Answer key · Unit 17

The Earth, the Moon and the Sun.

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

Day and night

The Earth spins on its axis: why half of it is always lit, how many degrees the Sun moves in an hour, why day and night are equal everywhere at the equinox, how high the noon Sun stands, and what the midnight Sun is.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
A turning globe
Practice · Choose
Question
Why are there day and night?

Correct answer
- Correct The Earth turns on its axis once a day, and every place takes its turn on the lit side and the shadowed side
Setup, choices and feedback
The Scene at this step
Day of the year
78 day
Axial tilt
23.44 °
Latitude
32 ° N
The orbit / Sunlight up close
The orbit
Choices
Shown in a random order.
- Correct The Earth turns on its axis once a day, and every place takes its turn on the lit side and the shadowed side
- Wrong The Earth goes round the Sun, and for half the orbit it is far from it
  Misconception: Explains day and night by the Earth's yearly trip round the Sun (or the Sun circling the Earth each day), not by the Earth spinning.
  Feedback from the catalogue
  Day and night come from the Earth spinning on its axis, about once every 24 h. At any moment half the Earth faces the Sun, and there it is day, while the other half is in its own shadow. The trip round the Sun is $365$ ⁠ times slower: it gives the year and the seasons, not the day. The Sun only seems to cross the sky, the way the landscape seems to move past a train window.
- Wrong The Sun goes round the Earth once a day
  Misconception: Explains day and night by the Earth's yearly trip round the Sun (or the Sun circling the Earth each day), not by the Earth spinning.
  Feedback from the catalogue
  Day and night come from the Earth spinning on its axis, about once every 24 h. At any moment half the Earth faces the Sun, and there it is day, while the other half is in its own shadow. The trip round the Sun is $365$ ⁠ times slower: it gives the year and the seasons, not the day. The Sun only seems to cross the sky, the way the landscape seems to move past a train window.
- Wrong At night the air and clouds block the sunlight
  Feedback
  Even on a perfectly clear night the Sun is not seen: it is below the horizon, on the other side of the Earth.
Feedback for a right answer
At any moment the Sun lights one half of the globe. The spin about the axis, once every 24 h, carries every place from the lit half to the dark half and back.

Hint
Look at the globe: which part of it is lit at any moment?

Worked solution
1. The Sun is very far away, so its rays arrive almost parallel and light exactly half the globe.
2. The line between the lit and dark halves is called the terminator.
3. The Earth turns on its axis once every 24 h, and every place crosses that line twice: at sunrise and at sunset.
The trip round the Sun takes a whole year, so it cannot give a daily cycle.
Step 2 of 5
Open this step
Degree by degree
Practice · Calculate
Question
The Earth turns $360^\circ$ ⁠ in 24 h. By how many degrees does the Sun seem to move across the sky each hour?

Correct answer
15 ° ± 0.1 °
Setup, choices and feedback
The Scene at this step
Day of the year
78 day
Axial tilt
23.44 °
Latitude
32 ° N
The orbit / Sunlight up close
Sunlight up close
Accepted within
± 0.1 °
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.99 ° ± 0.05 °
  Misconception: Explains day and night by the Earth's yearly trip round the Sun (or the Sun circling the Earth each day), not by the Earth spinning.
  Feedback from the catalogue
  Day and night come from the Earth spinning on its axis, about once every 24 h. At any moment half the Earth faces the Sun, and there it is day, while the other half is in its own shadow. The trip round the Sun is $365$ ⁠ times slower: it gives the year and the seasons, not the day. The Sun only seems to cross the sky, the way the landscape seems to move past a train window.
Feedback for a right answer
$\frac{360^\circ}{24} = 15^\circ$ ⁠ an hour. That is why time zones are about $15^\circ$ ⁠ of longitude wide: one hour of turning.

Hint
Divide the full turn by the number of hours.

Worked solution
1. A full turn: $360^\circ$ ⁠ in 24 h.
2. In one hour: $\frac{360^\circ}{24} = 15^\circ$ ⁠.
If you answered about $1^\circ$ ⁠, you divided by $365$ ⁠: that is how far the Earth moves along its orbit each day, not how far it turns.
Step 3 of 5
Open this step
Equal night
Practice · Predict, then look
Question
It is the March equinox, and we are in Singapore, almost on the equator, with a day of about 12 h. How long is the day on the same date in Tromsø in Norway, at latitude $69.6^\circ$ ⁠ north?

Correct answer
- Correct Also about 12 h
Setup, choices and feedback
The Scene at this step
Day of the year
78 dayfixed
Axial tilt
23.44 °fixed
Latitude
1.3 ° Nfixed
The orbit / Sunlight up close
Sunlight up closefixed
Choices
- Correct Also about 12 h
- Wrong Much longer, because it is near the pole
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Wrong Much shorter, because the Sun is low there
  Feedback
  The Sun is indeed low in Tromsø, but its height at noon and the length of the day are two different things.
Once the prediction is locked in, the Scene changes
Latitude
69.6 ° Nfixed
The orbit / Sunlight up close
Sunlight up closefixed
The Scene then shows: 12.27 h · Time from sunrise to sunset at the chosen latitude (0 to 24 hours), the top of the Sun clearing a horizon lifted 0.833° by the air/nodes/year/outputs/dayLength
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 12 h 16 min. At the equinox the terminator runs through both poles and cuts every latitude exactly in two. The extra minutes come from the air, which bends sunlight over the horizon.

Hint
At the equinox the Sun is over the equator. Where does the terminator run?

Worked solution
1. At the equinox the Sun stands over the equator, and the terminator runs through both poles.
2. Every latitude, $69.6^\circ$ ⁠ included, is cut into two equal halves: half in day and half in night.
3. So everywhere the day lasts about 12 h, hence the name “equinox”, equal night.
The air bends sunlight and adds a few minutes, more at high latitudes, where the Sun rises and sets at a shallow slant.
Step 4 of 5
Open this step
The noon Sun
Practice · Calculate
Question
On the same equinox, how high above the horizon does the noon Sun stand in Tel Aviv, at latitude $32^\circ$ ⁠ north?

