Answer key · Unit 9

Circular motion and gravity.

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

Acceleration towards the centre

A ball going round a circle at a steady speed is accelerating all the time. Which way, by how much, and how long does one turn take?

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Letting go
Practice · Predict, then look
Question
The ball is going round at 2 m/s on a string 0.6 m long. The string is cut. What path will the ball take?

Correct answer
- Correct A straight line along the tangent, the way it was moving when the string was cut
Setup, choices and feedback
The Scene at this step
Speed
2 m/s
String length
0.6 m
Forces, then cut
Whirlinghidden
Choices
- Correct A straight line along the tangent, the way it was moving when the string was cut
- Wrong A straight line outward, away from the peg, in line with the string
  Misconception: Thinks a ball let go from a circle flies straight outward, away from the centre.
  Feedback from the catalogue
  The moment the string is cut there is no sideways force on the ball at all, and by Newton’s first law it carries on with the velocity it had at that instant: along the tangent to the circle, at right angles to the string. Nothing was pushing it outward before, so nothing throws it outward now. It does move away from the peg, but only because it goes straight on while the circle would have kept curving.
- Wrong It keeps curving, on a widening circle
  Misconception: Thinks a ball let go from a circle keeps curving round, as if it remembered the circle.
  Feedback from the catalogue
  The ball has no memory of the circle. Its path curved only because the string pulled it inward at every moment. Without that pull nothing bends the path, and the ball goes straight on along the tangent, at a steady speed.
- Wrong It stops where it is
  Feedback
  Nothing stops the ball: the table is smooth, and with no force its velocity stays as it was.
Once the prediction is locked in, the Scene changes
Forces, then cut
Cut the stringhidden
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
With no pull, nothing bends the path. The ghosts, at equal spacing, show the ball carrying on along the tangent in a straight line and at the same speed.

Worked solution
1. Before the cut, the only sideways force on the ball is the string’s pull, towards the peg.
2. When the string is cut, that force is gone too. The weight and the table’s push balance each other, so the net force on the ball is zero.
3. By Newton’s first law, a body with no net force on it carries on at a constant velocity in a straight line. At the moment of the cut it is moving at 2 m/s along the tangent, so it carries on like that until the table’s rim stops it.
Choosing “outward” means imagining a force pushing the ball away from the centre; but no such force was acting on it before either.
Step 2 of 6
Open this step
The velocity turns
Practice · Choose
Question
The ball moves on a circle at 2 m/s, and the size of its velocity does not change. Which way does its acceleration point?

Correct answer
- Correct Towards the peg, at the centre of the circle
Setup, choices and feedback
The Scene at this step
Speed
2 m/s
String length
0.6 m
Forces, then cut
0.4 (Whirling → Cut the string)fixed
Choices
Shown in a random order.
- Correct Towards the peg, at the centre of the circle
- Wrong It has no acceleration, because its speed is constant
  Misconception: Thinks a body moving at a constant speed has no acceleration, even on a circle.
  Feedback from the catalogue
  Acceleration is the rate of change of velocity, and velocity is a vector: a size and a direction. On a circle the size stays the same, but the direction changes all the time, so there is an acceleration. It points towards the centre, and its size is $a = \dfrac{v^2}{r}$ ⁠.
- Wrong Forward, in the direction of motion
  Misconception: Thinks the acceleration, or a force, on a body moving on a circle points forward along its velocity.
  Feedback from the catalogue
  A force or an acceleration along the motion would change the size of the velocity, and here the size is constant. To change only the direction, the acceleration has to be at right angles to the motion: along the radius, towards the centre. Nothing needs to push the ball forward; without friction it keeps moving on its own.
- Wrong Outward, away from the peg
  Misconception: Believes an outward centrifugal force acts on the circling body and balances the inward pull.
  Feedback from the catalogue
  If an outward force balanced the inward pull, the net force would be zero and the body would move in a straight line. It curves precisely because there is a net force inward. What you feel as an outward pull is the ball pulling on the string and your hand (Newton’s third law), or your own body carrying straight on while the car turns underneath it.
Feedback for a right answer
The velocity changes direction at every moment, and that change points towards the centre, like the orange arrow. That is the centripetal acceleration.

Worked solution
1. Acceleration is the change in velocity per second, $\vec a = \dfrac{\Delta \vec v}{\Delta t}$ ⁠, and velocity is a vector.
2. The size of the velocity is constant, 2 m/s, so the acceleration has no part along the motion; only the direction changes.
3. Changing the direction alone takes an acceleration at right angles to the velocity. The velocity is along the tangent, so the acceleration is along the radius.
4. The string, the only force in the plane of the table, pulls towards the peg, and by $\vec F = m\vec a$ ⁠ the acceleration points the same way: towards the centre.
The common slip is “no acceleration because the speed is constant”: only the size of the velocity is constant here.
Step 3 of 6
Open this step
One turn
Practice · Calculate
Question
The ball moves at 2.5 m/s on a string 0.8 m long. How long does one turn take?

Correct answer
2.011 s ± 2%
Setup, choices and feedback
The Scene at this step
Speed
2.5 m/sfixed
String length
0.8 mfixed
Forces, then cut
Whirlinghidden
Accepted within
± 2%
Computed from the Scene at this step’s setting
Seconds for one turn, 2πr/v/nodes/whirl/outputs/period
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.32 s ± 3%
  Misconception: Divides the radius by the speed for the period, forgetting the circumference 2πr.
  Feedback from the catalogue
  In one turn the ball covers the whole circumference of the circle, $2\pi r$ ⁠, not one radius. So $T = \dfrac{2\pi r}{v}$ ⁠. Dividing $\dfrac{r}{v}$ ⁠ gives only the time to cover one radius, about a sixth of a turn.
Feedback for a right answer
$T = \dfrac{2\pi r}{v} = \dfrac{2\pi \cdot 0.8}{2.5} \approx 2.01\ \mathrm{s}$ ⁠: about half a turn a second.

Hint
In one turn the ball covers the circumference, $2\pi r$ ⁠, at speed $v$ ⁠.

Worked solution
1. The givens: $v = 2.5\ \mathrm{m/s}$ ⁠ and $r = 0.8\ \mathrm{m}$ ⁠.
2. In one turn the ball covers the circumference, so $T = \dfrac{2\pi r}{v}$ ⁠.
3. Put in the numbers: $T = \dfrac{2\pi \cdot 0.8}{2.5} \approx \dfrac{5.03}{2.5} \approx 2.01\ \mathrm{s}$ ⁠.
4. A check through the angular speed: $\omega = \dfrac{v}{r} = \dfrac{2.5}{0.8} = 3.125\ \mathrm{rad/s}$ ⁠, and $\dfrac{2\pi}{3.125} \approx 2.01\ \mathrm{s}$ ⁠.
If you got 0.32 s, you divided the radius by the speed and forgot that a turn covers the whole circumference, $2\pi$ ⁠ radii.
Step 4 of 6
Open this step
How fast the velocity turns
Practice · Calculate
Question
The ball moves at 3 m/s on a string 0.5 m long. What is the size of its acceleration?

Correct answer
18 m/s² ± 2%
Setup, choices and feedback
The Scene at this step
Speed
3 m/sfixed
String length
0.5 mfixed
Forces, then cut
Whirlinghidden
Accepted within
± 2%
Computed from the Scene at this step’s setting
Acceleration towards the peg, v²/r/nodes/whirl/outputs/acceleration
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 6 m/s² ± 3%
  Misconception: Uses v/r or v²·r instead of v²/r for the centripetal acceleration.
  Feedback from the catalogue
  Check the units. $\dfrac{v}{r}$ ⁠ gives $\dfrac{\mathrm{m/s}}{\mathrm{m}} = \mathrm{1/s}$ ⁠: that is the angular speed, in radians per second, not an acceleration. $v^2 r$ ⁠ gives $\mathrm{m^3/s^2}$ ⁠. Only $\dfrac{v^2}{r}$ ⁠ gives $\dfrac{\mathrm{m^2/s^2}}{\mathrm{m}} = \mathrm{m/s^2}$ ⁠, a unit of acceleration.
- 4.5 m/s² ± 3%
  Misconception: Uses v/r or v²·r instead of v²/r for the centripetal acceleration.
  Feedback from the catalogue
  Check the units. $\dfrac{v}{r}$ ⁠ gives $\dfrac{\mathrm{m/s}}{\mathrm{m}} = \mathrm{1/s}$ ⁠: that is the angular speed, in radians per second, not an acceleration. $v^2 r$ ⁠ gives $\mathrm{m^3/s^2}$ ⁠. Only $\dfrac{v^2}{r}$ ⁠ gives $\dfrac{\mathrm{m^2/s^2}}{\mathrm{m}} = \mathrm{m/s^2}$ ⁠, a unit of acceleration.
Feedback for a right answer
$a = \dfrac{3^2}{0.5} = 18\ \mathrm{m/s^2}$ ⁠, towards the centre: almost twice the acceleration of free fall.

