D.1 · IB Physics
Kepler’s third law
Planets on circular orbits around one star. Gravity provides the centripetal force, which gives v = √(GM/r) and T² = (4π²/GM)r³, so every planet lies on one straight line of T² against r³ and a heavier star makes that line less steep.
Predict first
Start from the default settings (1.00 M☉ star, P at 2.00 AU). If you move planet P out to 8.00 AU, its orbital period becomes…
T² ∝ r³, so T ∝ r^1.5. Quadrupling r multiplies T by 4^1.5 = 8 (2.83 yr becomes 22.6 yr).
Check it in the model: Move the Orbit radius of planet P slider from 2.00 to 8.00 and compare Period of P in the readout.
planets your planet P star
T = 2πr ÷ v → T² = (4π²/GM) r³
AU, years, M☉: T² = r³ ÷ M
G = 6.67 × 10⁻¹¹ N m² kg⁻²
1 AU = 1.50 × 10¹¹ m, M☉ = 1.99 × 10³⁰ kg
Try this
- Move planet P from 1.00 AU to 4.00 AU around a 1.00 M☉ star. By what factor does the period grow, and why is it not 4?
- Find the star mass that makes a planet at 1.00 AU orbit in 0.707 years, using the gradient and the T²/r³ readout.
- Switch to the log–log graph and the outer planets. Read the gradient and explain why it is 1.5 whatever the star mass.
Common error: Thinking an outer planet moves faster because its circle is bigger: its speed is lower (v ∝ 1/√r), its period is longer (T² ∝ r³), and the planet’s own mass cancels from every equation.