Answer key · Unit 18

Orbits and spaceflight.

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 25 minutes

Kepler's laws

Orbits are ellipses with the Sun at one focus, a planet is faster when it is close to the Sun, and the length of the year depends only on the size of the orbit.

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The 6 steps and their answers

Step 1 of 6
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Ellipse and focus
Practice · Choose
Question
The planet in the Scene moves on an elliptical orbit, with Eccentricity $0.5$ ⁠. Where is the star it goes round?

Correct answer
- Correct At one of the ellipse’s foci, well away from its centre
Setup, choices and feedback
The Scene at this step
Eccentricity
0.5
Size of the orbit
1 AU
Show equal areas
Orbit
Choices
Shown in a random order.
- Correct At one of the ellipse’s foci, well away from its centre
- Wrong At the centre of the ellipse
  Misconception: Thinks the star sits at the centre of an elliptical orbit, so the planet is always the same distance from it.
  Feedback from the catalogue
  The Sun is not at the centre of the ellipse but at one of its two foci, $ae$ ⁠ from the centre. So the planet’s distance from it changes along the orbit: at the closest point it is $a(1-e)$ ⁠, and at the farthest $a(1+e)$ ⁠. Only on a circle, $e = 0$ ⁠, do the focus and the centre coincide.
- Wrong At the end of the ellipse, on the orbit itself
  Feedback
  If the star were on the orbit, the planet would hit it once a year. It sits inside the ellipse, on its long axis.
- Wrong At both foci at once, because the orbit is symmetric
  Feedback
  An ellipse has two foci, but only one of them holds a star. The other focus is empty: just a point of the geometry.
Feedback for a right answer
The star sits at a focus, and the other focus is empty. So the planet’s distance from it changes through the year: here from half an astronomical unit to one and a half.

Worked solution
1. An ellipse has two foci on its long axis, one on each side of the centre, each $ae$ ⁠ from it, where $a$ ⁠ is the semi-major axis and $e$ ⁠ the eccentricity.
2. Kepler’s first law: a planet moves on an ellipse with the Sun at one focus. The other focus is empty.
3. Here $e = 0.5$ ⁠ and $a = 1\ \mathrm{AU}$ ⁠, so the star is half an astronomical unit from the centre.
Choosing the centre means picturing a circle, with the star in the middle and the distance fixed. In a circle, $e = 0$ ⁠, both foci merge with the centre.
Step 2 of 6
Open this step
Closest and farthest
Practice · Calculate
Question
The size of the orbit, the semi-major axis, is locked at 1.2 AU, and the eccentricity at $0.3$ ⁠. How close to the star does the planet come? One astronomical unit is $1.50 \times 10^{8}\ \mathrm{km}$ ⁠.

Correct answer
125700000 km ± 2%
Setup, choices and feedback
The Scene at this step
Eccentricity
0.3fixed
Size of the orbit
1.2 AUfixed
Show equal areas
Orbit
Accepted within
± 2%
Computed from the Scene at this step’s setting
Closest distance to the star, a(1 − e)/nodes/orbit/outputs/perihelion
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 179500000 km ± 2%
  Misconception: Thinks the star sits at the centre of an elliptical orbit, so the planet is always the same distance from it.
  Feedback from the catalogue
  The Sun is not at the centre of the ellipse but at one of its two foci, $ae$ ⁠ from the centre. So the planet’s distance from it changes along the orbit: at the closest point it is $a(1-e)$ ⁠, and at the farthest $a(1+e)$ ⁠. Only on a circle, $e = 0$ ⁠, do the focus and the centre coincide.
Feedback for a right answer
$a(1-e) = 0.84\ \mathrm{AU}$ ⁠, about $1.26 \times 10^8\ \mathrm{km}$ ⁠. On the far side the distance is $a(1+e) = 1.56\ \mathrm{AU}$ ⁠, and the average of the two is exactly $a$ ⁠: that is why $a$ ⁠ is also called the mean distance.

Hint
The star is at a focus, $ae$ ⁠ from the centre. The closest point is the end of the long axis on its side.

Worked solution
1. The givens: $a = 1.2\ \mathrm{AU}$ ⁠, $e = 0.3$ ⁠.
2. The focus is $ae = 1.2 \cdot 0.3 = 0.36\ \mathrm{AU}$ ⁠ from the centre, and the end of the long axis is $a$ ⁠ from the centre.
3. The least distance, at the point closest to the star (perihelion): $r_{\min} = a - ae = a(1-e) = 1.2 \cdot 0.7 = 0.84\ \mathrm{AU}$ ⁠, that is $0.84 \cdot 1.50 \times 10^8 \approx 1.26 \times 10^8\ \mathrm{km}$ ⁠.
4. In the same way, at the farthest point (aphelion): $r_{\max} = a(1+e) = 1.56\ \mathrm{AU}$ ⁠.
Answering 1.2 AU, about $1.8 \times 10^8\ \mathrm{km}$ ⁠, means assuming the star is at the centre and the distance never changes.
Step 3 of 6
Open this step
Equal areas in equal times
Practice · Predict, then look
Question
The planet moves on an orbit with eccentricity $0.5$ ⁠. On which part of the orbit does it move fastest?

Correct answer
- Correct At the point closest to the star
Setup, choices and feedback
The Scene at this step
Eccentricity
0.5
Size of the orbit
1 AUfixed
Show equal areas
Orbit
Choices
- Correct At the point closest to the star
- Wrong At the point farthest from the star
  Feedback
  Gravity is weakest there, but it has also been slowing the planet all the way out. That is where it is slowest.
- Wrong At the same speed all the way round
  Misconception: Thinks a planet moves at the same speed all the way round its orbit, whatever its shape.
  Feedback from the catalogue
  On an elliptical orbit the speed changes. On the way in gravity pulls the planet a little forward and speeds it up, and on the way out it holds it back. That is Kepler’s second law: the line joining the Sun to the planet sweeps out equal areas in equal times, so close to the Sun, where the line is short, the planet has to cover more path every second. Only on a circle is the speed constant.
- Wrong At the sides, halfway between the closest and farthest points
  Feedback
  Halfway in, it is still speeding up: gravity keeps pulling it a little forward until the closest point.
Once the prediction is locked in, the Scene changes
Show equal areas
0.35 (Orbit → Equal areas)
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Fastest at the closest point, slowest at the farthest. The shaded wedge shows why: in the same time it is wide and short near the star and thin and long far out, and always the same area.

Worked solution
1. On the way in, gravity points a little forward, along the motion, so it increases the speed. On the way out it points a little backward, and slows the planet.
2. So the speed is greatest at the end of the way in, at the closest point, and least at the farthest point.
3. Show equal areas shows Kepler’s second law: the line joining the star to the planet sweeps out equal areas in equal times. Close to the star the line is short, so to sweep the same area the planet covers more path in the same time.
Only on a circular orbit is the speed constant.
Step 4 of 6
Open this step
Fast and slow
Practice · Calculate
Question
The orbit is locked on Mercury’s: eccentricity $0.21$ ⁠ and size 0.39 AU. How many times faster is Mercury at its closest point to the Sun than at its farthest?

Correct answer
1.532 ± 0.02
Setup, choices and feedback
The Scene at this step
Eccentricity
0.21fixed
Size of the orbit
0.39 AUfixed
Show equal areas
0.35 (Orbit → Equal areas)
Accepted within
± 0.02
Computed from the Scene at this step’s setting
Fastest speed over slowest, (1 + e)/(1 − e)/nodes/orbit/outputs/speedRatio
Wrong answers it recognises
- 1 ± 0.01
  Misconception: Thinks a planet moves at the same speed all the way round its orbit, whatever its shape.
  Feedback from the catalogue
  On an elliptical orbit the speed changes. On the way in gravity pulls the planet a little forward and speeds it up, and on the way out it holds it back. That is Kepler’s second law: the line joining the Sun to the planet sweeps out equal areas in equal times, so close to the Sun, where the line is short, the planet has to cover more path every second. Only on a circle is the speed constant.
- 2.346 ± 0.04
  Misconception: Thinks the speed on an orbit falls with the square of the distance, as the pull of gravity does.
  Feedback from the catalogue
  Gravity weakens like $\dfrac{1}{r^2}$ ⁠, but the speed does not. Kepler’s second law keeps $r \cdot v$ ⁠ the same at the closest and farthest points, so the ratio of the speeds equals the ratio of the distances, $\dfrac{v_{\max}}{v_{\min}} = \dfrac{1+e}{1-e}$ ⁠, with no square.
Feedback for a right answer
$\dfrac{1+e}{1-e} = \dfrac{1.21}{0.79} \approx 1.53$ ⁠: about 59 km/s near the Sun and 39 km/s far from it.