Hint
$a = \dfrac{v^2}{r}$ ⁠: the speed squared, divided by the radius.

Worked solution
1. The givens: $v = 3\ \mathrm{m/s}$ ⁠ and $r = 0.5\ \mathrm{m}$ ⁠.
2. The centripetal acceleration: $a = \dfrac{v^2}{r}$ ⁠.
3. Put in the numbers: $a = \dfrac{3^2}{0.5} = \dfrac{9}{0.5} = 18\ \mathrm{m/s^2}$ ⁠, towards the peg.
That is almost twice the acceleration of free fall. If you got $6$ ⁠, you worked out $\dfrac{v}{r}$ ⁠, which is the angular speed in radians per second, not an acceleration; if you got $4.5$ ⁠, you multiplied by $r$ ⁠ instead of dividing.
Step 5 of 6
Open this step
Twice the speed
Practice · Predict, then look
Question
On this circle the acceleration is 3.75 m/s². The speed is doubled, from 1.5 m/s to 3 m/s, on the same string. What will the acceleration be?

Correct answer
- Correct 15 m/s², four times as much
Setup, choices and feedback
The Scene at this step
Speed
1.5 m/sfixed
String length
0.6 mfixed
Forces, then cut
0.4 (Whirling → Cut the string)fixed
Choices
- Wrong 3.75 m/s², as before
  Feedback
  The size of the velocity changed, so the acceleration changes too: the direction has to change faster.
- Wrong 7.5 m/s², twice as much
  Misconception: Thinks the centripetal acceleration, and the pull, grow in proportion to the speed: twice the speed, twice the pull.
  Feedback from the catalogue
  The speed comes in squared: $a = \dfrac{v^2}{r}$ ⁠. At twice the speed the direction has to change twice as fast, and there is twice as much velocity to turn, so the acceleration, and the pull with it, grow $2^2 = 4$ ⁠ times.
- Correct 15 m/s², four times as much
- Wrong 30 m/s², eight times as much
  Feedback
  The speed comes in squared, not cubed: $2^2 = 4$ ⁠.
Once the prediction is locked in, the Scene changes
Speed
3 m/sfixed
The Scene then shows: 15 m/s² · Acceleration towards the peg, v²/r/nodes/whirl/outputs/acceleration
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Four times: $\dfrac{3^2}{0.6} = 15\ \mathrm{m/s^2}$ ⁠. The pull on the orange arrow grew four times too, from 0.75 N to 3 N.

Worked solution
1. Before: $a = \dfrac{1.5^2}{0.6} = 3.75\ \mathrm{m/s^2}$ ⁠.
2. After: $a = \dfrac{3^2}{0.6} = \dfrac{9}{0.6} = 15\ \mathrm{m/s^2}$ ⁠.
3. The ratio: $\dfrac{15}{3.75} = 4 = 2^2$ ⁠, because $a$ ⁠ depends on $v^2$ ⁠.
At twice the speed the ball goes round the peg twice as fast, so the blue arrow turns twice as fast, and it is also twice as long. Together the two give four times, not twice.
Step 6 of 6
Open this step
An acceleration of ten
Practice · Reach the target
Question
Change the Speed and the String length until the ball’s acceleration is 10 m/s², then press Check.

Goal
- 10 m/s² ± 0.25 m/s²
  At the start of the step: 1.667 m/s²
Setup, choices and feedback
The Scene at this step
Speed
1 m/s
String length
0.6 m
Forces, then cut
0.4 (Whirling → Cut the string)fixed
Read from the Scene at
Acceleration towards the peg, v²/r/nodes/whirl/outputs/acceleration
One setting that reaches it
Speed
2 m/s
String length
0.4 m
Feedback for a right answer
There is more than one way: any speed and length with $v^2 = 10\,r$ ⁠. A shorter string needs a lower speed.

Feedback for a wrong answer
Not yet. If the acceleration is too small, raise the speed or shorten the string; if it is too big, do the opposite.

Hint
You need $v^2 = 10\,r$ ⁠. Pick a string length, and work out the speed that goes with it.

Worked solution
1. You want $a = \dfrac{v^2}{r} = 10\ \mathrm{m/s^2}$ ⁠, that is, $v^2 = 10\, r$ ⁠.
2. Pick a string length where $10\,r$ ⁠ is a handy square: at $r = 0.4\ \mathrm{m}$ ⁠ you need $v^2 = 4$ ⁠, so $v = 2\ \mathrm{m/s}$ ⁠.
3. Check: $\dfrac{2^2}{0.4} = 10\ \mathrm{m/s^2}$ ⁠.
$r = 0.9\ \mathrm{m}$ ⁠ with $v = 3\ \mathrm{m/s}$ ⁠ works too: $\dfrac{9}{0.9} = 10$ ⁠. Turning up only the speed on the 0.6 m string jumps past the target: at 2.4 m/s the acceleration is 9.6 m/s², and at 2.5 m/s it is already about 10.4 m/s².

Lesson 2 of 5 · About 25 minutes

The force that holds it on the circle

Moving on a circle takes a net force towards the centre, of size m·v²/r. What supplies it, how strong is it, and why is there no force pushing outward?

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
What pulls the ball
Practice · Choose
Question
Look at the ball from above while it goes round. Which sideways forces act on it?

Correct answer
- Correct Only the string’s pull, towards the peg
Setup, choices and feedback
The Scene at this step
Speed
2 m/s
String length
0.6 m
Forces, then cut
0.4 (Whirling → Cut the string)fixed
Choices
Shown in a random order.
- Correct Only the string’s pull, towards the peg
- Wrong The string’s pull inward, and an equal centrifugal force outward
  Misconception: Believes an outward centrifugal force acts on the circling body and balances the inward pull.
  Feedback from the catalogue
  If an outward force balanced the inward pull, the net force would be zero and the body would move in a straight line. It curves precisely because there is a net force inward. What you feel as an outward pull is the ball pulling on the string and your hand (Newton’s third law), or your own body carrying straight on while the car turns underneath it.
- Wrong The string’s pull inward, and a forward force that keeps the ball moving
  Misconception: Thinks the acceleration, or a force, on a body moving on a circle points forward along its velocity.
  Feedback from the catalogue
  A force or an acceleration along the motion would change the size of the velocity, and here the size is constant. To change only the direction, the acceleration has to be at right angles to the motion: along the radius, towards the centre. Nothing needs to push the ball forward; without friction it keeps moving on its own.
- Wrong The string’s pull inward, and a centripetal force inward as well
  Misconception: Adds the centripetal force as a force of its own, on top of the real forces on the body.
  Feedback from the catalogue
  Centripetal force is not an extra kind of force but the name of a job: the net force pointing at the centre. Here the string’s pull does that job; for a car going round a bend, the friction between tyres and road; for the Moon, gravity. In a free-body diagram you draw only the real forces, and check that their sum points at the centre and equals $\dfrac{m v^2}{r}$ ⁠.
Feedback for a right answer
Only the string. Its pull is the centripetal force: the net force, towards the centre, that turns the velocity.

