Q1 Gravitation, Gravitational Force, Force of Gravity, Cavendish 6 marks ▼
(a) What do you understand by the following: (i) gravitation (ii) gravitational force (iii) force of gravity (iv) duality equivalence principle. (b) Explain the contribution of British Scientist Henry Cavendish using the example of cans of dog food as illustration of superposition of forces.
Part (a)(i) – Gravitation
Gravitation is the universal natural phenomenon by which all objects with mass attract one another. It is a long-range, always-attractive interaction described by Newton's Law of Universal Gravitation.

Newton's Law states: every particle of mass m₁ attracts every other particle of mass m₂ with a force:

F = G·m₁·m₂ / r² where: G = 6.674 × 10⁻¹¹ N·m²/kg² (universal gravitational constant) r = distance between the centres of mass of the two bodies
Part (a)(ii) – Gravitational Force
Gravitational force is the attractive force between any two masses due to gravitation. It acts along the line joining the two masses, is always attractive (never repulsive), decreases with the square of distance, and has infinite range.

In vector form, the gravitational force on mass m₂ due to mass m₁ is:

F₁₂ = −G·m₁·m₂ / r² · r̂₁₂ The negative sign indicates attraction (force directed toward m₁).
Part (a)(iii) – Force of Gravity
Force of gravity is the specific gravitational force exerted by the Earth on a body of mass m near its surface. It is what we call weight.
W = mg where g ≈ 9.8 m/s² is the acceleration due to gravity at Earth's surface. From Newton's law: g = G·M_Earth / R_Earth² M_Earth = 5.97 × 10²⁴ kg, R_Earth = 6.371 × 10⁶ m → g = (6.674×10⁻¹¹ × 5.97×10²⁴) / (6.371×10⁶)² ≈ 9.8 m/s² ✓
Part (a)(iv) – Duality Equivalence Principle (Principle of Equivalence)
The Equivalence Principle states that inertial mass (resistance to acceleration: F = mᵢa) and gravitational mass (response to gravitational field: F = mɡg) are exactly equal: mᵢ = mɡ.

This means all objects fall with the same acceleration in a gravitational field, regardless of their mass — a result confirmed by Galileo's experiments and later underpinning Einstein's General Relativity.

Part (b) – Cavendish Experiment & Superposition

Henry Cavendish (1798) was the first to experimentally measure G using a torsion balance. Two small lead balls were attached to a rod; two large fixed lead balls nearby attracted them, twisting a fine wire. By measuring the twist angle, he calculated G.

Superposition of gravitational forces states that the total gravitational force on a mass due to multiple sources is the vector sum of the individual forces.

Dog food can analogy: Imagine three cans of dog food sitting on a shelf. A small ball in the middle is attracted to each can separately. The force from can 1 pulls it left; the force from can 2 pulls it up; the force from can 3 pulls it right. By superposition:

F_total = F₁ + F₂ + F₃ (vector addition) Just as three cans exert independent gravitational tugs that add as vectors, any number of masses contribute independently to the total gravitational force on a test mass. This is the Principle of Superposition for gravity.
Key result: G = 6.674 × 10⁻¹¹ N·m²·kg⁻² (Cavendish's great contribution — first measurement of G, which also allowed the mass of Earth to be determined for the first time).
Q2 Orbit, Central Orbit, Inertial & Gravitational Mass; Net Force Calculation 6 marks ▼
(a) Define: (i) orbit (ii) central orbit (iii) inertial mass (iv) gravitational mass. (b) Given masses m₁ = 4 kg, m₂ = 6 kg, m₃ = m₂ = 6 kg, placed with the center of attraction at a = 20 m = 0.02 μ. Find the net force due to m₂ and m₃ on m₁. Obtain the angle which the forces make with the x-axis.
Part (a)(i) – Orbit
An orbit is the curved path traced by a body moving under the influence of a central force (such as gravity). The shape depends on the body's total mechanical energy.
Part (a)(ii) – Central Orbit
A central orbit is an orbit in which the net force on the body always points toward (or away from) a fixed centre called the centre of force. For gravity, this centre is the more massive body (e.g., the Sun).

The key property: angular momentum L = mr²θ̇ is conserved, and motion is confined to a plane.

Part (a)(iii) – Inertial Mass
Inertial mass (mᵢ) measures a body's resistance to being accelerated. Defined by Newton's 2nd Law: F = mᵢ·a. The harder it is to accelerate, the greater the inertial mass.
Part (a)(iv) – Gravitational Mass
Gravitational mass (mɡ) measures how strongly a body interacts with a gravitational field. It appears in F = G·mɡ·M/r². Experiments confirm mᵢ = mɡ to very high precision.
Part (b) – Net Force Calculation

We interpret the problem as: m₁ = 4 kg is at the origin. m₂ = 6 kg is along the +y axis at distance r = 20 m. m₃ = 6 kg is along the +x axis at distance r = 20 m. (A common symmetric arrangement in such problems.)

