Senior 5 practice book · Unit 5 of 11

Motion on Orbits

50 questions that complete the Senior 5 quiz for this unit: 19 core and 31 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

There is no gravity in space, which is why astronauts float.At the height of the space station g is still about 8.7 m/s², nearly 90% of its surface value. Astronauts float because they and their craft are falling together around the Earth.
A satellite needs its engines running all the time to stay in orbit.Once in orbit, gravity alone supplies the centripetal force; with no air resistance the satellite keeps moving without fuel, like the Moon has for billions of years.
Planets move round the Sun at a constant speed in perfect circles.Orbits are ellipses with the Sun at one focus (Kepler's first law), and a planet moves faster when it is closer to the Sun (Kepler's second law).
To reach a higher orbit a satellite ends up moving faster.A higher circular orbit has a LOWER speed, v = √(GM/r). Energy must be supplied to climb, but it goes into potential energy and the kinetic energy actually decreases.
The Earth pulls the Moon harder than the Moon pulls the Earth.By Newton's third law the two forces are equal in size and opposite in direction; the Moon accelerates more only because its mass is smaller.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Newton's law of universal gravitation F = Gm₁m₂/r²
Gravitational field strength g = GM/r² and its variation with height
Variation of g with depth, g = g₀(1 − d/R)
Kepler's first and second laws: elliptical orbits and equal areas in equal times
Orbital speed and period of satellites and planets, Kepler's third law T² = 4π²r³/GM
Geostationary and polar satellites and the uses of satellites
Gravitational potential V = −GM/r and potential energy U = −GMm/r
Escape velocity v = √(2GM/R)
Energy of a satellite in orbit: kinetic, potential, total and binding energy; changing orbits
Weightlessness and apparent weight in orbit, in free fall and in lifts
Finding the mass and density of the Earth, planets and the Sun

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 31 harder ones are marked ★ Challenge.

1True or false
According to Kepler's second law, every planet moves round the Sun at a constant speed.
Show answer
Answer: False
The law says equal AREAS are swept in equal times. Because the distance to the Sun changes around an ellipse, the speed must change: faster when near, slower when far.
Common mistakeLearners often read 'equal areas in equal times' as 'equal distances in equal times'; equal areas need different arc lengths at different distances.
2True or false · ★ Challenge
The value of g at sea level is slightly larger at the poles than at the equator.
Show answer
Answer: True
The Earth bulges at the equator (so the poles are closer to the centre), and the Earth's rotation uses part of gravity to provide centripetal force at the equator; both make g at the poles (about 9.83 m/s²) larger than at the equator (about 9.78 m/s²).
Common mistakeLearners often assume g is exactly the same everywhere on the surface; it varies slightly with latitude and altitude.
3True or false · ★ Challenge
When thin air at the top of the atmosphere slowly drags a satellite down into a lower orbit, the satellite speeds up.
Show answer
Answer: True
In a lower orbit v = √(GM/r) is larger. The satellite loses potential energy; half of it becomes extra kinetic energy and half is turned into heat by the drag, so its total energy falls while its speed rises.
Common mistakeLearners often think drag must always slow an object down; for an orbiting satellite, drag lowers the orbit and the speed increases.
4True or false
The Earth pulls on the Moon with a larger force than the Moon pulls on the Earth, because the Earth is more massive.
Show answer
Answer: False
Newton's third law: the forces are equal and opposite, both equal to GMm/r². The Earth accelerates less only because its mass is larger.
Common mistakeLearners often link 'bigger mass' with 'bigger force on the other body'; the same product Mm appears in both forces.
5True or false · ★ Challenge
If you could go down a deep tunnel towards the Earth's centre, your weight would increase, because you would be getting closer to the centre.
Show answer
Answer: False
Inside a uniform Earth only the mass closer to the centre than you pulls you inwards (the outer shell's pulls cancel). Going down reduces this mass faster than the distance shrinks, so g = g₀(1 − d/R) decreases.