Worked solution
1. What touches the ball? The table underneath and the string. And from a distance the Earth pulls it down.
2. The weight and the table’s push are vertical and balance each other, because the ball neither rises nor sinks.
3. In the plane of the table only the string is left, and a string can only pull along itself: towards the peg.
4. Check: the ball accelerates towards the centre, $a = \dfrac{v^2}{r}$ ⁠, so the net force has to point at the centre with size $\dfrac{m v^2}{r} = \dfrac{0.2 \cdot 2^2}{0.6} \approx 1.33\ \mathrm{N}$ ⁠. That is exactly the number on the orange arrow.
A centrifugal force outward would cancel the net force, and the ball would move in a straight line. And the centripetal force is the name of the job the string’s pull does here, not an extra force.
Step 2 of 5
Open this step
How strong
Practice · Calculate
Question
A 0.5 kg ball goes round at 2 m/s on a string 0.8 m long. How hard does the string pull on it?

Correct answer
2.5 N ± 2%
Setup, choices and feedback
The Scene at this step
Speed
2 m/sfixed
String length
0.8 mfixed
Forces, then cut
Whirlinghidden
Other settings
Mass/nodes/whirl/props/mass = 0.5
Accepted within
± 2%
Computed from the Scene at this step’s setting
The string's pull on the ball, m·v²/r/nodes/whirl/outputs/pull
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1.25 N ± 3%
  Misconception: Uses v/r or v²·r instead of v²/r for the centripetal acceleration.
  Feedback from the catalogue
  Check the units. $\dfrac{v}{r}$ ⁠ gives $\dfrac{\mathrm{m/s}}{\mathrm{m}} = \mathrm{1/s}$ ⁠: that is the angular speed, in radians per second, not an acceleration. $v^2 r$ ⁠ gives $\mathrm{m^3/s^2}$ ⁠. Only $\dfrac{v^2}{r}$ ⁠ gives $\dfrac{\mathrm{m^2/s^2}}{\mathrm{m}} = \mathrm{m/s^2}$ ⁠, a unit of acceleration.
- 1.6 N ± 3%
  Misconception: Uses v/r or v²·r instead of v²/r for the centripetal acceleration.
  Feedback from the catalogue
  Check the units. $\dfrac{v}{r}$ ⁠ gives $\dfrac{\mathrm{m/s}}{\mathrm{m}} = \mathrm{1/s}$ ⁠: that is the angular speed, in radians per second, not an acceleration. $v^2 r$ ⁠ gives $\mathrm{m^3/s^2}$ ⁠. Only $\dfrac{v^2}{r}$ ⁠ gives $\dfrac{\mathrm{m^2/s^2}}{\mathrm{m}} = \mathrm{m/s^2}$ ⁠, a unit of acceleration.
Feedback for a right answer
$F = \dfrac{m v^2}{r} = \dfrac{0.5 \cdot 4}{0.8} = 2.5\ \mathrm{N}$ ⁠, towards the peg.

Hint
First the acceleration, $a = \dfrac{v^2}{r}$ ⁠. Then Newton’s second law, $F = m a$ ⁠.

Worked solution
1. The givens: $m = 0.5\ \mathrm{kg}$ ⁠, $v = 2\ \mathrm{m/s}$ ⁠ and $r = 0.8\ \mathrm{m}$ ⁠.
2. The acceleration: $a = \dfrac{v^2}{r} = \dfrac{4}{0.8} = 5\ \mathrm{m/s^2}$ ⁠, towards the peg.
3. The force, from Newton’s second law: $F = m a = 0.5 \cdot 5 = 2.5\ \mathrm{N}$ ⁠.
Or in one go: $F = \dfrac{m v^2}{r} = \dfrac{0.5 \cdot 2^2}{0.8} = 2.5\ \mathrm{N}$ ⁠. If you got 1.25 N you forgot to square the speed, and if you got 1.6 N you multiplied by $r$ ⁠ instead of dividing.
Step 3 of 5
Open this step
A pull of two newtons
Practice · Reach the target
Question
The 0.2 kg ball. Change the Speed and the String length until the string pulls with 2 N, then press Check.

Goal
- 2 N ± 0.05 N
  At the start of the step: 0.3333 N
Setup, choices and feedback
The Scene at this step
Speed
1 m/s
String length
0.6 m
Forces, then cut
0.4 (Whirling → Cut the string)fixed
Read from the Scene at
The string's pull on the ball, m·v²/r/nodes/whirl/outputs/pull
One setting that reaches it
Speed
2 m/s
String length
0.4 m
Feedback for a right answer
The string pulls with 2 N: the acceleration from the last lesson, 10 m/s², times the mass, 0.2 kg.

Feedback for a wrong answer
Not yet. Look at the number on the orange arrow: if it is too small, raise the speed or shorten the string.

Hint
$F = \dfrac{m v^2}{r}$ ⁠, so you need $\dfrac{v^2}{r} = \dfrac{2}{0.2} = 10$ ⁠.

Worked solution
1. You want $\dfrac{m v^2}{r} = 2\ \mathrm{N}$ ⁠ with $m = 0.2\ \mathrm{kg}$ ⁠, that is, an acceleration of $a = \dfrac{F}{m} = 10\ \mathrm{m/s^2}$ ⁠.
2. So you need $v^2 = 10\, r$ ⁠: for example $r = 0.4\ \mathrm{m}$ ⁠ and $v = 2\ \mathrm{m/s}$ ⁠.
3. Check: $\dfrac{0.2 \cdot 2^2}{0.4} = 2\ \mathrm{N}$ ⁠, which is what the orange arrow shows.
$r = 0.9\ \mathrm{m}$ ⁠ with $v = 3\ \mathrm{m/s}$ ⁠ also gives 2 N. At the start, at 1 m/s, the string pulls with only 0.33 N: to get six times as much, raising the speed $\sqrt{6} \approx 2.4$ ⁠ times is enough, because it comes in squared.
Step 4 of 5
Open this step
A longer string
Practice · Predict, then look
Question
The ball moves at 2 m/s on a string 0.5 m long, and the string pulls with 1.6 N. The string is lengthened to 1 m, at the same speed. What happens to the pull?

Correct answer
- Correct It halves, to 0.8 N
Setup, choices and feedback
The Scene at this step
Speed
2 m/sfixed
String length
0.5 mfixed
Forces, then cut
0.4 (Whirling → Cut the string)fixed
Choices
- Correct It halves, to 0.8 N
- Wrong It doubles, to 3.2 N
  Misconception: Thinks that at the same speed a bigger circle needs a bigger pull.
  Feedback from the catalogue
  At the same speed, a bigger circle is a gentler turn: the direction changes more slowly. So $F = \dfrac{m v^2}{r}$ ⁠ gets smaller as the radius grows, and a string twice as long needs only half the pull.
- Wrong It stays the same
  Feedback
  The speed is the same, but the turn is gentler: the direction changes more slowly, so less pull is needed.
- Wrong It drops to a quarter, 0.4 N
  Feedback
  The radius is underneath to the first power, not squared: $F = \dfrac{m v^2}{r}$ ⁠.
Once the prediction is locked in, the Scene changes
String length
1 mfixed
The Scene then shows: 0.8 N · The string's pull on the ball, m·v²/r/nodes/whirl/outputs/pull
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Half: $F = \dfrac{0.2 \cdot 2^2}{1} = 0.8\ \mathrm{N}$ ⁠. At the same speed, a big circle is a gentler turn.

Worked solution
1. Before: $F = \dfrac{m v^2}{r} = \dfrac{0.2 \cdot 4}{0.5} = 1.6\ \mathrm{N}$ ⁠.
2. After: $F = \dfrac{0.2 \cdot 4}{1} = 0.8\ \mathrm{N}$ ⁠.
3. The radius is in the denominator, so twice the radius gives half the force.
At the same speed, a bigger circle is a gentler turn: the period is twice as long, and the velocity turns half as fast. The feeling that a long string pulls harder comes from whirling at the same rate of turns: then the ball on the long string moves faster, $v = \omega r$ ⁠, and the pull $m \omega^2 r$ ⁠ really does grow.
Step 5 of 5
Open this step
Who does the string's job
Practice · Choose
Question
A car drives at a steady speed round a roundabout, on a level road. Which force keeps it on the circle?