1
Force on m₁ due to m₂ (along +y direction):
F₁₂ = G·m₁·m₂ / r² = (6.674×10⁻¹¹ × 4 × 6) / (20)² = (6.674×10⁻¹¹ × 24) / 400 = 1.6018×10⁻⁹ / 400 = 4.00 × 10⁻¹² N (in the +y direction)
2
Force on m₁ due to m₃ (along +x direction):
F₁₃ = G·m₁·m₃ / r² = (6.674×10⁻¹¹ × 4 × 6) / (20)² = 4.00 × 10⁻¹² N (in the +x direction)
3
Net force magnitude (vector addition, since forces are perpendicular):
|F_net| = √(F₁₂² + F₁₃²) = √((4.00×10⁻¹²)² + (4.00×10⁻¹²)²) = 4.00×10⁻¹² × √2 = 5.66 × 10⁻¹² N
4
Angle with x-axis:
θ = arctan(F_y / F_x) = arctan(4.00×10⁻¹² / 4.00×10⁻¹²) = arctan(1) = 45°
Net force on m₁ = 5.66 × 10⁻¹² N at 45° above the positive x-axis.
Q3 Kepler's Laws; Torque and Net Force on a Rod 6 marks ▼
(a) State the three laws of Kepler. (b) Given two bodies with masses m₁ = 1.10 kg and m₂ = 25.0 kg connected by a rod. The m₁ pairs have separation l = 30.0 cm, and m₁ and m₂ are 12.0 cm apart (centre-to-centre). (i) Find the net force. (ii) Find the net torque about the rotation axis. (iii) Does the torque in (ii) have the capacity to rotate the rod?
Part (a) – Kepler's Three Laws of Planetary Motion
First Law (Law of Ellipses): Every planet moves in an elliptical orbit with the Sun located at one focus of the ellipse.
Second Law (Law of Equal Areas): A line segment joining a planet to the Sun sweeps out equal areas in equal times. This is equivalent to conservation of angular momentum: L = mr²θ̇ = constant.
Third Law (Law of Periods): The square of the orbital period T of a planet is proportional to the cube of the semi-major axis a of its orbit:
T² ∝ a³ or equivalently T² = (4π²/GM)·a³
Part (b) – Force and Torque Calculation

Setup: Two masses m₁ = 1.10 kg, each at distance l/2 = 15.0 cm = 0.15 m from the axis. m₂ = 25.0 kg at distance d = 12.0 cm = 0.12 m from the axis. The gravitational force between them:

1
(i) Net gravitational force between m₁ and m₂:
F = G·m₁·m₂ / r² where r = 0.12 m F = (6.674×10⁻¹¹ × 1.10 × 25.0) / (0.12)² = (6.674×10⁻¹¹ × 27.5) / 0.0144 = 1.835×10⁻⁹ / 0.0144 = 1.27 × 10⁻⁷ N
2
(ii) Net torque about the rotation axis:

Torque = Force × perpendicular distance from axis. If the force acts along the line connecting m₁ and m₂, and the rod lies perpendicular to the rotation axis:

τ = F × d_perp Taking d_perp = 0.12 m (the separation distance): τ = 1.27 × 10⁻⁷ × 0.12 = 1.53 × 10⁻⁸ N·m
3
(iii) Can this torque rotate the rod?

Yes — any non-zero net torque will produce angular acceleration α = τ/I, where I is the moment of inertia of the rod system. Since τ ≠ 0, the rod will rotate. The direction of rotation is determined by which side the net torque acts.

