Year 12 Physics Module 5 · IQ2 + IQ3 consolidation 50 min Lessons 6–12 only Lesson 13 of 18

Circular Motion and Orbits Consolidation

Diagnose what you can retrieve, connect force diagrams to equations, and repair the sign and model-condition errors that cost marks across torque, circular motion, gravitation and circular orbits.

Today's diagnostic: For each problem, can you name the real forces, choose the radial or torque sign convention, state the model condition, and predict the sign of the answer before calculating?
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You’re here

Diagnose and orient

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

One printable cumulative worksheet covering the same Lessons 6–12 boundary as this lesson.

Build Foundations & guided practice
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 — Retrieve Lessons 6–12

  • Recall torque, circular-motion, gravitation and circular-orbit relationships
  • Attach each equation to its model conditions and symbol definitions

Understand — Diagnose errors

  • Identify and correct common errors in circular motion problems
  • Distinguish real forces, radial net force, signed torque and signed orbital energy

Can Do — Select and justify

  • Solve mixed circular motion and orbital problems under exam conditions
  • Use force diagrams, sign conventions, units and applicability statements
Key Terms, Lessons 6–12
vocab
Torque$\tau=rF_\perp=rF\sin\theta$; a signed moment of force measured in N·m.
Centripetal acceleration$a_c = v^2/r$, directed toward the instantaneous 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 $r$.
Gravitational field strength$g=GM/r^2$ outside a spherical mass; $r$ is centre-to-centre and the field points inward.
Gravitational potential energy$U = -GMm/r$ when $U=0$ at infinity.
Circular-orbit total energy$E = -GMm/(2r) = K + U$; this radius form is restricted to circular orbits.
Binding boundary: This is cumulative retrieval from Lessons 6–12: torque and equilibrium; uniform, horizontal and vertical circular motion; Newtonian gravitation; circular satellite motion; and circular-orbit energy. Escape speed begins in Lesson 16, while non-circular Kepler orbits begin in Lesson 17, so neither is assessed here.
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.

2

Map formulae to conditions

Lessons 6–12: formula and condition map

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Lessons 6–12 Formula and Condition Map
+5 XP

Six relationship families, when each applies, and the sign or model condition that must accompany it.

Lessons 6 to 12 summary linking torque, circular motion, gravitation and circular-orbit energy

Cumulative map for the exact Lessons 6–12 assessment boundary.

Torque and equilibrium

$\tau = rF_\perp = rF\sin\theta$   ·   $\Sigma \tau = 0$ and $\Sigma F = 0$ for static equilibrium

Use: Declare a clockwise/anticlockwise sign convention and use the perpendicular lever arm. Torque units are N·m, not J.

Centripetal acceleration & radial net force

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

Use: $F_c$ names the net inward requirement, supplied by real forces. Constant speed does not mean constant velocity.

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

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

Newtonian gravitation and circular satellites

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

Use: $F=GMm/r^2$ and $g=GM/r^2$ require centre-to-centre $r$ outside a spherical mass. The speed/period equations shown are circular-orbit results.

Gravitational energy

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

Use: $U=0$ at infinity. The displayed $K$ and $E$ radius forms apply to circular orbits; then $U<0$, $E<0$ and $E=U/2=-K$.

Signed energy change between circular orbits

$\Delta E = E_f-E_i = -\dfrac{GMm}{2r_f}+\dfrac{GMm}{2r_i}$

Use: Compare specified initial and final circular states. A higher final circular orbit has $\Delta E>0$ even though its final speed and kinetic energy are smaller.

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: $F_T + mg = mv^2/r$ (top). Orbits: $GMm/r^2 = mv^2/r$.
$$F_T + mg = \frac{mv^2}{r}$$
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, $F_T$ and $mg$ both point inward, so they add. At the bottom, $F_T$ is inward and $mg$ is outward, so $F_T-mg=mv^2/r$.
Connects to Minimum speed at top: set $F_T=0$ 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 Gravity supplies the radial net force: $GMm/r^2=mv^2/r$. Combining this with $v=2\pi r/T$ gives the circular-orbit period relation.
$$U = -\frac{GMm}{r}$$
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: $\Delta E = E_2 - E_1$.
$$\Delta E = -\frac{GMm}{2r_f}+\frac{GMm}{2r_i}$$
Circular-orbit energy change, click to reveal
Use when Comparing the total energy of two specified circular orbits about the same central mass.
Trap Using only $\Delta K$, or applying a circular-state formula to the intermediate transfer trajectory. Always calculate $E_f-E_i$ with signed values.
Connects to A prograde first burn immediately raises speed and energy; a separate burn is required to circularise at the higher radius.