Correct answer
- Correct The friction between the tyres and the road, sideways, towards the middle of the roundabout
Setup, choices and feedback
The Scene at this step
Speed
2 m/s
String length
0.6 m
Forces, then cut
0.4 (Whirling → Cut the string)
Choices
Shown in a random order.
- Correct The friction between the tyres and the road, sideways, towards the middle of the roundabout
- Wrong The centrifugal force
  Misconception: Believes an outward centrifugal force acts on the circling body and balances the inward pull.
  Feedback from the catalogue
  If an outward force balanced the inward pull, the net force would be zero and the body would move in a straight line. It curves precisely because there is a net force inward. What you feel as an outward pull is the ball pulling on the string and your hand (Newton’s third law), or your own body carrying straight on while the car turns underneath it.
- Wrong The engine’s force, pushing the car forward
  Misconception: Thinks the acceleration, or a force, on a body moving on a circle points forward along its velocity.
  Feedback from the catalogue
  A force or an acceleration along the motion would change the size of the velocity, and here the size is constant. To change only the direction, the acceleration has to be at right angles to the motion: along the radius, towards the centre. Nothing needs to push the ball forward; without friction it keeps moving on its own.
- Wrong The car’s weight
  Feedback
  The weight points down, and the road balances it. A vertical force cannot bend a horizontal path.
Feedback for a right answer
Friction points sideways, towards the middle of the roundabout, and it is the centripetal force here. On ice there is not enough friction, and the car carries straight on.

Worked solution
1. Like the ball, the car is accelerating towards the middle of the roundabout, $a = \dfrac{v^2}{r}$ ⁠, so it needs a net force towards the centre.
2. The weight and the road’s normal force are vertical and balance each other.
3. The only horizontal force that can point sideways is the friction between the tyres and the road. Here it does the string’s job.
4. That is why every bend has a top speed: when the most friction the tyres can get is less than $\dfrac{m v^2}{r}$ ⁠, on a wet road for instance, the car skids and moves straighter than the roundabout.
No centrifugal force acts on the car; it is what the passengers feel as their bodies carry straight on.

Lesson 3 of 5 · About 30 minutes

Universal gravitation

Any two masses pull on each other, with the same force, which grows with the masses and weakens with the square of the distance. From the pull between two bodies in the Scene to your own weight.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Who pulls whom
Practice · Predict, then look
Question
The second body has $12\%$ ⁠ of the first one’s mass. Which body pulls harder on the other?

Correct answer
- Correct They pull on each other with the same force
Setup, choices and feedback
The Scene at this step
Second mass
12 % of first
Distance apart
9
Show path
ondefault
Show the well
Offdefault
Choices
- Wrong The heavy body pulls harder on the light one
  Misconception: Thinks the heavier body pulls harder on the lighter one than the lighter pulls on it.
  Feedback from the catalogue
  Gravity is mutual. By Newton’s third law, and in the formula $F = G\dfrac{m_1 m_2}{r^2}$ ⁠, there is one number for both bodies: the Earth pulls on the Moon exactly as hard as the Moon pulls on the Earth. The difference is in the acceleration, $a = \dfrac{F}{m}$ ⁠: the same force moves the heavy body a little and the light one a lot.
- Correct They pull on each other with the same force
- Wrong The light body pulls harder on the heavy one
  Feedback
  The light body does move more, but because of its small mass, not because of a stronger force.
- Wrong Only the heavy body pulls; the light one does not pull at all
  Misconception: Thinks the heavier body pulls harder on the lighter one than the lighter pulls on it.
  Feedback from the catalogue
  Gravity is mutual. By Newton’s third law, and in the formula $F = G\dfrac{m_1 m_2}{r^2}$ ⁠, there is one number for both bodies: the Earth pulls on the Moon exactly as hard as the Moon pulls on the Earth. The difference is in the acceleration, $a = \dfrac{F}{m}$ ⁠: the same force moves the heavy body a little and the light one a lot.
Once the prediction is locked in, the Scene changes
Other settings
Forces/nodes/pair/props/forces = 1
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Both arrows are the same length and carry the same number. The same force, but the light body accelerates far more, so it looks as if only the heavy one pulls.

Worked solution
1. By Newton’s third law, the force the first body exerts on the second is equal in size, and opposite in direction, to the force the second exerts on the first.
2. In the law of gravitation too, $F = G\dfrac{m_1 m_2}{r^2}$ ⁠, the two masses appear in the same place: there is one force, and it acts on both.
3. What differs is the acceleration, $a = \dfrac{F}{m}$ ⁠. The second body’s mass is $\dfrac{100}{12} \approx 8.3$ ⁠ times smaller, so its acceleration is $8.3$ ⁠ times bigger: it does most of the moving.
The feeling that the big body pulls harder comes from seeing only the small one move.
Step 2 of 6
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Newton's law
No practice question in this step.
The Scene at this step
Second mass
100 % of first
Distance apart
10
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Forces/nodes/pair/props/forces = 1
Step 3 of 6
Open this step
The square of the distance
Practice · Predict, then look
Question
Two equal bodies $4$ ⁠ apart pull on each other with a force of $6.25$ ⁠, in the Scene’s unit. They are moved to $12$ ⁠ apart, three times as far. What will the pull be?

Correct answer
0.6944 ± 5%
A number prediction counts as right within ± 5%
Setup, choices and feedback
The Scene at this step
Second mass
100 % of firstfixed
Distance apart
4
Show path
ondefault
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Offdefault
Other settings
Forces/nodes/pair/props/forces = 1
Once the prediction is locked in, the Scene changes
Distance apart
12
The Scene then shows: 0.6944 · The pull between the bodies, in the Scene unit: 1 between two bodies like the first, 10 apart/nodes/pair/outputs/pull
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Three times as far, nine times as weak: $\dfrac{6.25}{9} \approx 0.69$ ⁠.

Hint
The pull goes as one over the square of the distance: $F \propto \dfrac{1}{r^2}$ ⁠.

Worked solution
1. The masses do not change, and $F \propto \dfrac{1}{r^2}$ ⁠.
2. The distance grows $3$ ⁠ times, so the pull shrinks $3^2 = 9$ ⁠ times.
3. $\dfrac{6.25}{9} \approx 0.69$ ⁠.
A check in the Scene’s unit: $F = 1 \cdot \left(\dfrac{10}{12}\right)^2 \approx 0.69$ ⁠. If you divided by $3$ ⁠ and got about $2.08$ ⁠, you forgot that the distance is squared in the law.
Step 4 of 6
Open this step
Mass and distance together
Practice · Choose
Question
The second body’s mass is doubled, and so is the distance between the bodies. What happens to the pull between them?

Correct answer
- Correct It halves
Setup, choices and feedback
The Scene at this step
Second mass
30 % of first
Distance apart
6
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ondefault
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Forces/nodes/pair/props/forces = 1
Choices
Shown in a random order.
- Correct It halves
- Wrong It stays the same
  Misconception: Thinks the pull falls in proportion to the distance (1/r), not to its square.
  Feedback from the catalogue
  The distance in the law of gravitation is squared: $F \propto \dfrac{1}{r^2}$ ⁠. Twice the distance gives a quarter of the pull, and three times the distance gives a ninth, not a third.
- Wrong It doubles
  Feedback
  Twice the mass doubles the pull, but twice the distance divides it by four.
- Wrong It drops to a quarter
  Feedback
  Twice the distance alone gives a quarter, but twice the mass doubles it back up.
Feedback for a right answer
The mass gives times $2$ ⁠, the distance divided by $4$ ⁠: half altogether. Try it in the Scene: from $30\%$ ⁠ at $6$ ⁠ to $60\%$ ⁠ at $12$ ⁠.

Worked solution
1. $F = G\dfrac{m_1 m_2}{r^2}$ ⁠. Twice the mass, on top: times $2$ ⁠.
2. Twice the distance, underneath and squared: divided by $2^2 = 4$ ⁠.
3. Together: $\dfrac{2}{4} = \dfrac12$ ⁠.
In the Scene: with the second mass at $30\%$ ⁠ and a distance of $6$ ⁠ the pull is about $0.83$ ⁠; at $60\%$ ⁠ and $12$ ⁠ it is about $0.42$ ⁠, half. If you answered “it stays the same”, you divided by the distance without squaring it.
Step 5 of 6
Open this step
Your weight
Practice · Calculate
Question
The Earth’s mass is $5.97 \times 10^{24}\ \mathrm{kg}$ ⁠ and its radius is 6370 km. How hard does it pull on a 70 kg person standing on its surface? Use $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ ⁠.