Correct answer
57.87 ° ± 0.5 °
Setup, choices and feedback
The Scene at this step
Day of the year
78 dayfixed
Axial tilt
23.44 °fixed
Latitude
32 ° Nfixed
The orbit / Sunlight up close
The orbitfixed
Accepted within
± 0.5 °
Computed from the Scene at this step’s setting
The noon Sun's height above the horizon at the chosen latitude, 90° − |latitude − declination|; negative when it stays below/nodes/year/outputs/noonAltitude
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 90 ° ± 0.5 °
  Misconception: Thinks the Sun stands straight overhead at noon everywhere, every day.
  Feedback from the catalogue
  The Sun stands exactly overhead (at the zenith) at only one latitude on any day, the one it is “over” that day, and always between $23.4^\circ$ ⁠ north and $23.4^\circ$ ⁠ south. Elsewhere the noon Sun is lower: at a height of $90^\circ - |\varphi - \delta|$ ⁠, where $\varphi$ ⁠ is the latitude and $\delta$ ⁠ the Sun’s latitude. In Israel, at latitude $32^\circ$ ⁠, it never reaches the zenith.
- 32 ° ± 0.5 °
  Misconception: Thinks the Sun stands straight overhead at noon everywhere, every day.
  Feedback
  $32^\circ$ ⁠ is the latitude, the angle between Tel Aviv and the equator. The Sun’s height is its complement.
Feedback for a right answer
About $58^\circ$ ⁠: $90^\circ - 32^\circ$ ⁠. The Sun is overhead only on the equator, and every degree of latitude north of it the Sun is a degree lower.

Hint
At the equinox the Sun is over the equator. How many degrees is Tel Aviv from the place where the Sun is overhead?

Worked solution
1. At the equinox the Sun is overhead at the equator: $\delta \approx 0^\circ$ ⁠.
2. The rays are parallel, so the Sun’s height drops a degree for every degree of latitude: $h = 90^\circ - |\varphi - \delta|$ ⁠.
3. In Tel Aviv: $h = 90^\circ - 32^\circ = 58^\circ$ ⁠. The Scene gives $57.9^\circ$ ⁠, because on this day the Sun is still slightly south of the equator.
The Sun is not overhead at noon everywhere: only at the latitude it is over that day.
Step 5 of 5
Open this step
The midnight Sun
Practice · Predict, then look
Question
Now it is the June solstice, and the latitude is $80^\circ$ ⁠ north, in Svalbard. How long will the Sun be above the horizon?

Correct answer
- Correct All 24 hours
Setup, choices and feedback
The Scene at this step
Day of the year
171 dayfixed
Axial tilt
23.44 °fixed
Latitude
32 ° Nfixed
The orbit / Sunlight up close
Sunlight up closefixed
Choices
- Correct All 24 hours
- Wrong About 12 h, as everywhere
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Wrong Not at all
  Misconception: Expects the whole Earth to have summer at the same time.
  Feedback from the catalogue
  When the north leans towards the Sun, the south leans away. So the seasons are opposite in the two hemispheres: in June it is summer in Israel and winter in Sydney, and Christmas in Australia is spent on the beach. That is more evidence the distance from the Sun is not the cause: both hemispheres are always the same distance from it.
- Wrong About 18 h
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Latitude
80 ° Nfixed
The orbit / Sunlight up close
Sunlight up closefixed
The Scene then shows: 24 h · Time from sunrise to sunset at the chosen latitude (0 to 24 hours), the top of the Sun clearing a horizon lifted 0.833° by the air/nodes/year/outputs/dayLength
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The midnight Sun: the whole parallel is on the lit side, and the Sun circles the sky without setting. In December, the other way round, it does not rise at all.

Hint
In June the North Pole leans towards the Sun. Does latitude $80^\circ$ ⁠ reach the dark side at all?

Worked solution
1. At the June solstice the axis leans $23.4^\circ$ ⁠ towards the Sun, and the terminator runs $23.4^\circ$ ⁠ beyond the North Pole.
2. Every latitude above $90^\circ - 23.4^\circ = 66.6^\circ$ ⁠, the Arctic Circle, stays entirely on the lit side.
3. $80^\circ$ ⁠ is above the circle, so the Sun does not set: 24 h of daylight.

Lesson 2 of 4 · About 30 minutes

The seasons

Why summer is hot and winter cold: not the distance from the Sun but the tilt of the axis. The noon Sun's height and the length of the day at latitude $32^\circ$, light spread over more ground, opposite seasons in the south, and what a different tilt would do.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
Closest to the Sun
Practice · Predict, then look
Question
The Earth’s orbit is an almost circular ellipse. When is the Earth closest to the Sun?

Correct answer
- Correct In early January
Setup, choices and feedback
The Scene at this step
Day of the year
184 day
Axial tilt
23.44 °fixed
Latitude
32 ° Nfixed
The orbit / Sunlight up close
The orbitfixed
Choices
- Correct In early January
- Wrong In early July, in summer
  Misconception: Thinks summer happens when the Earth is closest to the Sun and winter when it is farthest.
  Feedback from the catalogue
  The Earth is closest to the Sun in early January, in the northern winter, and about $3\%$ ⁠ farther in early July. And when it is summer in the north it is winter in the south, at exactly the same distance. What makes the seasons is the tilt of the axis: in the half of the year when the north leans towards the Sun, the noon Sun is high, the same light falls on less ground, and the days are long.
- Wrong At the equinoxes
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Day of the year
2 day
The Scene then shows: 147100000 km · The Earth–Sun distance on this day/nodes/year/outputs/distance
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
On $2$ ⁠ January the Earth is 147.1 million km away, and in early July 152.1 million. The north has winter just when the Sun is closest.

Hint
The distance in the Scene comes from the real orbit. What does that say about distance and the seasons?

Worked solution
1. The closest point of the orbit, perihelion, falls in early January, and the farthest, aphelion, in early July.
2. The distance differs by about $3\%$ ⁠, and the light by about $7\%$ ⁠, since light falls off as $\frac{1}{r^2}$ ⁠.
3. Yet January is winter in the north and summer in the south. So distance is not what makes the seasons.
Step 2 of 7
Open this step
The tilted axis
Practice · Choose
Question
So why is June summer in the northern hemisphere?

