The shape of the lesson
By the end of the lesson, learners can
- Gravity is the centripetal force: A satellite in a circular orbit has only one force on it: the planet's gravity, and that single force is the centripetal force.
- Field strength of a point mass, and near the surface: For a point mass the field strength is g = GM / r².
- Newton's law: F = Gm₁m₂ / r², with uniform spheres treated as point masses at their centres.
- Field strength and potential: g = GM / r² and ϕ = −GM / r, so g is the gradient of ϕ.
- Orbits: gravity is the centripetal force, GMm / r² = mv² / r, giving v = √(GM / r) and T = 2π√(r³ / GM).
One picture every idea returns to
Gravity supplies the centripetal force for an orbit, GMm / r² = mv² / r, giving v = √(GM / r) and T = 2π√(r³ / GM). The field strength of a point mass is g = GM / r², nearly constant close to a surface.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Held by gravity, nothing else.. An orbit is not a balance of forces. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Gravity is the centripetal force | A satellite in a circular orbit has only one force on it: the planet's gravity, and that single force is the centripetal force. Project tab 1 of the lab (the simulation) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Field strength of a point mass, and near the surface | For a point mass the field strength is g = GM / r². | Jigsaw |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (4 questions, 44 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Field strength of a point mass, and near the surface, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
No examiner trap is recorded for this lesson. Use the quiz key below to spot the distractor most of the class picks.
Orbits and field strength: topic worksheet
4 questions, 44 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
A satellite is in a geostationary orbit around the Earth, which has mass M = 6.0 × 10²⁴ kg. The orbit is shown in Fig. 2.1.
(a) State two features of a geostationary orbit. [2]
(b) Show that the radius of the orbit is about 4.2 × 10⁷ m. [3]
(c) Calculate the orbital speed of the satellite. [2]
(d) Calculate the gravitational field strength of the Earth at this orbital radius. [2]
(e) Explain why a satellite in a lower circular orbit cannot remain above the same point on the Earth's surface. [2]
(f) State one practical application that the geostationary orbit makes possible. [1]
A satellite orbits the Earth (M = 6.0 × 10²⁴ kg) in a circular orbit of radius r = 7.0 × 10⁶ m.
(a) State what provides the centripetal force on the satellite. [1]
(b) Show that the orbital speed is given by v = √(GM/r). [2]
(c) Calculate the orbital speed. [2]
(d) Calculate the orbital period. [2]
(e) State and explain how the orbital speed differs for a satellite in a higher orbit. [2]
(f) State the direction of the satellite’s acceleration. [1]
This question is about Kepler’s third law for circular orbits. The Moon orbits the Earth at radius r = 3.8 × 10⁸ m with a period of 27.3 days.
(a) State Kepler’s third law for circular orbits. [1]
(b) Show that for a circular orbit T² = 4π²r³ / GM. [3]
(c) Use the Moon’s orbital data to estimate the mass of the Earth. [3]
(d) State one assumption made in this calculation. [1]
(e) Two satellites orbit at radii r and 4r. State and explain the ratio of their periods. [2]
A satellite of mass 500 kg orbits the Earth (M = 6.0 × 10²⁴ kg) in a circular orbit of radius r = 7.0 × 10⁶ m. This question connects the orbital motion of Q6 with the energy ideas of Q8.
(a) Write expressions, in terms of G, M, m and r, for the kinetic energy and the gravitational potential energy of the satellite. [2]
(b) Show that the total energy of the satellite is E = −GMm / 2r. [2]
(c) Calculate the kinetic energy, the potential energy and the total energy of this satellite. [3]
(d) The satellite is moved to a higher orbit. State and explain what happens to its kinetic energy and to its total energy. [3]
(e) Explain how a satellite in a higher orbit can move more slowly yet have a greater total energy. [2]
Answer key and marking notes
- (a)Any two: period of 24 h (equal to Earth's rotation period) ✓; orbit lies in the equatorial plane ✓; moves in the same direction (west to east) as the Earth's rotation, so it stays above one fixed point.
