The shape of the lesson
By the end of the lesson, learners can
- A field is force per unit mass: A mass placed near a planet feels a pull even with nothing touching it.
- Newton's law, and treating spheres as points: Newton's law of gravitation gives the attraction between two point masses as F = Gm₁m₂ / r², directed along the line joining them.
- 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
A gravitational field is a region where a mass experiences a force; its strength is g = F / m. Between two point masses the attraction is F = Gm₁m₂ / r², with uniform spheres treated as point masses at their centres.
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, Mass makes a field.. A gravitational field is a region where a mass feels a force. 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: A field is force per unit mass | A mass placed near a planet feels a pull even with nothing touching it. Project tab 1 of the lab (Field around a mass (3D)) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Newton's law, and treating spheres as points | Newton's law of gravitation gives the attraction between two point masses as F = Gm₁m₂ / r², directed along the line joining them. | Mix-Pair-Share |
| 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, 45 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 Newton's law, and treating spheres as points, 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.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
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.
Gravitational field and Newton's law: topic worksheet
4 questions, 45 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 planet P has mass M = 4.8 × 10²⁴ kg and radius R = 5.6 × 10⁶ m. A satellite of mass m = 320 kg is in a circular orbit at a height h = 2.0 × 10⁶ m above the surface of P, as shown in Fig. 1.1.
(a) State Newton's law of gravitation. [2]
(b) Show that the gravitational field strength at the surface of P is about 10 N kg⁻¹. [2]
(c) For the orbiting satellite, calculate:
(i) the orbital radius r, [1]
(ii) the gravitational force acting on it, [2]
(iii) its orbital speed, [2]
(iv) its period of orbit. [2]
(d) State and explain whether the orbital speed would change if a satellite of larger mass were placed in the same orbit. [2]
This question is about gravitational field strength and Newton’s law of gravitation. Take G = 6.67 × 10⁻¹¹ N m² kg⁻², and for the Earth M = 6.0 × 10²⁴ kg and R = 6.4 × 10⁶ m.
(a) Define gravitational field strength. [2]
(b) State Newton’s law of gravitation. [2]
(c) Show that the field strength at distance r from a point mass M is g = GM / r². [2]
(d) Calculate the gravitational field strength at the Earth’s surface. [2]
(e) Calculate the weight of a 70 kg astronaut at the Earth’s surface. [1]
(f) State and explain how g changes if the astronaut moves to twice the distance from the Earth’s centre. [1]
Planet P has twice the mass of the Earth and twice its radius, as shown in Fig. 5.1. The Earth has surface field strength 9.8 N kg⁻¹.
(a) Write the expression for the gravitational field strength at the surface of a planet of mass M and radius R. [2]
(b) Show that the surface field strength of Planet P is half that of the Earth. [3]
(c) Calculate the surface field strength of Planet P. [1]
(d) A 5.0 kg mass is taken from the Earth’s surface to the surface of Planet P. Compare its weight in the two places. [2]
(e) Explain why surface field strength depends on both the mass and the radius of a planet. [2]
Fig. 10.1 shows how the gravitational potential φ near a planet varies with distance r from its centre. This question connects potential (Q8) with field strength (Q4).