α = τ / I (angular acceleration) Since τ = 1.53 × 10⁻⁸ N·m ≠ 0, the torque CAN rotate the rod.
(i) F ≈ 1.27 × 10⁻⁷ N  |  (ii) τ ≈ 1.53 × 10⁻⁸ N·m  |  (iii) Yes, the torque can rotate the rod.
Q4 Superposition, Comet Halley, Moon's Force on a Human 6 marks ▼
(a) Briefly explain: (i) superposition (ii) comet Halley (iii) energy conservation (iv) moon gravity. (b) A typical adult human has a mass of about 70 kg. (i) What force does a full moon exert on a human? (ii) Compare this force with that exerted on the human by the Earth.
Part (a)(i) – Superposition
The Principle of Superposition states that when multiple gravitational forces act on a body simultaneously, the total (net) force is the vector sum of all individual forces. Forces add independently and do not interfere with each other.
Part (a)(ii) – Comet Halley
Comet Halley is a famous short-period comet visible from Earth approximately every 75–76 years. It follows a highly elliptical orbit around the Sun (eccentricity e ≈ 0.967), with perihelion near 0.59 AU and aphelion near 35 AU. It perfectly illustrates Kepler's First and Second Laws — moving fastest near the Sun (perihelion) and slowest far from it (aphelion).
Part (a)(iii) – Energy Conservation
In an isolated gravitational system, total mechanical energy E = KE + PE is conserved:
E = ½mv² − GMm/r = constant.
As a planet moves closer to the Sun, PE decreases and KE increases, but their sum stays constant.
Part (a)(iv) – Moon Gravity
Moon gravity refers to the gravitational attraction between the Moon and objects on Earth. The Moon's surface gravity is about 1/6 of Earth's (g_moon ≈ 1.62 m/s²). The Moon's gravity causes ocean tides on Earth.
Part (b)(i) – Force of Full Moon on a 70 kg Human
1
Known data:
Mass of Moon: M_moon = 7.342 × 10²² kg Mass of human: m = 70 kg Distance Earth–Moon: r = 3.84 × 10⁸ m (centre to centre) G = 6.674 × 10⁻¹¹ N·m²/kg²
2
Apply Newton's Law:
F_moon = G · M_moon · m / r² = (6.674×10⁻¹¹ × 7.342×10²² × 70) / (3.84×10⁸)² = (6.674×10⁻¹¹ × 5.139×10²⁴) / (1.475×10¹⁷) = (3.430×10¹⁴) / (1.475×10¹⁷) = 2.33 × 10⁻³ N ≈ 0.0023 N
Part (b)(ii) – Compare with Earth's Force
3
F_Earth = m · g = 70 × 9.8 = 686 N Ratio = F_moon / F_Earth = 0.0023 / 686 ≈ 3.35 × 10⁻⁶
F_moon ≈ 0.0023 N  |  F_Earth = 686 N  |  The Moon's pull on you is about 300,000× weaker than Earth's. Despite being massive, the Moon is very far away; Earth wins by its proximity.
Q5/Q7 Equation of Motion Analysis via Integration; Total Energy Conservation 6 marks ▼
(a) With the aid of good illustrations, using a two-steps approach, explain the analysis of the equation of motion through integration. The total energy can be written as E = ½μṙ² + U(r). (b) Given the equation of motion for the relative coordinate: d/dt(∂L/∂ṙ) = ∂L/∂r, show that the total energy through the relative coordinate contribution is conserved.
Part (a) – Two-Step Analysis of Equation of Motion

In the two-body central force problem, we replace two bodies with one body of reduced mass μ moving in an effective potential.

1
Step 1: Reduction to one-body problem

Define the reduced mass:

μ = m₁·m₂ / (m₁ + m₂) The Lagrangian for the relative motion is: L = ½μṙ² + ½μr²θ̇² − U(r) = ½μ(ṙ² + r²θ̇²) − U(r) The angular momentum is conserved: l = μr²θ̇ = constant This allows us to eliminate θ and work only in r.
2
Step 2: Integrate the radial equation

The total energy (with centrifugal term from angular momentum):

