Physics • Year 12 • Module 5 • Lesson 10
Newton’s Law of Universal Gravitation
Lock in the formula, the meaning of every symbol, the inverse square law pattern, and the Cavendish experiment before tackling harder problems.
1. Formula recall card
Complete the table. For each formula write the SI units of every variable and a one-line note on when to use it. 6 marks (1 per completed row)
| Name | Formula | Variables & SI units | When to use it |
|---|---|---|---|
| Gravitational force | F = GMm / r2 | F = ______ (______) ; G = ______ ; M, m = ______ (______) ; r = ______ (______) | |
| Gravitational field strength (general) | g = GM / r2 | g in ______ ; M = ______ (______) ; r = ______ (______) | |
| Field strength at altitude | g′ = g (R / (R + h))2 | R = ______ (______) ; h = ______ (______) | |
| Centre-to-centre distance | r = R + h | R = ______ ; h = ______ | |
| Inverse square law | F ∝ 1 / r2 | No new variables | |
| Earth’s mass from surface g | M = gR2 / G | M in ______ ; R = ______ (______) |
2. Term–definition match
Write the matching term from this list into the right-hand column: gravitational field strength, inverse square law, universal gravitational constant (G), centre-to-centre distance, free fall, Newton’s Law of Universal Gravitation, microgravity, torsion balance, attractive force. 9 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 2.1 | Every particle of matter attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the separation between their centres. | |
| 2.2 | The proportionality constant in Newton’s gravitational equation; measured to be 6.67 × 1021 N m2 kg−2. (Note: correct the exponent on this line to −11 in your answer.) | |
| 2.3 | Gravitational force per unit mass at a point in space; SI units are N/kg or m/s2. | |
| 2.4 | A relationship where a quantity decreases as the square of distance increases; doubling distance reduces the quantity to one-quarter. | |
| 2.5 | The distance measured from the centre of one mass to the centre of the other; always used in gravitational calculations, not surface-to-surface. | |
| 2.6 | Motion under gravity alone; the correct description for astronauts in orbit who are falling toward Earth while also moving forward fast enough to miss it. | |
| 2.7 | The near-weightless environment experienced in an orbiting spacecraft, arising because the spacecraft and its occupants fall at the same rate under gravity. | |
| 2.8 | The instrument used by Henry Cavendish in 1797–1798 to measure the gravitational attraction between laboratory-scale lead masses, enabling the first determination of G. | |
| 2.9 | The nature of the gravitational force between any two masses — it always pulls the masses toward each other; it never repels. |
3. True or false — with correction
Circle T or F. If false, write the corrected statement on the line below. 12 marks (1 T/F + 1 correction each)
3.1 The gravitational force between two masses is repulsive when the masses are very large. T / F
3.2 If the distance between two masses doubles, the gravitational force halves. T / F
3.3 Gravitational field strength g at the surface of a planet depends on both the planet’s mass and its radius. T / F
3.4 Astronauts on the International Space Station experience zero gravity because there is no gravitational force at that altitude. T / F
3.5 The variable r in F = GMm/r2 is the centre-to-centre distance between the two masses, not the surface-to-surface distance. T / F
3.6 When Cavendish performed his experiment in 1797–1798, the forces he measured between his lead spheres were on the order of 103 N — large enough to feel by hand. T / F
4. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word or phrase is used once. 9 marks (1 per blank)
Word bank:
attractive · Cavendish · centre-to-centre · field strength · free fall · gravity · inverse square · product · torsion balance
Newton’s Law of Universal Gravitation states that every mass exerts an ___________ force on every other mass. The magnitude of this force is proportional to the ___________ of the two masses and obeys the ___________ law with respect to the ___________ distance between them. The constant of proportionality, G, was first measured by Henry ___________ in 1797–1798 using a sensitive ___________. The quantity g is called the gravitational ___________ and has the same numerical value as the acceleration due to ___________ at that location. Astronauts in orbit are in ___________, not zero gravity — g at ISS altitude is still about 88 % of its surface value.