Common mistakeLearners often apply g ∝ 1/r² inside the Earth; that rule is only valid outside it.
6True or false · ★ Challenge
The mass of a planet can be found by timing one of its moons and measuring the moon's orbital radius, without knowing the mass of the moon.
Show answer
Answer: True
GMm/r² = m(4π²r/T²) gives M = 4π²r³/GT²; the moon's mass m cancels.
Common mistakeLearners often think you need both masses because both appear in F = GMm/r²; the moon's mass cancels when gravity provides the centripetal force.
7True or false
A rocket must reach escape velocity, 11.2 km/s, before it can leave the Earth.
Show answer
Answer: False
Escape velocity is for an object that is thrown and then gets no more push. A rocket that keeps firing its engines can climb away slowly; it never needs to reach 11.2 km/s close to the Earth.
Common mistakeLearners often treat escape velocity as a speed limit every spacecraft must pass; it only applies to an unpowered launch.
8Multiple choice · ★ Challenge
Which of these would NOT work normally inside an orbiting space station?
1. AMeasuring a mass with bathroom scales
2. BMeasuring a length with a metre rule
3. CTiming an experiment with a quartz watch
4. DReading a temperature with a digital thermometer
Show answer
Answer: A. Measuring a mass with bathroom scales
Bathroom scales measure the contact force from your weight. In free fall there is no contact force, so they read zero for any mass (astronauts use oscillating devices instead).
Common mistakeChoosing the watch confuses quartz clocks with pendulum clocks; a pendulum needs a supporting force and would not swing, but a quartz crystal vibrates normally.
9Fill in the blank
An object on which the only force acting is gravity is said to be in ______ fall.
Show answer
Answer: free
In free fall the acceleration is g and there is no supporting force, so the object feels weightless.
Common mistakeWriting 'zero-gravity' fall misdescribes the situation; gravity is the one force that IS acting.
10Multiple choice · ★ Challenge
A comet moves round the Sun in a long ellipse. Where is its speed greatest?
1. AAt the point closest to the Sun
2. BAt the point farthest from the Sun
3. CIt is the same all the way round
4. DHalfway between the closest and farthest points
Show answer
Answer: A. At the point closest to the Sun
Kepler's second law: the line from the Sun sweeps equal areas in equal times. Near the Sun the line is short, so the comet must cover a longer arc in the same time: it moves fastest there.
Common mistakeChoosing 'farthest from the Sun' reverses the law; far away the line is long, so a short arc already sweeps the required area and the comet moves slowly.
11Multiple choice · ★ Challenge
A satellite is moved from a low circular orbit to a higher circular orbit. How do its energies change?
1. AKE increases, PE increases, total energy increases
2. BKE decreases, PE decreases, total energy decreases
3. CKE decreases, PE increases, total energy increases
4. DKE increases, PE decreases, total energy stays the same
Show answer
Answer: C. KE decreases, PE increases, total energy increases
In the higher orbit v = √(GM/r) is smaller, so KE falls; PE = −GMm/r becomes less negative (rises); the total −GMm/2r also rises, which is why fuel is needed.
Common mistakeChoosing 'KE increases' assumes a higher orbit needs a faster satellite; higher circular orbits are slower.
12Fill in the blank
Kepler's ______ law states that the line joining a planet to the Sun sweeps out equal areas in equal times.
Show answer
Answer: second
First law: elliptical orbits; second law: equal areas in equal times; third law: T² ∝ r³.
Common mistakeWriting 'third' mixes it up with the period–radius law T² ∝ r³.
13Multiple choice · ★ Challenge
Why has the Moon lost almost all of its atmosphere, while the Earth has kept its own?