Condition map: torque uses a declared rotational sign and perpendicular lever arm; radial equations sum real inward forces; vertical-circle signs change with position; gravitational and orbital radii are centre-to-centre; $v=\sqrt{GM/r}$, $K=GMm/(2r)$ and $E=-GMm/(2r)$ are circular-orbit results; calculate orbit changes with signed $\Delta E=E_f-E_i$.

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

Worked exemplar 1

Use a signed torque ledger

reveal by step

A 120 N downward force acts 0.40 m to the right of a pivot. What upward force 0.60 m to the right produces rotational equilibrium? Take anticlockwise as positive.

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Represent: the 120 N force produces clockwise torque, so $\tau_1=-(0.40)(120)=-48$ N·m.
2
Apply equilibrium: $\Sigma\tau=0$, so $-48+(0.60)F=0$.
3
Solve and interpret: $F=80$ N upward. The pivot reaction is included when checking $\Sigma F=0$, but it has zero moment about the pivot.
Worked exemplar 2

Compare two circular-orbit states

reveal by step

A 500 kg satellite moves from a circular orbit at $r_i=7.0\times10^6$ m to a final circular orbit at $r_f=1.40\times10^7$ m. Find $\Delta E$. Use $GM_E=3.986\times10^{14}$ m³/s².

1
Condition: both stated endpoints are circular, so $E=-GMm/(2r)$ applies to each endpoint, not to the transfer trajectory.
2
Initial: $E_i=-\dfrac{(3.986\times10^{14})(500)}{2(7.0\times10^6)}=-1.42\times10^{10}$ J.
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Final: $E_f=-\dfrac{(3.986\times10^{14})(500)}{2(1.40\times10^7)}=-7.12\times10^9$ J.
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Signed change: $\Delta E=E_f-E_i=+7.12\times10^9$ J. The positive sign means energy was added; the final state is less tightly bound.

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?

3

Repair common reasoning errors

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
Treating a prograde burn as an instant higher circular orbit
▾
Student working: “A brief prograde burn moves a satellite straight into a higher circular orbit, where it continues at the increased speed.”
Correction: The burn immediately increases speed and total energy at the original radius, placing the satellite on a transfer trajectory rather than a new circular orbit. A later burn is needed to circularise at the higher radius. The final higher circular orbit has lower speed because $v=\sqrt{GM/r}$, even though its total energy is greater (less negative).
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) no centrifugal force in an inertial FBD; (2) speed varies in a vertical circle when gravity does work; (3) use centre-to-centre $r=R+h$; (4) use $g=GM/r^2$ at altitude; (5) separate an immediate burn from the final circular state; (6) satellite mass cancels from circular-orbit 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).

4

Apply mixed calculations

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. Starting with gravity supplying the radial net force, derive $T^2=4\pi^2r^3/(GM)$ for a circular satellite and explain what it predicts for larger $r$.

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. For a circular satellite, derive $K=GMm/(2r)$ and hence $E=-GMm/(2r)$. State the condition that makes both radius forms valid.

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 approaches the top of a hill of radius 20 m at 15 m/s. Determine whether it can remain in contact with the road. The car later enters a valley of radius 15 m at the same speed; calculate the normal force at the bottom. Explain the contrast using the direction of radial 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?

5

Write structured responses

Extended response practice

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 Force 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. Use the component equations to show why the period is independent of the bob’s mass. (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.

6

Feedback and remediation

Activity 1, Circular Motion & Orbits Consolidation Drills
ApplyBand 3

Practise the key concepts from Lessons 6–12.

  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 circular-orbit 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: compared with a lower circular orbit, a higher circular orbit has total energy that is _____ negative.