Hint
At those two points the velocity is at right angles to the line to the Sun, and Kepler’s second law keeps $r \cdot v$ ⁠ the same.

Worked solution
1. At the closest and farthest points the motion is at right angles to the line to the Sun, and the area the line sweeps in a short time $\Delta t$ ⁠ is a thin triangle: $\tfrac12 r \cdot v\,\Delta t$ ⁠.
2. By the second law the areas are equal, so $r_{\min} v_{\max} = r_{\max} v_{\min}$ ⁠.
3. $\dfrac{v_{\max}}{v_{\min}} = \dfrac{r_{\max}}{r_{\min}} = \dfrac{a(1+e)}{a(1-e)} = \dfrac{1.21}{0.79} \approx 1.53$ ⁠.
The size $a$ ⁠ cancels, and the ratio depends only on the eccentricity. Getting about $2.35$ ⁠ means squaring the ratio of the distances, as in the law of gravity; the speed is kept through $r \cdot v$ ⁠, with no square.
Step 5 of 6
Open this step
The third law
Practice · Reach the target
Question
A year on Mars lasts $1.88$ ⁠ Earth years. Change Size of the orbit until the planet’s year in the Scene lasts $1.88$ ⁠ years, and press Check.

Goal
- 686.7 d ± 1%
  At the start of the step: 365.3 d
Setup, choices and feedback
The Scene at this step
Eccentricity
0.09fixed
Size of the orbit
1 AU
Show equal areas
Orbit
Read from the Scene at
One orbit, from Kepler's third law/nodes/orbit/outputs/period
One setting that reaches it
Size of the orbit
1.52 AU
Feedback for a right answer
About 1.52 AU: $1.52^3 \approx 3.5 \approx 1.88^2$ ⁠. Notice that the Earth in the Scene goes round almost twice while the planet goes round once.

Feedback for a wrong answer
Not yet. If the year is too short, make the orbit bigger; if it is too long, make it smaller.

Hint
In astronomical units and years, $T^2 = a^3$ ⁠. So $a = T^{2/3}$ ⁠.

Worked solution
1. Kepler’s third law, in years and astronomical units, round a star of one solar mass: $T^2 = a^3$ ⁠.
2. $T = 1.88$ ⁠, so $a^3 = 1.88^2 \approx 3.53$ ⁠.
3. $a = \sqrt[3]{3.53} \approx 1.52\ \mathrm{AU}$ ⁠.
That is exactly Mars’s distance from the Sun. This is how Kepler worked out the size of the Solar System: from the periods, which are easy to measure, to the distances, which are hard.
Step 6 of 6
Open this step
Shape and year
Practice · Predict, then look
Question
The orbit is now a circle of size 1 AU, and the planet’s year is the same as the Earth’s. Size of the orbit stays as it is, and the orbit is stretched to eccentricity $0.8$ ⁠. What will happen to the planet’s year?

Correct answer
- Correct It will stay exactly one year
Setup, choices and feedback
The Scene at this step
Eccentricity
0
Size of the orbit
1 AUfixed
Show equal areas
Orbit
Choices
- Correct It will stay exactly one year
- Wrong It will get shorter, because the planet passes close to the Sun and is fast there
  Misconception: Thinks stretching an orbit at the same size (semi-major axis) changes how long a year lasts.
  Feedback from the catalogue
  By Kepler’s third law, the period depends only on the semi-major axis $a$ ⁠: $T^2 \propto a^3$ ⁠. A stretched orbit with the same $a$ ⁠ comes closer to the Sun on one side and goes farther from it on the other, and the two effects cancel exactly. An ellipse with $e = 0.8$ ⁠ and a circle with the same $a$ ⁠ take the same year.
- Wrong It will get longer, because the way round an ellipse is longer
  Misconception: Thinks stretching an orbit at the same size (semi-major axis) changes how long a year lasts.
  Feedback from the catalogue
  By Kepler’s third law, the period depends only on the semi-major axis $a$ ⁠: $T^2 \propto a^3$ ⁠. A stretched orbit with the same $a$ ⁠ comes closer to the Sun on one side and goes farther from it on the other, and the two effects cancel exactly. An ellipse with $e = 0.8$ ⁠ and a circle with the same $a$ ⁠ take the same year.
Once the prediction is locked in, the Scene changes
Eccentricity
0.8
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The year stays a year: it depends only on $a$ ⁠. The planet simply spends most of the year far out and slow, and makes the time up on the fast run past the Sun.

Worked solution
1. Kepler’s third law: $T^2 = \dfrac{4\pi^2}{GM}\, a^3$ ⁠. The formula holds only the semi-major axis $a$ ⁠, not $e$ ⁠.
2. $a$ ⁠ stays 1 AU, so $T$ ⁠ stays one year.
3. What actually happens: close to the Sun the planet is very fast and gets through that part quickly, and far out it is slow and spends most of the year there. The two effects cancel exactly.
In the Scene, the planet and the Earth come back to the same place together on every orbit.

Lesson 2 of 4 · About 25 minutes

Escape velocity

How fast you have to go never to come back. Kinetic energy against the well of gravity, the escape speed from the Earth, the Sun and the Moon, and stretched orbits that come close to it.

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The 5 steps and their answers

Step 1 of 5
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Energy to leave
Practice · Choose
Question
A 500 kg probe and a 5000 kg probe are launched from the same place. To escape gravity, how fast does the heavy probe need to go?

Correct answer
- Correct At the same speed as the light one
Setup, choices and feedback
The Scene at this step
Eccentricity
0
Size of the orbit
1 AU
Orbit / Equal areas
Orbit
Choices
Shown in a random order.
- Correct At the same speed as the light one
- Wrong Faster, because gravity pulls it harder
  Misconception: Thinks a heavier probe needs a higher speed to escape a planet or a star.
  Feedback from the catalogue
  A heavier body is pulled harder, but at the same speed it also has more kinetic energy. In $\tfrac12 m v^2 = \dfrac{G M m}{r}$ ⁠ the mass $m$ ⁠ cancels: $v = \sqrt{\dfrac{2 G M}{r}}$ ⁠. The escape speed depends only on the mass of the planet or star and the distance from it. A heavy probe needs more energy, so more propellant, but not a higher speed.
- Wrong $\sqrt{10}$ ⁠ times faster, because its energy has to be $10$ ⁠ times as large
  Misconception: Thinks a heavier probe needs a higher speed to escape a planet or a star.
  Feedback from the catalogue
  A heavier body is pulled harder, but at the same speed it also has more kinetic energy. In $\tfrac12 m v^2 = \dfrac{G M m}{r}$ ⁠ the mass $m$ ⁠ cancels: $v = \sqrt{\dfrac{2 G M}{r}}$ ⁠. The escape speed depends only on the mass of the planet or star and the distance from it. A heavy probe needs more energy, so more propellant, but not a higher speed.
- Wrong Slower, because a heavy body is harder to stop
  Misconception: Thinks a heavier probe needs a higher speed to escape a planet or a star.
  Feedback from the catalogue
  A heavier body is pulled harder, but at the same speed it also has more kinetic energy. In $\tfrac12 m v^2 = \dfrac{G M m}{r}$ ⁠ the mass $m$ ⁠ cancels: $v = \sqrt{\dfrac{2 G M}{r}}$ ⁠. The escape speed depends only on the mass of the planet or star and the distance from it. A heavy probe needs more energy, so more propellant, but not a higher speed.
Feedback for a right answer
The escape speed is $\sqrt{\dfrac{2GM}{r}}$ ⁠, with no probe mass in it. The heavy probe needs more energy and more propellant, but not more speed.