Correct answer
687 N ± 2%
Setup, choices and feedback
The Scene at this step
Second mass
1 % of first
Distance apart
9
Show path
ondefault
Show the well
Offdefault
Other settings
Forces/nodes/pair/props/forces = 1
Accepted within
± 2%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 4380000000 N ± 5%
  Misconception: Thinks the pull falls in proportion to the distance (1/r), not to its square.
  Feedback from the catalogue
  The distance in the law of gravitation is squared: $F \propto \dfrac{1}{r^2}$ ⁠. Twice the distance gives a quarter of the pull, and three times the distance gives a ninth, not a third.
- 687000000 N ± 5%
  Misconception: Puts a distance in kilometres into G·M·m/r², whose G is in metres.
  Feedback from the catalogue
  $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ ⁠ is built on metres, so $r$ ⁠ has to be in metres too: $6370\ \mathrm{km} = 6.37 \times 10^6\ \mathrm{m}$ ⁠. With kilometres the denominator is $10^6$ ⁠ times too small, and the force comes out a million times too big.
Feedback for a right answer
$F = G\dfrac{M m}{r^2} \approx 687\ \mathrm{N}$ ⁠, the person’s weight: $\dfrac{687}{70} \approx 9.8\ \mathrm{m/s^2}$ ⁠, the acceleration of free fall.

Hint
The distance $r$ ⁠ is measured from the Earth’s centre, and it has to be in metres.

Worked solution
1. The givens: $M = 5.97 \times 10^{24}\ \mathrm{kg}$ ⁠, $m = 70\ \mathrm{kg}$ ⁠, and $r = 6.37 \times 10^6\ \mathrm{m}$ ⁠, the distance between the centres.
2. The law of gravitation: $F = G\dfrac{M m}{r^2}$ ⁠.
3. The top: $6.67 \times 10^{-11} \cdot 5.97 \times 10^{24} \cdot 70 \approx 2.79 \times 10^{16}$ ⁠.
4. The bottom: $r^2 = (6.37 \times 10^6)^2 \approx 4.06 \times 10^{13}\ \mathrm{m^2}$ ⁠.
5. $F \approx \dfrac{2.79 \times 10^{16}}{4.06 \times 10^{13}} \approx 687\ \mathrm{N}$ ⁠.
That is exactly the weight, $m g$ ⁠: divide by $70$ ⁠ and you get $g \approx 9.8\ \mathrm{m/s^2}$ ⁠. If you got an enormous number, you forgot to square $r$ ⁠, or put it in kilometres.
Step 6 of 6
Open this step
The centre of mass
Practice · Calculate
Question
The second body’s mass is $25\%$ ⁠ of the first one’s, and they are $10$ ⁠ apart. How far is the first, heavy body from their shared centre of mass?

Correct answer
2 ± 0.05
Setup, choices and feedback
The Scene at this step
Second mass
25 % of firstfixed
Distance apart
10fixed
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ondefault
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Forces/nodes/pair/props/forces = 1
Accepted within
± 0.05
Computed from the Scene at this step’s setting
The first (heavier) body's distance from the centre of mass, Scene units/nodes/pair/outputs/distance1
Wrong answers it recognises
- 0 ± 0.01
  Misconception: Thinks the heavier body sits fixed at the centre while the lighter one goes round it.
  Feedback from the catalogue
  Both bodies go round their shared centre of mass. It lies on the line between them, closer to the heavier body, so that $m_1 r_1 = m_2 r_2$ ⁠. The heavy body moves too, on a small circle. Only when the second body is negligible beside it does the centre of mass sit almost at the heavy body’s centre.
Feedback for a right answer
$r_1 = 10 \cdot \dfrac{0.25}{1.25} = 2$ ⁠. The same force on both, but the heavy body’s acceleration is $4$ ⁠ times smaller, so its circle is $4$ ⁠ times smaller.

Hint
The centre of mass satisfies $m_1 r_1 = m_2 r_2$ ⁠, and $r_1 + r_2 = 10$ ⁠.

Worked solution
1. Write $m_1 = 1$ ⁠ and $m_2 = 0.25$ ⁠, in units of the first mass, and $r_1 + r_2 = 10$ ⁠.
2. The centre of mass: $m_1 r_1 = m_2 r_2$ ⁠, so $r_1 = 0.25\, r_2$ ⁠.
3. Substitute: $0.25\, r_2 + r_2 = 10$ ⁠, so $r_2 = 8$ ⁠ and $r_1 = 2$ ⁠.
The heavy body moves on a circle of radius $2$ ⁠ and the light one on a circle of radius $8$ ⁠: four times bigger, like the ratio of the masses. If you answered $0$ ⁠, you assumed the heavy body stands still.

Lesson 4 of 5 · About 30 minutes

Orbits: a fall that never ends

A satellite is a body that keeps falling and keeps missing the Earth. Gravity is its centripetal force, and from that come its speed, its period, Kepler's third law, and why astronauts float.

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The 6 steps and their answers

Step 1 of 6
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Falling and missing
Practice · Choose
Question
The International Space Station orbits the Earth at a height of about 400 km. Why do the astronauts inside it float?

Correct answer
- Correct They and the station are falling together round the Earth, with the same acceleration
Setup, choices and feedback
The Scene at this step
Second mass
1 % of firstfixed
Distance apart
9
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ondefault
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Offdefault
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Forces/nodes/pair/props/forces = 1
Choices
Shown in a random order.
- Correct They and the station are falling together round the Earth, with the same acceleration
- Wrong There is no gravity that high up
  Misconception: Thinks there is no gravity in orbit, so astronauts float because nothing pulls on them.
  Feedback from the catalogue
  At the height of the space station, about 400 km, gravity is still about $89\%$ ⁠ of what it is on the ground. Without it the station would carry straight on into space instead of going round the Earth. The astronauts float because they and the station are falling together with the same acceleration, so nothing presses on them: that is free fall, not an absence of gravity.
- Wrong Gravity is balanced by an outward centrifugal force
  Misconception: Believes an outward centrifugal force acts on the circling body and balances the inward pull.
  Feedback from the catalogue
  If an outward force balanced the inward pull, the net force would be zero and the body would move in a straight line. It curves precisely because there is a net force inward. What you feel as an outward pull is the ball pulling on the string and your hand (Newton’s third law), or your own body carrying straight on while the car turns underneath it.
- Wrong They are too far from the Earth, and gravity there is negligible
  Misconception: Thinks there is no gravity in orbit, so astronauts float because nothing pulls on them.
  Feedback from the catalogue
  At the height of the space station, about 400 km, gravity is still about $89\%$ ⁠ of what it is on the ground. Without it the station would carry straight on into space instead of going round the Earth. The astronauts float because they and the station are falling together with the same acceleration, so nothing presses on them: that is free fall, not an absence of gravity.
Feedback for a right answer
Shared free fall: gravity accelerates the station and everyone in it equally, so nothing presses on them. It is called weightlessness, but gravity is there.

Worked solution
1. At a height of 400 km the distance from the Earth’s centre is $r = 6370 + 400 = 6770\ \mathrm{km}$ ⁠, only about $6\%$ ⁠ more than on the ground.
2. Gravity is weaker by a factor of $\left(\dfrac{6770}{6370}\right)^2 \approx 1.13$ ⁠: it is still about $89\%$ ⁠ of gravity on the ground.
3. Gravity is the only force on the station and the astronauts, and it gives them all the same acceleration, towards the Earth’s centre. They fall together, so the floor does not press on them, and there is no weight to feel.
Nor is there a centrifugal force balancing gravity: if the forces were balanced, the station would move in a straight line.
Step 2 of 6
Open this step
How much gravity there is up there
Practice · Calculate
Question
What is the acceleration due to gravity at the height of the space station, 400 km above the Earth’s surface? The Earth’s mass is $5.97 \times 10^{24}\ \mathrm{kg}$ ⁠, its radius 6370 km, and $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ ⁠.