Correct answer
- Correct The axis points the same way all year, and in June the Earth is on the side of its orbit where the north leans towards the Sun
Setup, choices and feedback
The Scene at this step
Day of the year
171 day
Axial tilt
23.44 °fixed
Latitude
32 ° Nfixed
The orbit / Sunlight up close
The orbit
Choices
Shown in a random order.
- Correct The axis points the same way all year, and in June the Earth is on the side of its orbit where the north leans towards the Sun
- Wrong In June the Earth tips its axis towards the Sun, and in December away from it
  Misconception: Thinks the Earth tips its axis towards the Sun in summer and away in winter, as if the axis rocked during the year.
  Feedback from the catalogue
  The Earth’s axis points the same way in space all year, roughly at the Pole Star. What changes is where the Earth is on its orbit. In June it is on the side where the north end of the axis faces the Sun; in December it is on the far side of the orbit, and the same axis, pointing the same way, now faces away from it.
- Wrong In June the north is nearer the Sun
  Misconception: Thinks summer happens when the Earth is closest to the Sun and winter when it is farthest.
  Feedback from the catalogue
  The Earth is closest to the Sun in early January, in the northern winter, and about $3\%$ ⁠ farther in early July. And when it is summer in the north it is winter in the south, at exactly the same distance. What makes the seasons is the tilt of the axis: in the half of the year when the north leans towards the Sun, the noon Sun is high, the same light falls on less ground, and the days are long.
- Wrong In June the whole Earth gets more light
  Misconception: Expects the whole Earth to have summer at the same time.
  Feedback from the catalogue
  When the north leans towards the Sun, the south leans away. So the seasons are opposite in the two hemispheres: in June it is summer in Israel and winter in Sydney, and Christmas in Australia is spent on the beach. That is more evidence the distance from the Sun is not the cause: both hemispheres are always the same distance from it.
Feedback for a right answer
The axis is fixed in space, tilted $23.4^\circ$ ⁠. In June its north end faces the Sun: the noon Sun is high in the north and the days are long. Half a year later the Earth is on the far side, and it is the south that leans towards the Sun.

Hint
Change Day of the year and watch the axis: does its direction change?

Worked solution
1. The Earth’s axis is tilted $23.4^\circ$ ⁠ and keeps its direction in space all the way round the orbit.
2. In June the Earth is on the side of the orbit where the north end of the axis faces the Sun. In December, on the opposite side, it faces away.
3. When the north faces the Sun: the noon Sun is high, the same light falls on less ground, and the day is longer than the night. All of these warm it.
The distance hardly changes, and it is the same for both hemispheres. That is why the seasons are opposite in the north and the south.
Step 3 of 7
Open this step
A high Sun
Practice · Calculate
Question
At the June solstice the Sun is overhead at $23.4^\circ$ ⁠ north, the Tropic of Cancer. How high does the noon Sun stand at latitude $32^\circ$ ⁠ north?

Correct answer
81.44 ° ± 0.5 °
Setup, choices and feedback
The Scene at this step
Day of the year
171 dayfixed
Axial tilt
23.44 °fixed
Latitude
32 ° Nfixed
The orbit / Sunlight up close
The orbitfixed
Accepted within
± 0.5 °
Computed from the Scene at this step’s setting
The noon Sun's height above the horizon at the chosen latitude, 90° − |latitude − declination|; negative when it stays below/nodes/year/outputs/noonAltitude
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 90 ° ± 0.5 °
  Misconception: Thinks the Sun stands straight overhead at noon everywhere, every day.
  Feedback from the catalogue
  The Sun stands exactly overhead (at the zenith) at only one latitude on any day, the one it is “over” that day, and always between $23.4^\circ$ ⁠ north and $23.4^\circ$ ⁠ south. Elsewhere the noon Sun is lower: at a height of $90^\circ - |\varphi - \delta|$ ⁠, where $\varphi$ ⁠ is the latitude and $\delta$ ⁠ the Sun’s latitude. In Israel, at latitude $32^\circ$ ⁠, it never reaches the zenith.
- 34.56 ° ± 0.5 °
  Misconception: Thinks the Earth tips its axis towards the Sun in summer and away in winter, as if the axis rocked during the year.
  Feedback
  $34.6^\circ$ ⁠ is the height in December, when the Sun is over $23.4^\circ$ ⁠ south. In June it is over $23.4^\circ$ ⁠ north, much closer to Tel Aviv.
Feedback for a right answer
$90^\circ - (32^\circ - 23.4^\circ) = 81.4^\circ$ ⁠: nearly overhead. In December the same formula gives only $34.6^\circ$ ⁠.

Feedback for a wrong answer
How many degrees is $32^\circ$ ⁠ north from the line where the Sun is overhead?

Hint
$h = 90^\circ - |\varphi - \delta|$ ⁠, with $\varphi = 32^\circ$ ⁠ and $\delta = 23.4^\circ$ ⁠.

Worked solution
1. The Sun is overhead at $\delta = 23.4^\circ$ ⁠ north.
2. Tel Aviv, $\varphi = 32^\circ$ ⁠, is $32^\circ - 23.4^\circ = 8.6^\circ$ ⁠ from there.
3. So the Sun is $8.6^\circ$ ⁠ below the zenith: $h = 90^\circ - 8.6^\circ = 81.4^\circ$ ⁠.
In December: $\delta = -23.4^\circ$ ⁠, the gap is $55.4^\circ$ ⁠, so $h = 34.6^\circ$ ⁠. The difference between summer and winter is $2 \times 23.4^\circ = 46.9^\circ$ ⁠: twice the tilt.
Step 4 of 7
Open this step
Long days
Practice · Predict, then look
Question
It is now the March equinox in Tel Aviv: about 12 h of daylight. Predict how long the day there lasts at the June solstice.

Correct answer
14.24 h ± 0.5 h
A number prediction counts as right within ± 0.5 h
Setup, choices and feedback
The Scene at this step
Day of the year
78 dayfixed
Axial tilt
23.44 °fixed
Latitude
32 ° Nfixed
The orbit / Sunlight up close
Sunlight up closefixed
Once the prediction is locked in, the Scene changes
Day of the year
171 dayfixed
The orbit / Sunlight up close
Sunlight up closefixed
The Scene then shows: 14.24 h · Time from sunrise to sunset at the chosen latitude (0 to 24 hours), the top of the Sun clearing a horizon lifted 0.833° by the air/nodes/year/outputs/dayLength
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 14 h 15 min. In December, the other way round, about 10 h. The higher the latitude, the bigger the gap between summer and winter.

Hint
At latitude $32^\circ$ ⁠ day and night differ by a few hours, not by $12$ ⁠.