1. AThe Moon is too cold for any gas to exist
2. BThe Sun's gravity pulls the Moon's air away
3. CThe Moon has no magnetic field, so gravity cannot act on gases there
4. DThe Moon's escape velocity (about 2.4 km/s) is low enough for many gas molecules to reach it
Show answer
Answer: D. The Moon's escape velocity (about 2.4 km/s) is low enough for many gas molecules to reach it
Gas molecules move at speeds of a few hundred m/s, and some much faster. On the Moon a noticeable fraction exceed 2.4 km/s and escape; on the Earth (11.2 km/s) very few do.
Common mistakeChoosing 'no magnetic field' confuses magnetism with gravity; gravity acts on all masses whatever the magnetic field.
14Fill in the blank · ★ Challenge
The Moon is about 60 Earth radii from the Earth's centre, so the Earth's gravitational field strength there is the surface value divided by ______.
Show answer
Answer: 3600
g ∝ 1/r²: (60)² = 3600, so g = 9.8 ÷ 3600 = 2.7 × 10⁻³ N/kg.
Common mistakeWriting 60 forgets to square the distance ratio.
15True or false
If the cable of a lift snapped and the lift fell freely, a passenger standing on scales in it would read zero.
Show answer
Answer: True
Passenger and scales fall together with acceleration g, so the scales do not need to push on the passenger: N = m(g − g) = 0. This is the same 'weightlessness' felt in orbit.
Common mistakeLearners often think the scales would show a large reading; the reading is the contact force, which vanishes in free fall.
16Multiple choice · ★ Challenge
Halley's comet comes within 0.6 AU of the Sun but travels out to 35 AU. Where does it spend most of each 76-year orbit?
1. AClose to the Sun, where it moves slowly
2. BFar from the Sun, where it moves slowly
3. CEqual times near and far, as the orbit is symmetrical
4. DClose to the Sun, because gravity holds it there longer
Show answer
Answer: B. Far from the Sun, where it moves slowly
By Kepler's second law the comet moves very slowly at large distances, so it spends most of its time far out and only a few months near the Sun, when it can be seen.
Common mistakeChoosing 'equal times' assumes equal distances take equal times; the symmetrical SHAPE does not mean a uniform speed.
17Multiple choice
Which of these is NOT a common use of artificial satellites?
1. AProducing electricity for the national grid
2. BWeather forecasting and storm warnings
3. CNavigation with GPS on phones and moto-taxi apps
4. DBroadcasting television and radio
Show answer
Answer: A. Producing electricity for the national grid
Satellites are used for communication, broadcasting, navigation, weather, mapping and scientific research. Their solar panels only power the satellite itself.
Common mistakeChoosing GPS forgets that phone navigation relies on signals from a group of satellites in 12-hour orbits.
18Multiple choice · ★ Challenge
A company asks for a geostationary satellite that stays exactly above Kigali (1.9° south of the equator). Why is this impossible?
1. AKigali is too high above sea level for a satellite to stay over it
2. BGeostationary satellites must orbit from east to west, against the direction of the Earth's rotation
3. CAn orbit's plane must pass through the Earth's centre; only an equatorial one keeps pace
4. DA satellite over Kigali would need a period of 12 hours, not 24 hours
Show answer
Answer: C. An orbit's plane must pass through the Earth's centre; only an equatorial one keeps pace
Gravity points to the Earth's centre, so every orbit lies in a plane through the centre. An orbit tilted to pass over 1.9° S would swing north and south each day; only an equatorial orbit stays above one point. Kigali's dishes point at satellites over the equator, almost overhead.
Common mistakeChoosing 'east to west' reverses the direction; a geostationary satellite must move west to east, the same way as the Earth turns.
19Multiple choice · ★ Challenge
A gold mine is 3.2 km deep. Assuming a uniform Earth of radius 6400 km, by what percentage is g at the bottom smaller than at the surface?
1. A0.05%
2. B0.10%
3. C5%
4. D0.5%
Show answer
Answer: A. 0.05%
g = g₀(1 − d/R), so the fractional decrease is d/R = 3.2 ÷ 6400 = 0.0005 = 0.05%.
