Skip to content
HSCScience Physics · Y12 · M5
⚡0 XP
🪙0
🔥0
Lv 1
Year 12 Physics Module 5 ⏱ ~50 min 5 MC · 3 Short Answer Lesson 13 of 18

Phase 2 Consolidation

On 16 July 1969, NASA Mission Control performed a 347-second engine burn at perigee to send Apollo 11 on its trans-lunar trajectory — increasing velocity from 7.79 km/s to 10.84 km/s. This single manoeuvre required simultaneous application of orbital energy (L12), Kepler's Third Law (L11), and gravitational field calculations (future L14). Today's checkpoint drills exactly those skills.

Today's hook: On 16 July 1969, Apollo 11 performed a 347-second trans-lunar injection burn, accelerating from 7.79 km/s to 10.84 km/s — a delta-v of 3.05 km/s. Before today's consolidation session, write down: what orbital energy change does this velocity increase represent, and is 10.84 km/s above Earth's escape velocity?
0/5TASKS
Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions
Before you begin — reflect

What is the difference between centripetal and centrifugal force? Write your answer before looking at any notes.

Warm-up — which formula gives the centripetal acceleration of an object in circular motion?

Learning Intentions
goals

Know — Recall all Phase 2 Formulae

  • Recall and apply all Phase 2 formulae for circular motion and orbits
  • Select the correct equation for each problem type

Understand — Identify and Correct Common Errors

  • Identify and correct common errors in circular motion problems
  • Explain why centrifugal force is fictitious

Can Do — Solve Under Exam Conditions

  • Solve mixed circular motion and orbital problems under exam conditions
  • Answer 3 timed extended responses accurately
Key Terms — Phase 2
vocab
Centripetal acceleration$a_c = v^2/r$, directed toward the centre of circular motion.
Centripetal force$F_c = mv^2/r$; the net force required for circular motion — not a separate force, but the net result of real forces.
Angular velocity$\omega = 2\pi f = 2\pi/T$; rate of angle swept out in rad/s.
Orbital velocity$v = \sqrt{GM/r}$; speed for stable circular orbit at radius $r$.
Kepler's Third Law$T^2 \propto r^3$; relates orbital period to orbital radius.
Gravitational PE$U = -GMm/r$; always negative, zero at infinity.
Total orbital energy$E = -\frac{1}{2}GMm/r = KE + U$; negative for bound orbits.
Escape velocity$v_e = \sqrt{2GM/r} = \sqrt{2} \times v_{orb}$.
Cross-lesson links: L11–L12 built orbital mechanics. L13 consolidates them — this checkpoint covers the exam-critical skills: calculating orbital speed from $M$ and $r$, applying Kepler's Third Law, and determining whether an orbit is bound or unbound from its total energy.
Misconceptions to fix before you review
✗ Wrong: Centrifugal force is a real force that pushes objects outward in circular motion.
✓ Right: Centrifugal force is fictitious — it only appears in rotating reference frames. In an inertial frame, only real forces exist; the net inward force is the centripetal force.
✗ Wrong: A heavier satellite at the same altitude orbits more slowly because it needs more centripetal force.
✓ Right: Orbital speed $v = \sqrt{GM/r}$ is independent of satellite mass — the mass cancels exactly. All satellites at the same radius orbit at the same speed.

Centrifugal force is a real force that acts outward on objects in circular motion (as seen in an inertial frame).

A more massive satellite at the same orbital radius has a higher orbital speed.

IQ2: Circular Motion & Orbits Formula Sprint

1
Phase 2 Formula Sheet & Sprint Cards
+5 XP

Six formulae — when to use each, key variables, and common traps.

Phase 2 Summary — all six formulae at a glance

Phase 2 Summary — all six formulae at a glance.

Centripetal acceleration & force

$a_c = \dfrac{v^2}{r} = \omega^2 r$   ·   $F_c = \dfrac{mv^2}{r} = m\omega^2 r$

Use: Any object moving in a circle at constant speed — $F_c$ is the net inward force, not a separate force.

Speed and angular frequency

$v = \dfrac{2\pi r}{T} = \omega r$   ·   $\omega = 2\pi f$

Use: Relating linear speed, period, frequency and angular velocity for circular motion.

Vertical circles

Top: $T + mg = \dfrac{mv^2}{r}$   ·   Bottom: $T - mg = \dfrac{mv^2}{r}$

Use: At the top, both tension and gravity point toward the centre (add). At the bottom, tension is up, gravity down (subtract).