(a) State the relationship between gravitational field strength and gravitational potential. [2]
(b) Explain how the field strength at a point can be obtained from the φ-r graph. [1]
(c) Compare how field strength and potential each vary with distance from a point mass, referring to their different powers of r. [3]
(d) State the relationship between gravitational field lines and equipotential surfaces. [2]
(e) The potential at the planet’s surface is −6.3 × 10⁷ J kg⁻¹. Calculate the energy needed to remove a 2.0 kg mass completely from the surface to infinity. [3]
(f) State the value of the gravitational potential at infinity. [1]
Answer key and marking notes
- (a)The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation ✓, directed along the line joining them (attractive) ✓
- (b)g = GM / R² = (6.67 × 10⁻¹¹ × 4.8 × 10²⁴) / (5.6 × 10⁶)² ✓ = 3.20 × 10¹⁴ / 3.14 × 10¹³ = 10.2 N kg⁻¹, about 10 N kg⁻¹ ✓
- (c)(i)r = R + h = 5.6 × 10⁶ + 2.0 × 10⁶ = 7.6 × 10⁶ m ✓
- (c)(ii)F = GMm / r² = (6.67 × 10⁻¹¹ × 4.8 × 10²⁴ × 320) / (7.6 × 10⁶)² ✓ = 1.8 × 10³ N ✓
- (c)(iii)v = √(GM / r) = √(3.20 × 10¹⁴ / 7.6 × 10⁶) ✓ = √(4.21 × 10⁷) = 6.5 × 10³ m s⁻¹ ✓
- (c)(iv)T = 2πr / v = (2π × 7.6 × 10⁶) / 6.5 × 10³ ✓ = 7.4 × 10³ s (about 2.0 h) ✓
- (d)No change ✓. Gravity provides the centripetal force: GMm / r² = mv² / r, so v = √(GM / r); the satellite mass cancels and v depends only on M and r ✓
- (a)The gravitational force per unit mass acting on a small test mass placed at the point ✓✓
- (b)The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation ✓✓
- (c)g = F/m = (GMm/r²)/m = GM/r² ✓✓
- (d)g = GM/R² = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴)/(6.4 × 10⁶)² ✓ = 9.8 N kg⁻¹ ✓
- (e)W = mg = 70 × 9.8 = 6.9 × 10² N ✓
- (f)g ∝ 1/r², so at twice the distance g falls to one quarter (about 2.4 N kg⁻¹) ✓
- (a)g = GM/R² ✓✓
- (b)gₚ = G(2M)/(2R)² = 2GM/4R² = ½(GM/R²) ✓✓ = ½ gₐ ✓
- (c)gₚ = ½ × 9.8 = 4.9 N kg⁻¹ ✓
- (d)On Earth W = 5.0 × 9.8 = 49 N ✓; on Planet P W = 5.0 × 4.9 = 25 N (half) ✓
- (e)g = GM/R²: a larger mass increases the field, while a larger radius places the surface further from the centre and reduces it through the inverse-square dependence ✓✓
- (a)The field strength is the negative gradient of the potential: g = −dφ/dr ✓✓
- (b)It is the magnitude of the gradient (steepness) of the φ-r graph at that point ✓
- (c)Field strength varies as 1/r² (an inverse-square fall-off, steeper) ✓; potential varies as 1/r (a gentler fall-off) ✓; so the potential reaches further before becoming negligible ✓
- (d)Field lines are perpendicular to equipotential surfaces ✓ and point from high to low potential (toward the planet) ✓
- (e)Energy = m(φ∞ − φₙ) = 2.0 × (0 − (−6.3 × 10⁷)) ✓✓ = 1.3 × 10⁸ J ✓
- (f)Zero ✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
A gravitational field strength of 9.8 N kg⁻¹ at a point means that:
- Aeach kilogram of mass there feels a force of 9.8 N
- Beach newton of force moves 9.8 kg
- Cthe potential there is 9.8 J kg⁻¹
- Dthe field does 9.8 J of work on any mass
The gravitational field lines around an isolated point mass:
- Apoint radially inward, toward the mass
- Bpoint radially outward, away from the mass
- Care parallel and evenly spaced
- Dform closed loops around the mass
In F = Gm₁m₂ / r² applied to two uniform spheres, the distance r is measured between:
- Atheir centres
- Btheir nearest surfaces
- Ctheir farthest surfaces
- Dany pair of surface points
Two point masses attract with force F. If both masses are doubled while the separation is unchanged, the force becomes:
- A4F
- B2F
- CF
- DF / 4
Quiz key
Field strength is force per unit mass, g = F/m, so a 1 kg mass feels 9.8 N.
Field lines show the direction of the force on a small test mass, which is toward the attracting mass, so they are radial and inward; their spreading shows the field weakening.
A uniform sphere acts as a point mass at its centre, so r is the centre-to-centre separation.
F is proportional to the product m₁m₂, so doubling each mass multiplies the force by 2 × 2 = 4.
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.