Circular motion, satellites & Kepler's law

Learning goals

What students should be able to explain, calculate and investigate after completing Circular Motion, Satellites and Kepler's Third Law.

Goals marked Higher Level are required only for Higher Level students; the remaining goals are for all students.

• Higher Level: Explain centripetal force and state why it is needed for circular motion.
• Higher Level: Use F = mv²/r to solve problems involving objects moving in a circle.
• Higher Level: Explain how gravity provides the centripetal force for satellite and planetary orbits.
• Higher Level: State and apply Kepler's Third Law.
• Distinguish near-Earth orbits from geostationary orbits and relate orbit type to satellite use.

Key definitions

Precise terms, symbols and SI units used in Circular Motion, Satellites and Kepler's Third Law.

Centripetal Force and Acceleration

An object moving at constant speed in a circle is accelerating because its direction is changing.

In circular motion, the velocity is always along the tangent to the circle. Even when the speed stays constant, the direction of the velocity is changing, so the object is accelerating. This acceleration points towards the centre of the circle. From F = ma, the resultant force must also point towards the centre. The inward resultant may be supplied by tension, friction, gravity or a normal reaction; it is not an extra force.

• The acceleration is directed towards the centre of the circle.
• The resultant force is also directed towards the centre.
• F = mv²/r

Gravity as the Centripetal Force for Orbits

For a satellite or planet in a stable circular orbit, gravity provides the inward resultant force needed for circular motion.

Setting these two expressions for F equal to each other allows the orbital speed, period, or radius to be calculated for a given central mass. This is strong evidence supporting Newton's law of gravitation: the same inverse-square gravitational force law that describes objects falling on Earth also correctly predicts the motion of orbiting satellites and planets.

Kepler's Third Law

Kepler's Third Law states that for any set of objects orbiting the same central mass, the square of the orbital period is proportional to the cube of the orbital radius.

This can be derived by combining the gravitational force equation with the centripetal force equation for circular orbits, giving T² = 4π²R³/(GM). This means more distant orbits (larger R) always have longer periods (larger T) — for example, outer planets in the solar system take much longer to orbit the Sun than inner planets.

• The displayed form applies to a circular orbit around a central mass M when the orbiting mass is much smaller than M.

Near-Earth and Geostationary Orbits

A near-Earth orbit is at relatively low altitude, giving a short orbital period (e.g. the International Space Station orbits roughly every 90 minutes).

For orbital calculations, distinguish altitude h above Earth’s surface from orbital radius r measured from Earth’s centre. They are related by r = R_E + h. Gravity supplies the centripetal acceleration, so the satellite is continuously falling around Earth. A geostationary orbit is a special circular, equatorial, prograde orbit whose period matches Earth’s rotation, so the satellite remains above the same longitude.

• A geostationary satellite has a 24-hour period and orbits above the equator in Earth’s direction of rotation.
• It remains above the same point on Earth and is useful for continuous communication coverage.
• A lower circular orbit has a shorter period and greater orbital speed.

• h is measured from Earth’s surface.
• r is measured from Earth’s centre and is the distance used in gravitational and orbital formulae.
• Always convert h to r using r = R_E + h before substituting into F = GMm/r² or orbital-period relationships.

Diagrams and formulae

Key diagrams and equations for reviewing Circular Motion, Satellites and Kepler's Third Law.

• Circular motion: the velocity is always tangential to the circle, while the centripetal force (and acceleration) always points towards the centre.
• Near-Earth orbits have relatively low altitude and short periods. A geostationary satellite follows a circular equatorial orbit in the direction of Earth's rotation, with a period of approximately 24 hours, so it remains above the same point on the equator.

Circular motion: the velocity is always tangential to the circle, while the centripetal force (and acceleration) always points towards the centre.

Near-Earth orbits have relatively low altitude and short periods. A geostationary satellite follows a circular equatorial orbit in the direction of Earth's rotation, with a period of approximately 24 hours, so it remains above the same point on the equator.

Worked Example

A step-by-step calculation applying the central relationships in Circular Motion, Satellites and Kepler's Third Law.

A satellite of mass 800 kg orbits the Earth at a radius of 7 × 10⁶ m with an orbital speed of 7.5 × 10³ m s⁻¹. Calculate the centripetal force acting on it.

• F = mv²/r
• F = 800 × (7.5 × 10³)² ÷ (7 × 10⁶)
• F = 800 × 5.625 × 10⁷ ÷ 7 × 10⁶ ≈ 6429 N

Practical / Investigation

Experimental planning, measurement and evaluation for Circular Motion, Satellites and Kepler's Third Law.