Orbital mechanics

$v_{\text{orb}} = \sqrt{\dfrac{GM}{r}}$   ·   $T^2 = \dfrac{4\pi^2}{GM} r^3$

Use: Speed and period of any satellite in stable circular orbit. Remember $r = R_{\text{planet}} + h$.

Gravitational energy

$KE = \dfrac{1}{2}\dfrac{GMm}{r}$   ·   $U = -\dfrac{GMm}{r}$   ·   $E = -\dfrac{1}{2}\dfrac{GMm}{r}$

Use: PE is always negative for bound orbits. Total orbital energy is half the PE in magnitude (virial theorem).

Escape velocity

$v_e = \sqrt{\dfrac{2GM}{r}} = \sqrt{2} \times v_{\text{orb}}$

Use: Minimum speed to escape from radius $r$. Exactly $\sqrt{2}$ times the orbital speed at the same radius.

Sprint Cards — click to reveal traps

$$a_c = \frac{v^2}{r}$$
Click to reveal when to use this
Use when Finding centripetal acceleration of any object in circular motion. Works for cars on curves, satellites, electrons — anything going in a circle.
Trap Using $a = v^2/r$ with $v$ as a velocity vector that is changing. This is magnitude only. Also: forgetting that $v$ may not be constant in vertical circles.
Connects to $F_c = mv^2/r$ (multiply by mass to get force). Also connects to $v = 2\pi r/T$ when period is given instead of speed.
$$F_c = \frac{mv^2}{r}$$
Click to reveal when to use this
Use when Finding the net force required to keep an object moving in a circle, or identifying which real force(s) provide that net force.
Trap Inventing "centrifugal force" as a real force. $F_c$ is not a separate force — it is the net result of real forces (tension, gravity, friction, normal).
Connects to Banked curves: $N\sin\theta = mv^2/r$. Vertical circles: $T + mg = mv^2/r$ (top). Orbits: $GMm/r^2 = mv^2/r$.
$$T + mg = \frac{mv^2}{r}$$
Top of vertical loop — click to reveal
Use when Analysing forces at the top of a vertical circle (roller coaster, bucket of water, pendulum). Both tension and gravity point toward the centre.
Trap Using a sign convention that makes both forces positive without thinking. At the top: $T$ and $mg$ both point down (toward centre), so they add. At the bottom: $T$ points up, $mg$ points down, so $T - mg = mv^2/r$.
Connects to Minimum speed at top: set $T = 0$ to find $v_{\min} = \sqrt{gr}$. Conservation of energy connects top and bottom speeds.
$$v = \sqrt{\frac{GM}{r}}$$
Orbital velocity — click to reveal
Use when Finding the speed of any satellite in stable circular orbit. Derived by equating gravitational force to required centripetal force.
Trap Forgetting $r = R + h$. The orbital radius is centre-to-centre. Also: using $g = 9.8$ m/s² instead of computing $g$ at altitude via $g = GM/r^2$.
Connects to Kepler's Third Law ($T^2 \propto r^3$), escape velocity ($v_e = \sqrt{2} \times v_{\text{orb}}$), and gravitational PE ($U = -GMm/r$).
$$U = -\frac{GMm}{r}$$
Gravitational potential energy — click to reveal
Use when Calculating gravitational potential energy in orbital mechanics. The negative sign indicates a bound orbit. Zero PE is defined at infinity.
Trap Dropping the negative sign. $U$ is always negative for bound orbits. Using $U = mgh$ (only valid near Earth's surface) for orbital problems.
Connects to Total energy $E = -\frac{1}{2}GMm/r = KE + U$. Energy changes when moving between orbits: $\Delta E = E_2 - E_1$.
$$v_e = \sqrt{\frac{2GM}{r}}$$
Escape velocity — click to reveal
Use when Finding the minimum speed needed to escape a planet's gravitational pull from a given distance $r$. At the surface, $r = R_{\text{planet}}$.
Trap Forgetting the $\sqrt{2}$ factor. $v_e = \sqrt{2} \times v_{\text{orb}} \approx 1.414 \times v_{\text{orb}}$. Many students use $v_e = \sqrt{GM/r}$ (orbital velocity) by mistake.
Connects to Total orbital energy $E = -\frac{1}{2}GMm/r$. Escape requires adding exactly $+\frac{1}{2}GMm/r$ of energy to bring total to zero.

