Feedback from the catalogue
  The Coriolis effect grows with the time something keeps moving. Water leaves a sink in seconds, over centimetres, and in that time the ground hardly turns. The swirl left from filling it and the shape of the plughole are thousands of times stronger. Only things that keep moving for hours and days, such as winds and storms, feel it strongly.
- Wrong It is true only near the equator, where the effect is strong
  Misconception: Thinks the Coriolis effect is strongest at the equator, where the ground moves fastest.
  Feedback from the catalogue
  At the equator the ground does move fastest round the Earth’s axis, about 1670 km/h, but it does not turn about its own upright. The Coriolis effect comes from that turning, so it grows with $\sin\varphi$ ⁠: zero at the equator and largest at the poles. That is why hurricanes almost never form within about 5° of the equator.
Feedback for a right answer
Too short: the effect builds up only over hours and days, as in winds and storms.

Worked solution
1. The bend grows with the time something keeps moving: a 3-hour throw in the Scene bends by tens of degrees.
2. Water leaves a sink within seconds. In one second the ground turns only about $\dfrac{360^\circ}{86400} \approx 0.004^\circ$ ⁠.
3. The swirl left from filling it and the shape of the plughole are thousands of times stronger.
In a huge, perfectly still tank emptied over hours the effect can be measured. In a sink at home it cannot.
Step 8 of 8
Open this step
On Jupiter
Practice · Calculate
Question
The Rotation rate is locked at Jupiter, 2.42 turns per Earth day, and the Latitude at 30° north. How many turns a day does the ground there make about its own upright?

Correct answer
1.21 ± 2%
Setup, choices and feedback
The Scene at this step
Rotation rate
2.42 turns/dayfixed
Latitude
30 ° Nfixed
Ride along
Riding along
Accepted within
± 2%
Computed from the Scene at this step’s setting
Turns a day the ground makes about its own upright, the planet’s rate times sin(latitude); negative clockwise seen from above/nodes/spin/outputs/groundTurn
Wrong answers it recognises
- 2.42 ± 2%
  Misconception: Thinks the ground turns about its own upright once a day everywhere on Earth, as it does at a pole.
  Feedback from the catalogue
  Only at a pole does the ground turn about its own upright a full turn a day. At the equator it does not turn about it at all, and in between it turns at the Earth’s rate times $\sin\varphi$ ⁠, the sine of the latitude. At 30°, $\sin 30^\circ = 0.5$ ⁠: half a turn a day.
Feedback for a right answer
1.21 turns a day: a Foucault pendulum on Jupiter would go round in under twenty hours.

Hint
The ground turns about its upright at the planet’s rate times the sine of the latitude.

Worked solution
1. $\sin 30^\circ = 0.5$ ⁠.
2. $2.42 \times 0.5 = 1.21$ ⁠ turns a day.
Only at the pole does the ground turn about itself at the planet’s full rate. At the equator it does not turn about its upright at all, which is why there are no hurricanes there.

Lesson 4 of 5 · About 25 minutes

The ozone layer

A layer of gas that would be a few millimetres thick if squeezed down shields life from ultraviolet light. Measure how thin it is, see how ozone absorbs light without running out, and how chlorine, a catalyst, tore a hole in it over Antarctica every spring.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
A thin shield
Practice · Calculate
Question
Before CFCs, in December. One Dobson unit is a thickness of 0.01 mm of ozone squeezed to sea-level pressure. The thinnest ozone column anywhere on the globe this month is about $288$ ⁠ Dobson units. How thick would that column be, squeezed like that?

Correct answer
2.882 mm ± 3%
Setup, choices and feedback
The Scene at this step
Chlorine in the stratosphere
0.6 ppbfixed
Month
Decemberfixed
Zoom to the molecules
Globe
Accepted within
± 3%
Computed from the Scene at this step’s setting
The thinnest column squeezed to sea-level pressure: a Dobson unit is 0.01 mm/nodes/earth/outputs/minimumThickness
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 2.9 mm: a shield as thin as a coin.

Hint
Multiply the number of Dobson units by 0.01 mm.