Correct answer
8.69 m/s² ± 2%
Setup, choices and feedback
The Scene at this step
Second mass
1 % of firstfixed
Distance apart
9
Show path
ondefault
Show the well
Offdefault
Other settings
Forces/nodes/pair/props/forces = 1
Accepted within
± 2%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 m/s² ± 0.01 m/s²
  Misconception: Thinks there is no gravity in orbit, so astronauts float because nothing pulls on them.
  Feedback from the catalogue
  At the height of the space station, about 400 km, gravity is still about $89\%$ ⁠ of what it is on the ground. Without it the station would carry straight on into space instead of going round the Earth. The astronauts float because they and the station are falling together with the same acceleration, so nothing presses on them: that is free fall, not an absence of gravity.
- 2490 m/s² ± 5%
  Misconception: Uses the height above the ground instead of the distance from the Earth's centre as r.
  Feedback from the catalogue
  In the law of gravitation $r$ ⁠ is the distance between the bodies’ centres. For a satellite at a height $h$ ⁠ above the ground, $r = R + h$ ⁠, where $R \approx 6.37 \times 10^6\ \mathrm{m}$ ⁠ is the Earth’s radius. A height of 400 km on its own is nowhere near the real distance to the centre.
Feedback for a right answer
$g = \dfrac{GM}{r^2} \approx 8.7\ \mathrm{m/s^2}$ ⁠, not much less than the $9.8$ ⁠ on the ground. Gravity is certainly there.

Hint
$g = \dfrac{G M}{r^2}$ ⁠, where $r$ ⁠ is the distance from the Earth’s centre.

Worked solution
1. The distance from the Earth’s centre: $r = R + h = 6.37 \times 10^6 + 0.40 \times 10^6 = 6.77 \times 10^6\ \mathrm{m}$ ⁠.
2. A body of mass $m$ ⁠ feels $F = G\dfrac{M m}{r^2}$ ⁠, so its acceleration is $g = \dfrac{F}{m} = \dfrac{G M}{r^2}$ ⁠.
3. $G M = 6.67 \times 10^{-11} \cdot 5.97 \times 10^{24} \approx 3.98 \times 10^{14}\ \mathrm{m^3/s^2}$ ⁠.
4. $g = \dfrac{3.98 \times 10^{14}}{(6.77 \times 10^6)^2} \approx \dfrac{3.98 \times 10^{14}}{4.58 \times 10^{13}} \approx 8.7\ \mathrm{m/s^2}$ ⁠.
That is about $89\%$ ⁠ of the $9.8$ ⁠ on the ground. If you put in $r = 400\ \mathrm{km}$ ⁠ you got thousands of metres per second squared: $r$ ⁠ is measured from the Earth’s centre, not from the ground.
Step 3 of 6
Open this step
Orbital speed
Practice · Calculate
Question
How fast does the space station move on a circular orbit 400 km up? Use the same data: $M = 5.97 \times 10^{24}\ \mathrm{kg}$ ⁠, $R = 6370\ \mathrm{km}$ ⁠ and $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ ⁠.

Correct answer
7670 m/s ± 1.5%
Setup, choices and feedback
The Scene at this step
Second mass
1 % of firstfixed
Distance apart
9
Show path
ondefault
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Offdefault
Other settings
Forces/nodes/pair/props/forces = 1
Accepted within
± 1.5%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 7910 m/s ± 1%
  Misconception: Uses the Earth's radius alone for r, leaving out the height of the orbit.
  Feedback from the catalogue
  The satellite is not on the ground: its distance from the Earth’s centre is $r = R + h$ ⁠. With $R$ ⁠ alone you get the speed of an orbit skimming the ground, about 7.9 km/s, not the speed at the station’s height.
- 31600 m/s ± 3%
  Misconception: Uses the height above the ground instead of the distance from the Earth's centre as r.
  Feedback from the catalogue
  In the law of gravitation $r$ ⁠ is the distance between the bodies’ centres. For a satellite at a height $h$ ⁠ above the ground, $r = R + h$ ⁠, where $R \approx 6.37 \times 10^6\ \mathrm{m}$ ⁠ is the Earth’s radius. A height of 400 km on its own is nowhere near the real distance to the centre.
Feedback for a right answer
$v = \sqrt{\dfrac{GM}{r}} \approx 7.7\ \mathrm{km/s}$ ⁠, about 27,600 km/h: a full lap of the Earth in about $92$ ⁠ minutes.

Hint
Gravity is the centripetal force: $G\dfrac{M m}{r^2} = \dfrac{m v^2}{r}$ ⁠.

Worked solution
1. On a circular orbit gravity is the centripetal force: $G\dfrac{M m}{r^2} = \dfrac{m v^2}{r}$ ⁠.
2. Cancel $m$ ⁠ and one $r$ ⁠: $v^2 = \dfrac{G M}{r}$ ⁠, so $v = \sqrt{\dfrac{G M}{r}}$ ⁠.
3. Put in $r = 6.77 \times 10^6\ \mathrm{m}$ ⁠: $v = \sqrt{\dfrac{3.98 \times 10^{14}}{6.77 \times 10^6}} = \sqrt{5.88 \times 10^7} \approx 7670\ \mathrm{m/s}$ ⁠.
4. The period: $T = \dfrac{2\pi r}{v} = \dfrac{2\pi \cdot 6.77 \times 10^6}{7670} \approx 5550\ \mathrm{s}$ ⁠, about $92$ ⁠ minutes.
If you got about 7910 m/s, you used the Earth’s radius without the height; if you got about 31,600 m/s, you used the height alone.
Step 4 of 6
Open this step
Kepler's third law
Practice · Predict, then look
Question
In the Scene, a small body ( $1\%$ ⁠) orbits a big one $4$ ⁠ away, and one orbit takes about 3.3 s. It is moved out to $16$ ⁠, four times as far. How many times longer will an orbit take?

Correct answer
- Correct $8$ ⁠ times
Setup, choices and feedback
The Scene at this step
Second mass
1 % of firstfixed
Distance apart
4
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ondefault
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Choices
- Wrong $2$ ⁠ times
  Feedback
  $2$ ⁠ is $\sqrt{4}$ ⁠, but the orbit itself is also $4$ ⁠ times longer.
- Wrong $4$ ⁠ times
  Misconception: Thinks the orbital period grows in proportion to the distance.
  Feedback from the catalogue
  Farther out, the orbit is both longer and slower, because $v = \sqrt{\dfrac{GM}{r}}$ ⁠ shrinks with distance. Together the two effects give Kepler’s third law, $T^2 \propto r^3$ ⁠: four times the distance gives $\sqrt{4^3} = 8$ ⁠ times the period.
- Correct $8$ ⁠ times
- Wrong $16$ ⁠ times
  Feedback
  $T^2 \propto r^3$ ⁠, not $T \propto r^2$ ⁠: $\sqrt{4^3} = 8$ ⁠ times.
Once the prediction is locked in, the Scene changes
Distance apart
16
The Scene then shows: 26.68 s · Seconds the Scene takes for one orbit/nodes/pair/outputs/period
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$8$ ⁠ times: the orbit is $4$ ⁠ times longer and the speed $2$ ⁠ times lower. That is $T^2 \propto r^3$ ⁠, Kepler’s third law.

Worked solution
1. On a circular orbit $v = \sqrt{\dfrac{G M}{r}}$ ⁠ and $T = \dfrac{2\pi r}{v}$ ⁠.
2. Substitute: $T = 2\pi\sqrt{\dfrac{r^3}{G M}}$ ⁠, that is, $T^2 = \dfrac{4\pi^2}{G M}\, r^3$ ⁠: Kepler’s third law.
3. Four times the distance: $T$ ⁠ grows $\sqrt{4^3} = \sqrt{64} = 8$ ⁠ times.
4. In the Scene: from about 3.3 s to about 26.7 s.
The orbit is $4$ ⁠ times longer, and the body also moves $2$ ⁠ times slower, because $v \propto \dfrac{1}{\sqrt{r}}$ ⁠. Answering $4$ ⁠ times counts only the length of the orbit.
Step 5 of 6
Open this step
On the same orbit
Practice · Choose
Question
A second satellite, ten times heavier than the first, is put on the same circular orbit round the Earth. How fast must it move?