Six key formulae: $a_c = v^2/r = \omega^2 r$; $F_c = mv^2/r$ (net inward force); $v = 2\pi r/T = \omega r$; vertical-circle top $T + mg = mv^2/r$; orbital $v = \sqrt{GM/r}$; $U = -GMm/r$; $E = -GMm/(2r)$; $v_e = \sqrt{2GM/r} = \sqrt{2}\,v_\text{orb}$. Always use $r = R + h$.

Pause — copy the highlighted formula summary into your book before moving on.

A satellite orbits at altitude 400 km. A student uses $r = 4.0 \times 10^5$ m in $v = \sqrt{GM/r}$. What is wrong?

Error Clinic

2
Six Common Errors — Find the Fix
+5 XP

Each card shows student working with an error. Identify the error and reveal the correction.

We just saw the six formulae that drive this module. That raises a question: where do students go wrong when applying them? This card answers it → six error types, each with a common mistake and the corrected physics.

1
Thinking "centrifugal force" is a real force
▾
Student working: "In the rotating frame of a spinning ride, the person is pushed outward by centrifugal force, which balances the tension in the chains."
Correction: Centrifugal force is a fictitious force that appears in rotating (non-inertial) reference frames. In an inertial frame, there is no outward force. The only real force is tension (and gravity), which provides the centripetal force toward the centre. "Centrifugal force" is the felt effect of inertia — the tendency to continue in a straight line — not a real force.
2
Using constant speed equations for vertical circles
▾
Student working: "A roller coaster loop has radius 5 m. The cart enters at 8 m/s. Find the normal force at the top." Student uses $v = 8$ m/s at the top and calculates $N + mg = mv^2/r$ with $v = 8$.
Correction: In a vertical circle, speed is not constant — gravity does work, converting between kinetic and gravitational PE. The entry speed (bottom) is the maximum; speed at the top is lower. Use conservation of energy: $\frac{1}{2}mv_{bot}^2 = \frac{1}{2}mv_{top}^2 + mg(2r)$ to find $v_{top}$, then substitute into $N + mg = mv_{top}^2/r$.
3
Forgetting $r = R + h$ for orbital problems
▾
Student working: "A satellite orbits at altitude 400 km. Find its orbital speed." Student uses $r = 400 \times 10^3$ m $= 4.0 \times 10^5$ m in $v = \sqrt{GM/r}$.
Correction: The orbital radius $r$ is always centre-to-centre. You must add the planet's radius: $r = R_{\text{Earth}} + h = 6.371 \times 10^6 + 0.400 \times 10^6 = 6.771 \times 10^6$ m. Using just the altitude gives a speed that is too high by a factor of $\sqrt{R_{\text{Earth}}/h} \approx 4$.
4
Using $g = 9.8$ m/s² for orbital altitude problems
▾
Student working: "A satellite at 1000 km altitude. Find the gravitational force on a 500 kg satellite." Student calculates $F = mg = 500 \times 9.8 = 4900$ N.
Correction: $g = 9.8$ m/s² is only valid at Earth's surface. At altitude $h$, gravitational field strength is $g' = GM/(R+h)^2$. For this satellite: $r = 7.371 \times 10^6$ m, so $g' = 7.33$ m/s². The correct force is $F = 500 \times 7.33 = 3665$ N. Alternatively, use Newton's law directly: $F = GMm/r^2$.
5
Confusing $v_{\text{orbital}}$ with $v_{\text{escape}}$ — forgetting $\sqrt{2}$
▾
Student working: "The orbital speed at Earth's surface is 7.9 km/s. Therefore the escape velocity is also 7.9 km/s."
Correction: Escape velocity is $\sqrt{2}$ times orbital velocity at the same radius: $v_e = \sqrt{2GM/r} = \sqrt{2} \times \sqrt{GM/r} = \sqrt{2} \times v_{\text{orb}} \approx 1.414 \times 7.9 = 11.2$ km/s. Orbital velocity gives enough speed to stay in a circular orbit; escape velocity gives enough kinetic energy to reach infinity. The factor of 2 inside the square root makes all the difference.
6
Thinking heavier satellites orbit slower
▾
Student working: "A 1000 kg satellite and a 2000 kg satellite are at the same altitude. The heavier one moves slower because it needs more force to stay in orbit."
Correction: From $v = \sqrt{GM/r}$, the satellite's own mass $m$ cancels out entirely. Both satellites orbit at exactly the same speed. The heavier satellite does experience twice the gravitational force, but it also has twice the inertia — and these effects exactly cancel. Orbital motion is independent of the orbiting body's mass.