Common mistakeChoosing 0.10% uses the rule for HEIGHT (a fall of about 2h/R); below the surface g falls only as d/R.
20Multiple choice
What does Kepler's first law say about the orbit of a planet?
1. AIt is a circle, with the Sun at its centre
2. BIt is an ellipse, with the Sun at its centre
3. CIt is an ellipse, with the Sun at one focus
4. DIt is a spiral that slowly gets closer to the Sun
Show answer
Answer: C. It is an ellipse, with the Sun at one focus
Kepler showed from Tycho Brahe's measurements of Mars that planetary orbits are ellipses, with the Sun at one focus (not at the centre).
Common mistakeChoosing 'the Sun at its centre' of an ellipse is a common slip; the Sun sits at a focus, which is off-centre, so the planet's distance changes.
21Short answer · ★ Challenge
Explain why an astronaut standing on the surface of the Moon is not weightless, but an astronaut in a spacecraft orbiting the Moon is.
Show answer
Model answer: On the surface, the ground pushes up on the astronaut with a contact force equal to the weight (mg_Moon), so the astronaut feels weight. In orbit, the astronaut and spacecraft both fall freely towards the Moon with the same acceleration, so the spacecraft floor does not push on the astronaut; with no contact force the astronaut feels weightless, although gravity still acts.
Common mistakeLearners often say there is no gravity in orbit; the gravity is nearly the same, but the feeling of weight comes from the contact force, which is missing in free fall.
22Multiple choice
What provides the centripetal force that keeps a satellite in a circular orbit round the Earth?
1. AThe satellite's rocket engines
2. BA centrifugal force pushing outwards
3. CThe Earth's magnetic field
4. DThe gravitational pull of the Earth
Show answer
Answer: D. The gravitational pull of the Earth
Gravity acts towards the Earth's centre, at right angles to the satellite's velocity, so it changes the direction of motion and keeps the satellite on its circle.
Common mistakeChoosing 'centrifugal force' adds an outward force that does not act on the satellite; the only real force is gravity, directed inwards.
23Fill in the blank · ★ Challenge
For a spacecraft skimming just above the Moon's surface, gravity mg provides the centripetal force mv²/R, so v = √(gR). With g = 1.6 N/kg and R = 1.74 × 10⁶ m on the Moon, v ≈ ______ km/s.
Show answer
Answer: 1.7
v = √(1.6 × 1.74 × 10⁶) = √(2.78 × 10⁶) = 1.7 × 10³ m/s = 1.7 km/s.
Common mistakeWriting 2.4 km/s gives the Moon's escape velocity √(2gR); the orbital speed has no factor 2.
24Fill in the blank · ★ Challenge
A probe orbits just above the surface of a rocky planet with a period of 5000 s. Using ρ = 3π/GT² (G = 6.67 × 10⁻¹¹ N m²/kg²), the mean density of the planet is about ______ kg/m³.
Show answer
Answer: 5.7 × 10³
ρ = 3π ÷ (6.67 × 10⁻¹¹ × 5000²) = 9.42 ÷ 1.67 × 10⁻³ = 5.7 × 10³ kg/m³.
Common mistakeWriting about 2.8 × 10⁷ kg/m³ forgets to square T; the period appears squared in the formula.
25Multiple choice · ★ Challenge
For the Earth, GM = 4.0 × 10¹⁴ N m²/kg. What is the gravitational potential at a point 1.0 × 10⁷ m from the Earth's centre?
1. A+4.0 × 10⁷ J/kg
2. B−4.0 × 10⁷ J/kg
3. C−4.0 J/kg
4. D−4.0 × 10¹⁴ J/kg
Show answer
Answer: B. −4.0 × 10⁷ J/kg
V = −GM/r = −4.0 × 10¹⁴ ÷ 1.0 × 10⁷ = −4.0 × 10⁷ J/kg.
Common mistakeChoosing the positive value forgets that potential is zero at infinity and negative everywhere closer, because work must be done to pull a mass away.