Worked solution
1. In June the north leans towards the Sun, and the terminator runs beyond the North Pole.
2. So latitude $32^\circ$ ⁠ is cut unevenly: more than half of it is on the lit side.
3. The lit part is about $0.59$ ⁠ of the circle, which is 14.2 h out of 24 h.
On the equator the day stays about 12 h all year, and above the Arctic Circle it reaches 24 h.
Step 5 of 7
Open this step
Light spread thin
Practice · Calculate
Question
The December solstice, $32^\circ$ ⁠ north. How much sunlight does each square metre of level ground get at noon, as a per cent of what it would get with the Sun overhead? Write the number only.

Correct answer
56.73 ± 1
Setup, choices and feedback
The Scene at this step
Day of the year
354 dayfixed
Axial tilt
23.44 °fixed
Latitude
32 ° Nfixed
The orbit / Sunlight up close
The orbitfixed
Accepted within
± 1
Computed from the Scene at this step’s setting
The noon sunlight on each square metre of level ground at the chosen latitude, as a share of the light with the Sun overhead: sin(altitude), 0 when it stays below/nodes/year/outputs/sunlightShare
Wrong answers it recognises
- 38.4 ± 1
  Misconception: Thinks the Sun stands straight overhead at noon everywhere, every day.
  Feedback
  $\frac{34.6^\circ}{90^\circ}$ ⁠ assumes the light falls in step with the angle. The beam is smeared over an area $\frac{1}{\sin h}$ ⁠ times larger, so the share is $\sin h$ ⁠.
Feedback for a right answer
$\sin 34.6^\circ \approx 0.57$ ⁠: about $57\%$ ⁠. In June, at $81.4^\circ$ ⁠, about $99\%$ ⁠. In winter the same beam is smeared over $1.76$ ⁠ times the area, so each square metre warms less.

Hint
A beam arriving at height $h$ ⁠ spreads over an area $\frac{1}{\sin h}$ ⁠ times larger.

Worked solution
1. In December, at $32^\circ$ ⁠ north, the noon Sun is at height $h = 90^\circ - (32^\circ + 23.4^\circ) = 34.6^\circ$ ⁠.
2. A beam one unit wide falls on level ground $\frac{1}{\sin h}$ ⁠ wide.
3. So each square metre gets a share $\sin h = \sin 34.6^\circ \approx 0.57$ ⁠, about $57\%$ ⁠.
This is the main reason winter is cold: not that the Sun is far, but that its light spreads over more ground, for fewer hours.
Step 6 of 7
Open this step
The other hemisphere
Practice · Predict, then look
Question
The June solstice. In Tel Aviv the day is long, about 14 h. What happens on the same date in Sydney, at latitude $33.9^\circ$ ⁠ south?

Correct answer
- Correct A short day, about 10 h: it is winter there
Setup, choices and feedback
The Scene at this step
Day of the year
171 dayfixed
Axial tilt
23.44 °fixed
Latitude
32 ° Nfixed
The orbit / Sunlight up close
Sunlight up closefixed
Choices
- Correct A short day, about 10 h: it is winter there
- Wrong A long day too, about 14 h: June is summer everywhere
  Misconception: Expects the whole Earth to have summer at the same time.
  Feedback from the catalogue
  When the north leans towards the Sun, the south leans away. So the seasons are opposite in the two hemispheres: in June it is summer in Israel and winter in Sydney, and Christmas in Australia is spent on the beach. That is more evidence the distance from the Sun is not the cause: both hemispheres are always the same distance from it.
- Wrong About 12 h
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Latitude
-33.9 ° Nfixed
The orbit / Sunlight up close
Sunlight up closefixed
The Scene then shows: 9.894 h · Time from sunrise to sunset at the chosen latitude (0 to 24 hours), the top of the Sun clearing a horizon lifted 0.833° by the air/nodes/year/outputs/dayLength
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 9 h 54 min, and the noon Sun only $33^\circ$ ⁠ up. Sydney has winter in June, at the same distance from the Sun as Tel Aviv.

Hint
When the north leans towards the Sun, which way does the south lean?

Worked solution
1. In June the north leans towards the Sun, so the south leans away.
2. In Sydney, $33.9^\circ$ ⁠ south, the parallel is cut so that less than half of it is on the lit side: a day of about 9.9 h.
3. The noon Sun: $h = 90^\circ - (33.9^\circ + 23.4^\circ) \approx 32.7^\circ$ ⁠.
Same date, same distance from the Sun, opposite seasons: the simplest proof that the tilt, not the distance, makes the seasons.
Step 7 of 7
Open this step
A different tilt
Practice · Reach the target
Question
On our Earth the Sun is never overhead in Tel Aviv. Change Axial tilt until the noon Sun at the June solstice stands straight overhead at latitude $32^\circ$ ⁠ north, then press Check.

Goal
- 90 ° ± 0.3 °
  At the start of the step: 81.44 °
Setup, choices and feedback
The Scene at this step
Day of the year
171 dayfixed
Axial tilt
23.44 °
Latitude
32 ° Nfixed
The orbit / Sunlight up close
Sunlight up closefixed
Read from the Scene at
The noon Sun's height above the horizon at the chosen latitude, 90° − |latitude − declination|; negative when it stays below/nodes/year/outputs/noonAltitude
One setting that reaches it
Axial tilt
31.71 °
Feedback for a right answer
A tilt of $32^\circ$ ⁠: the Tropic of Cancer now runs through Tel Aviv. A bigger tilt makes the seasons more extreme, and $0^\circ$ ⁠ removes them altogether: a 12 h day all year.

Feedback for a wrong answer
Not yet. How many degrees north does the line where the Sun is overhead need to move?

Hint
At the solstice the Sun is overhead at the latitude equal to the tilt.

Worked solution
1. At the June solstice the Sun’s latitude equals the axial tilt: $\delta = \varepsilon$ ⁠.
2. The Sun is overhead at $32^\circ$ ⁠ when $\delta = 32^\circ$ ⁠.
3. So set Axial tilt to $32^\circ$ ⁠.
On our Earth $\varepsilon = 23.4^\circ$ ⁠, so the Sun is overhead only between $23.4^\circ$ ⁠ north and $23.4^\circ$ ⁠ south, between the two tropics.