Six error rules: (1) centrifugal force is fictitious — never in an inertial FBD; (2) speed is NOT constant in vertical circles — use energy conservation; (3) always $r = R + h$; (4) use $g = GM/r^2$ at altitude, not 9.8; (5) $v_e = \sqrt{2}\,v_\text{orb}$, not equal; (6) satellite mass cancels — all satellites at same $r$ orbit at same speed.

Add the highlighted error checklist to your notes before the check below.

Three of these statements about circular motion and orbits are correct. Pick the odd one out (the incorrect statement).

Mixed Practice

3
Mixed Practice Questions
+5 XP

Ten questions from Band 3 to Band 6. Worked solutions in the Answers accordion at the bottom of the Practice phase.

We just saw the six common errors students make in this topic. That raises a question: can you now apply the correct approach across a full range of question types? This card answers it → ten mixed practice questions from Band 3 recall to Band 6 derivation.

Apply Band 3 2 marks

Q1. A car travels around a curve of radius 30 m at a constant speed of 12 m/s. Calculate the centripetal acceleration.

Apply Band 3 2 marks

Q2. A 0.40 kg ball on a 0.50 m string is whirled in a horizontal circle at 3.0 revolutions per second. Calculate the centripetal force on the ball.

Understand Band 3 2 marks

Q3. State Kepler's Third Law and explain what it tells us about satellites orbiting the same central body.

Analyse Band 4-5 3 marks

Q4. A banked curve has radius 60 m and banking angle $\theta = 10°$. (a) Calculate the design speed (no friction needed). (b) If $\mu_s = 0.20$, calculate the maximum speed before slipping up the bank.

Analyse Band 4-5 3 marks

Q5. A roller coaster loop has radius 4.0 m. (a) Find the minimum speed required at the bottom of the loop to just complete the circle. (b) If the entry speed at the bottom is 18 m/s, find the normal force at the bottom for a 60 kg rider. ($g = 9.8$ m/s²)

Apply Band 4-5 3 marks

Q6. A satellite orbits Earth at $r = 7.5 \times 10^6$ m from Earth's centre. Calculate its orbital speed and period. ($G = 6.67 \times 10^{-11}$ N m²/kg², $M_E = 5.97 \times 10^{24}$ kg)

Analyse Band 4-5 4 marks

Q7. A 0.30 kg bob on a 0.70 m string moves in a vertical circle. (a) Find the minimum speed at the top of the circle for the string to remain taut. (b) If the speed at the top is 4.0 m/s, find the tension at the top and at the bottom. ($g = 9.8$ m/s²)

Analyse Band 6 4 marks

Q8. Derive the relationship between orbital velocity and escape velocity at the same radius. Starting from the energy equations for each case, show clearly that $v_e = \sqrt{2} \times v_{\text{orb}}$.

Analyse Band 6 4 marks

Q9. A 500 kg satellite moves from low Earth orbit ($r_1 = 6.8 \times 10^6$ m) to a higher orbit ($r_2 = 1.2 \times 10^7$ m). Calculate the energy change required and the new orbital speed. ($G = 6.67 \times 10^{-11}$ N m²/kg², $M_E = 5.97 \times 10^{24}$ kg)

Evaluate Band 6 5 marks

Q10. A 1000 kg car travels over a hill of radius 20 m at 15 m/s. Calculate the normal force at the top of the hill. The car then descends into a valley of radius 15 m at the same speed. Calculate the normal force at the bottom of the valley. Explain the difference in normal forces in terms of centripetal force requirements and the direction of acceleration. ($g = 9.8$ m/s²)

Mixed practice method: (1) identify the net centripetal force direction; (2) apply $F_c = mv^2/r$ with the correct sign (top: forces add; bottom: forces subtract); (3) for orbital problems find $r = R + h$ first; (4) for energy changes use $\Delta E = -GMm/(2r_2) - (-GMm/(2r_1))$ — not just $\Delta KE$.

Add the highlighted problem-solving method to your notes before the check below.

Quick check — for Q2 above (ball on string at 3.0 rev/s), which is the correct first step?

