Applications of Circular Motion

Circular motion is not just an abstract physics concept — it governs the motion of satellites, the operation of centrifuges, the design of fairground rides, and much more. In this lesson, we connect the theory to real-world applications that frequently appear in Edexcel exam questions.

Satellites and Orbital Speed

A satellite in a circular orbit has its centripetal force provided by gravity. Setting the gravitational force equal to the centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

Where G is the gravitational constant (6.67 × 10⁻¹¹ N m² kg⁻²), M is the mass of the central body, m is the satellite mass, and r is the orbital radius (measured from the centre of the central body).

Simplifying (the satellite mass cancels):

$v = \sqrt{\frac{GM}{r}}$

Key insight: orbital speed depends only on the orbital radius and the mass of the central body — not on the satellite's mass. A feather and a bowling ball at the same orbital radius orbit at the same speed.

Worked Example 1

Calculate the orbital speed of the International Space Station (ISS), which orbits at approximately 400 km above Earth's surface. (Mass of Earth = 5.97 × 10²⁴ kg, radius of Earth = 6.37 × 10⁶ m)

Orbital radius: $r = 6.37 \times 10^6 + 400{,}000 = 6.77 \times 10^6 \text{ m}$

Orbital speed: $v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.77 \times 10^6}} = \sqrt{\frac{3.98 \times 10^{14}}{6.77 \times 10^6}} = \sqrt{5.88 \times 10^7} = 7670 \text{ m s}^{-1}$

The ISS travels at about 7.7 km s⁻¹ — roughly 27,600 km/h. It completes one orbit in about 92 minutes.

Orbital Period

From v = 2πr/T and v = √(GM/r):

$T = 2\pi\sqrt{\frac{r^3}{GM}}$

This is Kepler's third law: T² ∝ r³.

Orbital Data Table

Object	Central Body	Orbital Radius	Orbital Speed	Period
ISS	Earth	6.77 × 10⁶ m	7 670 m s⁻¹	92 min
Geostationary satellite	Earth	4.22 × 10⁷ m	3 070 m s⁻¹	24.0 h
Moon	Earth	3.84 × 10⁸ m	1 020 m s⁻¹	27.3 days
Mercury	Sun	5.79 × 10¹⁰ m	47 400 m s⁻¹	88 days
Earth	Sun	1.50 × 10¹¹ m	29 800 m s⁻¹	365.25 days
Mars	Sun	2.28 × 10¹¹ m	24 100 m s⁻¹	687 days
Jupiter	Sun	7.78 × 10¹¹ m	13 100 m s⁻¹	11.9 years
Neptune	Sun	4.50 × 10¹² m	5 430 m s⁻¹	165 years

Notice the clear trend: objects further from the central body orbit more slowly but have longer periods. Both relationships follow from Kepler's third law.

Geostationary Orbit

A geostationary satellite has a period of exactly 24 hours (86,400 s), so it stays above the same point on Earth's equator. Using Kepler's third law:

$r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}$

$r = \left(\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 86400^2}{4\pi^2}\right)^{1/3} = 4.22 \times 10^7 \text{ m}$

The geostationary altitude is 42,200 km from Earth's centre, or about 35,800 km above the surface. Communication satellites and weather satellites use this orbit.

Centrifuges

A centrifuge spins samples at high speed to separate components by density. The "effective gravity" experienced by the contents is the centripetal acceleration:

$a = \omega^2 r$

Type	Typical Speed	Radius	Acceleration	Application
Clinical centrifuge	3 000 rpm	0.15 m	1 500g	Separating blood cells from plasma
Lab centrifuge	10 000 rpm	0.10 m	11 200g	Separating cell fragments
Ultracentrifuge	60 000 rpm	0.08 m	320 000g	Separating proteins, DNA
Gas centrifuge (uranium)	50 000 rpm	0.15 m	420 000g	Isotope enrichment

Worked Example 2

A clinical centrifuge with radius 0.15 m spins at 3000 rpm. Calculate the centripetal acceleration.

$\omega = \frac{2\pi \times 3000}{60} = 314 \text{ rad s}^{-1}$ $a = \omega^2 r = 314^2 \times 0.15 = 98{,}600 \times 0.15 = 14{,}800 \text{ m s}^{-2} \approx 1{,}510g$

Red blood cells sediment 1,510 times faster than under normal gravity, separating from plasma in minutes.