Worked solution
1. To escape, the kinetic energy has to cover the work gravity does all the way out to infinity: $\tfrac12 m v^2 = \dfrac{G M m}{r}$ ⁠.
2. The probe’s mass $m$ ⁠ is on both sides and cancels: $v = \sqrt{\dfrac{2 G M}{r}}$ ⁠.
3. $M$ ⁠ is the mass of the planet or star, and $r$ ⁠ is the same distance, so the speed is the same.
The heavy probe does need $10$ ⁠ times the energy, so $10$ ⁠ times the propellant. But at the same speed it also has $10$ ⁠ times the kinetic energy.
Step 2 of 5
Open this step
From the Earth
Practice · Calculate
Question
What is the escape speed from the Earth’s surface, ignoring the air? The Earth’s mass is $5.97 \times 10^{24}\ \mathrm{kg}$ ⁠, its radius is 6370 km, and $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ ⁠.

Correct answer
11180 m/s ± 2%
Setup, choices and feedback
The Scene at this step
Eccentricity
0
Size of the orbit
1 AU
Orbit / Equal areas
Orbit
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
- 7906 m/s ± 1.5%
  Misconception: Uses the circular orbital speed √(GM/r) as the escape speed.
  Feedback from the catalogue
  The speed of a circular orbit, $\sqrt{\dfrac{GM}{r}}$ ⁠, only keeps the body on the circle: it falls and misses, but does not get any farther away. To escape it needs kinetic energy that covers the whole well of gravity, $\tfrac12 m v^2 = \dfrac{G M m}{r}$ ⁠, which gives $v = \sqrt{\dfrac{2GM}{r}}$ ⁠: $\sqrt 2 \approx 1.41$ ⁠ times the circular speed.
- 15810 m/s ± 2%
  Misconception: Thinks the escape speed is twice the circular orbital speed.
  Feedback from the catalogue
  The factor $2$ ⁠ is inside the square root: $v_{\text{escape}} = \sqrt{\dfrac{2GM}{r}} = \sqrt 2 \cdot \sqrt{\dfrac{GM}{r}}$ ⁠. Escaping takes twice the energy of a circular orbit at the same distance, and kinetic energy grows like $v^2$ ⁠, so the speed grows only $\sqrt 2 \approx 1.41$ ⁠ times, not $2$ ⁠.
Feedback for a right answer
About 11.2 km/s, more than 40,000 km/h. A real rocket does not have to reach that speed near the ground: it keeps pushing on the way up, and higher up the escape speed is lower.

Hint
$v = \sqrt{\dfrac{2GM}{r}}$ ⁠, with $r$ ⁠ in metres.

Worked solution
1. The givens: $GM = 6.67 \times 10^{-11} \cdot 5.97 \times 10^{24} \approx 3.98 \times 10^{14}\ \mathrm{m^3/s^2}$ ⁠, $r = 6.37 \times 10^6\ \mathrm{m}$ ⁠.
2. The formula: $v = \sqrt{\dfrac{2GM}{r}}$ ⁠.
3. Substituting: $v = \sqrt{\dfrac{7.96 \times 10^{14}}{6.37 \times 10^6}} = \sqrt{1.25 \times 10^8} \approx 11{,}200\ \mathrm{m/s}$ ⁠.
That is $\sqrt 2$ ⁠ times the speed of an orbit skimming the ground, about 7900 m/s. Answering $7900$ ⁠ means working out an orbit, not an escape; answering about $15{,}800$ ⁠ means doubling the speed instead of the energy.
Step 3 of 5
Open this step
From the Sun
Practice · Calculate
Question
The planet moves on a circle of radius 1 AU, like the Earth. How fast would a body at that distance have to move to escape the Sun altogether?

Correct answer
42120 m/s ± 2%
Setup, choices and feedback
The Scene at this step
Eccentricity
0fixed
Size of the orbit
1 AUfixed
Orbit / Equal areas
Orbit
Accepted within
± 2%
Computed from the Scene at this step’s setting
The speed that would escape the star from the closest point, √(2GM/r)/nodes/orbit/outputs/escapeSpeed
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 29790 m/s ± 1%
  Misconception: Uses the circular orbital speed √(GM/r) as the escape speed.
  Feedback from the catalogue
  The speed of a circular orbit, $\sqrt{\dfrac{GM}{r}}$ ⁠, only keeps the body on the circle: it falls and misses, but does not get any farther away. To escape it needs kinetic energy that covers the whole well of gravity, $\tfrac12 m v^2 = \dfrac{G M m}{r}$ ⁠, which gives $v = \sqrt{\dfrac{2GM}{r}}$ ⁠: $\sqrt 2 \approx 1.41$ ⁠ times the circular speed.
- 59570 m/s ± 2%
  Misconception: Thinks the escape speed is twice the circular orbital speed.
  Feedback from the catalogue
  The factor $2$ ⁠ is inside the square root: $v_{\text{escape}} = \sqrt{\dfrac{2GM}{r}} = \sqrt 2 \cdot \sqrt{\dfrac{GM}{r}}$ ⁠. Escaping takes twice the energy of a circular orbit at the same distance, and kinetic energy grows like $v^2$ ⁠, so the speed grows only $\sqrt 2 \approx 1.41$ ⁠ times, not $2$ ⁠.
Feedback for a right answer
About 42.1 km/s, $\sqrt 2$ ⁠ times the Earth’s 29.8 km/s. Voyager 1 passed that speed with swings past Jupiter and Saturn.

Hint
The Earth’s orbital speed is about 29.8 km/s. The escape speed at the same distance is $\sqrt 2$ ⁠ times as large.

Worked solution
1. On a circular orbit $v_{\text{orbit}} = \sqrt{\dfrac{GM}{r}}$ ⁠, and the Earth moves like that at about 29.8 km/s.
2. The escape speed at the same distance: $v_{\text{escape}} = \sqrt{\dfrac{2GM}{r}} = \sqrt 2 \cdot v_{\text{orbit}}$ ⁠.
3. $v_{\text{escape}} = 1.414 \cdot 29.8 \approx 42.1\ \mathrm{km/s}$ ⁠.
So to send a probe from the Earth out of the Solar System, it only has to be given about 12 km/s on top of the speed it already shares with the Earth, if it is launched the way the Earth is moving.
Step 4 of 5
Open this step
Stretched orbits
Practice · Reach the target
Question
The more stretched the orbit, the closer and faster the planet passes the Sun. Change Eccentricity until at the closest point its speed is $90\%$ ⁠ of the escape speed there, and press Check.

Goal
- 0.9 ± 0.0025
  At the start of the step: 0.866
Setup, choices and feedback
The Scene at this step
Eccentricity
0.5
Size of the orbit
1 AUfixed
Orbit / Equal areas
Orbit
Read from the Scene at
Speed at the closest point over the escape speed there, √((1 + e)/2)/nodes/orbit/outputs/escapeShare
One setting that reaches it
Eccentricity
0.62
Feedback for a right answer
At $e = 0.62$ ⁠, $\sqrt{\dfrac{1.62}{2}} = 0.9$ ⁠. Comets falling in from the edge of the Solar System move on far more stretched orbits, and near the Sun they are moving at almost the escape speed.

Feedback for a wrong answer
Not yet. To get closer to the escape speed, stretch the orbit more; if you have gone past $90\%$ ⁠, back off a little.

Hint
At the closest point $\dfrac{v}{v_{\text{escape}}} = \sqrt{\dfrac{1+e}{2}}$ ⁠.