E = ½μṙ² + l²/(2μr²) + U(r) [effective potential approach] or equivalently for purely radial motion: E = ½μṙ² + U(r) Solving for ṙ: ṙ = dr/dt = √[(2/μ)(E − U(r))] Separating variables and integrating: t = ∫ dr / √[(2/μ)(E − U(r))] This integral gives r(t), from which the full orbit can be reconstructed. Similarly: θ = ∫ (l/μr²) dt gives the angular position.
Part (b) – Showing Energy Conservation from the Euler-Lagrange Equation
1
Start with the Lagrangian for relative motion:
L = ½μṙ² − U(r) The Euler-Lagrange equation: d/dt(∂L/∂ṙ) = ∂L/∂r Computing each side: ∂L/∂ṙ = μṙ → d/dt(μṙ) = μr̈ ∂L/∂r = −dU/dr = F(r) (the central force) So: μr̈ = −dU/dr ✓ (Newton's 2nd Law in radial direction)
2
Show total energy E = ½μṙ² + U(r) is conserved:
dE/dt = d/dt[½μṙ² + U(r)] = μṙr̈ + (dU/dr)·ṙ = ṙ[μr̈ + dU/dr] = ṙ[μr̈ − F(r)] since F = −dU/dr = ṙ × 0 by the equation of motion (μr̈ = F) Therefore: dE/dt = 0 → E = constant ✓ The total energy is conserved.
Energy conservation follows directly from the Euler-Lagrange equation applied to the Lagrangian of the relative coordinate. Since dE/dt = ṙ(μr̈ − F) = 0 by Newton's second law, E is a constant of the motion.
Q6 Central Potential; Two-Body to One-Body Reduction; Properties of Gravitational Force 6 marks ▼
(a) What is central potential? Show that the two-body central force problem may always be reduced to two independent one-body problems by transforming to the centre of mass R and relative coordinate r. Given R = (m₁r₁ + m₂r₂)/(m₁+m₂), r = r₁ − r₂, with r₁ = R + m₂r/(m₁+m₂) and r₂ = R − m₁r/(m₁+m₂). (b) With the aid of good examples, explain properties of gravitational force.
Central Potential – Definition
A central potential U(r) is one that depends only on the distance r = |r| from a fixed centre, not on the direction. The corresponding force F = −∇U(r) always points radially (toward or away from the centre). Gravity is a central potential: U(r) = −GMm/r.
Two-Body Reduction – Full Derivation
1
Define coordinates:
R = (m₁r₁ + m₂r₂)/(m₁+m₂) [centre of mass] r = r₁ − r₂ [relative coordinate] Inverting: r₁ = R + (m₂/(m₁+m₂))·r r₂ = R − (m₁/(m₁+m₂))·r
2
Write the total kinetic energy:
T = ½m₁ṙ₁² + ½m₂ṙ₂² Substituting ṙ₁ = Ṙ + (m₂/M)ṙ, ṙ₂ = Ṙ − (m₁/M)ṙ where M = m₁+m₂: T = ½MṘ² + ½μṙ² where μ = m₁m₂/(m₁+m₂) is the reduced mass.
3
The total Lagrangian separates:
L = T − V = [½MṘ²] + [½μṙ² − U(r)] = L_CM + L_rel These are INDEPENDENT: • L_CM describes a free particle of mass M at the centre of mass → moves in straight line • L_rel describes a single particle of mass μ in potential U(r) → the orbit problem This is the reduction from 2-body to 2 independent 1-body problems. ✓
Part (b) – Properties of Gravitational Force
  • Always attractive: Unlike electric forces, gravity never repels. Two masses always pull toward each other (e.g., Earth pulls Moon toward it).
  • Inverse-square law: F ∝ 1/r². Doubling the distance reduces the force by a factor of 4 (e.g., force on a satellite at 2R_Earth is ¼ that at R_Earth).
  • Proportional to both masses: F ∝ m₁·m₂. The Earth pulls a 100 kg person twice as hard as a 50 kg person.
  • Acts at a distance (long range): Gravity has infinite range. It decreases with distance but never becomes exactly zero.
  • Conservative force: Work done by gravity is path-independent. A satellite returns to the same point with the same speed — energy is conserved.
  • Central force: It always acts along the line joining two masses, hence angular momentum is conserved in orbital motion.
  • Weakest fundamental force: G = 6.67×10⁻¹¹ N·m²/kg² makes gravity extremely weak between everyday objects but dominant at astronomical scales due to large masses.
Q8 Frames of Reference; Inertial vs Non-Inertial Frames 6 marks ▼
(b) Briefly explain: (i) frame (ii) frame of reference (iii) inertial reference frame (iv) non-inertial reference frame.
(i) Frame
A frame is a coordinate system (set of axes) used to measure position, velocity, and acceleration of objects. It provides a viewpoint from which physical events are described.
(ii) Frame of Reference
A frame of reference is a frame together with a clock, allowing measurement of both position and time. All physical measurements (position, velocity, force) depend on the frame chosen. Example: a person on a moving train uses the train as their frame of reference.
(iii) Inertial Reference Frame
An inertial reference frame is one in which Newton's Laws hold in their standard form — a body with no net force moves at constant velocity. An inertial frame is either at rest or moving at constant velocity (non-accelerating). Example: a laboratory on Earth (approximately), or a space station far from gravity.
Newton's 2nd Law in inertial frame: F = ma (no extra terms)
(iv) Non-Inertial Reference Frame
A non-inertial reference frame is one that is accelerating (including rotating). Newton's Laws do not hold in their standard form. To apply Newton's Laws, we must introduce fictitious (pseudo) forces that account for the frame's acceleration. Examples: a car accelerating, a rotating Earth, a merry-go-round.
In a rotating frame: F_apparent = F_real − mω×(ω×r) − 2m(ω×v) − mα×r [centrifugal] [Coriolis] [Euler force]
Q9 Rotating Frames; Relative Velocity; Zero Momentum Frame; Inelastic Collision 6 marks ▼
(a) Briefly explain: (i) rotating frame of reference (ii) show that the relative speed between two particles is the same in both frames. (b) Two cars travel in the same direction with velocities 20 m/s and 23 m/s in the road's frame. (i) What is the speed of the slower car in the rest frame of the faster one? (ii) By considering a collision in the zero momentum frame, explain why a perfectly inelastic collision results in maximum loss of kinetic energy.
Part (a)(i) – Rotating Frame of Reference
A rotating frame of reference is one that rotates with angular velocity ω relative to an inertial frame. Observers in this frame see fictitious forces: the centrifugal force (outward) and Coriolis force (perpendicular to velocity). Example: Earth's surface rotates with ω = 7.27×10⁻⁵ rad/s, making it a rotating non-inertial frame.
Part (a)(ii) – Relative Speed is Frame-Independent (Galilean)
In frame S: particle A has velocity vₐ, particle B has velocity v_b Relative velocity of A w.r.t B: v_rel = vₐ − v_b In frame S' moving at velocity V relative to S: vₐ' = vₐ − V v_b' = v_b − V Relative velocity in S': v_rel' = vₐ' − v_b' = (vₐ−V) − (v_b−V) = vₐ − v_b Therefore: v_rel' = v_rel ✓ The relative speed between two particles is the same in all inertial frames.
Part (b)(i) – Speed of Slower Car in Frame of Faster Car
1
v_slow = 20 m/s (in road frame) v_fast = 23 m/s (in road frame) Speed of slow car in fast car's frame: v_relative = v_slow − v_fast = 20 − 23 = −3 m/s The slower car appears to move at 3 m/s backward (in the −x direction) relative to the faster car's frame.
Part (b)(ii) – Zero Momentum Frame and Maximum KE Loss in Inelastic Collision
2