Timed Exam Block

4
Extended Response Practice
+5 XP

Three exam-style questions. Each: 4 marks, 8 minutes recommended. Worked solutions in the Answers accordion of the Practice phase.

We just saw ten mixed practice questions testing all formula areas. That raises a question: can you write extended, structured responses under timed conditions? This card answers it → three 4-mark questions with 8 minutes each — conical pendulum, banked curve, and orbital transfer.

08:00
Question 11 — Conical Pendulum and Energy Analysis (4 marks, ~8 min)

A conical pendulum consists of a 0.50 kg mass on a 1.2 m string, tracing out a horizontal circle of radius 0.40 m with constant speed.

  1. Draw a free-body diagram showing all forces on the mass. (1 mark)
  2. Calculate the tension in the string. (1 mark)
  3. Calculate the speed of the mass. (1 mark)
  4. If the string is shortened to 0.80 m while keeping the same radius, explain how the speed and period change, using energy concepts. (1 mark)
08:00
Question 12 — Banked Curve with Friction (4 marks, ~8 min)

A race track has a banked curve of radius 80 m with banking angle 15°. The coefficient of static friction between tyres and track is 0.35.

  1. Calculate the design speed for which no friction is required. (1 mark)
  2. Derive expressions for the maximum and minimum speeds the car can travel without slipping, considering friction acts both up and down the bank. (1 mark)
  3. Calculate the numerical values for $v_{\max}$ and $v_{\min}$. (1 mark)
  4. Explain what happens if the car travels faster than $v_{\max}$. (1 mark)
08:00
Question 13 — Orbital Mechanics: Geostationary Transfer (4 marks, ~8 min)

A 2000 kg satellite is to be moved from a parking orbit ($r_1 = 7.0 \times 10^6$ m) to geostationary orbit ($r_2 = 4.22 \times 10^7$ m).

  1. Calculate the orbital speed in the parking orbit and in geostationary orbit. (1 mark)
  2. Calculate the total energy change required for this transfer. (1 mark)
  3. Explain why the total energy change is positive even though the satellite slows down in the higher orbit. (1 mark)
  4. A common student error is to use $\Delta E = \frac{1}{2}m(v_2^2 - v_1^2)$. Explain why this is incorrect and what the correct approach is. (1 mark)

Extended response strategy: (1) draw and label a FBD before writing equations; (2) show force decomposition explicitly; (3) use $r = R + h$ for orbital radius; (4) for orbit transfer energy use $\Delta E = E_2 - E_1 = -GMm/(2r_2) + GMm/(2r_1)$ — not $\Delta KE$ alone. Explain the sign of $\Delta E$ in context.

Pause — write the highlighted exam strategy into your book before moving on.

Activity 1 — Circular Motion & Orbits Consolidation Drills
ApplyBand 3

Practise the key concepts from Phase 2.

  1. Define centripetal acceleration and state its SI unit. Give the formula in terms of both $v$ and $\omega$.
  2. A satellite orbits Earth at radius $r = 8.0 \times 10^6$ m. Calculate its orbital speed and period. ($G = 6.67 \times 10^{-11}$ N m²/kg², $M_E = 5.97 \times 10^{24}$ kg)
  3. Explain why a higher orbit requires more total energy, even though the satellite moves more slowly.
Activity 2 — Concept Check
UnderstandBand 5

Explain the reasoning behind a key Phase 2 principle.

A student claims that firing a satellite's thrusters to speed it up will always result in a faster-moving satellite in its final orbit. Assess this claim using your knowledge of orbital mechanics, referring to the relationship between orbital speed and radius.

Fill the gap: the escape velocity from any radius $r$ equals $\sqrt{\_\_\_}$ times the orbital velocity at the same radius. The missing number is _____.

Wrap-up — Misconceptions & Summary

Misconceptions — final check

✗ "Centrifugal force is real — I can feel it pushing me outward in a spinning ride."
✓ What you feel is your body's inertia trying to continue in a straight line. In an inertial frame, only the inward centripetal force (provided by the seat/harness) exists. Never include centrifugal force in a free-body diagram in an inertial frame.
✗ "To find $\Delta E$ for an orbit transfer, use $\Delta E = \frac{1}{2}m(v_2^2 - v_1^2)$."
✓ This ignores the change in gravitational potential energy. Use total orbital energy: $\Delta E = E_2 - E_1 = -\frac{1}{2}GMm/r_2 - (-\frac{1}{2}GMm/r_1)$.