Correct answer
- Correct At the same speed as the first
Setup, choices and feedback
The Scene at this step
Second mass
1 % of firstfixed
Distance apart
9
Show path
ondefault
Show the well
Offdefault
Choices
Shown in a random order.
- Correct At the same speed as the first
- Wrong Faster, because gravity pulls it harder
  Misconception: Thinks a heavier satellite needs a different speed, or period, on the same orbit.
  Feedback from the catalogue
  The satellite’s mass cancels out: $\dfrac{G M m}{r^2} = \dfrac{m v^2}{r}$ ⁠ gives $v = \sqrt{\dfrac{G M}{r}}$ ⁠, with no $m$ ⁠ in it. A heavier satellite is pulled harder, but it is also harder to turn by the same amount, just as all bodies fall with the same acceleration.
- Wrong Slower, because it is harder to move
  Misconception: Thinks a heavier satellite needs a different speed, or period, on the same orbit.
  Feedback from the catalogue
  The satellite’s mass cancels out: $\dfrac{G M m}{r^2} = \dfrac{m v^2}{r}$ ⁠ gives $v = \sqrt{\dfrac{G M}{r}}$ ⁠, with no $m$ ⁠ in it. A heavier satellite is pulled harder, but it is also harder to turn by the same amount, just as all bodies fall with the same acceleration.
- Wrong $\sqrt{10}$ ⁠ times faster
  Misconception: Thinks a heavier satellite needs a different speed, or period, on the same orbit.
  Feedback from the catalogue
  The satellite’s mass cancels out: $\dfrac{G M m}{r^2} = \dfrac{m v^2}{r}$ ⁠ gives $v = \sqrt{\dfrac{G M}{r}}$ ⁠, with no $m$ ⁠ in it. A heavier satellite is pulled harder, but it is also harder to turn by the same amount, just as all bodies fall with the same acceleration.
Feedback for a right answer
The satellite’s mass cancels, and $v = \sqrt{\dfrac{GM}{r}}$ ⁠ depends only on the Earth’s mass and the distance. That is why a bolt that comes loose from the space station carries on beside it, on the same orbit.

Worked solution
1. Gravity on the satellite is its centripetal force: $G\dfrac{M m}{r^2} = \dfrac{m v^2}{r}$ ⁠.
2. The mass $m$ ⁠ appears on both sides and cancels: $v = \sqrt{\dfrac{G M}{r}}$ ⁠.
3. $M$ ⁠ is the Earth’s mass and $r$ ⁠ is the same orbit, so $v$ ⁠ is the same for both satellites.
The heavy satellite is pulled $10$ ⁠ times harder, but it also takes $10$ ⁠ times the force to turn it: the same sum that makes a heavy ball and a light ball fall together.
Step 6 of 6
Open this step
Twice the period
Practice · Reach the target
Question
The small body orbits $6$ ⁠ away, and one orbit takes about 6.1 s in the Scene. Change the Distance apart until one orbit takes twice as long, about 12.3 s, then press Check.

Goal
- 12.26 s ± 3%
  At the start of the step: 6.128 s
Setup, choices and feedback
The Scene at this step
Second mass
1 % of firstfixed
Distance apart
6
Show path
ondefault
Show the well
Offdefault
Read from the Scene at
Seconds the Scene takes for one orbit/nodes/pair/outputs/period
One setting that reaches it
Distance apart
9.5
Show path
off
Feedback for a right answer
$\sqrt[3]{4} \approx 1.59$ ⁠ times the distance gives twice the period. That is how we know that Mars, whose year is nearly two of ours, is about $1.5$ ⁠ times as far from the Sun as the Earth.

Feedback for a wrong answer
Not yet. If the orbit is too quick, move the body out; if it is too slow, bring it in.

Hint
$T^2 \propto r^3$ ⁠, so $r \propto T^{2/3}$ ⁠: twice the time needs $2^{2/3}$ ⁠ times the distance.

Worked solution
1. From Kepler’s third law: $\dfrac{r_2^3}{r_1^3} = \dfrac{T_2^2}{T_1^2} = 2^2 = 4$ ⁠.
2. So $\dfrac{r_2}{r_1} = \sqrt[3]{4} \approx 1.59$ ⁠.
3. $r_2 \approx 1.59 \cdot 6 \approx 9.5$ ⁠.
At $9.5$ ⁠ an orbit takes about 12.2 s in the Scene. Doubling the distance to $12$ ⁠ gives about 17.3 s, nearly three times: $2^{3/2} \approx 2.8$ ⁠.

Lesson 5 of 5 · About 25 minutes

Angular momentum: a skater and a gyroscope

A spinning body keeps its angular momentum, L = Iω, as long as no torque acts on it: a skater who pulls in her arms and spins faster, the energy she adds, and a gyroscope that circles its stand instead of falling.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Pulling in
Practice · Predict, then look
Question
The skater spins with her arms out, about one turn a second. The ice is smooth and nobody pushes her. What will happen when she pulls her arms in to her body?

Correct answer
- Correct She will spin faster
Setup, choices and feedback
The Scene at this step
Arm reach
0.8 m
Gyroscope spin
3000 rpm
Show angular momentum
Off
Choices
- Correct She will spin faster
- Wrong She will keep the same pace: nothing outside pushed her
  Misconception: Thinks a skater can spin faster only if something outside pushes her, so pulling her arms in changes nothing.
  Feedback from the catalogue
  What is kept without a torque from outside is the angular momentum $L = I\omega$ ⁠, not the angular velocity. As the arms come in, the moment of inertia $I$ ⁠ falls, so $\omega$ ⁠ rises by the same factor, with no push at all.
- Wrong She will slow down, because the arms brake the spin
  Feedback
  The arms do not rub against anything. As they come in, their mass circles closer to the axis, and the skater has to turn faster, not slower.
Once the prediction is locked in, the Scene changes
Arm reach
0.15 m
The Scene then shows: 2.601 Hz · The skater’s turns per second, L / (2π I)/nodes/spin/outputs/spin
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
With her arms tucked she turns about $2.6$ ⁠ times a second, more than twice as fast, with no push at all. Her angular momentum is kept, and what changed is her moment of inertia.

Worked solution
1. A spinning body carries angular momentum $L = I\omega$ ⁠. It changes only when something outside applies a torque to it.
2. The ice is smooth, so the skater’s $L$ ⁠ is fixed.
3. As the arms come in towards the axis, $I$ ⁠ falls, so $\omega$ ⁠ must rise by the same factor.
If you expected her to keep the same pace, you took a faster spin to need a push. What is kept is the product $I\omega$ ⁠, not $\omega$ ⁠.
Step 2 of 6
Open this step
The moment of inertia
Practice · Reach the target
Question
With her arms out, 0.8 m from the axis, the skater turns once a second. Change Arm reach until she turns exactly twice as fast, 2 Hz, and press Check.

Goal
- 2 Hz ± 0.03 Hz
  At the start of the step: 1.001 Hz
Setup, choices and feedback
The Scene at this step
Arm reach
0.8 m
Gyroscope spin
3000 rpmfixed
Show angular momentum
On
Read from the Scene at
The skater’s turns per second, L / (2π I)/nodes/spin/outputs/spin
One setting that reaches it
Arm reach
0.35 m
Feedback for a right answer
At about 0.35 m the moment of inertia is about 1.30 kg·m², half of the 2.59 kg·m² with her arms out, and the spin is doubled.

Feedback for a wrong answer
Not yet. With her arms closer to the axis she spins faster. Read the turns per second on the board.

Hint
Twice the spin means half the moment of inertia. Mass far from the axis counts by the square of its distance, so the arms need not come all the way in.