The zero-momentum (CM) frame is the frame where total momentum = 0, i.e., P_total = 0. Objects move with equal and opposite momenta.

In the CM frame before collision: m₁v₁' = −m₂v₂' → total momentum = 0 In a perfectly inelastic collision, the objects STICK TOGETHER. Since momentum must still be zero (conserved) in the CM frame: (m₁ + m₂)·V_final' = 0 → V_final' = 0 So both objects are at REST in the CM frame after a perfectly inelastic collision. KE_after (CM frame) = 0 This means ALL kinetic energy in the CM frame is lost as heat/deformation. No other type of collision produces zero final KE in the CM frame. Hence perfectly inelastic → MAXIMUM kinetic energy loss. ✓
Relative speed of slow car in fast car's frame = 3 m/s (backward). Perfectly inelastic collision destroys all KE in the CM frame → maximum energy loss.
Q10/Q11 Rotating Frame Equations; Fictitious Forces; Foucault Pendulum 6 marks ▼
(a) Write down the resulting equation from the transformation of Newton's Laws to a uniformly rotating frame of reference, with identification of Centrifugal and Coriolis forces. (b) Why is Centrifugal force in a rotating frame called a "fictitious force"? Write a short note on the Foucault Pendulum.
Part (a) – Newton's Law in a Rotating Frame

Let S be an inertial frame and S' rotate with angular velocity ω relative to S. For any vector A:

(dA/dt)_inertial = (dA/dt)_rotating + ω × A

Applying this to position r twice to get acceleration, and substituting into F = ma:

m·a_rotating = F − 2m(ω × v') − mω × (ω × r) − m(dω/dt × r) where: F = real physical force (gravity, normal, etc.) −2m(ω × v') = Coriolis force [depends on velocity in rotating frame] −mω×(ω×r) = Centrifugal force [outward, depends on position] −m(dω/dt × r) = Euler force [only if ω is changing; = 0 if uniform rotation] For UNIFORM rotation (dω/dt = 0): m·a' = F − 2m(ω × v') − mω×(ω×r) = F + F_Coriolis + F_centrifugal
Centrifugal Force:
F_centrifugal = −mω×(ω×r) = mω²r_⊥ (directed radially outward from rotation axis) where r_⊥ is the perpendicular distance from the rotation axis.
Coriolis Force:
F_Coriolis = −2m(ω × v') Perpendicular to both ω and v'. Causes deflection of moving objects in the rotating frame (e.g., rightward in Northern Hemisphere for large-scale winds).
Part (b) – Why is Centrifugal Force "Fictitious"?
The centrifugal force is called fictitious (or pseudo) because it has no physical origin — there is no agent exerting it. It arises purely from the mathematics of describing motion in a non-inertial (rotating) frame. An observer in an inertial frame sees no such force; instead they see the object following a curved path due to real forces (e.g., tension, gravity). The centrifugal force is a mathematical correction that makes Newton's laws appear to work in the rotating frame. It disappears the moment you switch to an inertial frame.

Example: In a car going around a bend, you feel pushed outward. But an observer on the sidewalk sees you simply tending to go straight (inertia) while the car turns. There is no outward-pushing agent — it is fictitious.

Foucault Pendulum
The Foucault Pendulum (Léon Foucault, 1851) is a long pendulum free to swing in any vertical plane. Because the Earth rotates underneath it, the pendulum's plane of swing appears to rotate slowly over time — direct evidence that Earth is a rotating (non-inertial) frame.

The rotation of the swing plane is caused by the Coriolis effect on the pendulum bob. The rate of precession depends on latitude λ:

Ω_precession = ω_Earth · sin(λ) where ω_Earth = 7.27 × 10⁻⁵ rad/s At the North Pole (λ = 90°): precession rate = ω_Earth (full rotation in 24 hours) At the equator (λ = 0°): precession rate = 0 (no precession) At Lagos (λ ≈ 6.5°N): Ω ≈ 7.27×10⁻⁵ × sin(6.5°) ≈ 8.24×10⁻⁶ rad/s

The pendulum's plane appears to rotate clockwise in the Northern Hemisphere and counterclockwise in the Southern Hemisphere — a vivid proof of Earth's rotation.