Worked solution
1. The thinnest column: about $288$ ⁠ Dobson units.
2. Each unit is 0.01 mm.
3. $288 \times 0.01 \approx 2.9$ ⁠ millimetres.
All the ozone overhead, brought down to the ground, would be a layer only a few millimetres thick. Yet it stops most of the harmful ultraviolet.
Step 2 of 7
Open this step
What ozone stops
Practice · Choose
Question
Choose Molecules under Zoom to the molecules. An ultraviolet photon hits an ozone molecule, and it splits with an orange flash. What happens next?

Correct answer
- Correct The loose atom joins an $\mathrm{O_2}$ ⁠ again, the ozone is back, and the light’s energy stays in the air as heat
Setup, choices and feedback
The Scene at this step
Chlorine in the stratosphere
0.6 ppb
Month
December
Zoom to the molecules
Globe
Choices
Shown in a random order.
- Correct The loose atom joins an $\mathrm{O_2}$ ⁠ again, the ozone is back, and the light’s energy stays in the air as heat
- Wrong The ozone is destroyed for good: every photon absorbed thins the layer
  Misconception: Thinks each ozone molecule is used up for good when it absorbs ultraviolet.
  Feedback from the catalogue
  When ozone absorbs an ultraviolet photon it does break apart, into $\mathrm{O_2}$ ⁠ and a loose oxygen atom, but the two soon join again into $\mathrm{O_3}$ ⁠. Nothing is used up: the light’s energy stays in the air as heat, which is what warms the stratosphere. Ozone is truly destroyed only when something else, such as chlorine, takes atoms away from it.
- Wrong The photon carries on to the ground, only weaker
  Feedback
  The photon is absorbed whole: its energy is what split the molecule, and it stays in the air as heat.
Feedback for a right answer
The ozone is back: nothing is used up, and the light has become heat.

Worked solution
1. Ozone, $\mathrm{O_3}$ ⁠, absorbs the photon and splits into $\mathrm{O_2}$ ⁠ and an $\mathrm{O}$ ⁠ atom.
2. The atom soon meets an $\mathrm{O_2}$ ⁠ molecule and joins it again into $\mathrm{O_3}$ ⁠.
3. At the end of the loop there is the same amount of ozone, and the light’s energy has warmed the air. That is what warms the stratosphere.
So absorbing alone does not thin the layer. What thins it is something that takes oxygen atoms out of play, such as chlorine.
Step 3 of 7
Open this step
Polar spring
Practice · Predict, then look
Question
Today, in June: polar night over Antarctica, and icy clouds have already freed the chlorine. But there is no hole. Choose October under Month. What will happen to the ozone over the pole?

Correct answer
- Correct It will collapse, and a hole will open
Setup, choices and feedback
The Scene at this step
Chlorine in the stratosphere
3.2 ppb
Month
June
Zoom to the molecules
Globe
Choices
- Correct It will collapse, and a hole will open
- Wrong It will grow, because the cold winter is over
  Misconception: Thinks the ozone hole opens over the countries that released the CFCs, or appears in the darkest, coldest months.
  Feedback from the catalogue
  CFCs mix through the whole atmosphere over years, so the chlorine reaches everywhere. The hole opens over Antarctica because only there does the stratosphere get cold enough in the polar night for icy clouds, on which the chlorine is freed. And it opens only in spring, in September and October, when sunlight returns and splits the freed chlorine into atoms.
- Wrong Nothing, because the hole opens only over the countries that release CFCs
  Misconception: Thinks the ozone hole opens over the countries that released the CFCs, or appears in the darkest, coldest months.
  Feedback from the catalogue
  CFCs mix through the whole atmosphere over years, so the chlorine reaches everywhere. The hole opens over Antarctica because only there does the stratosphere get cold enough in the polar night for icy clouds, on which the chlorine is freed. And it opens only in spring, in September and October, when sunlight returns and splits the freed chlorine into atoms.
Once the prediction is locked in, the Scene changes
Month
October
The Scene then shows: 131.4 · The thinnest ozone column anywhere that month, Dobson units/nodes/earth/outputs/minimumOzone
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A hole: the returning light set to work the chlorine that waited in the dark.