Lesson 3 of 4 · About 25 minutes

Phases of the Moon

The Sun always lights half the Moon, and through the month we see a different part of that half. First quarter, full moon, how much of the disc is lit each day, and why the Moon rises about $50$ minutes later every day.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Always half lit
Practice · Choose
Question
At any moment, how much of the Moon’s surface does the Sun light?

Correct answer
- Correct Always half, the half facing the Sun
Setup, choices and feedback
The Scene at this step
Day of the month
7.4
Show tides
off
Eclipse season
off
Sunlight / Shadows
Sunlight
Choices
Shown in a random order.
- Correct Always half, the half facing the Sun
- Wrong It depends on the phase: all of it at full moon, none at new moon
  Misconception: Thinks the Moon's phases are the Earth's shadow covering part of it.
  Feedback from the catalogue
  The Earth’s shadow falls on the Moon only in a lunar eclipse, a few times a year. The phases are something else: the Sun always lights the half of the Moon facing it, and we see a different part of that lit half depending on where the Moon is on its orbit. At full moon, precisely when the Earth is between the Moon and the Sun, we see the whole lit half. If the shadow made the phases, the line between light and dark would curve the wrong way.
- Wrong Only the side facing us, so the far side is always dark
  Misconception: Thinks the far side of the Moon, the side we never see, is always dark.
  Feedback from the catalogue
  The Moon turns on its axis once every orbit, so the same side always faces us. But the far side gets sunlight just like the near side, only at other times: at new moon, when we see a dark Moon, the far side is fully lit. “The far side” is not “the dark side”.
Feedback for a right answer
The Sun always lights half the Moon, exactly as it lights half the Earth. The phases are how much of that lit half we see from the Earth.

Hint
The Moon is a ball in the light of a distant Sun, like the Earth.

Worked solution
1. The Moon is a ball, and the Sun is far away, so at any moment it lights exactly half of it.
2. As the Moon goes round the Earth, the lit half always faces the Sun, but we see it from different angles.
3. At new moon the lit half faces away from us, and at full moon it faces us.
At new moon the far side of the Moon, which we never see, is fully lit.
Step 2 of 5
Open this step
First quarter
Practice · Calculate
Question
It is day $7.4$ ⁠ of the month, first quarter. What share of the Moon’s disc looks lit from the Earth? Give it as a per cent, the number only.

Correct answer
50.27 ± 1.5
Setup, choices and feedback
The Scene at this step
Day of the month
7.4fixed
Show tides
offfixed
Eclipse season
offfixed
Sunlight / Shadows
Sunlight
Accepted within
± 1.5
Computed from the Scene at this step’s setting
The share of the Moon's face seen from Earth that is lit, (1 − cos elongation) / 2, 0 to 1/nodes/pair/outputs/illuminated
Wrong answers it recognises
- 25 ± 1.5
  Misconception: Takes "first quarter" to mean a quarter of the Moon's disc is lit.
  Feedback from the catalogue
  The “quarter” is a quarter of the orbit, not of the disc: at first quarter the Moon has gone a quarter of the way from new moon to new moon, $90^\circ$ ⁠ from the Sun. From Earth you then see half the disc lit, a half circle. At full moon it has gone half its orbit.
Feedback for a right answer
Half: $\frac{1 - \cos 90^\circ}{2} = 50\%$ ⁠. We see the Moon side-on, and the line between light and dark runs down the middle of the disc.

Hint
The Moon is $90^\circ$ ⁠ from the Sun. How much of the lit half faces us?

Worked solution
1. After $7.4$ ⁠ days the Moon has gone a quarter of the month, $\frac{7.4}{29.5} \times 360^\circ \approx 90^\circ$ ⁠ from the Sun.
2. The lit share we see is $\frac{1 - \cos \theta}{2}$ ⁠, where $\theta$ ⁠ is the angle between the Sun and the Moon as seen from the Earth.
3. At $\theta = 90^\circ$ ⁠: $\frac{1 - 0}{2} = 50\%$ ⁠.
The “quarter” is a quarter of the way round the orbit, not a quarter of the disc.
Step 3 of 5
Open this step
Full moon
Practice · Predict, then look
Question
Where is the Moon at full moon?

Correct answer
- Correct On the side of the Earth opposite the Sun: Sun, Earth and Moon almost in a line
Setup, choices and feedback
The Scene at this step
Day of the month
7.4fixed
Show tides
offfixed
Eclipse season
offfixed
Sunlight / Shadows
Sunlight
Choices
- Correct On the side of the Earth opposite the Sun: Sun, Earth and Moon almost in a line
- Wrong Between the Earth and the Sun
  Feedback
  That is where it is at new moon: its lit half faces away from us.
- Wrong Out of the Earth’s shadow, which covers it the rest of the month
  Misconception: Thinks the Moon's phases are the Earth's shadow covering part of it.
  Feedback from the catalogue
  The Earth’s shadow falls on the Moon only in a lunar eclipse, a few times a year. The phases are something else: the Sun always lights the half of the Moon facing it, and we see a different part of that lit half depending on where the Moon is on its orbit. At full moon, precisely when the Earth is between the Moon and the Sun, we see the whole lit half. If the shadow made the phases, the line between light and dark would curve the wrong way.
- Wrong At a right angle to the line from the Earth to the Sun
  Misconception: Takes "first quarter" to mean a quarter of the Moon's disc is lit.
  Feedback from the catalogue
  The “quarter” is a quarter of the orbit, not of the disc: at first quarter the Moon has gone a quarter of the way from new moon to new moon, $90^\circ$ ⁠ from the Sun. From Earth you then see half the disc lit, a half circle. At full moon it has gone half its orbit.
Once the prediction is locked in, the Scene changes
Day of the month
14.75fixed
The Scene then shows: 180 ° · The angle Sun–Earth–Moon, round from new moon: 0 new, 90 first quarter, 180 full, 270 last quarter/nodes/pair/outputs/elongation
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$180^\circ$ ⁠ from the Sun: we are between the Sun and the Moon, looking straight at its lit half. That is why a full moon rises at sunset and sets at sunrise.

Hint
From where would you have to look at the Moon to see all of its lit half?