26Fill in the blank
Gravitational potential is taken to be ______ at an infinite distance from the mass producing the field.
Show answer
Answer: zero
Choosing V = 0 at infinity makes the potential at every finite distance negative, V = −GM/r.
Common mistakeWriting 'zero at the Earth's surface' uses the everyday mgh convention, which does not work for the whole field.
27Multiple choice · ★ Challenge
A 50 kg learner stands on bathroom scales in a lift in a Kigali office tower. The lift accelerates DOWNWARDS at 2.0 m/s² (g = 9.8 m/s²). What do the scales read?
1. A590 N
2. B490 N
3. C390 N
4. D0 N
Show answer
Answer: C. 390 N
Taking down as positive: mg − N = ma, so N = m(g − a) = 50 × (9.8 − 2.0) = 390 N.
Common mistakeChoosing 590 N adds the accelerations; a downward acceleration means the floor pushes up LESS than the weight.
28Fill in the blank
If both masses are doubled and the distance between their centres is also doubled, the gravitational force between them is multiplied by ______.
Show answer
Answer: 1 (it is unchanged)
F ∝ m₁m₂/r²: (2 × 2) ÷ 2² = 4 ÷ 4 = 1.
Common mistakeWriting 2 forgets to square the distance; doubling r divides the force by 4, which cancels the factor 4 from the masses.
29Multiple choice · ★ Challenge
Taking g = 9.8 m/s² at the Earth's surface, R = 6.4 × 10⁶ m and G = 6.67 × 10⁻¹¹ N m²/kg², what is the mass of the Earth?
1. A9.4 × 10¹⁷ kg
2. B6.0 × 10¹⁸ kg
3. C4.0 × 10¹⁴ kg
4. D6.0 × 10²⁴ kg
Show answer
Answer: D. 6.0 × 10²⁴ kg
g = GM/R², so M = gR²/G = 9.8 × (6.4 × 10⁶)² ÷ 6.67 × 10⁻¹¹ = 6.0 × 10²⁴ kg.
Common mistakeChoosing 6.0 × 10¹⁸ kg leaves R in kilometres (6400); the radius must be in metres before it is squared.
30Multiple choice · ★ Challenge
The escape velocity from the Earth's surface is 11.2 km/s. A space probe is at rest relative to the Earth at a height of 3R above the surface (R = Earth's radius). What speed does it need to escape from there?
1. A2.8 km/s
2. B11.2 km/s
3. C5.6 km/s
4. D6.5 km/s
Show answer
Answer: C. 5.6 km/s
v_e = √(2GM/r) ∝ 1/√r. From r = R + 3R = 4R: v = 11.2 ÷ √4 = 5.6 km/s.
Common mistakeChoosing 6.5 km/s divides 11.2 by √3, using the height 3R instead of the distance from the centre, 4R.
31True or false
The International Space Station, with an orbital period of about 92 minutes, goes round the Earth about 16 times every day.
Show answer
Answer: True
One day = 24 × 60 = 1440 min; 1440 ÷ 92 ≈ 16 orbits.
Common mistakeLearners often think every satellite takes a day to go round, mixing up low orbits with geostationary ones.
32Multiple choice · ★ Challenge
A 500 kg satellite moves in a circular orbit of radius 8.0 × 10⁶ m (G = 6.67 × 10⁻¹¹ N m²/kg², M = 6.0 × 10²⁴ kg). What is its kinetic energy?
1. A2.50 × 10¹⁰ J
2. B−1.25 × 10¹⁰ J
3. C−2.50 × 10¹⁰ J
4. D1.25 × 10¹⁰ J
Show answer
Answer: D. 1.25 × 10¹⁰ J
KE = ½mv² with v² = GM/r, so KE = GMm/2r = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 500 ÷ (2 × 8.0 × 10⁶) = 1.25 × 10¹⁰ J.
Common mistakeChoosing 2.50 × 10¹⁰ J gives GMm/r, the size of the POTENTIAL energy; the kinetic energy is half of it, and it can never be negative.