Copy into your books

Circular Motion

  • $a_c = v^2/r = \omega^2 r$
  • $F_c = mv^2/r = m\omega^2 r$
  • $v = 2\pi r/T = \omega r$
  • Vertical top: $T + mg = mv^2/r$
  • Vertical bottom: $T - mg = mv^2/r$

Orbital Mechanics

  • $v = \sqrt{GM/r}$
  • $T^2 = (4\pi^2/GM)r^3$
  • $r = R_{\text{planet}} + h$ always
  • $KE = \frac{1}{2}GMm/r$
  • $U = -GMm/r$ (negative!)
  • $E = -\frac{1}{2}GMm/r$

Escape Velocity

  • $v_e = \sqrt{2GM/r}$
  • $v_e = \sqrt{2} \times v_{\text{orb}}$
  • Escape energy: $+\frac{1}{2}GMm/r$

Six Common Errors

  • No "centrifugal force" in inertial frames
  • Speed varies in vertical circles
  • Use $r = R + h$, not just $h$
  • Use $g = GM/r^2$, not 9.8
  • $v_e$ has $\sqrt{2}$ factor
  • Mass cancels in orbit speed

For a satellite moving to a higher orbit, which correctly describes the energy change?

Quick recall — Phase 2 Consolidation
+5 XP

A fresh five-question set drawn from this lesson's bank — feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence — that tells the system what to drill next.

Short Answer — 10 marks
+5 XP

ApplyBand 4-5(3 marks) 1. A conical pendulum has a bob of mass 0.40 kg on a 0.90 m string, tracing a horizontal circle of radius 0.30 m. Calculate the tension in the string and the period of the motion.

1 mark: tension · 1 mark: speed · 1 mark: period

AnalyseBand 5(3 marks) 2. The Moon orbits Earth with period 27.3 days at a mean distance of $3.84 \times 10^8$ m. Use this information to calculate the mass of Earth.

1 mark: period conversion · 1 mark: Kepler's Third rearrangement · 1 mark: correct answer

EvaluateBand 6(4 marks) 3. A satellite in circular orbit fires its thrusters briefly in the direction of motion. Evaluate what happens to its orbit, explaining whether it moves to a higher or lower orbit and how its speed and period change.

1 mark: initial speed increase · 1 mark: transfer to higher orbit · 1 mark: final speed decreases, period increases · 1 mark: energy reasoning

Show all answers

Mixed Practice Answers (Q1–Q10)

Q1 (2 marks): $a_c = v^2/r = (12)^2/30 = \mathbf{4.8 \text{ m/s}^2}$ (1 mark correct substitution, 1 mark answer with units).

Q2 (2 marks): $\omega = 2\pi \times 3.0 = 6\pi$ rad/s. $F_c = m\omega^2 r = 0.40 \times (6\pi)^2 \times 0.50 = \mathbf{71 \text{ N}}$ (1 mark correct $\omega$, 1 mark final answer).

Q3 (2 marks): Kepler's Third Law: $T^2 = \frac{4\pi^2}{GM}r^3$, i.e. $T^2/r^3 = \text{const}$ for all bodies orbiting the same central mass. This tells us that satellites with larger orbital radii have longer periods, and the ratio $T^2/r^3$ is the same for all satellites regardless of their own mass (1 mark statement, 1 mark explanation).

Q4 (3 marks): (a) $v_0 = \sqrt{gr\tan\theta} = \sqrt{9.8 \times 60 \times \tan 10°} = \sqrt{103.5} = \mathbf{10.2 \text{ m/s}}$ (1 mark). (b) $v_{\max} = \sqrt{gr\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}} = \sqrt{588 \times \frac{0.376}{0.965}} = \sqrt{229} = \mathbf{15.1 \text{ m/s}}$ (1 mark formula, 1 mark answer).

Q5 (3 marks): (a) Minimum speed: $v_{top}^2 = gr = 9.8 \times 4.0 = 39.2$ m²/s². Energy: $v_{bot}^2 = v_{top}^2 + 4gr = 196$, so $v_{bot} = \mathbf{14.0 \text{ m/s}}$ (1 mark energy method). (b) $N - mg = mv^2/r$, $N = 60 \times 81/4.0 + 60 \times 9.8 = 4860 + 588 = \mathbf{5448 \text{ N}}$ (1 mark formula, 1 mark answer).