Worked solution
1. To see the whole lit half you have to stand on the Sun’s side, between the Sun and the Moon.
2. So at full moon the Moon is on the side opposite the Sun, $180^\circ$ ⁠ from it.
3. It is almost never exactly in line, because the orbit is tilted, so the Earth’s shadow usually misses it.
The link to the shadow is backwards: precisely when the Earth is in the middle, we see the most light.
Step 4 of 5
Open this step
Three quarters
Practice · Reach the target
Question
Change Day of the month until $75\%$ ⁠ of the disc looks lit while the Moon is waxing (between first quarter and full), then press Check.

Goal
- 0.75 ± 0.02
  At the start of the step: 0
- 3 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Day of the month
0
Show tides
offfixed
Eclipse season
offfixed
Sunlight / Shadows
Sunlight
Read from the Scene at
The share of the Moon's face seen from Earth that is lit, (1 − cos elongation) / 2, 0 to 1/nodes/pair/outputs/illuminated
Read from the Scene at
The named phase nearest: 0 new, 1 waxing crescent, 2 first quarter, 3 waxing gibbous, 4 full, 5 waning gibbous, 6 last quarter, 7 waning crescent/nodes/pair/outputs/phase
One setting that reaches it
Day of the month
10
Feedback for a right answer
About day $10$ ⁠: the Moon is $120^\circ$ ⁠ from the Sun, a waxing gibbous Moon. After full moon the same share comes back around day $19.5$ ⁠, but then the Moon is waning.

Feedback for a wrong answer
Not yet. Between first quarter ( $50\%$ ⁠) and full ( $100\%$ ⁠), where is $75\%$ ⁠?

Hint
$\frac{1 - \cos \theta}{2} = 0.75$ ⁠ means $\cos \theta = -0.5$ ⁠.

Worked solution
1. You want $\frac{1 - \cos \theta}{2} = 0.75$ ⁠, so $\cos \theta = -0.5$ ⁠ and $\theta = 120^\circ$ ⁠.
2. Each day the Moon moves $\frac{360^\circ}{29.5} \approx 12.2^\circ$ ⁠ relative to the Sun.
3. $\frac{120^\circ}{12.2^\circ} \approx 9.8$ ⁠ days: set Day of the month to $10$ ⁠.
Notice: day $5$ ⁠ is a third of the way to full, but only about a quarter of the disc is lit. The lit share does not grow in step with the days.
Step 5 of 5
Open this step
Rising later
Practice · Calculate
Question
New moon comes round every $29.5$ ⁠ days. Each day the Moon moves along its orbit, so it rises a little later than yesterday. By roughly how many minutes?

Correct answer
50 min ± 3 min
Setup, choices and feedback
The Scene at this step
Day of the month
14.75
Show tides
off
Eclipse season
off
Sunlight / Shadows
Sunlight
Accepted within
± 3 min
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 4 min ± 1 min
  Misconception: Explains day and night by the Earth's yearly trip round the Sun (or the Sun circling the Earth each day), not by the Earth spinning.
  Feedback
  About $4$ ⁠ minutes a day is how much the stars shift, from the Earth’s orbit round the Sun. The Moon goes round the Earth far faster.
Feedback for a right answer
About $49$ ⁠ minutes: the Moon moves $12.2^\circ$ ⁠ a day, and the Earth needs another $\frac{12.2^\circ}{15^\circ}$ ⁠ of an hour to “catch up”. That is why a full moon rises at sunset, but a last-quarter Moon only at midnight.

Hint
How many degrees does the Moon move in a day, and how long does the Earth take to turn one degree?

Worked solution
1. In one day the Moon moves $\frac{360^\circ}{29.5} \approx 12.2^\circ$ ⁠ relative to the Sun.
2. The Earth turns $15^\circ$ ⁠ an hour, which is $\frac{1}{4}^\circ$ ⁠ a minute.
3. The extra time: $\frac{12.2^\circ}{15^\circ} \times 60 \approx 49$ ⁠ minutes, and a little more, since the Moon keeps moving meanwhile: about $50$ ⁠ minutes.
Over $29.5$ ⁠ days the delays add up to a whole day, and the Moon rises with the Sun again: new moon.

Lesson 4 of 4 · About 30 minutes

Eclipses and tides

Why not every full moon brings a lunar eclipse, when a solar eclipse happens, and how the gravity of the Moon and the Sun lifts the ocean: two high tides a day, spring and neap tides, and why the Sun's tide is small.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
Every full moon?
Practice · Predict, then look
Question
A full moon in an ordinary month. Will the Moon pass into the Earth’s shadow, that is, will there be a lunar eclipse?

Correct answer
- Correct No: the Moon passes above or below the shadow
Setup, choices and feedback
The Scene at this step
Day of the month
14.75fixed
Show tides
offfixed
Eclipse season
offfixed
Sunlight / Shadows
Sunlight
Choices
- Correct No: the Moon passes above or below the shadow
- Wrong Yes: every full moon brings a lunar eclipse
  Misconception: Expects a lunar eclipse at every full moon and a solar eclipse at every new moon.
  Feedback from the catalogue
  The Moon’s orbit is tilted by about $5^\circ$ ⁠ to the plane in which the Earth goes round the Sun. In most months the full moon passes above or below the Earth’s shadow, and the new moon’s shadow misses the Earth. An eclipse happens only when a new or full moon falls near the points where the orbit crosses that plane, in an eclipse season, about twice a year.
- Wrong The full moon is always in the shadow, which is why it is not truly full
  Misconception: Thinks the Moon's phases are the Earth's shadow covering part of it.
  Feedback from the catalogue
  The Earth’s shadow falls on the Moon only in a lunar eclipse, a few times a year. The phases are something else: the Sun always lights the half of the Moon facing it, and we see a different part of that lit half depending on where the Moon is on its orbit. At full moon, precisely when the Earth is between the Moon and the Sun, we see the whole lit half. If the shadow made the phases, the line between light and dark would curve the wrong way.
Once the prediction is locked in, the Scene changes
Sunlight / Shadows
Shadows
The Scene then shows: 0 · 0 none; 1 lunar, the Moon at least partly inside Earth's shadow cone; 2 solar, the Moon in front of some of the Earth's disc seen from the Sun/nodes/pair/outputs/eclipse
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The orbit is tilted, and the full moon passes above the shadow. In the Scene the tilt is exaggerated, $25^\circ$ ⁠, because the distances are squeezed; really it is $5^\circ$ ⁠, but the Moon is $60$ ⁠ Earth radii away, so even $5^\circ$ ⁠ is enough to miss.