Worked solution
1. In the polar night the chlorine is freed on the icy clouds, but stays in a form that does not destroy ozone, $\mathrm{Cl_2}$ ⁠, and waits.
2. In September the Sun returns, and its light splits the $\mathrm{Cl_2}$ ⁠ into chlorine atoms.
3. So much chlorine is freed at once that within weeks the column falls from about $280$ ⁠ to about $130$ ⁠ Dobson units.
That is why the hole belongs to the Antarctic spring, not to the dark winter itself.
Step 4 of 7
Open this step
How much chlorine it takes
Practice · Reach the target
Question
Month: October. The ozone hole is defined as everywhere the column falls below $220$ ⁠ Dobson units. Raise the Chlorine in the stratosphere from its natural level until the hole just begins to open, and press Check.

Goal
- 212 ± 10
  At the start of the step: 288.5
Setup, choices and feedback
The Scene at this step
Chlorine in the stratosphere
0.6 ppb
Month
Octoberfixed
Zoom to the molecules
Globe
Read from the Scene at
The thinnest ozone column anywhere that month, Dobson units/nodes/earth/outputs/minimumOzone
One setting that reaches it
Chlorine in the stratosphere
2.5 ppb
Feedback for a right answer
About 2.5 ppb: the chlorine of 1980.

Feedback for a wrong answer
Not yet. If there is no hole, add chlorine; if the column is already far below $220$ ⁠, take some away.

Hint
Raise the chlorine in small steps and watch the hole line on the chart: look for the first step at which a hole appears.

Worked solution
1. At 0.6 ppb, the natural level, there is no hole even in October.
2. The thinnest column falls fast as chlorine is added: at 2.2 ppb about $261$ ⁠, at 2.4 ppb about $224$ ⁠.
3. At 2.5 ppb it is about $208$ ⁠, below $220$ ⁠: the hole opens.
That is just where things stood around 1980, when the hole began to show in the measurements.
Step 5 of 7
Open this step
A catalyst
Practice · Choose
Question
October, Molecules. In the parcel of air a chlorine atom takes an atom from an ozone molecule, becomes $\mathrm{ClO}$ ⁠, and loses it again. Why are a few parts per billion of chlorine enough to destroy much of the ozone?

Correct answer
- Correct The chlorine atom comes out of each loop unchanged, and runs it again and again, thousands of times
Setup, choices and feedback
The Scene at this step
Chlorine in the stratosphere
3.2 ppb
Month
October
Zoom to the molecules
Molecules
Choices
Shown in a random order.
- Correct The chlorine atom comes out of each loop unchanged, and runs it again and again, thousands of times
- Wrong Each chlorine atom destroys one ozone molecule, and there is far more chlorine than it looks
  Misconception: Thinks each chlorine atom destroys one ozone molecule and is then used up.
  Feedback from the catalogue
  Chlorine is a catalyst: it takes an atom from an ozone molecule and becomes $\mathrm{ClO}$ ⁠, then loses that atom again and comes out of the loop unchanged. One chlorine atom can run the loop thousands of times before it is locked up in a harmless form. That is why a tiny amount, a few parts per billion, is enough to tear a hole.
- Wrong Chlorine is poisonous, and the ozone dies of it
  Feedback
  Ozone is a molecule, not a living thing. Chlorine reacts with it chemically, and comes out of the reaction as it went in.
Feedback for a right answer
A catalyst: one atom, thousands of ozone molecules.

Worked solution
1. $\mathrm{Cl} + \mathrm{O_3} \to \mathrm{ClO} + \mathrm{O_2}$ ⁠: the chlorine takes an atom.
2. Then the $\mathrm{ClO}$ ⁠ loses that atom, and the chlorine atom is free again, unchanged.
3. In each loop ozone turns into ordinary $\mathrm{O_2}$ ⁠, and the same chlorine atom starts a new loop, thousands of times before it is locked away.
That is how a catalyst works: it takes part in a reaction without being used up.
Step 6 of 7
Open this step
Fixing it
Practice · Choose
Question
The Montreal Protocol banned CFCs in 1987, and the chlorine in the stratosphere has been falling slowly since the late 1990s. Is the ozone hole what causes global warming?