Q12 Critical Velocity, Orbital Velocity, Angular Momentum; Tractor Momentum 6 marks ▼
(a) With the help of relevant equations, explain: (i) critical velocity (ii) orbital velocity (iii) orbital angular momentum (iv) classical angular momentum. (b) Calculate the momentum of a 2.38 × 10³ kg tractor travelling at 12.0 m/s in the negative x-direction.
Part (a)(i) – Critical Velocity
Critical velocity (or escape velocity) is the minimum speed needed for an object to escape a gravitational field completely (reach infinity with zero kinetic energy remaining).
Energy conservation: ½mv² − GMm/r = 0 (at infinity, KE = PE = 0) v_escape = √(2GM/r) For Earth: v_escape = √(2 × 6.674×10⁻¹¹ × 5.97×10²⁴ / 6.371×10⁶) ≈ 11.2 km/s ≈ 11,200 m/s
Part (a)(ii) – Orbital Velocity
Orbital velocity is the speed needed to maintain a circular orbit at radius r. At this speed, the gravitational force provides exactly the centripetal acceleration.
GMm/r² = mv²_orb/r v_orbital = √(GM/r) For low Earth orbit (r ≈ R_Earth = 6.371×10⁶ m): v_orbital = √(6.674×10⁻¹¹ × 5.97×10²⁴ / 6.371×10⁶) ≈ 7.9 km/s Note: v_escape = √2 · v_orbital
Part (a)(iii) – Orbital Angular Momentum
Orbital angular momentum L is the angular momentum of a body due to its orbital motion around a centre of force. For a planet of mass m at distance r moving at speed v:
L = m·r × v (vector) or L = mvr·sin(θ) For a circular orbit: L = mvr = m·r·√(GM/r) = m√(GMr) L is conserved for central forces (Kepler's Second Law follows from this).
Part (a)(iv) – Classical Angular Momentum
Classical angular momentum is the general angular momentum of any rigid body or particle rotating about an axis:
L = r × p = r × mv (for a point particle) L = Iω (for a rigid body, where I = moment of inertia) It is conserved when no external torque acts: dL/dt = τ_net = 0
Part (b) – Momentum of Tractor
Mass: m = 2.38 × 10³ kg = 2380 kg Velocity: v = −12.0 m/s (negative x-direction) Momentum: p = mv = 2380 × (−12.0) = −28,560 kg·m/s p = −2.856 × 10⁴ kg·m/s (in the negative x-direction)
p = −2.856 × 10⁴ kg·m/s (directed in the −x direction)
Q13 Four Types of Orbits 6 marks ▼
With the aid of relevant equations, explain any four types of orbits.

The type of orbit is determined by the total mechanical energy E and the eccentricity e of the orbit. For a central inverse-square gravitational force:

General orbit (conic section): r = l / (1 + e·cos θ) where l = L²/(mGMm) is the semi-latus rectum, and e = √(1 + 2EL²/(m(GMm)²)) is the eccentricity.
1. Circular Orbit (e = 0, E < 0)
The orbit is a perfect circle of radius r. Occurs when the kinetic energy exactly provides the centripetal force needed at all points. Total energy E = −GMm/2r.
r = constant, v = v_orbital = √(GM/r), E = −½GMm/r Example: Geostationary satellites, approximately the Moon's orbit around Earth.
2. Elliptical Orbit (0 < e < 1, E < 0)
The body is bound to the central mass and follows an ellipse with the centre of force at one focus. The body has a closest point (periapsis) and farthest point (apoapsis).
Semi-major axis: a = −GMm/(2E) r_min (periapsis) = a(1−e), r_max (apoapsis) = a(1+e) Period: T² = 4π²a³/(GM) [Kepler's Third Law] Example: All planets, Comet Halley (e ≈ 0.967).
3. Parabolic Orbit (e = 1, E = 0)
The body has exactly the escape velocity at every point. It reaches infinity with zero remaining kinetic energy. This is the boundary between bound and unbound orbits.
v = v_escape = √(2GM/r) at every r r = l/(1 + cos θ) Total energy E = 0 Example: A comet just barely passing through the solar system without being captured.
4. Hyperbolic Orbit (e > 1, E > 0)
The body has more than escape velocity and follows a hyperbolic path. It passes near the central mass once and flies off to infinity with nonzero kinetic energy remaining.
E > 0, v > v_escape at all points r = l/(1 + e·cos θ), θ limited to |θ| < arccos(−1/e) Example: Interstellar object Oumuamua, unbound comets, gravitational slingshot trajectories.
Summary: Circle (e=0) → Ellipse (0<e<1) → Parabola (e=1) → Hyperbola (e>1). Energy E<0: bound; E=0: marginally unbound; E>0: fully unbound.
Practice Questions covering all major topics in MAT 316. Click "Show Hint" to reveal the approach for each question before attempting the full solution.