Worked solution
1. At the closest point, $r = a(1-e)$ ⁠, conservation of energy gives the speed as $v^2 = \dfrac{GM}{a}\cdot\dfrac{1+e}{1-e}$ ⁠, and the escape speed there is $v_{\text{escape}}^2 = \dfrac{2GM}{a(1-e)}$ ⁠.
2. The ratio: $\dfrac{v^2}{v_{\text{escape}}^2} = \dfrac{1+e}{2}$ ⁠.
3. We want $0.9^2 = 0.81 = \dfrac{1+e}{2}$ ⁠, so $e = 0.62$ ⁠.
On a circle the ratio is $\sqrt{1/2} \approx 0.71$ ⁠, and for any eccentricity below $1$ ⁠ it stays below $1$ ⁠: the planet stays bound to the Sun. At $e = 1$ ⁠ the orbit opens into a parabola, and the body escapes.
Step 5 of 5
Open this step
The Moon
Practice · Calculate
Question
The Apollo spacecraft took off from the Moon without a giant rocket. What is the escape speed from the Moon’s surface? Its mass is $7.35 \times 10^{22}\ \mathrm{kg}$ ⁠, and its radius is 1737 km.

Correct answer
2376 m/s ± 2%
Setup, choices and feedback
The Scene at this step
Eccentricity
0.85
Size of the orbit
1 AU
Orbit / Equal areas
Orbit
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
- 1680 m/s ± 2%
  Misconception: Uses the circular orbital speed √(GM/r) as the escape speed.
  Feedback from the catalogue
  The speed of a circular orbit, $\sqrt{\dfrac{GM}{r}}$ ⁠, only keeps the body on the circle: it falls and misses, but does not get any farther away. To escape it needs kinetic energy that covers the whole well of gravity, $\tfrac12 m v^2 = \dfrac{G M m}{r}$ ⁠, which gives $v = \sqrt{\dfrac{2GM}{r}}$ ⁠: $\sqrt 2 \approx 1.41$ ⁠ times the circular speed.
- 3360 m/s ± 2%
  Misconception: Thinks the escape speed is twice the circular orbital speed.
  Feedback from the catalogue
  The factor $2$ ⁠ is inside the square root: $v_{\text{escape}} = \sqrt{\dfrac{2GM}{r}} = \sqrt 2 \cdot \sqrt{\dfrac{GM}{r}}$ ⁠. Escaping takes twice the energy of a circular orbit at the same distance, and kinetic energy grows like $v^2$ ⁠, so the speed grows only $\sqrt 2 \approx 1.41$ ⁠ times, not $2$ ⁠.
Feedback for a right answer
About 2.4 km/s. That is why one small engine was enough for the Apollo ascent stage, while leaving the Earth takes a three-stage rocket.

Hint
The same formula as for the Earth: $v = \sqrt{\dfrac{2GM}{r}}$ ⁠, with $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ ⁠.

Worked solution
1. $GM = 6.67 \times 10^{-11} \cdot 7.35 \times 10^{22} \approx 4.90 \times 10^{12}\ \mathrm{m^3/s^2}$ ⁠, $r = 1.737 \times 10^6\ \mathrm{m}$ ⁠.
2. $v = \sqrt{\dfrac{2 \cdot 4.90 \times 10^{12}}{1.737 \times 10^6}} = \sqrt{5.65 \times 10^6} \approx 2380\ \mathrm{m/s}$ ⁠.
Less than a quarter of the Earth’s escape speed, and with no air to slow you. Getting about $1680$ ⁠ means working out the speed of an orbit skimming the Moon’s surface.

Lesson 3 of 4 · About 30 minutes

Lagrange points

Five points where a probe goes round the Sun in step with the Earth. Why L1 and L2 are so close to the Earth, why they are unstable, and why L4 and L5 hold asteroids.

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The 7 steps and their answers

Step 1 of 7
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Turning with the Earth
Practice · Choose
Question
A probe at L1, between the Sun and the Earth, goes round the Sun in exactly one year, like the Earth, even though it is closer to the Sun. On an orbit that close it ought to go round faster. What slows it down?

Correct answer
- Correct The Earth pulls it outward, making the net force towards the Sun smaller
Setup, choices and feedback
The Scene at this step
Park the probe at
L1 · between
Nudge away from the Sun
0 % of Sun–Earth distance
Earth's mass
1 % of the Sun's
Show effective gravity
Off
Choices
Shown in a random order.
- Correct The Earth pulls it outward, making the net force towards the Sun smaller
- Wrong The two forces cancel at the point, so nothing pulls on it
  Misconception: Thinks the forces on a probe at a Lagrange point cancel, so nothing pulls on it there.
  Feedback from the catalogue
  At a Lagrange point the forces do not cancel. The probe is going round the Sun, so it needs a net force towards the centre of its turning, just like the Earth. What is special about the point is that the net force of the Sun and the Earth together is exactly the force needed to go round in one year, at the same pace as the Earth. Only in a frame turning with the Earth does it look like a balance.
- Wrong At L1 the Earth’s pull equals the Sun’s
  Misconception: Thinks L1 (or L2) is where the Sun's pull and the Earth's pull are equal.
  Feedback from the catalogue
  Where the pulls are equal the net force would be zero, and the probe would carry on in a straight line instead of going round the Sun. L1 is farther from the Earth: there the Sun’s pull is still the stronger, and the Earth’s pull only weakens it until just the centripetal force of a one-year orbit is left. So L1 and L2 are roughly $\sqrt[3]{\mu/3}$ ⁠ of the distance to the Sun from the Earth, where $\mu$ ⁠ is the Earth’s share of the total mass, and not at the distance $\sqrt{\mu}$ ⁠ where the pulls are equal.
- Wrong Nothing: close to the Sun every body goes round more slowly
  Feedback
  The other way round: by Kepler’s third law, a closer body goes round faster. Something has to weaken the pull inward.
Feedback for a right answer
The Earth’s pull weakens the Sun’s by just enough for the probe to go round in a year, so it stays in line with the Sun and the Earth.

Worked solution
1. To go round the Sun on a circle of radius $r$ ⁠ in a time $T$ ⁠ takes a net inward force of $m\dfrac{4\pi^2 r}{T^2}$ ⁠.
2. At L1 the probe is closer to the Sun than the Earth is, so the Sun’s pull alone is more than an orbit of one year needs: on its own, it would go round in less than a year.
3. The Earth, on the other side, pulls it outward. The net force, the Sun’s pull minus the Earth’s, is exactly the force an orbit of one year needs.
The net force is not zero: the probe is going round, so it needs a force inward. Where the pulls are equal, the probe would carry on in a straight line.
Step 2 of 7
Open this step
How far
Practice · Calculate
Question
Earth’s mass in the Scene is locked at $1\%$ ⁠ of the Sun’s, and the probe is parked at L1. How far is L1 from the Earth, as a percentage of the Sun–Earth distance? Look at the Scene, then use the estimate $\sqrt[3]{\mu/3}$ ⁠, where $\mu$ ⁠ is the Earth’s share of the total mass.

Correct answer
14.15 ± 1
Setup, choices and feedback
The Scene at this step
Park the probe at
L1 · betweenfixed
Nudge away from the Sun
0 % of Sun–Earth distance
Earth's mass
1 % of the Sun'sfixed
Show effective gravity
Off
Accepted within
± 1
Computed from the Scene at this step’s setting
Distance from Earth to the chosen point, as a share of the Sun–Earth distance/nodes/system/outputs/fromEarth
Wrong answers it recognises
- 50 ± 3
  Misconception: Thinks L1 lies halfway between the Sun and the Earth.
  Feedback from the catalogue
  The Sun is more than $300{,}000$ ⁠ times heavier than the Earth, so its pull dominates almost all the way. The Earth changes the picture only close to it: L1 and L2 are about one per cent of the way to the Sun, about $1.5$ ⁠ million kilometres from the Earth, and not halfway.
- 9.1 ± 0.8
  Misconception: Thinks L1 (or L2) is where the Sun's pull and the Earth's pull are equal.
  Feedback from the catalogue
  Where the pulls are equal the net force would be zero, and the probe would carry on in a straight line instead of going round the Sun. L1 is farther from the Earth: there the Sun’s pull is still the stronger, and the Earth’s pull only weakens it until just the centripetal force of a one-year orbit is left. So L1 and L2 are roughly $\sqrt[3]{\mu/3}$ ⁠ of the distance to the Sun from the Earth, where $\mu$ ⁠ is the Earth’s share of the total mass, and not at the distance $\sqrt{\mu}$ ⁠ where the pulls are equal.
Feedback for a right answer
About $14\%$ ⁠. The real Earth is $3000$ ⁠ times lighter than this, so its L1 is $\sqrt[3]{3000} \approx 14$ ⁠ times closer: about one per cent of the way to the Sun.