5. Inverse square law — complete the pattern table
Two masses experience a gravitational force F0 when separated by distance r0. Complete the table. 8 marks (1 per row)
| New separation | Factor change in r | New force (in terms of F0) | Brief reasoning |
|---|---|---|---|
| 2r0 | × 2 | ||
| 3r0 | × 3 | ||
| 4r0 | × 4 | ||
| 10r0 | × 10 | ||
| r0 / 2 | ÷ 2 (× ½) | ||
| r0 / 3 | ÷ 3 | ||
| r0 / 10 | ÷ 10 | ||
| 5r0 / 2 | × 2.5 |
Q1 — Formula recall card (sample entries)
Row 1 (Gravitational force): F = force (N) ; G = 6.67 × 10−11 N m2 kg−2 ; M, m = masses (kg) ; r = centre-to-centre distance (m). When: finding the gravitational force between any two masses.
Row 2 (Field strength): g in N/kg or m/s2 ; M = mass of attracting body (kg) ; r = distance from centre of mass (m). When: finding the field strength at distance r from a large mass.
Row 3 (Field strength at altitude): R = planet radius (m) ; h = altitude above surface (m). When: you already know the surface value of g and want g at altitude h.
Row 4 (Centre-to-centre): R = planet radius ; h = altitude. When: setting up any gravitational calculation — always convert altitude to r = R + h first.
Row 5 (Inverse square): When: predicting how force changes when separation changes without calculating absolute values.
Row 6 (Earth’s mass): M in kg ; R = Earth’s radius (m). When: deducing a planet’s mass from its surface g and radius — as Cavendish effectively did.
Marking criteria: 1 mark per row for correct units on all variables AND a sensible “when to use” note.
Q2 — Term–definition match
2.1 Newton’s Law of Universal Gravitation • 2.2 universal gravitational constant (G) • 2.3 gravitational field strength • 2.4 inverse square law • 2.5 centre-to-centre distance • 2.6 free fall • 2.7 microgravity • 2.8 torsion balance • 2.9 attractive force.
Note for 2.2: the exponent in the definition is deliberately wrong (21 instead of −11) so students must identify and correct it — full credit for writing “universal gravitational constant, G” and noting the correct value 6.67 × 10−11 N m2 kg−2.
Q3 — True / false with correction
3.1 False. Gravitational force is always attractive, regardless of the size of the masses. There is no gravitational repulsion.
3.2 False. If distance doubles, force becomes one-quarter (not one-half) because F ∝ 1/r2. Doubling r means r2 increases by a factor of 4.
3.3 True. g = GM/R2 depends on both the planet’s mass M and its radius R.
3.4 False. At ISS altitude (~400 km), g is approximately 8.7 m/s2 — about 89% of the surface value. Astronauts experience microgravity because they are in free fall (orbiting), not because there is no gravity.
3.5 True.
3.6 False. The forces Cavendish measured were approximately 10−7 N — comparable to the weight of a single bacterium — far too small to feel by hand. That is why a sensitive torsion balance was required.
Marking criteria: 1 mark for the correct T/F; 1 mark for the correction (required for false statements only).
Q4 — Cloze paragraph (in order)
attractive / product / inverse square / centre-to-centre / Cavendish / torsion balance / field strength / gravity / free fall.
Q5 — Inverse square law pattern table
Rule: new force = F0 / (factor)2. For distance halved / thirded, force multiplies by the square of the reciprocal.
| New separation | New force | Reasoning |
|---|---|---|
| 2r0 | F0 / 4 | r × 2 ⇒ r2 × 4 ⇒ F ÷ 4 |
| 3r0 | F0 / 9 | r × 3 ⇒ r2 × 9 ⇒ F ÷ 9 |
| 4r0 | F0 / 16 | r × 4 ⇒ r2 × 16 ⇒ F ÷ 16 |
| 10r0 | F0 / 100 | r × 10 ⇒ r2 × 100 ⇒ F ÷ 100 |
| r0 / 2 | 4F0 | r halved ⇒ r2 one-quarter ⇒ F × 4 |
| r0 / 3 | 9F0 | r thirded ⇒ r2 one-ninth ⇒ F × 9 |
| r0 / 10 | 100F0 | r ÷ 10 ⇒ r2 ÷ 100 ⇒ F × 100 |
| 5r0/2 = 2.5r0 | F0 / 6.25 | r × 2.5 ⇒ r2 × 6.25 ⇒ F ÷ 6.25 |
Marking criteria: 1 mark per row for correct new force AND correct brief reasoning (must reference squaring the factor).