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4.1.4Theme A — Space, Time and Motion

Kepler’s Laws of Planetary Motion — IB Physics SL Revision Notes & Topical Questions

5 min read 5 parts Lesson 41 of 41 Unit quiz

📚 IB Physics SL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 5The Three Laws

Kepler summarised the motion of the planets in three laws that were later explained by gravity.

First law
  • First law Each planet moves in an ellipse with the Sun at one of the two foci.
Second law
  • Second law A line from the Sun to a planet sweeps out equal areas in equal times, so planets move faster when nearer the Sun.
Third law
  • Third law The square of the orbital period is proportional to the cube of the average orbital radius, T² ∝ r³.
Scope
  • Scope The same laws describe moons orbiting planets and satellites orbiting the Earth.

Part 2 of 5Deriving the Third Law

For a circular orbit the gravitational force provides the centripetal force.

1
Equate the gravitational force to the centripetal force: GMm/r² = mv²/r, where M is the mass of the central body.
2
The orbital speed for one complete circle is v = 2πr/T, where T is the period.
3
Substitute and cancel m and one factor of r to get GM/r² = 4π²r/T².
4
Rearrange to obtain T² = (4π²/GM) r³, so T² / r³ is a constant for all bodies orbiting the same central mass.
5
The mass of the orbiting body cancels, so the period does not depend on the mass of the satellite.

Part 3 of 5Applications

Applications
Comparing planets
  • Comparing planets Ratios let one orbit be found from another, such as T₂/T₁ = (r₂/r₁)1.5.
Mass of central body
  • Mass of central body Measuring the period and radius of a moon gives M = 4π²r³/GT².
Geostationary orbit
  • Geostationary orbit A satellite with a 24-hour period orbits above the equator and stays above one point on Earth.
Ellipses
  • Ellipses Orbits are slightly elliptical, though many planetary orbits are almost circular.
PlanetOrbital radius (AU)Period (years)
Earth1.001.00
Mars1.521.87
Jupiter5.2011.9

Part 4 of 5Worked Example

Question

  • Mars has an average orbital radius of 1.52 AU.
  • Use Kepler's third law to find its orbital period in years, given that Earth's radius is 1.00 AU and its period is 1.00 year.

  1. T² ∝ r³, so T₂² / T₁² = (r₂ / r₁)³.
  2. T₂² = 1.00² × (1.52)³ = 3.51.
  3. T₂ = √3.51 = 1.87 years.

Answer The orbital period of Mars is about 1.87 years.

Part 5 of 5Exam Checklist & Tips

4.1.4 – Key Points for the Exam

  • Orbital period The time taken for one complete orbit.
  • Ellipse A closed oval curve with two foci.
  • Kepler's third law T² is proportional to r³ for bodies orbiting the same central mass.
  • Geostationary orbit An equatorial orbit with a period of 24 hours.

Common Mistakes

  • Using T ∝ r³ rather than T² ∝ r³.
  • Putting the Sun at the centre of the ellipse rather than at one focus.
  • Thinking a heavier satellite orbits more quickly; period does not depend on its mass.
  • Mixing units, such as using the period in years with the radius in metres.

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Physics 0625 & 9702 SUVAT Mastery Graph Interpretation

Constant Acceleration Equations (SUVAT) & Slopes

v = u + at  |  s = ut + ½at2  |  v2 = u2 + 2as  |  s = ½(u + v)t

  • Displacement-Time Graph: Gradient = Velocity (curve = accelerating).
  • Velocity-Time Graph: Gradient = Acceleration; Area under curve = Distance travelled / Displacement.
  • Free Fall in vacuum: a = g ≈ 9.81 m/s2 downwards (air resistance negligible).
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