Hint
$\mu = \dfrac{0.01}{1.01} \approx 0.0099$ ⁠, and $\sqrt[3]{0.0033} \approx 0.15$ ⁠. The estimate gets better the smaller $\mu$ ⁠ is.

Worked solution
1. $\mu = \dfrac{0.01}{1 + 0.01} \approx 0.0099$ ⁠.
2. The estimate for L1 and L2: $\dfrac{r}{R} \approx \sqrt[3]{\dfrac{\mu}{3}} = \sqrt[3]{0.0033} \approx 0.149$ ⁠, about $15\%$ ⁠.
3. The Scene’s exact calculation gives about $14\%$ ⁠ for L1 and about $16\%$ ⁠ for L2: the estimate is their average.
Answering $50\%$ ⁠ means thinking the point is halfway; answering about $9\%$ ⁠ means looking for where the pulls are equal, $\sqrt{\mu}/(1 + \sqrt{\mu})$ ⁠. There the net force is zero, and no body could go round.
Step 3 of 7
Open this step
The real Earth
Practice · Calculate
Question
The James Webb Space Telescope is near the Sun–Earth L2. The Earth’s mass is $3.0 \times 10^{-6}$ ⁠ of the Sun’s, and the distance to the Sun is $1.50 \times 10^{11}\ \mathrm{m}$ ⁠. Using the estimate $\sqrt[3]{\mu/3}$ ⁠, how far is L2 from the Earth?

Correct answer
1500000000 m ± 3%
Setup, choices and feedback
The Scene at this step
Park the probe at
L2 · beyond Earth
Nudge away from the Sun
0 % of Sun–Earth distance
Earth's mass
0.1 % of the Sun's
Show effective gravity
Off
Accepted within
± 3%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 260000000 m ± 5%
  Misconception: Thinks L1 (or L2) is where the Sun's pull and the Earth's pull are equal.
  Feedback from the catalogue
  Where the pulls are equal the net force would be zero, and the probe would carry on in a straight line instead of going round the Sun. L1 is farther from the Earth: there the Sun’s pull is still the stronger, and the Earth’s pull only weakens it until just the centripetal force of a one-year orbit is left. So L1 and L2 are roughly $\sqrt[3]{\mu/3}$ ⁠ of the distance to the Sun from the Earth, where $\mu$ ⁠ is the Earth’s share of the total mass, and not at the distance $\sqrt{\mu}$ ⁠ where the pulls are equal.
- 75000000000 m ± 5%
  Misconception: Thinks L1 lies halfway between the Sun and the Earth.
  Feedback from the catalogue
  The Sun is more than $300{,}000$ ⁠ times heavier than the Earth, so its pull dominates almost all the way. The Earth changes the picture only close to it: L1 and L2 are about one per cent of the way to the Sun, about $1.5$ ⁠ million kilometres from the Earth, and not halfway.
Feedback for a right answer
About $1.5$ ⁠ million kilometres, four times the distance to the Moon. There the telescope goes round the Sun in a year, always behind the Earth.

Hint
$\dfrac{\mu}{3} = 1.0 \times 10^{-6}$ ⁠, and its cube root is easy to work out.

Worked solution
1. $\dfrac{\mu}{3} = \dfrac{3.0 \times 10^{-6}}{3} = 1.0 \times 10^{-6}$ ⁠.
2. $\sqrt[3]{10^{-6}} = 10^{-2} = 0.01$ ⁠: one per cent of the distance to the Sun.
3. $r = 0.01 \cdot 1.50 \times 10^{11} = 1.5 \times 10^9\ \mathrm{m}$ ⁠, about $1.5$ ⁠ million kilometres.
That is four times the distance to the Moon. Getting about $2.6 \times 10^8\ \mathrm{m}$ ⁠ means looking for where the pulls are equal, $\sqrt{\mu}$ ⁠ of the way.
Step 4 of 7
Open this step
A saddle
Practice · Predict, then look
Question
The probe is parked at L2 and nudged slightly, half a per cent of the distance to the Sun, away from the Sun. What will it do once it is let go?

Correct answer
- Correct Drift farther and farther from L2
Setup, choices and feedback
The Scene at this step
Park the probe at
L2 · beyond Earthfixed
Nudge away from the Sun
0.5 % of Sun–Earth distance
Earth's mass
1 % of the Sun's
Show effective gravity
Off
Choices
- Correct Drift farther and farther from L2
- Wrong Come back to L2 and rock about it, like a marble in a bowl
  Misconception: Thinks a Lagrange point is a dip that pulls a nudged probe back, like a marble in a bowl.
  Feedback from the catalogue
  On the effective-gravity surface, L1, L2 and L3 are saddles: in one direction the surface rises and in another it falls. A probe that moves slightly along the line of the Sun and the Earth rolls downhill and drifts off, faster and faster. That is why satellites at L1 and L2 need small orbit corrections every few weeks.
- Wrong Stay at its new place, because the forces balance there too
  Misconception: Thinks the forces on a probe at a Lagrange point cancel, so nothing pulls on it there.
  Feedback from the catalogue
  At a Lagrange point the forces do not cancel. The probe is going round the Sun, so it needs a net force towards the centre of its turning, just like the Earth. What is special about the point is that the net force of the Sun and the Earth together is exactly the force needed to go round in one year, at the same pace as the Earth. Only in a frame turning with the Earth does it look like a balance.
Once the prediction is locked in, the Scene changes
Show effective gravity
On
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It slides off and drifts away: L2 is a saddle, and nothing brings it back. L1 and L3 are the same.

Worked solution
1. At L2 the net force is exactly what an orbit of one year needs. Farther from the Sun, the pull of the Sun and the Earth is too weak for an orbit of one year, so the probe starts to move outward, relative to the Earth.
2. The farther it goes, the bigger the mismatch, and the faster it moves away.
3. Show effective gravity shows the effective-gravity surface: L2 is a saddle. In one direction the surface rises, and along the line to the Sun it falls away on both sides.
So L1, L2 and L3 are unstable, and spacecraft there have to correct their orbits every few weeks.
Step 5 of 7
Open this step
Trojans
Practice · Predict, then look
Question
Now the probe is parked at L4, sixty degrees ahead of the Earth, and nudged $1.5\%$ ⁠ of the distance to the Sun. Show effective gravity will show that L4 is a hilltop of the effective-gravity surface. What will the probe do?

Correct answer
- Correct Move in loops round L4, and stay nearby
Setup, choices and feedback
The Scene at this step
Park the probe at
L4 · 60° aheadfixed
Nudge away from the Sun
1.5 % of Sun–Earth distance
Earth's mass
1 % of the Sun's
Show effective gravity
Off
Choices
- Correct Move in loops round L4, and stay nearby
- Wrong Slide off the hilltop and drift away, as at L2
  Misconception: Thinks a nudged probe must drift away from L4 or L5 because they are hilltops of the effective gravity.
  Feedback from the catalogue
  L4 and L5 really are hilltops of the effective-gravity surface, but in a turning frame there is also the Coriolis force, which pushes every moving body sideways. A probe that starts down the hill is turned into a circuit round it, and stays in loops round the point. This is how thousands of Trojan asteroids have gathered at Jupiter’s L4 and L5. It works as long as the small body is light enough, below about $4\%$ ⁠ of the Sun’s mass.
- Wrong Fall into the Sun
  Feedback
  The Sun pulls on it, but the probe keeps going round it just as the Earth does. It does not lose that motion.
Once the prediction is locked in, the Scene changes
Show effective gravity
On
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Loops round L4: a hilltop, and stable all the same, because the turning of the frame steers the probe round the point instead of down the slope.