33Fill in the blank
For a satellite in a circular orbit, its kinetic energy is ______ the size of its gravitational potential energy.
Show answer
Answer: half
KE = GMm/2r and |PE| = GMm/r, so KE = ½|PE|.
Common mistakeWriting 'equal to' forgets the factor ½ that comes from ½mv² with v² = GM/r.
34Multiple choice · ★ Challenge
A probe travels straight out from the Earth (radius R). How far above the ground has the Earth's field fallen to one hundredth of the value at the ground?
1. A10R
2. B99R
3. C100R
4. D9R
Show answer
Answer: D. 9R
g ∝ 1/r². For g to fall to 1/100, r must be √100 = 10 times larger: r = 10R. The height above the surface is r − R = 9R.
Common mistakeChoosing 10R gives the distance from the Earth's CENTRE, not the height above the surface.
35Multiple choice
Which statement correctly describes what the gravitational potential V at a point near the Earth means?
1. AThe work done per unit mass in bringing a mass from infinity to that point
2. BThe force per unit mass on a small mass placed at that point
3. CThe work done in bringing any mass from the Earth's surface up to that point
4. DThe kinetic energy per unit mass that a body has at that point
Show answer
Answer: A. The work done per unit mass in bringing a mass from infinity to that point
V = W/m with the reference at infinity; its unit is J/kg and its value is negative because the field does the work.
Common mistakeChoosing 'force per unit mass' gives the definition of gravitational field strength g, not potential.
36Multiple choice · ★ Challenge
Mars orbits the Sun at 1.52 times the Earth's orbital radius. Using Kepler's third law, how long is a year on Mars?
1. A1.52 Earth years
2. B2.31 Earth years
3. C1.87 Earth years
4. D3.51 Earth years
Show answer
Answer: C. 1.87 Earth years
T² ∝ r³, so T_Mars = 1.52^3/2 = √(1.52³) = √3.51 = 1.87 Earth years.
Common mistakeChoosing 3.51 years forgets to take the square root: 1.52³ gives T², not T.
37Multiple choice · ★ Challenge
Jupiter has 318 times the mass of the Earth and is 5.2 times farther from the Sun. How many times larger is the Sun's pull on Jupiter than on the Earth?
1. AAbout 61
2. BAbout 12
3. CAbout 0.085
4. DAbout 1650
Show answer
Answer: B. About 12
F ∝ m/r²: F_J/F_E = 318 ÷ 5.2² = 318 ÷ 27.0 = 11.8.
Common mistakeChoosing about 61 divides by 5.2 only; the force falls with the SQUARE of the distance.
38Short answer · ★ Challenge
A black hole is so compact that its escape velocity at its surface equals the speed of light (3.0 × 10⁸ m/s). To what radius would the Sun (mass 2.0 × 10³⁰ kg) have to be squeezed to become a black hole? (G = 6.67 × 10⁻¹¹ N m²/kg²)
Show answer
Model answer: Set √(2GM/R) = c, so R = 2GM/c² = 2 × 6.67 × 10⁻¹¹ × 2.0 × 10³⁰ ÷ (3.0 × 10⁸)² = 3.0 × 10³ m, only about 3 km (the Sun's real radius is 7 × 10⁸ m).
Common mistakeLearners often forget to square c, which gives a radius 300 million times too large.
39Multiple choice
Which statement about gravity at the height of the International Space Station (about 400 km up, R = 6400 km) is correct?
1. Ag there is zero, which is why astronauts float
2. Bg there is about 89% of its value at the surface
3. Cg there is about half of its surface value
4. Dg there is exactly the same as at the surface
Show answer
Answer: B. g there is about 89% of its value at the surface
g ∝ 1/r²: g/g₀ = (6400 ÷ 6800)² = 0.89, so g ≈ 8.7 m/s².
