Lesson 5: Circular Motion and Gravitation

Motion in a circle is accelerated motion even at constant speed, and the same inverse-square law that holds planets in orbit also describes the gravity field near Earth. This lesson joins Theme A (circular motion) to Theme D.1 (gravitational fields) at SL depth; HL field theory returns in Lesson 22.

Circular motion

For motion at speed v in a circle of radius r: angular speed ω = 2π/T = 2πf; centripetal acceleration a = v²/r = ω²r, directed towards the centre; centripetal force F = mv²/r. "Centripetal" is not a new kind of force — it is the role played by tension, friction, gravity or a normal reaction. The first step in any problem is: identify which real force points towards the centre.

Horizontal circles: a car on a flat curve relies on friction, Fmax = μsN; a conically swinging pendulum relies on the horizontal component of tension. Vertical circles add gravity: at the top, mg and the reaction both point towards the centre, and the minimum speed to stay on the circle satisfies mg = mv²/r.

Worked problem 1 — maximum speed on a curve

A 950 kg car takes a flat curve of radius 45 m. The maximum static friction coefficient between tyres and road is 0.55. Find the maximum speed before sliding.

Required centripetal force is provided by friction: mv²/r ≤ μs mg
vmax = √(μs g r) = √(0.55 × 9.81 × 45) = √242.8 = 16 m s⁻¹ (≈ 56 km h⁻¹)

If the road is wet and μs halves, vmax falls by √2, not by 2 — proportional reasoning of this kind is a Paper 1 favourite.

Newton's law of gravitation

Every mass attracts every other mass with F = GMm/r², where G = 6.67 × 10⁻¹¹ N m² kg⁻² and r is the separation of the centres. The gravitational field strength at distance r from a mass M is g = GM/r²: the force per unit test mass. Near Earth's surface this reproduces 9.81 N kg⁻¹ — verify: g = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)/(6.37 × 10⁶)² = 3.98 × 10¹⁴/4.06 × 10¹³ = 9.81 ✓. For a circular orbit, gravity provides the centripetal force: GMm/r² = mv²/r, so v = √(GM/r) and Kepler's third law follows: T² = 4π²r³/(GM).

Worked problem 2 — the Moon's orbital speed

The Moon orbits at r = 3.84 × 10⁸ m from Earth's centre; Earth's mass is 5.97 × 10²⁴ kg.

v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 3.84 × 10⁸)
  = √(1.037 × 10⁶) = 1.02 × 10³ m s⁻¹
T = 2πr / v = 2π × 3.84 × 10⁸ / 1018 = 2.37 × 10⁶ s ≈ 27.4 days ✓ (the sidereal month)

An answer that reproduces a known quantity is a strong self-check in the exam.

Practical worked example — conical pendulum determination of g

A string of length L = 0.850 ± 0.005 m swings a bob in a horizontal circle; the period is measured for several cone angles. Plotting T² against the vertical height h of the cone gives a straight line through the origin with gradient 4π²/g — because for a conical pendulum T = 2π√(h/g). A measured gradient of 4.05 ± 0.06 s² m⁻¹ gives g = 4π²/4.05 = 9.75 m s⁻² with an uncertainty of 1.5% from the gradient alone: g = 9.75 ± 0.15 m s⁻². Using h instead of L is the subtle point most students miss, and it is exactly the kind of linearisation Lesson 18 develops.

Nature of Science

Newton's gravitation was revolutionary because one law explained falling apples and orbiting planets — unification. Its prediction of Neptune's position from irregularities in Uranus's orbit (1846) is a classic case of a theory making a risky, testable prediction and being confirmed.

Real-world application

Geostationary satellites must orbit in the equatorial plane with T = 24 h. Setting T² = 4π²r³/(GM) gives r = 4.22 × 10⁷ m from Earth's centre — an altitude of 3.58 × 10⁷ m — which is why satellite TV dishes point at a fixed patch of sky and why there is only one geostationary "ring" to allocate.

Common mistakes (examiner style)

• Measuring r from the surface instead of the centre of the planet.
• In vertical circles, forgetting gravity at the top or adding it at the side position.
• Using degrees in ω = 2π/T calculations or confusing ω (rad s⁻¹) with f (Hz).
• Writing g = GM/r² with r in kilometres.
• Believing astronauts in orbit are "beyond gravity": at ISS altitude g is still ≈ 8.7 N kg⁻¹; they are in continuous free fall.