Correct answer
- Correct No: the hole lets in more ultraviolet, but the warming comes from greenhouse gases that catch infrared
Setup, choices and feedback
The Scene at this step
Chlorine in the stratosphere
3.7 ppb
Month
October
Zoom to the molecules
Globe
Choices
Shown in a random order.
- Correct No: the hole lets in more ultraviolet, but the warming comes from greenhouse gases that catch infrared
- Wrong Yes: the Sun’s heat comes in through the hole and warms the Earth
  Misconception: Thinks the ozone hole is what causes global warming.
  Feedback from the catalogue
  These are two different problems. The ozone hole lets in more ultraviolet, which harms skin, eyes and plants, but it is a very small part of the Sun’s energy and is not what warms the Earth. The warming comes from greenhouse gases, such as carbon dioxide, which catch the infrared rising from the ground. The only link: CFCs are also strong greenhouse gases.
- Wrong The opposite: the hole cools the Earth, because heat escapes through it
  Feedback
  The hole is not an opening in anything: the air is still there, only with less ozone. The heat escaping to space is infrared, which ozone hardly affects.
Feedback for a right answer
No: two problems, two kinds of light.

Worked solution
1. Ozone mostly absorbs ultraviolet. In the hole more of it reaches the ground, which harms skin, eyes and plants.
2. But ultraviolet is a small part of the Sun’s energy, and a hole over Antarctica in spring does not change the Earth’s heat balance.
3. The warming comes from greenhouse gases, such as carbon dioxide, which catch the infrared rising from the ground.
There is one link: CFCs are also strong greenhouse gases, so banning them slowed the warming too.
Step 7 of 7
Open this step
A hole the size of a continent
Practice · Reach the target
Question
Month: October. Antarctica covers about 14.2 million km². Raise the Chlorine in the stratosphere until the ozone hole, everywhere the column falls below $220$ ⁠ Dobson units, is as big as the whole continent. Press Check.

Goal
- 1 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Chlorine in the stratosphere
0.6 ppb
Month
Octoberfixed
Zoom to the molecules
Globe
Read from the Scene at
Area under 220 DU, in areas of Antarctica (14.2 million km²)/nodes/earth/outputs/holeAntarcticas
One setting that reaches it
Chlorine in the stratosphere
2.6 ppb
Feedback for a right answer
At about 2.6–2.7 ppb: a little above the threshold where the hole opens.

Hint
The hole opens only above a certain level of chlorine, and then grows fast.

Worked solution
1. Below about 2.5 ppb of chlorine there is no hole at all.
2. Above the threshold the hole grows fast: at 2.6–2.7 ppb it is already the size of Antarctica.
3. Today, at 3.2 ppb, it is about 1.4 times as big.
Because of this threshold the hole appeared suddenly in the early 1980s, though the chlorine had been rising slowly for decades.

Lesson 5 of 5 · About 25 minutes

Auroras

The Earth's magnetic field holds off the solar wind and steers electrons into rings round the poles. See how a storm brings the aurora nearer the equator, and why it glows green and red.

Open the lesson

The 8 steps and their answers

Step 1 of 8
Open this step
Lights over the poles
Practice · Choose
Question
Solar wind (Kp index) is at 5, a storm. Rings of green light circle both poles, on the night side too. What is glowing there?

Correct answer
- Correct Oxygen atoms and nitrogen molecules in the thin air, hit by fast electrons
Setup, choices and feedback
The Scene at this step
Solar wind (Kp index)
5
Show the magnetic field
Off
Choices
Shown in a random order.
- Correct Oxygen atoms and nitrogen molecules in the thin air, hit by fast electrons
- Wrong Sunlight reflected off the polar ice
  Misconception: Thinks auroras are sunlight reflected off polar ice or clouds.
  Feedback from the catalogue
  Auroras glow at night, when there is no sunlight to reflect. The light is made 100–300 km up: fast electrons that slid down the magnetic field lines hit oxygen atoms and nitrogen molecules, excite them, and they give off light in their own colours: green and red from oxygen, violet from nitrogen.
- Wrong The solar wind itself, glowing on its way from the Sun
  Feedback
  The solar wind is far too thin to glow like that, and it does not enter the air everywhere. The light is made only where electrons hit the air, near the poles.
Feedback for a right answer
The air itself: oxygen and nitrogen excited by electrons.

