Physics • Year 12 • Module 5 • Lesson 18
Phase 3 Consolidation
Lock in every Phase 3 formula, the zero-reference convention, and the six common errors before attempting calculation or exam questions.
1. Formula–description match
Match each formula (A–F) to its correct description (1–6). Write the letter next to the matching description. 6 marks (1 each)
| Letter | Formula | Match | Description |
|---|---|---|---|
| A | F = GMm / r² | 1. Gravitational potential energy of mass m at distance r; always negative; zero at infinity. | |
| B | g = GM / r² | 2. Minimum speed for a mass to permanently escape a gravitational field from distance r. | |
| C | U = −GMm / r | 3. Gravitational potential; energy per unit mass (J/kg); related to field strength by g = −dV/dr. | |
| D | V = −GM / r | 4. Attractive force between two point masses M and m separated by centre-to-centre distance r. | |
| E | v₁ = √(2GM/r) | 5. Gravitational field strength at distance r; force per unit mass; units m/s². | |
| F | T² = (4π²/GM) r³ | 6. Kepler’s Third Law; relates orbital period T to orbital radius r for any orbiting body. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 14 marks (1 T/F + 1 correction each)
2.1 The gravitational field strength g = 9.8 m/s² can be used for any altitude calculation involving Earth-orbiting satellites. T / F
2.2 The gravitational potential energy U = −GMm/r is always negative for any finite separation. T / F
2.3 For a satellite orbiting Earth at altitude h, the distance to use in formulae is r = h (the height above the surface). T / F
2.4 Gravitational potential V (J/kg) and gravitational potential energy U (J) are the same quantity. T / F
2.5 The escape velocity formula is v₁ = √(GM/r); the factor of 2 under the square root is optional. T / F
2.6 Kepler’s Third Law T² ∝ r³ is valid for elliptical orbits if the semi-major axis is used in place of r. T / F
2.7 The zero-reference point for gravitational potential energy is defined at the surface of the planet. T / F
3. 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 · centre-to-centre · escape velocity · field lines · gradient · infinity · negative · potential · semi-major axis
Gravity is always ___________: two masses always attract each other. In the formula F = GMm/r², the distance r is the ___________ distance between the two masses, not the surface-to-surface gap. The gravitational potential energy U = −GMm/r is always ___________ for bound systems, with the zero reference chosen at ___________ because the gravitational force approaches zero there. The gravitational ___________ V = −GM/r is the potential energy per unit mass in J/kg; the field strength g equals the negative of its distance ___________ (g = −dV/dr). Equipotential surfaces are perpendicular to gravitational ___________. The minimum speed to escape a gravitational field is called the ___________, given by √(2GM/r). For elliptical orbits, Kepler’s Third Law uses the ___________ of the ellipse in place of the orbital radius r.
4. Formula function recall
Answer each question in 1–2 sentences. Use precise terminology and include the relevant formula symbol. 10 marks (2 each)
4.1 What quantity does g = GM/r² calculate, and what are its SI units?
4.2 Why must the negative sign be included in U = −GMm/r? What happens to U as r increases?
4.3 State the relationship between gravitational potential V and gravitational field strength g.
4.4 What is the total mechanical energy of a satellite in a circular orbit of radius r? Is this value positive, negative, or zero?
4.5 For a satellite at height h above Earth’s surface (radius Rₐ), write the expression for the distance r to use in gravitational calculations.
5. Error clinic — identify and fix the mistake
Each student working below has made one of the six common errors (E1–E6) from the lesson. Identify the error code (E1–E6), state what is wrong, and write the correct version. 9 marks (1 error code + 1 explanation + 1 fix each)
5.1 Student A calculates the gravitational field strength 800 km above Earth’s surface as: “g = 9.8 m/s² because that is Earth’s surface gravity.”
Error code: ________ What is wrong:
Correct approach:
5.2 Student B writes the gravitational potential energy as: “U = GMm/r = +3.5 × 10¹⁰ J.”
Error code: ________ What is wrong:
Correct version:
5.3 Student C calculates escape velocity as: “v₁ = √(GM/r) = 7900 m/s.”
Error code: ________ What is wrong:
Correct formula and consequence:
Q1 — Formula–description match
A–4 (Newton’s law, force between two masses) • B–5 (field strength, force per unit mass) • C–1 (potential energy U, always negative) • D–3 (gravitational potential V, J/kg) • E–2 (escape velocity) • F–6 (Kepler’s Third Law).
Q2 — True / false with correction
2.1 False. g = 9.8 m/s² is valid only at Earth’s surface. At altitude h, use g = GMₐ/(Rₐ + h)². (Error E1)
2.2 True. U = −GMm/r is always negative for any finite r; it approaches zero as r → ∞.
2.3 False. The correct distance is r = Rₐ + h (centre-to-centre from Earth’s centre to the satellite). (Error E4)
2.4 False. V (J/kg) is gravitational potential (energy per unit mass); U (J) is gravitational potential energy of a specific mass m. They are related by U = mV. (Error E3)
2.5 False. The factor of 2 is mandatory: v₁ = √(2GM/r). Omitting it underestimates v₁ by ~29%. (Error E5)
2.6 True. Kepler’s Third Law is exact for all closed orbits using the semi-major axis a.
2.7 False. The zero-reference point for gravitational potential energy is defined at r = ∞ (infinite separation), not at the planet’s surface. This is because the gravitational force approaches zero only at infinity.
Q3 — Cloze paragraph
In order: attractive / centre-to-centre / negative / infinity / potential / gradient / field lines / escape velocity / semi-major axis.
Q4.1 — g = GM/r²
It calculates the gravitational field strength: the gravitational force per unit mass at a point distance r from the centre of mass M. SI units are m/s² (equivalent to N/kg).
Q4.2 — Negative sign in U
The negative sign reflects that gravity is attractive and work must be done against the field to move a mass to infinity. The zero of U is at r = ∞; for any finite r the system is bound (U < 0). As r increases, U becomes less negative, approaching zero from below.
Q4.3 — V and g relationship
The gravitational field strength equals the negative rate of change of potential with distance: g = −dV/dr. The field points in the direction of decreasing potential (toward the mass).
Q4.4 — Total orbital energy
For a circular orbit: Eₚₘₜₙ = −GMm/(2r). This is negative, meaning the satellite is gravitationally bound. Eₚₘₜₙ = ½ U = −K, where K is the kinetic energy.
Q4.5 — Distance r for orbital calculations
r = Rₐ + h, where Rₐ = 6.37 × 10&sup6; m is Earth’s radius and h is the altitude above the surface. Always write this explicitly before substituting into any formula.
Q5.1 — Error E1
Error E1: Using the surface value g = 9.8 m/s² for a non-surface calculation. At 800 km altitude: r = 6.37 × 10&sup6; + 8.0 × 10&sup5; = 7.17 × 10&sup6; m. Correct: g = GMₐ/r² = (6.67 × 10⁻¹¹)(5.97 × 10²&sup4;)/(7.17 × 10&sup6;)² ≈ 7.75 m/s².
Q5.2 — Error E2
Error E2: Omitting the mandatory negative sign from U. Gravitational potential energy is always negative for a bound system. Correct: U = −GMm/r = −3.5 × 10¹⁰ J.
Q5.3 — Error E5
Error E5: Missing the factor of 2 inside the square root. The correct formula is v₁ = √(2GM/r). Using √(GM/r) gives the circular orbital speed, not the escape speed. The escape velocity is √2 × vₚₗᵇᵭₜₕₗ ≈ 1.41 times the orbital speed. Student C’s answer would be underestimated by factor √2.