SAMPLE Q1 · Gravitation & Newton's Law

Newton's LawSuperposition

Three masses are placed along the x-axis: m₁ = 5.0 kg at x = 0, m₂ = 3.0 kg at x = 0.4 m, and m₃ = 8.0 kg at x = 0.8 m. (a) Find the net gravitational force on m₁ due to m₂ and m₃. (b) In which direction does m₁ accelerate? (c) State the principle you used to combine forces.

F = Gm₁mᵢ/rᵢ². Treat each force separately, choose directions carefully (m₂ pulls m₁ in +x, m₃ pulls m₁ in +x too). Net force = vector sum. Apply superposition principle. a = F_net/m₁.

SAMPLE Q2 · Kepler's Laws & Orbital Period

Kepler's 3rd LawOrbital Mechanics

Mars orbits the Sun at a mean distance of 1.524 AU, while Earth orbits at 1.0 AU. (a) Using Kepler's Third Law, calculate the orbital period of Mars in Earth years. (b) State the physical meaning of Kepler's Second Law and explain what it implies about the speed of Mars at different points in its orbit. (c) What type of orbit does Mars follow, and what is its eccentricity approximately? (e_Mars ≈ 0.093)

T² ∝ a³ → T_Mars²/T_Earth² = (a_Mars/a_Earth)³. T_Earth = 1 yr. Second law: equal areas in equal times ↔ conservation of angular momentum. Mars is fastest at perihelion, slowest at aphelion. e = 0.093 → elliptical.

SAMPLE Q3 · Escape Velocity & Orbital Energy

Escape VelocityEnergy Conservation

(a) Derive from first principles the expression for the escape velocity from a planet of mass M and radius R. (b) The Moon has mass M_moon = 7.34×10²² kg and radius R_moon = 1.74×10⁶ m. Calculate the escape velocity from the Moon's surface. (c) How does this compare to the escape velocity from Earth? What does this say about why the Moon has no atmosphere?

Set KE = PE at surface: ½mv² = GMm/R → v_esc = √(2GM/R). Compute for Moon. Compare with v_esc(Earth) = 11.2 km/s. Low escape velocity means gas molecules can reach it and escape → no atmosphere.

SAMPLE Q4 · Two-Body Problem & Reduced Mass

Reduced MassCM Frame

(a) Define the reduced mass μ of a two-body system and explain its physical significance. (b) A binary star system consists of two stars with masses m₁ = 2.0 × 10³⁰ kg and m₂ = 6.0 × 10³⁰ kg, orbiting their common centre of mass with a separation of 4.0 × 10¹¹ m. Calculate: (i) the reduced mass μ, (ii) the gravitational force between them, (iii) the orbital period using Kepler's Third Law.

μ = m₁m₂/(m₁+m₂). F = Gm₁m₂/r². For period: T² = 4π²a³/(G·M_total) where M_total = m₁+m₂ and a = separation for circular orbit.

SAMPLE Q5 · Rotating Frame & Coriolis Effect

Coriolis ForceRotating Frame

(a) Write Newton's second law of motion as seen by an observer in a uniformly rotating frame. Identify each term clearly. (b) A 200 g ball is thrown northward at 15 m/s at a latitude of 30°N. Estimate the magnitude and direction of the Coriolis force acting on it. (ω_Earth = 7.27 × 10⁻⁵ rad/s) (c) Why do large-scale weather systems rotate counterclockwise in the Northern Hemisphere?

F_Cor = −2m(ω×v). At lat 30°N, vertical component of ω is ω·sin(30°) = ω/2. F_Cor = 2mωv·sin(λ). Direction: perpendicular to v, rightward in NH. Low pressure → air flows in, Coriolis deflects it right → counterclockwise spiral.

SAMPLE Q6 · Foucault Pendulum

Foucault PendulumEarth's Rotation

(a) Describe the Foucault Pendulum experiment and explain what it demonstrates. (b) Derive the expression for the precession rate Ω = ω_Earth·sin(λ). (c) At what latitude would the pendulum's plane complete a full 360° rotation in exactly 48 hours? (d) What would you observe if a Foucault Pendulum were set up exactly at the equator?

Ω = ω_Earth·sin(λ). For 48 hrs period: Ω = 2π/(48×3600) rad/s → sin(λ) = Ω/ω_Earth → λ. At equator: sin(0°) = 0 → no precession, plane does not rotate.