Fairground Rides

The Rotor (Gravitron)

In this ride, people stand inside a spinning drum. When the drum spins fast enough, the floor drops away but the riders do not fall. The drum wall pushes them inward (providing centripetal force), and friction between their back and the wall supports their weight.

For the rider not to slide down: friction ≥ weight $\mu N \geq mg$

The normal force from the wall is the centripetal force: N = mω²r

$\mu m\omega^2 r \geq mg$ $\omega \geq \sqrt{\frac{g}{\mu r}}$

Worked Example 3

A Gravitron has radius 3.0 m and μ = 0.40 between riders' clothing and the wall. What minimum angular velocity is needed?

$\omega_{\min} = \sqrt{\frac{9.81}{0.40 \times 3.0}} = \sqrt{\frac{9.81}{1.20}} = \sqrt{8.18} = 2.86 \text{ rad s}^{-1}$

$f = \frac{\omega}{2\pi} = \frac{2.86}{6.28} = 0.455 \text{ Hz}$

That is about 27 rpm — the drum must complete at least 27 revolutions per minute.

Washing Machine Spin Cycle

During the spin cycle, the drum rotates rapidly. Clothes are pressed against the drum wall by the normal force (which provides centripetal force). Water, however, can escape through the small holes in the drum.

The water leaves the drum because it cannot follow the circular path — there is no centripetal force acting on water droplets once they pass through the holes. They fly off tangentially (Newton's first law).

Planetary Motion and Kepler's Third Law

Circular motion provides a simplified model for planetary orbits. The same principles apply:

The Sun's gravity provides the centripetal force
Planets closer to the Sun orbit faster (v = √(GM/r) — smaller r means larger v)
T² ∝ r³ means doubling the orbital radius increases the period by factor 2√2 ≈ 2.83

Worked Example 4: Using Kepler's Third Law

The Earth orbits the Sun at r₁ = 1.50 × 10¹¹ m with T₁ = 365.25 days. Mars orbits at r₂ = 2.28 × 10¹¹ m. What is Mars's orbital period?

$\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}$ $T_2^2 = 365.25^2 \times \left(\frac{2.28}{1.50}\right)^3 = 133{,}408 \times 3.511 = 468{,}300$ $T_2 = 684 \text{ days}$

The actual value is 687 days — close agreement, confirming Kepler's law.

Worked Example 5: Unknown Planet Mass

A satellite orbits an unknown planet with period T = 8000 s at radius r = 4.0 × 10⁶ m. Find the planet's mass.

$T = 2\pi\sqrt{\frac{r^3}{GM}} \implies M = \frac{4\pi^2 r^3}{GT^2}$ $M = \frac{4\pi^2 \times (4.0 \times 10^6)^3}{6.67 \times 10^{-11} \times 8000^2} = \frac{4\pi^2 \times 6.4 \times 10^{19}}{6.67 \times 10^{-11} \times 6.4 \times 10^7}$ $M = \frac{2.528 \times 10^{21}}{4.27 \times 10^{-3}} = 5.92 \times 10^{23} \text{ kg}$

This is close to Mars's actual mass (6.42 × 10²³ kg).

Common Exam Mistakes

Mistake 1: Using the altitude instead of the orbital radius. Orbital radius = planet radius + altitude. Forgetting to add the planet's radius is one of the most frequent errors.

Mistake 2: Thinking heavier satellites orbit slower. The satellite mass cancels — all objects at the same radius orbit at the same speed, regardless of mass.

Mistake 3: Confusing geostationary orbit with geosynchronous orbit. Geostationary must be above the equator in a circular orbit. Geosynchronous has the same period but can be inclined or elliptical.

Mistake 4: In centrifuge problems, using diameter instead of radius in a = ω²r.

Summary

Satellite orbital speed: v = √(GM/r) — independent of satellite mass
Kepler's third law: T² ∝ r³, so T = 2π√(r³/GM)
Geostationary orbit: T = 24 hours, r = 42,200 km from Earth's centre
Centrifuges create large centripetal accelerations (a = ω²r) to separate mixtures
Fairground rides and washing machines rely on centripetal force and the absence thereof
Planetary motion follows the same circular motion equations with gravity as the centripetal force
Always use orbital radius (not altitude) in satellite calculations

A-Level Deep Dive: Applications of Circular Motion

Spec mapping

Edexcel 9PH0 specification, Topic 6 — Further Mechanics addresses the application of circular motion equations to bodies undergoing uniform motion in a circle, including motion under a centripetal force such as gravity in orbital systems and tension in vertical and banked-track problems (refer to the official specification document for exact wording). The applications strand draws explicitly on v = ωr, a = v²/r = ω²r and F = mv²/r, and joins with Topic 5 (gravitational fields) wherever orbital motion is discussed. Although applications appear in both 9PH0 Paper 1 and Paper 2, the synoptic Paper 3 routinely tests them alongside thermodynamics-style data analysis or particle-physics charged-particle motion. The Edexcel formula booklet provides the centripetal expressions and Newton's gravitational law, but candidates are expected to combine them — Kepler's third law is not given and must be derived in answer space.

