Physics • Year 12 • Module 5 • Lesson 10
Newton’s Law of Universal Gravitation
Apply Newton’s Law of Universal Gravitation and the inverse square law to calculations, data interpretation and contextualised scenarios involving Earth, Mars and spacecraft.
1. Eight single-formula calculations
Use G = 6.67 × 10−11 N m2 kg−2 throughout. Show every substitution with units. Report answers to 3 significant figures unless the question says otherwise. 2 marks each — 16 marks
Foundation — direct substitution (Q1.1–Q1.3)
Q1.1 Two steel spheres of mass 8.00 kg and 12.0 kg have their centres 0.500 m apart. Calculate the gravitational force between them. 2 marks
Q1.2 Calculate the gravitational field strength at Earth’s surface. Use MEarth = 5.97 × 1024 kg, REarth = 6.37 × 106 m. 2 marks
Q1.3 A 75.0 kg astronaut orbits at altitude h = 400 km above Earth. Using r = R + h, calculate the gravitational force on the astronaut. Use MEarth = 5.97 × 1024 kg. 2 marks
Standard — field strength at altitude (Q1.4–Q1.6)
Q1.4 Calculate the gravitational field strength at 1000 km above Earth’s surface using the ratio method: g′ = g × (R/(R + h))2. Use R = 6371 km, g = 9.80 m/s2. 2 marks
Q1.5 A geostationary orbit is at altitude h = 35 786 km. Calculate g at this altitude. Use R = 6371 km, gsurface = 9.80 m/s2. 2 marks
Q1.6 Calculate the gravitational field strength at the surface of Mars. Use MMars = 6.42 × 1023 kg, RMars = 3.40 × 106 m. Express your answer as a percentage of Earth’s surface gravity. 2 marks
Extension — rearranging and multi-step (Q1.7–Q1.8)
Q1.7 A planet has surface gravity g = 14.0 m/s2 and radius R = 7.50 × 106 m. Rearranging g = GM/R2, calculate the mass of the planet. 2 marks
Q1.8 A satellite experiences gravitational force F = 4.50 × 103 N at altitude h = 500 km. The satellite’s mass is 250 kg. Calculate the gravitational field strength at that altitude, then use g′ = g (R/(R + h))2 to verify it. Use R = 6371 km. 2 marks
2. Interpret gravitational data — Solar System comparison
The table below gives the mass and radius of five Solar System bodies. 8 marks
| Body | M (kg) | R (m) | gsurface (m/s2) | % of Earth g |
|---|---|---|---|---|
| Earth | 5.97 × 1024 | 6.37 × 106 | given: 9.80 | 100% |
| Moon | 7.35 × 1022 | 1.74 × 106 | ||
| Mars | 6.42 × 1023 | 3.40 × 106 | ||
| Jupiter | 1.90 × 1027 | 7.15 × 107 | ||
| Neutron star* | 2.00 × 1030 | 1.00 × 104 |
* Illustrative values; a typical neutron star has M ≈ 1–2 solar masses and R ≈ 10 km.
2.1 Calculate gsurface and percentage of Earth’s gravity for the Moon, Mars, Jupiter and the neutron star. Complete the table. 4 marks (1 per body calculated)
2.2 Jupiter has a mass about 318 times Earth’s and a radius about 11.2 times Earth’s. Without using your table values, use these ratios and the inverse square law to predict gJupiter/gEarth. Show your reasoning. Does your prediction match the table? 2 marks
2.3 Explain why the neutron star’s surface gravity is so extreme despite having a mass comparable to the Sun. Reference the formula g = GM/R2 in your answer. 2 marks
3. Graph interpretation — gravitational force vs separation
A student plots gravitational force F against two different horizontal variables for a pair of masses: (Graph A) F vs r, and (Graph B) F vs 1/r2. 6 marks
Figure 3. (Left) F vs r: hyperbolic decrease. (Right) F vs 1/r2: straight line through origin with gradient GMm. Illustrative.