Worked solution
1. $L$ ⁠ is fixed, so $\omega$ ⁠ doubles when $I$ ⁠ halves: from 2.59 kg·m² to 1.30 kg·m².
2. Part of $I$ ⁠ is the body itself, 0.8 kg·m², which does not change. Only the arms can make up the difference.
3. On the board that happens at about 0.35 m: set Arm reach there.
Halving the reach, to 0.4 m, cuts the arms’ share about four times, but the body’s share stays, so the whole $I$ ⁠ falls by less than half, and she turns only about 1.86 Hz.
Step 3 of 6
Open this step
The energy of the spin
Practice · Calculate
Question
Her arms are tucked, and the board shows $I$ ⁠ and $L$ ⁠. The energy of a spin is $E = \dfrac{L^2}{2I}$ ⁠. How much energy does the skater’s spin have now?

Correct answer
133.2 J ± 2%
Setup, choices and feedback
The Scene at this step
Arm reach
0.15 mfixed
Gyroscope spin
3000 rpmfixed
Show angular momentum
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
The energy of her spin, L² / 2I/nodes/spin/outputs/energy
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 51.25 J ± 2%
  Misconception: Thinks the energy of a spin stays the same when the skater pulls her arms in, as the angular momentum does.
  Feedback from the catalogue
  The angular momentum is kept, but the energy $E = \dfrac{L^2}{2I}$ ⁠ grows as $I$ ⁠ falls. The extra energy comes from her muscles: the arms keep trying to fly outward, and she does work pulling them in.
Feedback for a right answer
$E = \dfrac{16.3^2}{2 \times 1.00} \approx 133$ ⁠ joules. With her arms out she had $\dfrac{16.3^2}{2 \times 2.59} \approx 51$ ⁠ joules: the angular momentum is kept, but the energy grew by about 80 J.

Hint
On the board $L \approx 16.3$ ⁠ and $I \approx 1.00$ ⁠, in SI units.

Worked solution
1. Given: L ≈ 16.3 kg·m²/s, I ≈ 1.00 kg·m².
2. $E = \dfrac{L^2}{2I} = \dfrac{16.3^2}{2 \times 1.00} = \dfrac{265.7}{2} \approx 133$ ⁠.
3. The energy is about 133 J.
If you answered about 51 J, that is the energy with her arms out. The energy is not kept: the skater does work pulling in arms that keep trying to fly outward, and that work becomes faster spin.
Step 4 of 6
Open this step
The gyroscope
Practice · Predict, then look
Question
The gyroscope is not spinning, with one end of its axle resting on the stand: it has fallen against the post. What will happen if the wheel spins at 3000 rpm and the gyroscope is let go with its axle level?

Correct answer
- Correct The axle will stay almost level and swing slowly round the stand
Setup, choices and feedback
The Scene at this step
Arm reach
0.8 mfixed
Gyroscope spin
0 rpm
Show angular momentum
Off
Choices
- Correct The axle will stay almost level and swing slowly round the stand
- Wrong It will stay where it is, level: the spin cancels gravity
  Misconception: Thinks a fast spin cancels or cuts gravity, so a gyroscope stays up because its weight is gone.
  Feedback from the catalogue
  The spin does not cancel gravity. The weight keeps acting, and the stand keeps holding it up. What the spin changes is what the weight does: its torque turns the angular momentum sideways, so the axle swings round instead of falling.
- Wrong It will fall again, only more slowly
  Feedback
  The spin does not slow the fall: it turns it sideways. The torque of the weight changes the angular momentum in a level direction, so the axle moves round, not down.
Once the prediction is locked in, the Scene changes
Gyroscope spin
3000 rpm
The Scene then shows: 3.084 s · Seconds for the gyroscope to circle its post once, once the wobble has settled; NaN when it has fallen/nodes/spin/outputs/precessionPeriod
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The axle stays almost level and goes round the stand about once every 3.1 s. That motion is called precession. Gravity still acts: without the stand the gyroscope would fall.

Worked solution
1. The weight pulls the wheel down, and the stand pushes up at the pivot. Together they make a torque, pointing level, at right angles to the axle.
2. A torque changes the angular momentum in its own direction. The spinning wheel’s angular momentum points along the axle, so at each moment it gains a little sideways.
3. The axle swings sideways, not down, and keeps doing so as long as the wheel spins.
The spin does not cancel gravity: the weight keeps acting, and the stand keeps holding it up.
Step 5 of 6
Open this step
The rate of precession
Practice · Predict, then look
Question
At 3000 rpm the gyroscope goes round its stand once every 3.1 s. How long will one circle take at 6000 rpm?

Correct answer
6.163 s ± 0.3 s
A number prediction counts as right within ± 0.3 s
Setup, choices and feedback
The Scene at this step
Arm reach
0.8 mfixed
Gyroscope spin
3000 rpm
Show angular momentum
On
Once the prediction is locked in, the Scene changes
Gyroscope spin
6000 rpm
The Scene then shows: 6.163 s · Seconds for the gyroscope to circle its post once, once the wobble has settled; NaN when it has fallen/nodes/spin/outputs/precessionPeriod
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 6.2 s: twice the spin, half the rate of precession. A faster wheel has more angular momentum, and the same torque turns it less.

Worked solution
1. A fast wheel precesses at $\Omega = \dfrac{Mg\ell}{I\omega}$ ⁠: the torque divided by the angular momentum.
2. The torque $Mg\ell$ ⁠ has not changed, and $I\omega$ ⁠ has doubled. So $\Omega$ ⁠ halves.
3. The time per circle, $\dfrac{2\pi}{\Omega}$ ⁠, doubles: $2 \times 3.1 = 6.2$ ⁠ seconds.
If you predicted a shorter time, you expected a faster wheel to circle faster. It is the other way round: the more angular momentum, the harder it is to turn.
Step 6 of 6
Open this step
A slowing wheel
Practice · Choose
Question
In a real gyroscope, friction in the bearings slowly slows the wheel. What happens to the precession as the wheel slows?

Correct answer
- Correct It circles faster and faster and dips more, until it falls
Setup, choices and feedback
The Scene at this step
Arm reach
0.8 mfixed
Gyroscope spin
800 rpm
Show angular momentum
On
Choices
Shown in a random order.
- Correct It circles faster and faster and dips more, until it falls
- Wrong It circles more and more slowly, until it stops
  Misconception: Thinks a gyroscope precesses more slowly as its wheel slows, as if the circling followed the spin.
  Feedback from the catalogue
  The rate of precession is $\dfrac{Mg\ell}{I\omega}$ ⁠: the torque divided by the angular momentum. A slower wheel carries less angular momentum, so the same torque turns it faster. A slowing gyroscope circles faster, dips, and in the end falls.
- Wrong The precession stays the same until the wheel stops
  Feedback
  The rate of precession depends on the wheel’s angular momentum, so it changes as soon as the wheel slows. Try 800 rpm: it already circles in under a second.
Feedback for a right answer
As the wheel slows it has less angular momentum, and the same torque turns it faster. At 800 rpm it already dips about $19^\circ$ ⁠ and circles in under a second, and below about 325 rpm it falls.

Hint
Compare 3000 rpm with 800 rpm on the board.

Worked solution
1. $\Omega = \dfrac{Mg\ell}{I\omega}$ ⁠: as $\omega$ ⁠ falls, $\Omega$ ⁠ rises.
2. At 3000 rpm a circle takes 3.1 s, and at 800 rpm about 0.9 s.
3. At slow spins the formula is no longer enough: the axle dips and wobbles (nutation), and in the end it rests against the post.
If you chose that it slows, you took the precession to follow the spin. It goes the opposite way.

Across the unit · 23 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

Acceleration towards the centre

Letting go

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The velocity turns

Question

Correct answer

The Scene at this step

Choices

Worked solution

One turn

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

How fast the velocity turns

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Twice the speed

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

An acceleration of ten

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The force that holds it on the circle

What pulls the ball

Question

Correct answer

The Scene at this step

Choices

Worked solution

How strong

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A pull of two newtons

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

A longer string

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Who does the string's job

Question

Correct answer

The Scene at this step

Choices

Worked solution

Universal gravitation

Who pulls whom

Question

Correct answer