Worked example with full mark scheme

Question (8 marks):

A satellite of mass m orbits a planet of mass M in a circular orbit of radius r.

(a) By equating the gravitational and centripetal expressions, derive an equation for the orbital period T in terms of G, M and r. (4)

(b) A satellite orbits an unknown planet with period T = 7200 s at radius r = 7.5 × 10⁶ m. Calculate the mass of the planet, giving your answer to three significant figures. (4)

Solution with mark scheme:

(a) Step 1 — equate the forces.

The gravitational pull on the satellite supplies the centripetal force:

$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$

M1 — correct identification that gravity is the centripetal force, written as an equation. A common slip is to set them as a sum or to add a separate "centripetal force"; that loses M1 because it implies a second force exists.

Step 2 — eliminate satellite mass.

The orbiting mass m cancels:

$v^2 = \dfrac{GM}{r}$

A1 — correct elimination, leading to the orbital-speed expression.

Step 3 — substitute for v using the period.

For uniform circular motion, v = 2πr/T, hence v² = 4π²r²/T². Substituting:

$\dfrac{4\pi^2 r^2}{T^2} = \dfrac{GM}{r}$

M1 — correct substitution v = 2πr/T and rearrangement to expose T².

Step 4 — rearrange for T.

$T^2 = \dfrac{4\pi^2 r^3}{GM} \quad\Longrightarrow\quad T = 2\pi\sqrt{\dfrac{r^3}{GM}}$

A1 — correct final form (Kepler's third law). Both T² and T forms are acceptable provided the algebra is shown.

(b) Step 1 — rearrange Kepler's third law for M.

$M = \dfrac{4\pi^2 r^3}{GT^2}$

M1 — correct rearrangement.

Step 2 — substitute and evaluate the numerator and denominator separately.

r³ = (7.5 × 10⁶)³ = 4.219 × 10²⁰ m³. Hence 4π² × r³ = 1.665 × 10²² m³.

GT² = 6.67 × 10⁻¹¹ × (7200)² = 6.67 × 10⁻¹¹ × 5.184 × 10⁷ = 3.458 × 10⁻³ m³ kg⁻¹.

M1 — substitution of correct values into a correctly-rearranged formula.

Step 3 — divide.

$M = \dfrac{1.665 \times 10^{22}}{3.458 \times 10^{-3}} = 4.81 \times 10^{24} \text{ kg}$

A1 — numerical answer correct to 3 s.f. A1 — units (kg) and an answer in standard form. Total: 8 marks (M4 A4).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A geostationary satellite orbits the Earth (M = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m) with a period equal to one sidereal day, T = 86 164 s.

(a) Calculate the orbital radius from the centre of the Earth. (3)

(b) Hence find the altitude above the equator and explain why the orbit must lie in the equatorial plane. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1) — quoting Kepler's third law in the rearranged form r = (GMT²/4π²)^(1/3).
M1 (AO2) — substituting G = 6.67 × 10⁻¹¹, M = 5.97 × 10²⁴, T = 86 164 correctly.
A1 (AO2) — r ≈ 4.22 × 10⁷ m to 3 s.f.

(b)

B1 (AO1) — altitude h = r − R_E ≈ 3.59 × 10⁷ m (≈ 36 000 km).
M1 (AO3) — argument that the centripetal force must point through the Earth's centre, so the orbital plane must contain that centre.
A1 (AO3) — completion: only an equatorial orbit keeps the satellite stationary above one fixed point on the rotating Earth, since any inclined orbit would trace a ground-track that oscillates in latitude.

Total: 6 marks split AO1 = 2, AO2 = 2, AO3 = 2 — a balanced AO load typical of synoptic Paper 2 questions linking Topic 5 (fields) and Topic 6 (further mechanics).

Synoptic links

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