Investigation: Verifying Kepler's Third Law using secondary data

• Task: using secondary data (e.g. orbital radius and period for the planets of the solar system, or for several moons of Jupiter, from a reliable source), calculate T² and R³ for each and plot T² (y-axis) against R³ (x-axis).
• A straight line through the origin confirms Kepler's Third Law, T² ∝ R³; the slope of this line equals 4π²/(GM) for the central mass being orbited.
• Nature of Science link: this is a good example of using large secondary datasets to verify a physical law, and of how a slope can be used to extract a physical constant (here, related to the central mass M) from real data.

• Use bodies orbiting the same central mass and use one consistent set of units for orbital period T and orbital radius R.
• Calculate T² and R³, then plot T² on the vertical axis against R³ on the horizontal axis. A straight line through the origin supports T² ∝ R³.
• Quote the source and uncertainty/precision of the data. Do not mix planets orbiting the Sun with moons orbiting a planet on one graph.

Quick-Check Questions

Short retrieval questions covering the essential ideas in Circular Motion, Satellites and Kepler's Third Law.

1. Why is an object moving in a circle at constant speed still accelerating?

2. State the formula for centripetal force.

3. What provides the centripetal force for a satellite orbiting Earth?

4. State Kepler's Third Law.

5. Why is a geostationary orbit useful for communications satellites?

Common mistakes and exam tips

Common misconceptions and exam-technique checks for Circular Motion, Satellites and Kepler's Third Law.

• Avoid: Adding a separate “centripetal force” to the real forces. Correct idea: Centripetal force is the centre-directed resultant supplied by real forces such as gravity, tension, friction or a normal force.
• Avoid: Thinking constant speed means zero acceleration in circular motion. Correct idea: Velocity changes direction continuously, so the object has acceleration towards the centre.
• Avoid: Using height above a planet as the orbital radius. Correct idea: In circular-motion and orbit equations, r is measured from the centre of the circular path or orbited body.

• Identify the real force or resultant that points towards the centre.
• For a circular orbit, equate gravitational force to the required centripetal force.
• Measure orbital radius from the centre of the Earth or other body.

Exam-style questions

Five exam-style questions with worked answers on Circular Motion, Satellites and Kepler's Third Law.

Higher Level — Required Content

Required Higher Level content for this topic.

• The resultant force is directed towards the centre of the circular path.
• For an orbit, gravity can supply this centripetal force.

For circular orbits around the same central body, equating gravitational and centripetal force and using v = 2πr/T gives T² = 4π²r³/(GM). Use orbital radius measured from the centre.

End-of-unit revision

Formulae, definitions, experiments and mixed revision questions for Circular Motion, Satellites and Kepler's Third Law.

• F = mv²/r (centripetal force)
• a = v²/r (centripetal acceleration)
• T² = 4π²R³/(GM) (Kepler's Third Law)

• Uniform circular motion
• Centripetal force and acceleration
• Geostationary orbit
• Kepler's Third Law

• Circular motion showing tangential velocity and centripetal force/acceleration towards the centre
• Near-Earth vs geostationary orbit comparison

• Verifying Kepler's Third Law using secondary data (T² vs R³ graph)

• Thinking centripetal force is a separate, extra force
• Forgetting constant-speed circular motion still involves acceleration
• Believing in an outward 'centrifugal force'
• Assuming all satellites are geostationary

1. A car of mass 1200 kg turns a corner of radius 25 m at 10 m s⁻¹. Calculate the centripetal force required.

2. Explain why a satellite in a stable circular orbit does not need any engine thrust to maintain its orbit.

3. Two planets orbit the same star; planet A has an orbital radius twice that of planet B. Compare their orbital periods using Kepler's Third Law.

4. State one advantage of a near-Earth orbit over a geostationary orbit for an Earth-observation satellite.

1. F = mv²/r = 1200 × 10² ÷ 25 = 1200 × 100 ÷ 25 = 4800 N.

2. Gravity alone already provides exactly the centripetal force needed for the orbit (F = mv²/r = GMm/r²); no additional force is needed to 'keep it going' in the absence of resistance, since there is effectively no air resistance in orbit to overcome.

3. Since T² ∝ R³, doubling R gives T² ∝ 2³ = 8, so planet A's period is √8 ≈ 2.83 times longer than planet B's.

4. A near-Earth orbit is much closer to the surface, giving higher-resolution images/data and the ability to pass over most of the Earth's surface over time, whereas a geostationary satellite stays fixed above one point and is much further away.

The table below shows which equations from this unit appear in the official State Examinations Commission Formulae and Tables book, and where.

Satellite applications are relevant for all students; quantitative circular motion and Kepler's Third Law are identified as Higher Level content.