3.1 Describe the shape of Graph A and explain why it has this shape, referring to the formula F = GMm/r2. 2 marks
3.2 Explain why plotting F against 1/r2 gives a straight line through the origin (Graph B). State what the gradient of Graph B equals. 2 marks
3.3 A student measures gradient = 2.67 × 10−3 N m2 for Graph B with one mass fixed at M = 4.00 × 1010 kg. Calculate the other mass m. 2 marks
4. Predict and justify — the Moon versus a mountain
Mount Kosciuszko (2228 m) is Australia’s highest peak. The Moon’s average distance from Earth is 3.84 × 108 m. A 70 kg person stands at the top of Kosciuszko. 5 marks
4.1 The person stands 2228 m above sea level, but Earth’s radius is 6.37 × 106 m. Explain why the change in gravitational force going from sea level to the top of Kosciuszko is negligibly small. Use percentage change reasoning rather than a full calculation. 2 marks
4.2 Without calculating it, predict whether the gravitational force exerted by the Moon on the 70 kg person is larger or smaller than the gravitational force between two 70 kg people standing 1 m apart. Justify your prediction. Then calculate both forces to check. 3 marks
Q1.1 — Force between two steel spheres
F = GMm/r2 = (6.67 × 10−11 × 8.00 × 12.0) / (0.500)2 = (6.67 × 10−11 × 96.0) / 0.250 = 6.404 × 10−9 / 0.250 = 2.56 × 10−8 N.
Marking criteria: 1 mark for correct substitution with units; 1 mark for correct final answer (accept 2.56–2.57 × 10−8 N).
Q1.2 — Earth’s surface gravity
g = GM/R2 = (6.67 × 10−11 × 5.97 × 1024) / (6.37 × 106)2 = 3.982 × 1014 / 4.058 × 1013 = 9.81 m/s2 (accept 9.80–9.83 m/s2).
Marking criteria: 1 mark for setup; 1 mark for answer in range 9.80–9.83 m/s2.
Q1.3 — Force on astronaut at ISS
r = 6.37 × 106 + 4.00 × 105 = 6.77 × 106 m.
F = GMm/r2 = (6.67 × 10−11 × 5.97 × 1024 × 75.0) / (6.77 × 106)2 = 2.987 × 1016 / 4.583 × 1013 = 652 N (accept 648–656 N).
Marking criteria: 1 mark for correctly computing r = R + h; 1 mark for final answer in acceptable range.
Q1.4 — Field strength at 1000 km
r = 6371 + 1000 = 7371 km. g′ = 9.80 × (6371/7371)2 = 9.80 × (0.8643)2 = 9.80 × 0.7470 = 7.32 m/s2.
Marking criteria: 1 mark for correct ratio; 1 mark for final answer (accept 7.31–7.33).
Q1.5 — Field strength at geostationary orbit
r = 6371 + 35 786 = 42 157 km. g′ = 9.80 × (6371/42157)2 = 9.80 × (0.15111)2 = 9.80 × 0.02283 = 0.224 m/s2.
Marking criteria: 1 mark for correct r; 1 mark for final answer (accept 0.22–0.23 m/s2).
Q1.6 — Mars surface gravity
gMars = (6.67 × 10−11 × 6.42 × 1023) / (3.40 × 106)2 = 4.282 × 1013 / 1.156 × 1013 = 3.70 m/s2. Percentage: 3.70/9.80 × 100% = 37.8% of Earth’s gravity.
Marking criteria: 1 mark for g = 3.70–3.72 m/s2; 1 mark for percentage 37–38%.
Q1.7 — Planet mass from surface g
M = gR2/G = (14.0 × (7.50 × 106)2) / (6.67 × 10−11) = (14.0 × 5.625 × 1013) / (6.67 × 10−11) = 7.875 × 1014 / 6.67 × 10−11 = 1.18 × 1025 kg.
Marking criteria: 1 mark for rearranging correctly; 1 mark for final answer (accept 1.17–1.19 × 1025 kg).
Q1.8 — Field strength by two methods
Method 1 (definition): g′ = F/m = 4500/250 = 18.0 N/kg.
Method 2 (ratio): r = 6371 + 500 = 6871 km. g′ = 9.80 × (6371/6871)2 = 9.80 × (0.9272)2 = 9.80 × 0.8597 = 8.42 m/s2.
Note: the two methods give different answers (18.0 vs 8.42 N/kg) because the given force is not physically consistent with g = 9.80 at Earth’s surface for this satellite mass. Award full marks for correctly applying both methods and noting the discrepancy. This illustrates why you must always verify given data for internal consistency.
Marking criteria: 1 mark for g′ = F/m correctly; 1 mark for ratio method correctly applied.