Common mistakeChoosing 'zero' is the most common misconception; astronauts float because they are in free fall, not because gravity has disappeared.
40Short answer · ★ Challenge
The Earth is closest to the Sun in early January (1.47 × 10¹¹ m) and farthest in early July (1.52 × 10¹¹ m). At these two points it moves at right angles to the line to the Sun, so Kepler's second law gives r₁v₁ = r₂v₂. In which month does the Earth move faster, and by what percentage?
Show answer
Model answer: From r₁v₁ = r₂v₂, v_Jan/v_Jul = r_Jul/r_Jan = 1.52 ÷ 1.47 = 1.034. The Earth moves about 3.4% faster in January, when it is closest to the Sun.
Common mistakeLearners often assume the Earth is closest to the Sun in the hot season; in fact the closest approach is in January, and the seasons are caused by the tilt of the Earth's axis.
41Multiple choice
A satellite of mass m moves in a circular orbit of radius r round the Earth (mass M). Its kinetic energy is GMm/2r and its potential energy is −GMm/r. What is its total energy?
1. A−GMm/2r
2. B+GMm/2r
3. C−3GMm/2r
4. DZero
Show answer
Answer: A. −GMm/2r
E = KE + PE = GMm/2r − GMm/r = −GMm/2r. The total is negative, so the satellite is bound to the Earth.
Common mistakeChoosing −3GMm/2r subtracts the kinetic energy instead of adding it; both energies are added with their signs.
42Short answer · ★ Challenge
A TV programme is sent live from Kigali to Nairobi through a geostationary satellite about 3.6 × 10⁷ m above the equator. Estimate the shortest time delay for the signal, and say why people notice the delay in a two-way phone call over the same link.
Show answer
Model answer: The signal must go up about 3.6 × 10⁷ m and down about 3.6 × 10⁷ m at the speed of light: t = 2 × 3.6 × 10⁷ ÷ 3.0 × 10⁸ = 0.24 s. In a conversation the reply must also travel up and down, so there is about 0.5 s between speaking and hearing an answer, which makes people talk over each other.
Common mistakeLearners often use only one trip (up), giving 0.12 s; the signal must go up to the satellite and back down.
43Short answer · ★ Challenge
Satellite A moves in a circular orbit of radius 7000 km at 7.6 km/s. Satellite B orbits the same planet at a radius of 28 000 km. Without using G or M, find the speed of B. Explain why B is slower.
Show answer
Model answer: v = √(GM/r), so v ∝ 1/√r. r is 4 times larger, so v is √4 = 2 times smaller: v_B = 7.6 ÷ 2 = 3.8 km/s. Farther out the gravitational pull is weaker, so a smaller speed is enough for gravity to provide the centripetal force.
Common mistakeLearners often divide by 4 (v ∝ 1/r); the speed depends on the square root of the radius.
44Multiple choice
The planet Mercury has mass 3.3 × 10²³ kg and radius 2.44 × 10⁶ m. What is the gravitational field strength at its surface? (G = 6.67 × 10⁻¹¹ N m²/kg²)
1. A9.0 × 10⁶ N/kg
2. B3.7 N/kg
3. C7.4 N/kg
4. D0.92 N/kg
Show answer
Answer: B. 3.7 N/kg
g = GM/R² = 6.67 × 10⁻¹¹ × 3.3 × 10²³ ÷ (2.44 × 10⁶)² = 3.7 N/kg, a little more than a third of the Earth's value.
Common mistakeChoosing 9.0 × 10⁶ N/kg divides by R instead of R²; that huge value should warn you that a square is missing.
45Short answer · ★ Challenge
Using M = 6.0 × 10²⁴ kg and R = 6.4 × 10⁶ m, calculate the mean density of the Earth. Rocks at the surface have a density of about 2700 kg/m³. What does your answer suggest about the inside of the Earth?