IB Exam Experience

Paper 1-style questions. 1. An object moves in a horizontal circle at constant speed. Which quantity is constant? A. velocity B. momentum C. kinetic energy D. acceleration. Key: C — speed constant means Ek constant; velocity, momentum and acceleration all change direction continuously. 2. A planet's orbit radius is quadrupled. Its orbital period increases by a factor of: A. 4 B. 8 C. 16 D. 64. Key: B — T ∝ r^1.5, so 4^1.5 = 8. 3. The gravitational field strength at a distance 2R from a planet's centre is g. At the surface it is: A. g/4 B. g/2 C. 2g D. 4g. Key: D — inverse square: halving the distance quadruples g.

Paper 2-style structured question (7 marks). A 0.12 kg ball on a 0.90 m string is swung in a vertical circle. At the top of the circle the ball's speed is 6.0 m s⁻¹. (a) Calculate the tension in the string at the top. [3] (b) Calculate the minimum speed at the top for the string to remain taut. [2] (c) At the bottom the speed is 9.5 m s⁻¹. Calculate the tension there. [2] *Marking notes:* (a) At the top both weight and tension point to the centre: T + mg = mv²/r → T = 0.12 × 36/0.90 − 0.12 × 9.81 = 4.80 − 1.18 = 3.6 N. (b) Set T = 0: v = √(gr) = √(9.81 × 0.90) = 3.0 m s⁻¹. (c) At the bottom tension points to the centre, weight away: T − mg = mv²/r → T = 0.12 × 90.25/0.90 + 1.18 = 12.03 + 1.18 = 13.2 N. The asymmetry between (a) and (c) is the point of the question; candidates who give equal tensions lose both marks.

IA idea: Measure the orbital period of a simulated satellite (tethered ball on a force sensor) as a function of radius and test Kepler's third law with a log–log plot.

Banked tracks: circular motion without friction

On a track banked at angle θ, the normal reaction has a horizontal component towards the centre. With no friction needed at the design speed: N sin θ = mv²/r and N cos θ = mg, dividing gives tan θ = v²/(rg). For r = 45 m and a design speed of 20 m s⁻¹, tan θ = 400/(45 × 9.81) = 0.906, so θ = 42° — steep, which is why velodromes and bobsleigh runs look so dramatic. Below the design speed the car tends to slide *down* the bank (friction acts up the slope); above it, friction acts down the slope. Explaining these friction reversals with diagrams is a standard HL-style discussion at SL-accessible physics.

How g changes away from the surface

At altitude h, g = GM/(R + h)². At the ISS (h ≈ 400 km, so r = 6.77 × 10⁶ m): g = 3.98 × 10¹⁴/(6.77 × 10⁶)² = 3.98 × 10¹⁴/4.58 × 10¹³ = 8.7 N kg⁻¹ — nearly 90% of the surface value. Astronauts float not because gravity is absent but because station, crew and contents all fall together around the Earth. Below the surface (inside a uniform planet), g decreases linearly with depth, g ∝ r, because only the mass inside your radius pulls net-ward (the shell outside cancels — Newton's shell theorem).

Deriving Kepler's third law — and using it backwards

Setting gravity equal to the centripetal requirement for a circular orbit, GMm/r² = m(2π/T)²r, gives T² = 4π²r³/(GM). Read left to right it predicts periods; read right to left it weighs the central body: M = 4π²r³/(GT²). Every mass ever quoted for a planet, star or black hole comes from watching something orbit it. The moons of Jupiter (Galileo's observations) give Jupiter's mass; stars orbiting the galactic centre give the Milky Way's; the 4-million-solar-mass black hole at our centre was weighed this way from a single star's 16-year orbit.

Why the Moon keeps one face to us

The Moon rotates once per orbit (tidal locking) because Earth's gravity raised tides in the once-molten Moon, and the friction of those tides dissipated rotational energy until the same hemisphere always faced Earth — energy loss driving a dynamical system into its minimum-energy configuration for its (conserved) angular momentum. The Earth is slowing the same way: the day lengthens by about 2.3 ms per century and the Moon recedes 3.8 cm per year (measured by laser ranging to Apollo retroreflectors) as angular momentum transfers from Earth's spin to the Moon's orbit. Circular motion, gravitation and energy — this lesson's three strands — acting over geological time.

Geosynchronous versus geostationary: a precision point

Any orbit with a 24-hour period is geosynchronous, but only a circular orbit in the equatorial plane is geo*stationary* — appearing fixed in the sky. An inclined geosynchronous satellite traces a figure-of-eight (analemma) in the sky each day as its latitude oscillates. Polar orbits, used for Earth observation, exploit a different frame trick: the planet rotates beneath a fixed orbital plane, so a satellite at ~800 km with a ~100 minute period eventually passes over every point on the surface — global coverage from a single spacecraft, at the price of never lingering over one place.