- (b)Gravity is the centripetal force: GMm / r² = mr(2π/T)², so r³ = GMT² / 4π² ✓. With T = 24 h = 8.64 × 10⁴ s, r³ = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × (8.64 × 10⁴)²) / 4π² ✓ = 7.57 × 10²², r = 4.2 × 10⁷ m ✓
- (c)v = 2πr / T = (2π × 4.23 × 10⁷) / 8.64 × 10⁴ ✓ = 3.1 × 10³ m s⁻¹ ✓
- (d)g = GM / r² = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (4.23 × 10⁷)² ✓ = 0.22 N kg⁻¹ ✓
- (e)A smaller radius gives a shorter period (T² ∝ r³), so T < 24 h ✓; the satellite then completes an orbit faster than the Earth turns, so it moves relative to the ground and cannot stay above one point ✓
- (f)Communications or television relay using a fixed (non-tracking) ground aerial ✓
- (a)The gravitational attraction of the Earth on the satellite ✓
- (b)Gravity provides the centripetal force: GMm/r² = mv²/r ✓, so v² = GM/r and v = √(GM/r) ✓
- (c)v = √(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / 7.0 × 10⁶) ✓ = 7.6 × 10³ m s⁻¹ ✓
- (d)T = 2πr/v = (2π × 7.0 × 10⁶)/(7.56 × 10³) ✓ = 5.8 × 10³ s (about 97 min) ✓
- (e)v ∝ 1/√r, so a satellite in a higher orbit moves more slowly ✓✓
- (f)Directed toward the centre of the Earth (centripetal) ✓
- (a)The square of the orbital period is proportional to the cube of the orbital radius (T² ∝ r³) ✓
- (b)GMm/r² = mrω² = mr(2π/T)² ✓; rearranging, T² = 4π²r³/GM ✓✓
- (c)M = 4π²r³/(GT²); T = 27.3 × 86 400 = 2.36 × 10⁶ s ✓; M = 4π² × (3.8 × 10⁸)³ / (6.67 × 10⁻¹¹ × (2.36 × 10⁶)²) ✓ = 5.8 × 10²⁴ kg ✓
- (d)The orbit is circular, or the Earth’s mass greatly exceeds the Moon’s ✓
- (e)T ∝ r³⁄², so the ratio is 4³⁄² = 8 ✓: the outer satellite has a period eight times the inner one ✓
- (a)KE = ½mv² = GMm/2r (using v² = GM/r) ✓; Eₙ = −GMm/r ✓
- (b)E = KE + Eₙ = GMm/2r − GMm/r = −GMm/2r ✓✓
- (c)KE = +1.4 × 10¹⁰ J ✓; Eₙ = −2.9 × 10¹⁰ J ✓; total E = −1.4 × 10¹⁰ J ✓
- (d)At larger r the speed is smaller, so the kinetic energy decreases ✓; the total energy E = −GMm/2r increases (becomes less negative) ✓✓
- (e)The gain in potential energy is larger than the loss in kinetic energy, so the total energy rises toward zero even though the satellite is slower ✓✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
For a satellite in a stable circular orbit, the centripetal force is provided by:
- Athe gravitational pull of the planet
- Ba small onboard motor
- Cair resistance in the upper atmosphere
- Dan outward reaction force
Setting GMm / r² equal to mv² / r shows that the orbital speed is:
- Av = √(GM / r)
- Bv = GM / r
- Cv = √(GMr)
- Dv = GM / r²
A geostationary satellite must have an orbit that is:
- Aperiod 24 h, in the equatorial plane, moving with Earth's spin
- Bperiod 1 h, in any plane
- Cany period, directly above the poles
- Dperiod 12 h, moving against Earth's spin
Compared with sea level, the field strength g = GM / r² at the top of a mountain is:
- Aslightly smaller, because r is a little larger
- Bslightly larger
- Cexactly the same
- Dzero
Quiz key
Gravity is the only force acting, and it is the centripetal force: GMm / r² = mv² / r.
Cancel m and rearrange: v² = GM / r, so v = √(GM / r). A larger orbital radius gives a smaller speed.
To stay above one point it matches the Earth's rotation: a 24 h period in the equatorial plane, in the same sense as the spin.
g falls as 1/r²; a few kilometres of height barely change r, so g is very slightly smaller, which is why g is taken as constant near the surface.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.