Worked solution
1. At L4 the probe, the Sun and the Earth form an equilateral triangle, and the net force of the two pulls points exactly at the centre of the turning.
2. On the effective surface L4 is a hilltop: a probe that moves off it starts to go downhill.
3. But in the turning frame every moving body is pushed sideways by the Coriolis force. The probe that starts downhill is turned into a circuit round the hilltop, and does not get away.
This is how Jupiter holds thousands of Trojan asteroids at its L4 and L5.
Step 6 of 7
Open this step
Routh's limit
Practice · Reach the target
Question
L4 holds a probe only while the small body is light enough. The probe is at L4, nudged slightly. Increase Earth’s mass until L4 stops holding it, and press Check.

Goal
- 0 ± 0.1
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Park the probe at
L4 · 60° aheadfixed
Nudge away from the Sun
0.5 % of Sun–Earth distance
Earth's mass
1 % of the Sun's
Show effective gravity
Off
Read from the Scene at
1 if the chosen point holds a nudged probe (L4 or L5 below Routh's limit), else 0/nodes/system/outputs/stable
One setting that reaches it
Nudge away from the Sun
0 % of Sun–Earth distance
Earth's mass
4.1 % of the Sun's
Feedback for a right answer
Above about $4\%$ ⁠ of the Sun’s mass, Routh’s limit, L4 no longer holds. In every real system of a star and a planet, the small body is far lighter than that.

Feedback for a wrong answer
Not yet. The probe is still looping round L4: increase the mass further.

Hint
Routh’s limit: about $4\%$ ⁠ of the Sun’s mass.

Worked solution
1. Whether L4 and L5 are stable depends on the mass ratio. Edward Routh’s calculation, in 1875, gives a limit: the small body’s share of the total mass must be below $\mu \approx 0.0385$ ⁠.
2. $\dfrac{m}{M} = \dfrac{\mu}{1 - \mu} \approx \dfrac{0.0385}{0.9615} \approx 0.040$ ⁠, that is $4\%$ ⁠ of the Sun’s mass.
3. Above about $4\%$ ⁠ the probe at L4 moves off in growing loops, and in the end it leaves.
Jupiter is about $0.1\%$ ⁠ of the Sun’s mass, and the Moon about $1.2\%$ ⁠ of the Earth’s, so the L4 and L5 of both hold Trojan bodies or clouds of dust.
Step 7 of 7
Open this step
A telescope at L2
Practice · Choose
Question
The James Webb telescope sees very faint infrared light, so it has to stay cold, below $-230^\circ\mathrm{C}$ ⁠. Why was it sent to L2 in particular?

Correct answer
- Correct From there the Sun, the Earth and the Moon are always on the same side, and one shield blocks the heat of all three
Setup, choices and feedback
The Scene at this step
Park the probe at
L2 · beyond Earth
Nudge away from the Sun
0 % of Sun–Earth distance
Earth's mass
1 % of the Sun's
Show effective gravity
On
Choices
Shown in a random order.
- Correct From there the Sun, the Earth and the Moon are always on the same side, and one shield blocks the heat of all three
- Wrong There is no gravity at L2, so the telescope does not have to move
  Misconception: Thinks the forces on a probe at a Lagrange point cancel, so nothing pulls on it there.
  Feedback from the catalogue
  At a Lagrange point the forces do not cancel. The probe is going round the Sun, so it needs a net force towards the centre of its turning, just like the Earth. What is special about the point is that the net force of the Sun and the Earth together is exactly the force needed to go round in one year, at the same pace as the Earth. Only in a frame turning with the Earth does it look like a balance.
- Wrong L2 is stable, so the telescope stays there without fuel
  Misconception: Thinks a spacecraft parked at L2 stays there for ever without firing its engines.
  Feedback from the catalogue
  L2 is a saddle, not a dip: every small drift grows. The James Webb telescope does not sit exactly on L2 but circles it on a wide orbit, and it fires its thrusters every few weeks to keep from sliding off. The propellant for those corrections is what sets how many years it can work.
- Wrong L2 is several times farther from the Sun than the Earth, so it is cold there
  Feedback
  L2 is only about one per cent farther from the Sun than the Earth. Sunlight there is almost as strong as it is here.
Feedback for a right answer
All the sources of heat on one side, and one shield to block them all. Gravity is there, and the orbit needs small corrections, but the shade is permanent.

Worked solution
1. L2 is on the line from the Sun through the Earth, beyond the Earth, and goes round the Sun with it.
2. So from L2 the Sun, the Earth and the Moon are always in roughly the same direction. One sunshield, the size of a tennis court, blocks all of them, and the telescope behind it cools down by itself.
3. L2 is not stable: the telescope circles it and corrects its orbit every few weeks.
Gravity is certainly there: it is what swings the telescope round the Sun once a year.

Lesson 4 of 4 · About 30 minutes

Rockets and the rocket equation

A rocket throws mass backward and gets momentum forward. Tsiolkovsky's rocket equation, why each last per cent of propellant is worth more, the exhaust speed, stages, and what it takes to reach orbit.

Open the lesson

The 8 steps and their answers

Step 1 of 8
Open this step
Nothing to push on
Practice · Choose
Question
Beyond the atmosphere there is no air. What pushes a rocket forward when its engine runs in space?

Correct answer
- Correct The gas it throws backward: the gas takes momentum backward, and the rocket gets an equal momentum forward
Setup, choices and feedback
The Scene at this step
Exhaust speed
3.2 km/s
Propellant share
88 %
Stages
1
Show the equation
Off
Choices
Shown in a random order.
- Correct The gas it throws backward: the gas takes momentum backward, and the rocket gets an equal momentum forward
- Wrong The gas pushes against the air behind the rocket
  Misconception: Thinks a rocket moves by pushing its exhaust against the air or the ground.
  Feedback from the catalogue
  A rocket needs nothing to lean on. It pushes the gas backward, and the gas pushes it forward with an equal and opposite force, by Newton’s third law. In terms of momentum: the total momentum of the rocket and the gas stays as it was, so when the gas gets momentum backward, the rocket gets an equal momentum forward. In a vacuum this works even better, because there is no air to hold the rocket back.
- Wrong Nothing: in space an engine cannot speed anything up, and the rocket only coasts on its inertia
  Misconception: Thinks a rocket engine cannot speed a spacecraft up in empty space.
  Feedback from the catalogue
  There is no air in space, but momentum is still conserved. When the engine throws mass backward at high speed, the spacecraft gets an equal momentum forward. That is how probes are sped up on the way to other planets, far from any atmosphere.
- Wrong The push from the launch pad, which stays in the rocket all the way
  Misconception: Thinks a rocket moves by pushing its exhaust against the air or the ground.
  Feedback from the catalogue
  A rocket needs nothing to lean on. It pushes the gas backward, and the gas pushes it forward with an equal and opposite force, by Newton’s third law. In terms of momentum: the total momentum of the rocket and the gas stays as it was, so when the gas gets momentum backward, the rocket gets an equal momentum forward. In a vacuum this works even better, because there is no air to hold the rocket back.
Feedback for a right answer
Conservation of momentum: whatever the gas takes backward, the rocket gets forward. It needs nothing to lean on.

Worked solution
1. Before ignition the total momentum of the rocket and its propellant is zero, in the frame in which the rocket is at rest.
2. No outside force acts on the system, so the total momentum stays zero: $m_{\text{gas}} v_{\text{gas}} + m_{\text{rocket}} v_{\text{rocket}} = 0$ ⁠.
3. When the gas is thrown backward, the rocket must move forward. In terms of forces: the rocket pushes the gas backward, and the gas pushes it forward (Newton’s third law).
Air does not help, it only slows the rocket. Rocket engines actually work better in a vacuum.
Step 2 of 8
Open this step
Momentum out of the back
Practice · Calculate
Question
A spacecraft of mass 1000 kg, propellant included, is at rest in space. Its engine throws 10 kg of gas backward at 3000 m/s. How fast will the spacecraft move afterwards?