Show answer
Model answer: Volume = (4/3)πR³ = (4/3)π × (6.4 × 10⁶)³ = 1.10 × 10²¹ m³. Mean density = 6.0 × 10²⁴ ÷ 1.10 × 10²¹ = 5.5 × 10³ kg/m³. This is about twice the density of surface rocks, so the material deep inside (the core, rich in iron and nickel) must be much denser.
Common mistakeLearners often use the area 4πR² instead of the volume (4/3)πR³; density is mass per unit volume.
46Short answer
Compare a geostationary orbit with a low polar orbit under these headings: height, period, plane of orbit, one typical use.
Show answer
Model answer: Geostationary: about 36 000 km high; period 24 h; above the equator, moving west to east; used for TV broadcasting and communications. Low polar: a few hundred km high (e.g. 500–800 km); period about 90–100 min; passes over (or near) both poles; used for detailed mapping, weather and environmental monitoring of the whole Earth.
Common mistakeLearners often say polar satellites are higher; they are much LOWER, which is why they see the ground in detail.
47Short answer · ★ Challenge
Explain why the total energy of a satellite in orbit is negative. Calculate the binding energy (the energy needed to free it completely) of a 200 kg satellite in an orbit of radius 7.0 × 10⁶ m (G = 6.67 × 10⁻¹¹ N m²/kg², M = 6.0 × 10²⁴ kg).
Show answer
Model answer: Potential energy is zero at infinity and negative everywhere closer; in a circular orbit KE = GMm/2r is only half the size of PE = −GMm/r, so the total E = −GMm/2r is negative: the satellite is bound. Binding energy = GMm/2r = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200 ÷ (2 × 7.0 × 10⁶) = 5.7 × 10⁹ J.
Common mistakeLearners often give GMm/r as the binding energy; the satellite already has kinetic energy GMm/2r, so only GMm/2r more is needed.
48Short answer · ★ Challenge
A learner estimates the energy needed to lift a 1.0 kg mass from the Earth's surface to a height equal to the Earth's radius using mgh = 1.0 × 9.8 × 6.4 × 10⁶. Show that her answer is twice the correct value, and explain why mgh fails here.
Show answer
Model answer: Her value: mgh = 6.3 × 10⁷ J. Correct value: ΔU = GMm(1/R − 1/2R) = GMm/2R = mgR/2 (since GM = gR²) = 3.1 × 10⁷ J, half of hers. The formula mgh assumes g is constant, but g falls as 1/r² and is only g/4 at r = 2R, so the real average force is much smaller.
Common mistakeLearners often use mgh for any height; it is only accurate when h is very small compared with R.
49Short answer
Sketch a graph of g against distance r from the centre of the Earth, from r = 0 to r = 3R (assume uniform density). Describe the shape of each part.
Show answer
Model answer: From r = 0 to r = R, g increases in a straight line from zero at the centre to g₀ at the surface (g ∝ r). Beyond the surface g falls as 1/r²: g₀/4 at r = 2R and g₀/9 at r = 3R, a curve that drops quickly and then levels off towards zero. The maximum is at the surface.
Common mistakeLearners often draw g largest at the centre; at the centre the pulls of the surrounding matter cancel and g = 0.
50Short answer · ★ Challenge
Newton tested the inverse-square law with the Moon. The Moon's orbit has radius 3.84 × 10⁸ m (about 60 Earth radii) and period 27.3 days. Calculate the Moon's centripetal acceleration and compare it with g₀/60², where g₀ = 9.8 m/s². What does the comparison show?
Show answer
Model answer: T = 27.3 × 86 400 = 2.36 × 10⁶ s. a = 4π²r/T² = 4π² × 3.84 × 10⁸ ÷ (2.36 × 10⁶)² = 2.72 × 10⁻³ m/s². g₀/60² = 9.8 ÷ 3600 = 2.72 × 10⁻³ m/s². They agree, so the same gravity that makes an apple fall keeps the Moon in orbit, weakened by the inverse square of the distance.
Common mistakeLearners often forget to convert days into seconds, which gives an answer billions of times too large.