Worked solution
1. The light comes from 100–300 km up, far above the clouds and the ice.
2. Fast electrons slide down the magnetic field lines and hit oxygen atoms and nitrogen molecules.
3. The atoms are excited, and as they settle back they give off light at their own wavelengths: green and red from oxygen, violet and blue from nitrogen.
It is the same principle as a neon sign: a thin gas excited by electrons glows in its own colour.
Step 2 of 8
Open this step
The magnetic bubble
Practice · Calculate
Question
Solar wind (Kp index) is locked at 1, quiet, and Show the magnetic field is On. The Earth’s radius is 6371 km. The nose of the magnetosphere, the boundary where the solar wind is held off, is about ten radii from the centre. How many kilometres is that, by the model?

Correct answer
64140 km ± 5%
Setup, choices and feedback
The Scene at this step
Solar wind (Kp index)
1fixed
Show the magnetic field
On
Accepted within
± 5%
Computed from the Scene at this step’s setting
Distance from the Earth’s centre to the magnetopause’s nose, towards the Sun/nodes/earth/outputs/standoff
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 64,000 km: about a sixth of the way to the Moon.

Hint
In the model, on a quiet day the nose is about $10.1$ ⁠ radii from the centre.

Worked solution
1. On a quiet day, $\mathrm{Kp} = 1$ ⁠, the Shue model puts the nose about $10.1$ ⁠ Earth radii from the centre.
2. $10.1 \times 6371 \approx 64{,}000$ ⁠ kilometres.
3. That is about a sixth of the way to the Moon, which is about $60$ ⁠ radii from us.
The number comes from a balance: the solar wind’s pressure equals the pressure of the magnetic field, which weakens with distance.
Step 3 of 8
Open this step
A solar storm
Practice · Predict, then look
Question
Show the magnetic field is On, and the solar wind is quiet, $\mathrm{Kp} = 1$ ⁠. Raise it to an extreme storm, $\mathrm{Kp} = 9$ ⁠. What will happen to the nose of the magnetosphere?

Correct answer
- Correct It will be squeezed in, to about half the distance
Setup, choices and feedback
The Scene at this step
Solar wind (Kp index)
1
Show the magnetic field
On
Choices
- Correct It will be squeezed in, to about half the distance
- Wrong It will move out, because the storm inflates the field
  Feedback
  A stronger wind pushes harder, and the Earth’s field does not get stronger. The boundary moves to where the field is strong enough to resist: nearer in.
- Wrong It will stay put, because the Earth’s magnetic field has not changed
  Feedback
  The field is the same field, but the boundary is where its pressure equals the wind’s. When the wind pushes harder, the balance point moves in.
Once the prediction is locked in, the Scene changes
Solar wind (Kp index)
9
The Scene then shows: 5.347 · The same distance in Earth radii/nodes/earth/outputs/standoffRadii
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Squeezed to about $5.3$ ⁠ radii: about half.

Worked solution
1. In a storm the solar wind’s pressure grows about $20$ ⁠ times, and the field it carries points south, against the Earth’s.
2. The Earth’s field weakens with distance, so the boundary moves in until the field is strong enough to resist the push.
3. In the model the nose goes from about $10$ ⁠ radii to about $5.3$ ⁠.
In real extreme storms, such as 1859 and 2024, the magnetosphere was squeezed to a few radii, and satellites found themselves outside it.
Step 4 of 8
Open this step
The oval moves
Practice · Reach the target
Question
Near 50° of magnetic latitude lie central Europe, southern Canada and southern Australia. Raise the Solar wind (Kp index) until the edge of the auroral oval nearest the equator, on the night side, reaches 50°, and press Check.

Goal
- 50 ° ± 1.2 °
  At the start of the step: 61.2 °
Setup, choices and feedback
The Scene at this step
Solar wind (Kp index)
1
Show the magnetic field
Off
Read from the Scene at
Magnetic latitude of the oval’s edge nearest the equator, on the night side/nodes/earth/outputs/ovalEdge
One setting that reaches it
Solar wind (Kp index)
5
Feedback for a right answer
Around $\mathrm{Kp} = 5$ ⁠: a geomagnetic storm.

Feedback for a wrong answer
Not yet. If the edge is still far from the equator, strengthen the solar wind; if it has passed 50°, weaken it.

Hint
On a quiet day the edge is at about 61°, and it moves towards the equator by about 2.5° per Kp step.