Correct answer
30.3 m/s ± 3%
Setup, choices and feedback
The Scene at this step
Exhaust speed
3.2 km/s
Propellant share
88 %
Stages
1
Show the equation
Off
Accepted within
± 3%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 3000 m/s ± 5%
  Misconception: Thinks a rocket gains the exhaust speed itself, whatever the masses.
  Feedback from the catalogue
  What is conserved is momentum, not speed. If $10\ \mathrm{kg}$ ⁠ of gas leaves at 3000 m/s, it takes $30{,}000\ \mathrm{kg \cdot m/s}$ ⁠ of momentum backward, and a spacecraft of about $1000\ \mathrm{kg}$ ⁠ gets the same momentum forward: a speed of only about 30 m/s. To reach a high speed, a rocket has to throw away a great deal of mass.
- 0 m/s ± 0.5 m/s
  Misconception: Thinks a rocket engine cannot speed a spacecraft up in empty space.
  Feedback from the catalogue
  There is no air in space, but momentum is still conserved. When the engine throws mass backward at high speed, the spacecraft gets an equal momentum forward. That is how probes are sped up on the way to other planets, far from any atmosphere.
Feedback for a right answer
About 30 m/s: $\dfrac{10 \cdot 3000}{990}$ ⁠. To reach kilometres per second, the spacecraft has to throw away most of its mass.

Hint
The total momentum before the throw is zero, and it has to stay zero after it.

Worked solution
1. Before: everything at rest, total momentum $0$ ⁠.
2. After: the gas, $10\ \mathrm{kg}$ ⁠, takes $10 \cdot 3000 = 30{,}000\ \mathrm{kg \cdot m/s}$ ⁠ backward.
3. The spacecraft, $1000 - 10 = 990\ \mathrm{kg}$ ⁠, gets the same momentum forward: $v = \dfrac{30{,}000}{990} \approx 30.3\ \mathrm{m/s}$ ⁠.
The speed the spacecraft gets is $100$ ⁠ times smaller than the gas’s, because it is $99$ ⁠ times heavier. Answering $3000$ ⁠ means thinking the speed passes over as it is; what passes over is the momentum.
Step 3 of 8
Open this step
The rocket equation
Practice · Calculate
Question
A single-stage rocket, Exhaust speed 3.2 km/s (kerosene and oxygen), and Propellant share $88\%$ ⁠ of the liftoff mass. How much speed can it gain in empty space, with no gravity and no air?

Correct answer
6785 m/s ± 2%
Setup, choices and feedback
The Scene at this step
Exhaust speed
3.2 km/sfixed
Propellant share
88 %fixed
Stages
1fixed
Show the equation
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
The ideal speed gained by the whole design, in empty space (the rocket equation, stage by stage)/nodes/rocket/outputs/deltaV
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 2816 m/s ± 3%
  Misconception: Thinks the speed a rocket gains grows in proportion to its propellant share, so each extra kilogram adds the same.
  Feedback from the catalogue
  Each kilogram of propellant pushes what is left of the rocket, and near the end of the burn there is very little left to push, so each kilogram adds more speed. That is the rocket equation: $\Delta v = v_e \ln\dfrac{m_0}{m_f}$ ⁠. What counts is the mass ratio, so going from $85\%$ ⁠ to $95\%$ ⁠ propellant, a mass ratio three times as large, adds far more than going from $50\%$ ⁠ to $60\%$ ⁠.
- 26670 m/s ± 3%
  Misconception: Thinks the speed gained is in proportion to the mass ratio, Δv = vₑ·(m₀/m_f), or that the mass ratio is Δv/vₑ.
  Feedback from the catalogue
  The mass ratio goes in through a logarithm: $\Delta v = v_e \ln\dfrac{m_0}{m_f}$ ⁠. To double $\Delta v$ ⁠ the mass ratio has to be squared, so turned round it is a power: $\dfrac{m_0}{m_f} = e^{\Delta v / v_e}$ ⁠. That is what makes rockets so heavy: a little more speed takes a great deal more propellant.
Feedback for a right answer
$3.2 \ln 8.33 \approx 6.8\ \mathrm{km/s}$ ⁠: more than the exhaust speed itself, but still short of the 9.4 km/s it takes to reach orbit.

Hint
At burnout $12\%$ ⁠ of the mass is left, so $\dfrac{m_0}{m_f} = \dfrac{1}{0.12}$ ⁠.

Worked solution
1. The givens: $v_e = 3200\ \mathrm{m/s}$ ⁠; the liftoff mass is $m_0$ ⁠, and once all the propellant has burnt, $m_f = 0.12\, m_0$ ⁠ is left.
2. The rocket equation: $\Delta v = v_e \ln\dfrac{m_0}{m_f}$ ⁠.
3. Substituting: $\Delta v = 3200 \cdot \ln\dfrac{1}{0.12} = 3200 \cdot \ln 8.33 \approx 3200 \cdot 2.12 \approx 6780\ \mathrm{m/s}$ ⁠.
Answering about $2800$ ⁠ means multiplying $v_e$ ⁠ by the propellant share, as if every kilogram added the same; answering about $26{,}700$ ⁠ means multiplying by the mass ratio instead of its logarithm.
Step 4 of 8
Open this step
The last kilograms
Practice · Predict, then look
Question
Going from $50\%$ ⁠ to $60\%$ ⁠ propellant adds about 0.7 km/s to the rocket. Now Propellant share is $85\%$ ⁠. How much will going to $95\%$ ⁠, ten per cent more, add?

Correct answer
- Correct Much more, about 3.5 km/s
Setup, choices and feedback
The Scene at this step
Exhaust speed
3.2 km/sfixed
Propellant share
85 %
Stages
1fixed
Show the equation
On
Choices
- Wrong Also about 0.7 km/s
  Misconception: Thinks the speed a rocket gains grows in proportion to its propellant share, so each extra kilogram adds the same.
  Feedback from the catalogue
  Each kilogram of propellant pushes what is left of the rocket, and near the end of the burn there is very little left to push, so each kilogram adds more speed. That is the rocket equation: $\Delta v = v_e \ln\dfrac{m_0}{m_f}$ ⁠. What counts is the mass ratio, so going from $85\%$ ⁠ to $95\%$ ⁠ propellant, a mass ratio three times as large, adds far more than going from $50\%$ ⁠ to $60\%$ ⁠.
- Wrong Less, because a rocket full of propellant is heavier to speed up
  Feedback
  The extra propellant is heavy, but it burns. What counts is how much mass is left at the end, and that is three times smaller.
- Correct Much more, about 3.5 km/s
Once the prediction is locked in, the Scene changes
Propellant share
95 %
The Scene then shows: 9.586 km/s · The ideal speed gained by the whole design, in empty space (the rocket equation, stage by stage)/nodes/rocket/outputs/deltaV
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From 6.1 to 9.6 km/s: the mass ratio grew $3$ ⁠ times. That is why engineers fight for every kilogram of structure.

Worked solution
1. The rocket equation depends on the mass ratio: $\Delta v = v_e \ln\dfrac{m_0}{m_f}$ ⁠, and $\dfrac{m_0}{m_f} = \dfrac{1}{1-f}$ ⁠ for a propellant share $f$ ⁠.
2. From $50\%$ ⁠ to $60\%$ ⁠: the mass ratio rises from $2$ ⁠ to $2.5$ ⁠, $1.25$ ⁠ times. The gain is $3.2 \ln 1.25 \approx 0.7\ \mathrm{km/s}$ ⁠.
3. From $85\%$ ⁠ to $95\%$ ⁠: the mass ratio rises from $6.7$ ⁠ to $20$ ⁠, $3$ ⁠ times. The gain is $3.2 \ln 3 \approx 3.5\ \mathrm{km/s}$ ⁠.
The same ten per cent, five times the speed: near the end of the burn very little is left to push, so each kilogram of propellant adds more.
Step 5 of 8
Open this step
Exhaust speed
Practice · Reach the target
Question
Propellant share is locked at $85\%$ ⁠ and Stages at one. Change Exhaust speed until the rocket gains 8 km/s, and press Check.

