Physics • Year 12 • Module 5 • Lesson 13
Phase 2 Consolidation
Lock in all six Phase 2 formulae, their conditions of use, and the six common errors before tackling harder questions.
1. Formula–condition match
Draw a line (or write the letter) from each formula in Column A to its correct description in Column B. 6 marks (1 each)
| Column A — Formula | Column B — When and what it describes |
|---|---|
| A. \(a_c = v^2/r\) | i. The net radial force required to keep any mass moving in a circle; always directed toward the centre. |
| B. \(F_c = mv^2/r\) | ii. Centripetal acceleration; magnitude only, directed toward the centre; use whenever speed and radius are known. |
| C. \(T + mg = mv^2/r\) | iii. Speed of a satellite in stable circular orbit at distance \(r\) from the planet's centre. |
| D. \(v = \sqrt{GM/r}\) | iv. Gravitational PE of an orbiting body; always negative; zero defined at infinity. |
| E. \(U = -GMm/r\) | v. Net force equation at the top of a vertical loop, where both tension and weight point toward the centre. |
| F. \(v_e = \sqrt{2GM/r}\) | vi. Minimum launch speed needed to escape a planet's gravity from distance \(r\); equals \(\sqrt{2}\,v_{\text{orb}}\). |
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. 12 marks (1 T/F + 1 correction each)
2.1 Centrifugal force is a real outward force that acts on any object moving in a circle, balancing the centripetal force. T / F
2.2 In a vertical circle, the speed of a mass on a string is constant because the string length does not change. T / F
2.3 For a satellite orbiting at altitude \(h\) above Earth, the correct orbital radius to substitute into \(v = \sqrt{GM/r}\) is \(r = R_E + h\). T / F
2.4 At an altitude of 400 km above Earth's surface, the gravitational field strength is still 9.8 m/s² and \(F = mg\) can be used with \(g = 9.8\). T / F
2.5 The escape velocity from Earth's surface is approximately \(\sqrt{2}\) times larger than the circular orbital speed at the same radius. T / F
2.6 A heavier satellite placed at the same orbital altitude as a lighter satellite will orbit with a greater speed because gravity exerts a larger force on it. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word or phrase is used once. 8 marks (1 per blank)
Word bank:
centre · conservation of energy · fictitious · gravitational potential energy · inertial · negative · orbital radius · \(\sqrt{2}\)
Centripetal force is always directed toward the ___________ of the circular path. Centrifugal force is a ___________ force that appears only when analysing motion from a rotating, non-___________ reference frame. In a vertical circle the speed varies because ___________ converts between kinetic energy and ___________. The gravitational PE of a satellite is always ___________, becoming zero only at infinity. When a satellite moves to a higher orbit, its ___________ increases, its speed decreases, and the energy required for the transfer is positive. The escape velocity is ___________ times the circular orbital velocity at the same distance from the planet.
4. Function recall
Answer each question in 1–2 sentences using precise physics terms. 8 marks (2 each)
4.1 Why is centripetal force not listed as a separate force type in a free-body diagram?
4.2 Why must conservation of energy be used to find the speed at the top of a vertical loop if only the entry speed at the bottom is given?
4.3 State Kepler’s Third Law and explain what it means for two satellites orbiting the same planet at different altitudes.
4.4 Why does total orbital energy \(E = -\tfrac{1}{2}GMm/r\) have a negative value, and what does this tell us about the satellite?
5. Build a concept map
Draw labelled arrows between the six terms below to show how they are related. Each arrow must carry a linking phrase (e.g. “is provided by”, “equals \(\sqrt{2}\times\)”, “requires using”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Terms: centripetal force · gravity · orbital velocity · escape velocity · orbital radius · total orbital energy.
Q1 — Formula–condition match
A → ii (centripetal acceleration) • B → i (centripetal force) • C → v (top of vertical loop) • D → iii (orbital speed) • E → iv (gravitational PE) • F → vi (escape velocity).
Q2 — True / false with correction
2.1 False. Centrifugal force is a fictitious (pseudo) force that appears only in rotating reference frames. In an inertial frame, there is no real outward force; the net force on an object in circular motion is the centripetal force, directed inward.
2.2 False. In a vertical circle, speed is not constant. Gravity does work on the mass as it moves up and down, converting kinetic energy to gravitational potential energy and back. Use conservation of energy to relate speeds at different points.
2.3 True.
2.4 False. At altitude 400 km, the gravitational field strength is significantly less than 9.8 m/s². Use \(g' = GM/(R_E+h)^2\) or Newton’s law of gravitation directly: \(F = GMm/r^2\).
2.5 True.
2.6 False. From \(v = \sqrt{GM/r}\), the satellite’s own mass cancels entirely. All satellites at the same orbital radius orbit at exactly the same speed, regardless of mass.
Q3 — Cloze paragraph
In order: centre / fictitious / inertial / conservation of energy / gravitational potential energy / negative / orbital radius / \(\sqrt{2}\).
Q4.1 — Centripetal force in FBDs
Centripetal force is the net result of real forces (tension, gravity, friction, normal force) rather than a separate force. Listing it as a force in addition to the real forces would double-count those forces. The centripetal force equation gives the required net inward force, which is then equated to whatever real forces provide it.
Q4.2 — Why use energy conservation for vertical circles
In a vertical circle, gravity acts on the object throughout its path, doing positive work as it descends and negative work as it ascends. This changes the kinetic energy (and hence speed). Conservation of energy relates the kinetic and potential energies at any two points, allowing the speed at the top to be found from the entry speed at the bottom.
Q4.3 — Kepler’s Third Law
Kepler’s Third Law: \(T^2 = \frac{4\pi^2}{GM} r^3\), or equivalently \(T^2/r^3 = \text{constant}\) for all bodies orbiting the same central mass. For two satellites orbiting the same planet, the one at a greater altitude has a longer period. The ratio \(T^2/r^3\) is the same for both, depending only on the planet’s mass.
Q4.4 — Negative total orbital energy
The total orbital energy \(E = -\tfrac{1}{2}GMm/r\) is negative because the zero of energy is defined at infinity (where \(U = 0\) and a satellite at rest would have \(KE = 0\)). A negative total energy means the satellite is bound — it does not have enough energy to escape to infinity. Energy must be added to move it to a higher orbit or to escape.
Q5 — Sample concept map
Valid arrows include:
- gravity — provides → centripetal force (for satellites and vertical circles)
- orbital velocity — increases as → orbital radius decreases
- escape velocity — equals \(\sqrt{2}\times\) → orbital velocity
- orbital radius — determines → total orbital energy
- total orbital energy — equals \(-\tfrac{1}{2}GMm/r\) → orbital radius
- centripetal force — equals \(mv^2/r\), balanced by → gravity
Award 1 mark per valid labelled arrow (minimum 6, maximum 6 marked).