Hint
Look at the Moon’s orbit: is it flat in the same plane as the Sun?

Worked solution
1. The Earth’s shadow is a cone pointing away from the Sun, in the plane of the Earth’s orbit.
2. The Moon’s orbit is tilted to that plane, so in most months the full moon is above or below it.
3. At a distance of 60 Earth radii, $5^\circ$ ⁠ is about $5$ ⁠ Earth radii, far more than the width of the shadow.
That is why a lunar eclipse happens only a few times a year, not every month.
Step 2 of 7
Open this step
Eclipse season
Practice · Predict, then look
Question
In the eclipse season the orbit’s crossing line points at the Sun. Which eclipse happens at new moon, on day $0$ ⁠?

Correct answer
- Correct A solar eclipse: the Moon’s shadow falls on the Earth
Setup, choices and feedback
The Scene at this step
Day of the month
14.75fixed
Show tides
offfixed
Eclipse season
onfixed
Sunlight / Shadows
Shadowsfixed
Choices
- Correct A solar eclipse: the Moon’s shadow falls on the Earth
- Wrong A lunar eclipse: the Earth’s shadow falls on the Moon
  Feedback
  At new moon the Moon is between the Earth and the Sun, so the Earth cannot cast a shadow on it.
- Wrong There is no eclipse at new moon
  Misconception: Expects a lunar eclipse at every full moon and a solar eclipse at every new moon.
  Feedback from the catalogue
  The Moon’s orbit is tilted by about $5^\circ$ ⁠ to the plane in which the Earth goes round the Sun. In most months the full moon passes above or below the Earth’s shadow, and the new moon’s shadow misses the Earth. An eclipse happens only when a new or full moon falls near the points where the orbit crosses that plane, in an eclipse season, about twice a year.
Once the prediction is locked in, the Scene changes
Day of the month
0fixed
Sunlight / Shadows
Shadowsfixed
The Scene then shows: 2 · 0 none; 1 lunar, the Moon at least partly inside Earth's shadow cone; 2 solar, the Moon in front of some of the Earth's disc seen from the Sun/nodes/pair/outputs/eclipse
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A solar eclipse: the Moon hides the Sun, and its patch of shadow sweeps across the Earth. A lunar eclipse happens in the same season, but at full moon, two weeks before or after.

Hint
At new moon, who stands in the way of whose sunlight?

Worked solution
1. At new moon the order is Sun, Moon, Earth.
2. In an eclipse season the new moon sits exactly on the line, so its shadow reaches the Earth: a solar eclipse.
3. At full moon the order is reversed, Sun, Earth, Moon, and the Moon enters the Earth’s shadow: a lunar eclipse.
The Moon’s shadow is small and barely reaches the Earth, so a total solar eclipse is seen only along a narrow track.
Step 3 of 7
Open this step
Into the shadow
Practice · Reach the target
Question
The eclipse season goes on. Change Day of the month until the Moon enters the Earth’s shadow, then press Check.

Goal
- 1 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Day of the month
7.5
Show tides
offfixed
Eclipse season
onfixed
Sunlight / Shadows
Shadows
Read from the Scene at
0 none; 1 lunar, the Moon at least partly inside Earth's shadow cone; 2 solar, the Moon in front of some of the Earth's disc seen from the Sun/nodes/pair/outputs/eclipse
One setting that reaches it
Day of the month
14.5
Feedback for a right answer
Near full moon, day $14.5$ ⁠ to $15$ ⁠: the Moon enters the shadow and turns red. It does not vanish, because the Earth’s air bends red light onto it, the light of every sunset and sunrise at once.

Feedback for a wrong answer
Not yet. The eclipse is brief: look very close to full moon.

Hint
The Earth’s shadow points away from the Sun. At which phase is the Moon there?

Worked solution
1. The Earth’s shadow is on the side opposite the Sun, so a lunar eclipse is possible only at full moon.
2. In an eclipse season the full moon passes through the shadow.
3. Set Day of the month to $14.5$ ⁠ or $15$ ⁠; half a day further and the Moon is out of the shadow.
Step 4 of 7
Open this step
Two bulges
Practice · Choose
Question
How many times a day does the water rise to high tide on an open seashore?

Correct answer
- Correct Twice
Setup, choices and feedback
The Scene at this step
Day of the month
7.4
Show tides
off
Eclipse season
off
Sunlight / Shadows
Sunlight
Choices
Shown in a random order.
- Correct Twice
- Wrong Once, when the Moon is overhead
  Misconception: Thinks the Moon raises one bulge of water, on the side facing it, so there is one high tide a day.
  Feedback from the catalogue
  The Moon pulls harder on what is nearer to it. The ocean on the near side is pulled more than the Earth’s centre, and the Earth’s centre more than the ocean on the far side. So the water stretches both ways: one bulge towards the Moon and one on the opposite side. The Earth turns beneath both, and every shore passes through two bulges a day: two high tides and two low ones.
- Wrong Four times, twice from the Moon and twice from the Sun
  Feedback
  The Sun’s and the Moon’s tides are not separate: they add into one tide, bigger or smaller with the phase.
Feedback for a right answer
Two bulges, one towards the Moon and one on the far side, and every shore passes through both as the Earth turns.

Hint
Switch on Show tides: how many bulges does the water have?

Worked solution
1. The Moon’s pull weakens with distance from it.
2. The water on the near side is pulled more than the Earth’s centre, and piles up towards the Moon.
3. The Earth’s centre is pulled more than the water on the far side, and “moves away” from it: it piles up in a second bulge.
4. The Earth turns beneath both bulges, so every shore has two high tides and two low tides a day.
Step 5 of 7
Open this step
Tide after tide
Practice · Calculate
Question
The Moon returns to the same place in the sky every 24 h 50 min, because meanwhile it moves along its orbit. How long is it from one high tide to the next? Write it in minutes.