Q6 (3 marks): $v = \sqrt{GM/r} = \sqrt{(6.67 \times 10^{-11} \times 5.97 \times 10^{24})/(7.5 \times 10^6)} = \mathbf{7.29 \text{ km/s}}$ (1 mark). $T = 2\pi r/v = \mathbf{6466 \text{ s} = 108 \text{ min}}$ (1 mark each).

Q7 (4 marks): (a) $v_{\min} = \sqrt{gr} = \sqrt{9.8 \times 0.70} = \mathbf{2.62 \text{ m/s}}$ (1 mark). (b) $T_{top} = mv_{top}^2/r - mg = 0.30 \times 16/0.70 - 0.30 \times 9.8 = 6.86 - 2.94 = \mathbf{3.92 \text{ N}}$ (1 mark). Find $v_{bot}^2 = 16 + 4gr = 43.44$ (1 mark). $T_{bot} = 0.30 \times 43.44/0.70 + 0.30 \times 9.8 = 18.62 + 2.94 = \mathbf{21.6 \text{ N}}$ (1 mark).

Q8 (4 marks): Orbital: $GMm/r^2 = mv_{orb}^2/r$, giving $v_{orb} = \sqrt{GM/r}$ (1 mark). Escape (total energy = 0): $\frac{1}{2}mv_e^2 - GMm/r = 0$, giving $v_e = \sqrt{2GM/r}$ (1 mark). Ratio: $v_e/v_{orb} = \sqrt{2GM/r}/\sqrt{GM/r} = \sqrt{2}$, so $v_e = \sqrt{2}\,v_{orb}$ (1 mark clear mathematics, 1 mark reasoning).

Q9 (4 marks): $E_1 = -GMm/(2r_1) = -1.463 \times 10^{10}$ J (1 mark). $E_2 = -GMm/(2r_2) = -8.29 \times 10^9$ J (1 mark). $\Delta E = +6.34 \times 10^9$ J (1 mark). $v_2 = \sqrt{GM/r_2} = 5.76$ km/s (1 mark).

Q10 (5 marks): Hill top: $mg - N = mv^2/r$, $N = 1000 \times 9.8 - 1000 \times 225/20 = 9800 - 11250$. Since $mv^2/r > mg$ the car leaves the road at this speed. (1 mark formula and recognition). Valley bottom: $N - mg = mv^2/r$, $N = 9800 + 1000 \times 225/15 = 9800 + 15000 = \mathbf{24\,800 \text{ N}}$ (1 mark formula, 1 mark answer). Explanation (2 marks): At the top of the hill, centripetal acceleration is directed downward, so gravity assists and $N$ is reduced. At the valley bottom, centripetal acceleration is upward, so $N$ must overcome gravity AND provide centripetal force, greatly increasing $N$.

Timed Exam Answers (Q11–Q13)

Q11 (4 marks): FBD: tension at angle $\theta$ to vertical, weight $mg$ downward (1 mark). $\cos\theta = \sqrt{L^2 - r^2}/L = \sqrt{1.44 - 0.16}/1.2 = 0.943$. $T = mg/\cos\theta = 0.50 \times 9.8/0.943 = \mathbf{5.20 \text{ N}}$ (1 mark). $\sin\theta = r/L = 0.40/1.2 = 0.333$. $v = \sqrt{rT\sin\theta/m} = \sqrt{0.40 \times 5.20 \times 0.333/0.50} = \mathbf{1.18 \text{ m/s}}$ (1 mark). Shortening the string (same $r$) increases angle, tension and centripetal force — speed increases, period decreases (1 mark).

Q12 (4 marks): $v_0 = \sqrt{9.8 \times 80 \times \tan 15°} = \sqrt{9.8 \times 80 \times 0.268} = \mathbf{14.5 \text{ m/s}}$ (1 mark). $v_{\max} = \sqrt{gr(\tan\theta+\mu_s)/(1-\mu_s\tan\theta)} = \mathbf{23.1 \text{ m/s}}$ (1 mark). $\tan\theta < \mu_s$ so $v_{\min}$ is imaginary — the car won't slide down at any speed (1 mark). Above $v_{\max}$: required centripetal force exceeds what gravity and friction can provide; the car slides up the bank (1 mark).