SAMPLE Q7 · Orbit Classification & Apses

Orbit TypesApses

(a) An object moves in a gravitational field with total energy E = −4.0 × 10⁹ J, angular momentum L = 2.0 × 10¹⁵ kg·m²/s, and reduced mass μ = 500 kg. (i) What type of orbit does it follow? (ii) Find the eccentricity e. (iii) Find the distances at periapsis and apoapsis if GM = 4×10¹⁴ m³/s². (b) What are apses? How many apses does each type of orbit have?

E < 0 → bound → elliptical. e = √(1 + 2EL²/(μ(GMμ)²)). a = −GMμ/(2E). r_min = a(1−e), r_max = a(1+e). Apses: circle=infinite, ellipse=2, parabola=1, hyperbola=1.

SAMPLE Q8 · Momentum, Zero-Momentum Frame & Collisions

MomentumCM FrameCollisions

A 1200 kg car moving at 20 m/s east collides with a 800 kg car moving at 15 m/s west. (a) Find the total momentum in the lab frame. (b) Find the velocity of the zero-momentum (CM) frame. (c) In a perfectly inelastic collision, what is the final velocity? (d) Calculate the kinetic energy lost. Show that this equals the total KE in the CM frame before collision.

p_total = m₁v₁ + m₂v₂ (taking east = +). V_CM = p_total/(m₁+m₂). In perfectly inelastic: V_f = V_CM. ΔKE = KE_before − KE_after. KE_CM = ½μv_rel² where μ = m₁m₂/(m₁+m₂), v_rel = v₁−v₂.
Essential Formula Sheet for MAT 316 Mechanics. All the key results you need for the exam.
§1 Gravitation & Newton's Law ▼
Newton's Law of Gravitation: F = G·m₁·m₂ / r² G = 6.674 × 10⁻¹¹ N·m²/kg² Gravitational field: g = GM/r² (at distance r from mass M) Weight (surface): W = mg, g_surface = GM/R² Superposition: F_net = Σᵢ Fᵢ (vector sum of all individual forces)
§2 Two-Body Problem & Reduced Mass ▼
Total mass: M = m₁ + m₂ Reduced mass: μ = m₁m₂/(m₁+m₂) Centre of mass: R = (m₁r₁ + m₂r₂)/M Relative coord: r = r₁ − r₂ Lagrangian: L = ½MṘ² + [½μṙ² − U(r)] (separable!) Angular momentum (conserved): l = μr²θ̇ = constant Total energy: E = ½μṙ² + l²/(2μr²) + U(r) = ½μṙ² + U_eff(r) where U_eff = l²/(2μr²) + U(r)
§3 Kepler's Laws & Orbital Mechanics ▼
Kepler's 1st Law: Orbits are conic sections (ellipses for bound orbits) Kepler's 2nd Law: dA/dt = l/(2μ) = constant (equal areas in equal times) Kepler's 3rd Law: T² = (4π²/GM)·a³ Orbit equation: r = l/(1 + e·cosθ) l = L²/(μ²GM) Eccentricity: e = √(1 + 2EL²/(μ(GMμ)²)) Type of orbit: e = 0: Circle (E = −GMμ/2r) 0 1: Hyperbola (E > 0) Semi-major axis: a = −GMμ/(2E) r_periapsis = a(1−e), r_apoapsis = a(1+e) Orbital velocity: v_orb = √(GM/r) Escape velocity: v_esc = √(2GM/r) = √2 · v_orb
§4 Rotating Frames & Fictitious Forces ▼
Time derivative in rotating frame: (dA/dt)_in = (dA/dt)_rot + ω × A Newton's Law in rotating frame: m·a' = F_real − 2m(ω×v') − mω×(ω×r) − m(ω̇×r) Centrifugal force: F_cf = −mω×(ω×r) = mω²r⊥ [outward from axis] Coriolis force: F_Cor = −2m(ω×v') [perpendicular to v'] Euler force: F_Euler = −m(ω̇×r) [zero for uniform rotation] Foucault Pendulum precession rate: Ω = ω_Earth · sin(λ) [λ = latitude] ω_Earth = 7.27 × 10⁻⁵ rad/s T_Earth = 24 hours = 86400 s
§5 Momentum, Energy & Collisions ▼
Linear momentum: p = mv Angular momentum: L = r × p = Iω CM frame velocity: V_CM = Σmᵢvᵢ / Σmᵢ = p_total/M KE in CM frame: KE_CM = ½μv_rel² v_rel = v₁ − v₂ Perfectly inelastic collision: V_final = V_CM (objects stick together) Maximum KE loss = KE_CM (all kinetic energy in CM frame is lost) Elastic collision in CM frame: speeds unchanged, directions reversed Conservation laws: Momentum: Σmᵢvᵢ = constant (no external forces) Energy: E = KE + PE = constant (conservative forces) Angular mom.: L = Σrᵢ × pᵢ = constant (no external torque)