Worked solution
1. On a quiet day, $\mathrm{Kp} = 1$ ⁠, the oval on the night side sits around 63° magnetic latitude, its edge nearest the equator about 61°.
2. With each Kp step the oval moves about 2.2° towards the equator and widens too, so the edge moves about 2.5°.
3. The edge reaches 50° around $\mathrm{Kp} = 5$ ⁠: a storm.
In the storm of May 2024, $\mathrm{Kp} = 9$ ⁠, auroras were photographed across southern Europe and Mexico.
Step 5 of 8
Open this step
Down the field lines
Practice · Choose
Question
Show the magnetic field is On. The solar wind strikes the day side of the Earth, yet the aurora is a ring round each pole, brightest on the night side. Why?

Correct answer
- Correct Charged particles move along field lines, which converge on the poles, and they come mostly from the tail on the night side
Setup, choices and feedback
The Scene at this step
Solar wind (Kp index)
3
Show the magnetic field
On
Choices
Shown in a random order.
- Correct Charged particles move along field lines, which converge on the poles, and they come mostly from the tail on the night side
- Wrong The wind enters the air straight on the day side, and blows from there to the poles
  Misconception: Thinks the solar wind pours straight into the air wherever it hits, on the side facing the Sun.
  Feedback from the catalogue
  Charged particles cannot easily cross magnetic field lines, so the solar wind flows round the magnetosphere. The particles reach the air only along field lines, which converge near the magnetic poles, and mostly from the tail on the night side, when it is stretched and snaps back. So the aurora is a ring round each pole, brightest on the night side, not a patch on the side facing the Sun.
- Wrong Only in the cold polar air can the light be seen
  Feedback
  The light is made hundreds of kilometres up, far above the cold polar air, and in storms it is seen far from the poles.
Feedback for a right answer
The field lines: they guide the electrons into a ring round each pole.

Worked solution
1. The Lorentz force turns a charged particle round a field line, so it does not easily cross field lines but slides along them.
2. So the solar wind flows round the magnetosphere, and stretches the night side into a long tail.
3. When the stretched tail snaps back, electrons are accelerated along field lines that reach the air only round the magnetic poles: a ring round each pole.
It is the same spiralling round field lines met in the Unit on magnetism.
Step 6 of 8
Open this step
Green and red
Practice · Choose
Question
An extreme storm, $\mathrm{Kp} = 9$ ⁠: above the green curtains red appears. Both colours come from oxygen. Why is the red always on top?

Correct answer
- Correct The red light leaves the atom only after about two minutes, and only in the very thin air high up does the atom not hit a neighbour first
Setup, choices and feedback
The Scene at this step
Solar wind (Kp index)
9
Show the magnetic field
Off
Choices
Shown in a random order.
- Correct The red light leaves the atom only after about two minutes, and only in the very thin air high up does the atom not hit a neighbour first
- Wrong The air there is hotter, and hot air glows red
  Misconception: Thinks an aurora’s colours show how hot the gas is, red being cooler or hotter than green.
  Feedback from the catalogue
  Each colour is an emission line of a particular gas, not a temperature. Oxygen gives off green at 557.7 nm and red at 630.0 nm. The red comes out only after about two minutes, so it survives only in the very thin air above about 200 km, where the atom does not hit a neighbour first. That is why red sits on top and green below it.
- Wrong Light from the sunset still reaches it there
  Misconception: Thinks auroras are sunlight reflected off polar ice or clouds.
  Feedback from the catalogue
  Auroras glow at night, when there is no sunlight to reflect. The light is made 100–300 km up: fast electrons that slid down the magnetic field lines hit oxygen atoms and nitrogen molecules, excite them, and they give off light in their own colours: green and red from oxygen, violet from nitrogen.
Feedback for a right answer
Time: the red is slow, and survives only in thin air.

Worked solution
1. An excited oxygen atom can give off green at 557.7 nm, within under a second, or red at 630.0 nm, after almost two minutes.
2. At 100–200 km the atom hits a neighbour within two minutes and loses the energy without red light.
3. Only above about 200 km is the air thin enough for the red to get out.
In stronger storms electrons also strike very high up, which is why red auroras are seen far from the poles.
Step 7 of 8
Open this step
How hard the wind pushes
Practice · Calculate
Question
Solar wind (Kp index) is locked at an extreme storm, $\mathrm{Kp} = 9$ ⁠. Each Kp step multiplies the pressure the wind puts on the magnetic bubble by about $1.4$ ⁠. How many times harder does it push now than when quiet, $\mathrm{Kp} = 1$ ⁠?