Goal
- 8 km/s ± 1%
  At the start of the step: 6.071 km/s
Setup, choices and feedback
The Scene at this step
Exhaust speed
3.2 km/s
Propellant share
85 %fixed
Stages
1fixed
Show the equation
On
Read from the Scene at
The ideal speed gained by the whole design, in empty space (the rocket equation, stage by stage)/nodes/rocket/outputs/deltaV
One setting that reaches it
Exhaust speed
4.2 km/s
Feedback for a right answer
About 4.2 km/s: $\Delta v$ ⁠ is in proportion to the exhaust speed, so a faster engine is worth as much as a lot of extra propellant.

Feedback for a wrong answer
Not yet. If the rocket is too slow, raise the exhaust speed; if it is too fast, lower it.

Hint
The mass ratio here is $\dfrac{1}{0.15} \approx 6.67$ ⁠, so $v_e = \dfrac{\Delta v}{\ln 6.67}$ ⁠.

Worked solution
1. $\Delta v = v_e \ln\dfrac{m_0}{m_f}$ ⁠, with $\dfrac{m_0}{m_f} = \dfrac{1}{0.15} \approx 6.67$ ⁠ and $\ln 6.67 \approx 1.90$ ⁠.
2. $v_e = \dfrac{8.0}{1.90} \approx 4.2\ \mathrm{km/s}$ ⁠.
Almost a hydrogen and oxygen engine, 4.4 km/s, the fastest exhaust chemical propellant gives. The exhaust speed goes straight into the equation, so it is worth a lot of propellant: reaching the same 8 km/s with kerosene takes $92\%$ ⁠ propellant.
Step 6 of 8
Open this step
Stages
Practice · Predict, then look
Question
The same rocket, $88\%$ ⁠ propellant and exhaust speed 3.2 km/s, gains about 6.8 km/s with one stage. The same propellant and the same structure are split into three rockets stacked on top of each other. What will happen to the final speed?

Correct answer
- Correct It will grow
Setup, choices and feedback
The Scene at this step
Exhaust speed
3.2 km/sfixed
Propellant share
88 %fixed
Stages
1
Show the equation
Off
Choices
- Correct It will grow
- Wrong It will not change, because it is the same propellant
  Misconception: Thinks splitting the same propellant into stages cannot add speed, since the total propellant is the same.
  Feedback from the catalogue
  Stages do not add propellant, they shed dead weight. In a single-stage rocket the empty tanks and engines of the whole rocket ride to the end, and the last of the propellant has to speed them up too. In a staged rocket each stage that runs dry is dropped, and the next one starts lighter, with a better mass ratio.
- Wrong It will shrink, because three rockets waste more
  Misconception: Thinks splitting the same propellant into stages cannot add speed, since the total propellant is the same.
  Feedback from the catalogue
  Stages do not add propellant, they shed dead weight. In a single-stage rocket the empty tanks and engines of the whole rocket ride to the end, and the last of the propellant has to speed them up too. In a staged rocket each stage that runs dry is dropped, and the next one starts lighter, with a better mass ratio.
Once the prediction is locked in, the Scene changes
Stages
3
The Scene then shows: 10.19 km/s · The ideal speed gained by the whole design, in empty space (the rocket equation, stage by stage)/nodes/rocket/outputs/deltaV
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 10.2 km/s: the same propellant, but without hauling the empty tanks. That is why every rocket that reaches orbit has more than one stage.

Worked solution
1. In a single-stage rocket the structure, the tanks and engines, $10\%$ ⁠ of the mass, rides all the way, and at burnout it is most of what is left: $m_f = 12\%$ ⁠.
2. In a three-stage rocket, each stage whose propellant runs out is dropped with its structure. The next stage starts without that dead weight.
3. Each stage adds its own $v_e \ln\dfrac{m_0}{m_f}$ ⁠, and the sum of the three logarithms is bigger than the one-stage logarithm: about 10.2 km/s instead of 6.8.
That is the difference between a rocket that does not reach orbit and one that does.
Step 7 of 8
Open this step
To orbit
Practice · Reach the target
Question
Reaching low Earth orbit takes about 9.4 km/s: 7.8 km/s of orbital speed, and about 1.5 km/s more that gravity and the air take on the way. Change the rocket, any way you like, until it gets there, and press Check.

Goal
- 1 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Exhaust speed
3.2 km/s
Propellant share
88 %
Stages
1
Show the equation
On
Read from the Scene at
1 if the design gains the 9.4 km/s low Earth orbit takes, else 0/nodes/rocket/outputs/reachesOrbit
One setting that reaches it
Exhaust speed
4.5 km/s
Propellant share
88 %
Stages
1
Feedback for a right answer
The rocket reaches orbit. Real rockets combine all the ways: fast exhaust, light structure and two or three stages.

Feedback for a wrong answer
Not 9.4 km/s yet. Try adding a stage.

Hint
There are three ways: more propellant, faster exhaust, or more stages.

Worked solution
1. As it is: one stage, $88\%$ ⁠ propellant, 3.2 km/s: $\Delta v \approx 6.8\ \mathrm{km/s}$ ⁠.
2. Two stages with the same propellant give about 9.4 km/s: just on the line.
3. Other ways: one stage with $95\%$ ⁠ propellant, about 9.6 km/s, or three stages, about 10.2 km/s.
In practice it is very hard to build a rocket whose structure is only $3\%$ ⁠ of its mass, so real rockets use two or three stages.
Step 8 of 8
Open this step
The mass ratio
Practice · Calculate
Question
A single-stage rocket with a hydrogen and oxygen engine, exhaust speed 4.4 km/s, has to gain 9.4 km/s. What must its mass ratio, $\dfrac{m_0}{m_f}$ ⁠, be?

Correct answer
8.47 ± 2%
Setup, choices and feedback
The Scene at this step
Exhaust speed
4.4 km/s
Propellant share
88 %
Stages
1
Show the equation
On
Accepted within
± 2%
A fixed value in the question
Wrong answers it recognises
- 2.14 ± 0.05
  Misconception: Thinks the speed gained is in proportion to the mass ratio, Δv = vₑ·(m₀/m_f), or that the mass ratio is Δv/vₑ.
  Feedback from the catalogue
  The mass ratio goes in through a logarithm: $\Delta v = v_e \ln\dfrac{m_0}{m_f}$ ⁠. To double $\Delta v$ ⁠ the mass ratio has to be squared, so turned round it is a power: $\dfrac{m_0}{m_f} = e^{\Delta v / v_e}$ ⁠. That is what makes rockets so heavy: a little more speed takes a great deal more propellant.
Feedback for a right answer
About $8.5$ ⁠: nearly $90\%$ ⁠ of the rocket is propellant, even with the fastest exhaust there is. That is why a rocket looks like a fuel tank with a small payload on the end.

Hint
Turn the rocket equation round: $\dfrac{m_0}{m_f} = e^{\Delta v / v_e}$ ⁠.

Worked solution
1. The rocket equation: $\Delta v = v_e \ln\dfrac{m_0}{m_f}$ ⁠, so $\ln\dfrac{m_0}{m_f} = \dfrac{\Delta v}{v_e} = \dfrac{9.4}{4.4} \approx 2.14$ ⁠.
2. $\dfrac{m_0}{m_f} = e^{2.14} \approx 8.5$ ⁠.
3. So at burnout only $\dfrac{1}{8.5} \approx 12\%$ ⁠ of the mass is left: about $88\%$ ⁠ of the rocket is propellant.
Answering $2.14$ ⁠ means stopping at the logarithm; the mass ratio itself is $e$ ⁠ to that power.

Across the unit · 19 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Kepler's laws

Ellipse and focus

Question

Correct answer

The Scene at this step

Choices

Worked solution

Closest and farthest

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Equal areas in equal times

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Fast and slow

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The third law

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Shape and year

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Escape velocity

Energy to leave

Question

Correct answer

The Scene at this step

Choices

Worked solution

From the Earth

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

From the Sun

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Stretched orbits

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The Moon

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Lagrange points

Turning with the Earth

Question