Q13 (4 marks): $v_1 = \sqrt{GM/r_1} = \mathbf{7.54 \text{ km/s}}$; $v_2 = \sqrt{GM/r_2} = \mathbf{3.07 \text{ km/s}}$ (1 mark). $E_1 = -GMm/(2r_1) = -5.69 \times 10^{10}$ J; $E_2 = -GMm/(2r_2) = -9.44 \times 10^9$ J; $\Delta E = +\mathbf{4.75 \times 10^{10} \text{ J}}$ (1 mark). Although KE decreases, PE increases by a larger amount (becomes less negative) — the satellite climbs out of Earth's gravitational well (1 mark). The formula $\Delta E = \frac{1}{2}m(v_2^2-v_1^2)$ ignores the PE change — the correct approach uses $\Delta E = E_2 - E_1 = -GMm/2r_2 + GMm/2r_1$ (1 mark).

Multiple Choice — Key

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer — Model Answers

SA1 (3 marks): $\cos\theta = \sqrt{0.81 - 0.09}/0.90 = 0.943$. $T = mg/\cos\theta = (0.40 \times 9.8)/0.943 = \mathbf{4.16 \text{ N}}$ (1 mark). $\sin\theta = 0.30/0.90 = 0.333$; $v = \sqrt{rT\sin\theta/m} = \sqrt{0.30 \times 4.16 \times 0.333/0.40} = \mathbf{1.02 \text{ m/s}}$ (1 mark). $T_{\text{period}} = 2\pi r/v = 2\pi \times 0.30/1.02 = \mathbf{1.85 \text{ s}}$ (1 mark).

SA2 (3 marks): $T = 27.3 \times 24 \times 3600 = 2.36 \times 10^6$ s (1 mark). From $T^2 = 4\pi^2 r^3/(GM)$: $M = 4\pi^2 r^3/(GT^2) = 4\pi^2 \times (3.84 \times 10^8)^3/(6.67 \times 10^{-11} \times (2.36 \times 10^6)^2) = \mathbf{6.01 \times 10^{24} \text{ kg}}$ (1 mark rearrangement, 1 mark answer).

SA3 (4 marks): Thrusters fire forward, increasing speed momentarily (1 mark). The satellite is now moving too fast for its current circular orbit — it climbs to a higher orbit (apoapsis increases), making the orbit elliptical with thrust point as perigee (1 mark). In the new stable circular orbit (if a second burn is applied at apoapsis), orbital speed is lower ($v \propto 1/\sqrt{r}$) and period is longer ($T \propto r^{3/2}$) (1 mark). The apparent paradox (speeding up to end up slower) is resolved because the added energy goes mostly into increasing gravitational PE — the satellite is less tightly bound to Earth (1 mark).

Boss Battle — Module Quiz
boss

Five timed questions on Phase 2 consolidation — circular motion, vertical circles, banked curves, orbital mechanics, and gravitational energy. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).

⚔ Enter the arena
Arcade Practice · Asteroid Blaster

Orbital trajectories — keep your circular orbit stable and intercept targets. A quick session between study blocks keeps concepts fresh.

Revisit — How did your thinking change?

At the start you were asked about the 16 July 1969 Apollo 11 trans-lunar injection burn — a 347-second burn accelerating from 7.79 km/s to 10.84 km/s.

Earth's escape velocity at Earth's surface is 11.2 km/s. At the altitude where the burn occurred (~185 km orbital altitude, $r \approx 6.556 \times 10^6$ m), escape velocity is $v_e = \sqrt{2GM/r} = \sqrt{2 \times 3.98 \times 10^{14} / 6.556 \times 10^6} = 11.0 \text{ km/s}$. Apollo 11's post-burn speed of 10.84 km/s was just below this — the spacecraft was on a transfer trajectory, not yet free of Earth's gravity. It coasted to the Moon using the combined gravitational fields of both Earth and Moon. After working through this consolidation lesson, reflect again:

  • Which question type do you find easiest? Which is hardest?
  • Which error from the Error Clinic have you made before?
  • What is your plan to avoid that error in the exam?
← Previous: Lesson 12 Next: Lesson 14 →

Module overview · Physics · Checkpoint 2 · Module quiz

🎓
Want help with Lesson 13: Phase 2 Consolidation — HSC Physics?

Work through this topic 1-on-1 with an experienced HSC tutor.

Book a free session →