Correct answer
14.76 ± 4%
Setup, choices and feedback
The Scene at this step
Solar wind (Kp index)
9fixed
Show the magnetic field
On
Accepted within
± 4%
A fixed value in the question
Wrong answers it recognises
- 11.2 ± 3%
  Misconception: Treats a quantity that is multiplied at each step as if the same amount were added each step.
  Feedback from the catalogue
  Each Kp step multiplies the wind’s pressure by about $1.4$ ⁠; it does not add a fixed amount. Eight steps are $1.4 \times 1.4 \times \dots$ ⁠ eight times, almost 15 times, not $1.4 \times 8$ ⁠.
Feedback for a right answer
Almost 15 times. Yet the bubble’s nose comes in by not even a factor of two, because the magnetic field grows strong fast nearer the Earth.

Hint
From $1$ ⁠ to $9$ ⁠ there are eight steps, and each multiplies by $1.4$ ⁠.

Worked solution
1. The number of steps: $9 - 1 = 8$ ⁠.
2. The pressure grows $1.4^8$ ⁠ times: $1.4^2 = 1.96$ ⁠, $1.4^4 \approx 3.84$ ⁠, $1.4^8 \approx 14.8$ ⁠.
3. About 2.1 nanopascals when quiet become about 31 in the storm.
If you answered $11.2$ ⁠, you multiplied $1.4 \times 8$ ⁠. But each step multiplies what is already there, so the growth is much faster.
Step 8 of 8
Open this step
Where the oval goes
Practice · Predict, then look
Question
On a quiet day, $\mathrm{Kp} = 1$ ⁠, the middle of the auroral oval on the night side is at about $63^\circ$ ⁠ of magnetic latitude, over northern Norway. Predict: where will it be when the Solar wind (Kp index) rises to 9, an extreme storm?

Correct answer
- Correct Nearer the equator, about $45^\circ$ ⁠
Setup, choices and feedback
The Scene at this step
Solar wind (Kp index)
1
Show the magnetic field
Off
Choices
- Wrong Nearer the pole, about $75^\circ$ ⁠, because the strong wind pushes it back
  Misconception: Thinks a stronger solar wind pushes the auroral oval back towards the pole.
  Feedback from the catalogue
  In a storm the wind does squeeze the bubble, but that is exactly why the aurora moves to lower latitudes. More field lines open to the wind and are stretched into the tail, and those field lines leave the Earth farther from the poles. The oval spreads towards the equator.
- Wrong In the same place, only brighter
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Correct Nearer the equator, about $45^\circ$ ⁠
Once the prediction is locked in, the Scene changes
Solar wind (Kp index)
9
The Scene then shows: 45.2 ° · Magnetic latitude of the middle of the auroral oval on the night side/nodes/earth/outputs/ovalLatitude
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About $45^\circ$ ⁠: almost $18$ ⁠ degrees nearer the equator, over northern Italy or the northern United States.

Hint
In a storm the wind squeezes the bubble, and the field lines that lead electrons down into the air start farther from the poles.

Worked solution
1. In a storm the wind squeezes the bubble on the day side and stretches the tail on the night side, and more of the field lines open up to it.
2. The field lines that reach into the tail leave the Earth at lower latitudes.
3. So the oval widens and moves down, to about $45^\circ$ ⁠ at $\mathrm{Kp} = 9$ ⁠.
In the storm of May 2024 auroras were seen in southern Europe and Mexico.

Across the unit · 22 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

Layers of the atmosphere

Not colder all the way up

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

How fast the air cools

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Where the air is

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

High over the tropics, low near the poles

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A layer that warms

Question

Correct answer

The Scene at this step

Choices

Worked solution

Hot, but it feels cold

Question

Correct answer

The Scene at this step

Choices

Worked solution

Storms and weather

Thermals

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The surrounding air's gradient

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The cloud base

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Without moisture

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The fuel

Question

Correct answer

The Scene at this step

Choices

Worked solution

Wind shear

Question

Correct answer

The Scene at this step