Orbits of planets and satellites (A-level only) — AQA A-Level Physics

This topic explores how gravitational force governs the motion of planets and satellites in circular and elliptical orbits. You will learn to apply Newton's law of gravitation and centripetal force equations to derive relationships such as orbital speed, period, and Kepler's third law. Understanding these principles is essential for explaining phenomena like geostationary satellites, polar orbits, and the motion of planets in our solar system.

In the wider context of A-Level Physics, this topic bridges mechanics and fields, reinforcing concepts of circular motion and gravitational fields. It also has practical applications in satellite technology, space exploration, and astronomy, making it a rich area for exam questions that test both calculation and explanation skills.

• →Gravitational force provides the centripetal force for a satellite in circular orbit: GMm/r^2 = mv^2/r, leading to v = sqrt(GM/r).
• →Kepler's third law: T^2 is proportional to r^3 for any object orbiting a much more massive body, derived from equating gravitational and centripetal forces.
• →Geostationary satellites have a period of 24 hours, orbit above the equator in the same direction as Earth's rotation, and are used for communication and weather monitoring.
• →Polar orbits are low-altitude orbits that pass over the poles, allowing a satellite to scan the entire Earth's surface as it rotates beneath, useful for mapping and reconnaissance.
• →Total energy of a satellite in orbit is negative and consists of kinetic energy (positive) and gravitational potential energy (negative), with E = -GMm/(2r) for circular orbits.

• States that gravitational force provides the centripetal force, leading to mv²/r = GMm/r² and hence v² = GM/r.
• Derives kinetic energy Eₖ = ½mv² = GMm/(2r) for a circular orbit.
• Uses gravitational potential energy Eₚ = −GMm/r, measured with zero at infinity.
• Adds kinetic and potential energy to obtain total energy E = −GMm/(2r).
• Explains that total energy is negative for a bound orbit and increases towards zero as orbital radius increases.
• Calculates the energy required to move a satellite between circular orbits using the difference in total energy.
• States that escape occurs when the total energy (kinetic plus gravitational potential) of the projectile is zero or greater, so it can reach infinity with zero kinetic energy.
• Derives or quotes the energy condition ½mv² = GMm/r and cancels the mass m to show escape speed is independent of the projectile mass.
• Rearranges correctly to v = √(2GM/r) and substitutes G, planetary mass M and distance r from the centre, not the surface altitude.
• Evaluates a numerical escape speed with consistent SI units, for example about 1.1 × 10⁴ m s⁻¹ for Earth, and states the unit m s⁻¹.
• Explains that escape speed is a minimum speed and that a larger launch speed still allows escape, while a smaller speed leads to a bound orbit or return.
• Uses gravitational potential energy as negative, −GMm/r, and recognises that zero total energy is the boundary between bound and unbound motion.
• Defines a synchronous orbit as one whose orbital period equals the rotation period of the central body, so the satellite repeats its position relative to the surface.
• Equates gravitational force GMm/r² to centripetal force mω²r or mv²/r for a circular orbit and cancels the satellite mass m.
• Uses ω = 2π/T to obtain r³ = GMT²/(4π²) or an equivalent relationship linking orbital radius and period.
• Calculates the orbital radius or altitude for a given period, for example about 4.2 × 10⁷ m from Earth's centre for a 24-hour period.
• Distinguishes geostationary from geosynchronous: geostationary must be equatorial, circular and in the same rotational sense as the planet.
• Explains that only one orbital radius satisfies a chosen synchronous period for a given central mass, so synchronous orbits form a single shell.
• Gravitational force provides the centripetal force: GMm/r² = mv²/r, so v = √(GM/r).
• Period relation: T² = 4π²r³/(GM), so r³ = GMT²/(4π²).
• Low orbit: smaller r gives larger v and shorter T; typical periods are about 90 minutes, used for Earth observation and imaging.
• Geostationary orbit: T = 24 h = 8.64 × 10⁴ s, equatorial plane, same sense as Earth's rotation.
• Radius from centre r ≈ 4.22 × 10⁷ m; height above surface ≈ 3.6 × 10⁷ m.
• Geostationary satellites are used for communication and weather monitoring because the ground antenna can be fixed.