Correct answer
745 min ± 5 min
Setup, choices and feedback
The Scene at this step
Day of the month
7.4
Show tides
on
Eclipse season
off
Sunlight / Shadows
Sunlight
Accepted within
± 5 min
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1490 min ± 10 min
  Misconception: Thinks the Moon raises one bulge of water, on the side facing it, so there is one high tide a day.
  Feedback from the catalogue
  The Moon pulls harder on what is nearer to it. The ocean on the near side is pulled more than the Earth’s centre, and the Earth’s centre more than the ocean on the far side. So the water stretches both ways: one bulge towards the Moon and one on the opposite side. The Earth turns beneath both, and every shore passes through two bulges a day: two high tides and two low ones.
Feedback for a right answer
Half of 24 h 50 min: 12 h 25 min, which is $745$ ⁠ minutes. That is why high tide comes about $50$ ⁠ minutes later every day.

Hint
Each lunar day has two high tides.

Worked solution
1. A lunar day: 24 h 50 min $= 24 \times 60 + 50 = 1490$ ⁠ minutes.
2. In each lunar day the shore passes through two bulges.
3. From high tide to high tide: $\frac{1490}{2} = 745$ ⁠ minutes, which is 12 h 25 min.
If you answered $1490$ ⁠, you counted one high tide a day, as if there were only one bulge.
Step 6 of 7
Open this step
Spring and neap tides
Practice · Predict, then look
Question
The Sun raises a tide too, a weaker one. It is first quarter now. What happens to the gap between high and low tide when the Moon reaches new moon?

Correct answer
- Correct It grows: the Sun’s tide adds to the Moon’s
Setup, choices and feedback
The Scene at this step
Day of the month
7.4fixed
Show tides
onfixed
Eclipse season
offfixed
Sunlight / Shadows
Sunlight
Choices
- Correct It grows: the Sun’s tide adds to the Moon’s
- Wrong It shrinks: the Sun pulls a different way from the Moon
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Wrong It stays the same: only the Moon makes tides
  Misconception: Expects the Sun, whose pull on the Earth is far stronger than the Moon's, to raise the bigger tides.
  Feedback from the catalogue
  A tide comes not from the pull itself but from the difference in pull between the near and the far side of the Earth. That difference falls off like $\frac{M}{r^3}$ ⁠, faster than the pull, which falls off like $\frac{M}{r^2}$ ⁠. The Sun pulls about $180$ ⁠ times harder than the Moon, but it is about $390$ ⁠ times farther away, so its tide is only about $0.46$ ⁠ of the Moon’s.
Once the prediction is locked in, the Scene changes
Day of the month
0fixed
Show tides
onfixed
The Scene then shows: 1.46 · The tidal range compared with the Moon's tide alone: the Sun's tide, 0.46 of the Moon's, adds to it at new and full moon (spring tides) and takes from it at the quarters (neap tides)/nodes/pair/outputs/tideRange
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At new and full moon the Sun’s and the Moon’s bulges line up and add: a spring tide, $1.46$ ⁠ times the Moon’s tide alone. At the quarters they are at a right angle and partly cancel: a neap tide, only $0.54$ ⁠ times.

Hint
At new moon the Sun and the Moon are in the same direction. At first quarter they are at a right angle.

Worked solution
1. The Sun raises two bulges of its own, $0.46$ ⁠ the size of the Moon’s, along the line to the Sun.
2. At new and full moon the two lines coincide, and the bulges add: $1 + 0.46 = 1.46$ ⁠.
3. At the quarters the lines are at right angles, and one’s bulge falls on the other’s low: $1 - 0.46 = 0.54$ ⁠.
4. So the gap between high and low tide at new moon is $\frac{1.46}{0.54} \approx 2.7$ ⁠ times that at first quarter.
A spring tide has nothing to do with spring: it comes twice every month.
Step 7 of 7
Open this step
The Sun's tide
Practice · Choose
Question
The Sun pulls on the Earth about $180$ ⁠ times harder than the Moon does. So why is its tide only about half the Moon’s?

Correct answer
- Correct A tide comes from the difference in pull between the near and far sides of the Earth, and that difference shrinks fast with distance
Setup, choices and feedback
The Scene at this step
Day of the month
0
Show tides
on
Eclipse season
off
Sunlight / Shadows
Sunlight
Choices
Shown in a random order.
- Correct A tide comes from the difference in pull between the near and far sides of the Earth, and that difference shrinks fast with distance
- Wrong The claim is wrong: if the Sun pulls harder, its tide is bigger
  Misconception: Expects the Sun, whose pull on the Earth is far stronger than the Moon's, to raise the bigger tides.
  Feedback from the catalogue
  A tide comes not from the pull itself but from the difference in pull between the near and the far side of the Earth. That difference falls off like $\frac{M}{r^3}$ ⁠, faster than the pull, which falls off like $\frac{M}{r^2}$ ⁠. The Sun pulls about $180$ ⁠ times harder than the Moon, but it is about $390$ ⁠ times farther away, so its tide is only about $0.46$ ⁠ of the Moon’s.
- Wrong The Sun acts only on the day side
  Feedback
  Gravity acts on the whole Earth, day and night. The Sun has two bulges too, one on each side.
Feedback for a right answer
The difference in pull across the Earth falls off like $\frac{M}{r^3}$ ⁠. The Sun is $2.7 \times 10^7$ ⁠ times as massive as the Moon but $390$ ⁠ times as far, and $\frac{2.7 \times 10^7}{390^3} \approx 0.46$ ⁠.

Hint
The pull falls off as $\frac{1}{r^2}$ ⁠. How much does it change across the Earth’s diameter when $r$ ⁠ is very large?

Worked solution
1. The pull of a body of mass $M$ ⁠ at distance $r$ ⁠ is $\frac{GM}{r^2}$ ⁠ per kilogram.
2. On the near side the distance is $r - R$ ⁠ and on the far side $r + R$ ⁠, so the difference is about $\frac{2GMR}{r^3}$ ⁠ each side: it falls off like $\frac{1}{r^3}$ ⁠.
3. The ratio of masses is $2.7 \times 10^7$ ⁠, of distances $390$ ⁠, so the ratio of tides is $\frac{2.7 \times 10^7}{390^3} \approx 0.46$ ⁠.
The ratio of the pulls themselves: $\frac{2.7 \times 10^7}{390^2} \approx 180$ ⁠. The third power is what makes the difference.

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

Day and night

A turning globe

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Degree by degree

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Equal night

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

The noon Sun

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The midnight Sun

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

The seasons

Closest to the Sun

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

The tilted axis

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

A high Sun

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Long days

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Hint

Worked solution

Light spread thin

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The other hemisphere

Question

Correct answer

The Scene at this step

Choices