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: $F_T + mg = mv^2/r$
  • Vertical bottom: $F_T - mg = mv^2/r$

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!)
  • $E = -\frac{1}{2}GMm/r$

Torque & Gravitation

  • $\tau=rF_\perp$; declare the torque sign
  • Static equilibrium: $\Sigma F=0$ and $\Sigma\tau=0$
  • $F=GMm/r^2$, $g=GM/r^2$; use centre-to-centre $r$

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
  • Separate an immediate burn from the final circular state
  • Mass cancels in orbit speed

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

7

Independent cumulative assessment

Quick recall, Circular Motion and Orbits 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: circular-orbit period relation 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): For a circular orbit, $GMm/r^2=mv^2/r$. Substituting $v=2\pi r/T$ and cancelling $m$ gives $T^2=4\pi^2r^3/(GM)$ (1 mark). Therefore a circular satellite at larger $r$ has a longer period; for the same central mass, $T^2/r^3$ is constant (1 mark).

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) At the just-contact threshold, $v_{\rm top}^2=gr=39.2$ m²/s². Conservation of energy gives $v_{\rm bottom}^2=v_{\rm top}^2+4gr=196$, so $v_{\rm bottom}=\mathbf{14.0\text{ m/s}}$ (1 mark). (b) At 18 m/s, $N-mg=mv^2/r$, so $N=60(18^2/4.0)+60(9.8)=4860+588=\mathbf{5448\text{ N}}$ (1 mark equation, 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) $v_{\min}=\sqrt{gr}=\mathbf{2.62\text{ m/s}}$ (1 mark). (b) $F_{T,\rm top}=mv_{\rm top}^2/r-mg=\mathbf{3.92\text{ N}}$ (1 mark). Energy gives $v_{\rm bottom}^2=16+4gr=43.44$ (1 mark). $F_{T,\rm bottom}=0.30(43.44)/0.70+0.30(9.8)=\mathbf{21.6\text{ N}}$ (1 mark).

Q8 (4 marks): For a circular orbit, $GMm/r^2=mv^2/r$, so $mv^2=GMm/r$ (1 mark). Hence $K=\tfrac12mv^2=GMm/(2r)$ (1 mark). With $U=-GMm/r$, $E=K+U=-GMm/(2r)$ (1 mark). These radius forms for $K$ and $E$ require a circular orbit about a point or spherically symmetric central mass; they are not applied to an intermediate transfer trajectory (1 mark).

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: while contact holds, $mg-N=mv^2/r$. Substitution would require $N=9800-11250=-1450$ N, which a road cannot provide; equivalently, $15\text{ m/s}>\sqrt{gr}=14.0\text{ m/s}$. The car therefore loses contact and $N=0$ once airborne (2 marks). Valley bottom: $N-mg=mv^2/r$, so $N=9800+1000(225/15)=\mathbf{24\,800\text{ N}}$ (2 marks). At the hilltop gravity supplies the downward radial requirement and contact force is reduced; at the valley bottom the road force must both overcome weight and provide the upward radial net force (1 mark).

Timed Exam Answers (Q11–Q13)

Q11 (4 marks): FBD: tension along the string toward the pivot and weight $mg$ downward (1 mark). $\cos\theta = \sqrt{L^2-r^2}/L = 0.943$, so $F_T=mg/\cos\theta=\mathbf{5.20\text{ N}}$ (1 mark). With $\sin\theta=r/L=0.333$, $F_T\sin\theta=mv^2/r$ gives $v=\mathbf{1.18\text{ m/s}}$ (1 mark). Dividing the radial equation by $F_T\cos\theta=mg$ cancels $m$, so the resulting $v$ and $T_{\rm period}=2\pi r/v$ do not depend on bob mass (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): $\cos\theta = \sqrt{0.81 - 0.09}/0.90 = 0.943$. $F_T = mg/\cos\theta = (0.40 \times 9.8)/0.943 = \mathbf{4.16 \text{ N}}$ (1 mark). $\sin\theta = 0.30/0.90 = 0.333$; $v = \sqrt{rF_T\sin\theta/m} = \mathbf{1.02 \text{ m/s}}$ (1 mark). $T_{\text{period}} = 2\pi r/v = \mathbf{1.85 \text{ s}}$ (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).

8

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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?

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