Q2.1 — Solar System g values
Moon: g = (6.67 × 10−11 × 7.35 × 1022) / (1.74 × 106)2 = 4.902 × 1012 / 3.028 × 1012 = 1.62 m/s2 ≈ 16.5% of Earth.
Mars: 3.70 m/s2, 37.8% (see Q1.6).
Jupiter: g = (6.67 × 10−11 × 1.90 × 1027) / (7.15 × 107)2 = 1.267 × 1017 / 5.112 × 1015 = 24.8 m/s2 ≈ 253% of Earth.
Neutron star: g = (6.67 × 10−11 × 2.00 × 1030) / (1.00 × 104)2 = 1.334 × 1020 / 108 = 1.33 × 1012 m/s2 ≈ 1.36 × 1011% of Earth.
Marking criteria: 1 mark per correctly calculated body (4 bodies = 4 marks).
Q2.2 — Jupiter ratio prediction
gJ/gE = (MJ/ME) × (RE/RJ)2 = 318 × (1/11.2)2 = 318 × 0.007972 = 2.53 (i.e. ~253%). This matches the table value of ~2.53. This is the ratio method in action — no need to know G.
Marking criteria: 1 mark for correct ratio formula; 1 mark for answer ~2.5–2.6 with comment it matches the table.
Q2.3 — Neutron star explanation
In g = GM/R2, R appears as the denominator squared. A neutron star compresses about 1–2 solar masses into a radius of only ~10 km (vs the Sun’s radius of ~700 000 km), so R2 is many orders of magnitude smaller than for the Sun. Even with a comparable mass, the tiny R2 drives g to extreme values (~1012 m/s2).
Marking criteria: 1 mark for identifying that R2 in the denominator is responsible; 1 mark for quantitative or qualitative reasoning about how tiny R produces extreme g.
Q3.1 — Shape of Graph A
Graph A is a hyperbolic curve that starts very high at small r and decreases rapidly, approaching zero asymptotically as r → ∞. This shape arises because F = GMm/r2 — F is proportional to 1/r2, not to 1/r, so the decrease is faster than a simple inverse relationship.
Marking criteria: 1 mark for describing the shape as hyperbolic/asymptotic; 1 mark for linking to F ∝ 1/r2.
Q3.2 — Graph B explanation
Rewrite the formula: F = GMm × (1/r2). If we let x = 1/r2, then F = (GMm)x — this is y = mx form, a straight line through the origin. The gradient equals GMm (N m2).
Marking criteria: 1 mark for algebraic argument; 1 mark for stating gradient = GMm.
Q3.3 — Calculate m from gradient
Gradient = GMm ⇒ m = gradient / (GM) = (2.67 × 10−3) / (6.67 × 10−11 × 4.00 × 1010) = (2.67 × 10−3) / (2.668 × 100) = 1.00 × 10−3 kg = 1.00 g.
Marking criteria: 1 mark for correct rearrangement; 1 mark for correct answer (accept 0.99–1.01 × 10−3 kg).
Q4.1 — Kosciuszko altitude effect
h/R = 2228 / (6.37 × 106) = 3.5 × 10−4 = 0.035%. Since g′ = g (R/(R+h))2 ≈ g(1 − 2h/R) for small h/R, the change in g is only ~0.07% — negligible for most purposes.
Marking criteria: 1 mark for computing h/R ratio; 1 mark for concluding the change is negligible with quantitative support.
Q4.2 — Moon’s force vs two-person force
Prediction: The Moon exerts a larger force. Even though it is very far away, its mass (~7.35 × 1022 kg) is vastly greater than a 70 kg person, which should more than compensate for the large distance.
Moon on person: F = (6.67 × 10−11 × 7.35 × 1022 × 70.0) / (3.84 × 108)2 = 3.432 × 1014 / 1.475 × 1017 = 2.33 × 10−3 N.
Two 70 kg people, 1 m apart: F = (6.67 × 10−11 × 70 × 70) / 12 = (6.67 × 10−11 × 4900) / 1 = 3.27 × 10−7 N.
The Moon exerts a force ~7000 times larger than two nearby people. The prediction was correct.
Marking criteria: 1 mark for correct Moon force calculation; 1 mark for correct two-person force; 1 mark for comparison and comment on the prediction.