• 💡Start from the force equation mv²/r = GMm/r² whenever you need to link speed, radius and mass.
• 💡Keep the negative sign throughout the potential and total energy calculations, and interpret it as a bound state.
• 💡When calculating an energy change, subtract the initial total energy from the final total energy and state whether energy is absorbed or released.
• 💡Start from the energy condition and show the cancellation of m explicitly; this demonstrates understanding rather than quoting a formula.
• 💡Check that r is measured from the centre of the planet and convert kilometres to metres before calculating.
• 💡Give the final answer to an appropriate number of significant figures and include the unit m s⁻¹, since a speed without a unit is incomplete.
• 💡Convert the period to seconds before substituting into r³ = GMT²/(4π²), since G is in SI base units.
• 💡Show the force equation and the substitution ω = 2π/T so the derivation is clear and the relationship between r and T is visible.
• 💡When asked about geostationary satellites, state all three conditions: equatorial plane, circular orbit and same rotational sense as the planet.
• 💡Start orbital calculations by writing the force equation before substituting numbers, so the physics is clear.
• 💡Convert 24 hours to seconds before using T² = 4π²r³/(GM).
• 💡When asked for height, subtract Earth's radius from the orbital radius and state the unit.
• 💡Use a clear sketch to show r measured from Earth's centre and the equatorial plane.
• 💡Always start orbital calculations by equating gravitational force to centripetal force. Show clear algebraic manipulation before substituting numbers to avoid rounding errors.
• 💡When explaining satellite orbits, use precise language: 'the gravitational force provides the centripetal force' rather than vague statements like 'gravity keeps it in orbit'.
• 💡For 6-mark questions, structure your answer with a clear derivation or explanation, including equations, substitution, and a final statement with units. Use bullet points or numbered steps if it helps clarity.

• Using Eₚ = +GMm/r; the correction is that gravitational potential energy is negative when the zero is at infinity.
• Forgetting the factor of ½ and writing total energy as −GMm/r; the correction is that E = Eₖ + Eₚ = GMm/(2r) − GMm/r = −GMm/(2r).
• Assuming a more distant orbit has more kinetic energy; the correction is that Eₖ decreases as r increases, even though total energy becomes less negative.
• Using the surface altitude above the planet instead of the distance from the planet's centre; correct this by adding the planet's radius to the altitude before substituting into v = √(2GM/r).
• Treating escape velocity as dependent on the mass of the escaping object; correct this by cancelling m in the energy equation, which shows the mass cancels and the speed depends only on M and r.
• Forgetting that gravitational potential energy is negative and using +GMm/r; correct this by writing the potential energy as −GMm/r so that the zero total energy condition gives ½mv² = GMm/r.
• Confusing orbital radius with altitude above the surface; correct this by calculating r from r³ = GMT²/(4π²) and then subtracting the planet's radius to find altitude.
• Assuming any 24-hour orbit is geostationary; correct this by stating that a geostationary orbit must also be equatorial, circular and in the same sense as the planet's rotation.
• Using the satellite's mass in the force equation and failing to cancel it; correct this by cancelling m from both sides, showing the orbital radius depends only on M and T.
• Using the height above the surface as r in GMm/r² or T² = 4π²r³/(GM); correct by adding Earth's radius, 6.37 × 10⁶ m, to the height.
• Treating a geostationary orbit as merely any 24-hour orbit; correct by stating it must also be equatorial and in the same sense as Earth's rotation.
• Assuming a polar low-Earth orbit is geostationary; correct by noting that a polar orbit does not keep a fixed position over the equator.
• Students often think that a satellite in a higher orbit moves faster. In fact, orbital speed decreases with increasing radius (v proportional to 1/sqrt(r)), while the period increases.
• Many believe that geostationary satellites are completely stationary in space. They are only stationary relative to a point on Earth's equator; they are still moving at about 3 km/s.
• Some confuse the terms 'geostationary' and 'geosynchronous'. Geosynchronous satellites have a 24-hour period but can be in an inclined or elliptical orbit, whereas geostationary must be circular, equatorial, and moving eastward.

1. 1Day 1-2: Review circular motion and gravitational force equations. Derive v = sqrt(GM/r) and T^2 proportional to r^3 from first principles.
2. 2Day 3-4: Practice calculations for orbital speed, period, and radius using standard values. Check units and significant figures.
3. 3Day 5-6: Learn the differences between geostationary and polar orbits, including their uses and required conditions. Create a comparison table.
4. 4Day 7-8: Attempt exam-style questions, focusing on 6-mark explanations and derivations. Mark your answers using the mark scheme and note common pitfalls.
5. 5Day 9-10: Review energy of orbiting satellites, including kinetic, potential, and total energy. Practice questions involving energy changes in orbit transfers.

• 📋Calculation of orbital speed, period, or radius using GMm/r^2 = mv^2/r or Kepler's third law. Advice: Rearrange equations before substituting, and keep values in standard form.
• 📋Explanation of geostationary and polar orbits, including their properties and applications. Advice: List all conditions and uses, and link to the satellite's purpose.
• 📋Derivation of Kepler's third law from Newton's law of gravitation and circular motion. Advice: Show each algebraic step clearly and state assumptions (e.g., circular orbit, mass of satellite negligible).
• 📋Analysis of energy changes when a satellite moves between orbits. Advice: Use E = -GMm/(2r) and remember that total energy becomes less negative (increases) as r increases.