D.1 · IB Physics
Orbital motion
A planet on an elliptical orbit sweeps out equal areas in equal times, so it moves fastest at its closest approach to the star. Change the shape and size of the orbit and watch the speed, the swept area and the period.
Predict first
Keep the eccentricity at its default of 0.45. If you double the semi-major axis a from 4 AU to 8 AU, what happens to the orbital period T?
T² = a³, so T = a3/2. Doubling a multiplies T by 23/2 ≈ 2.83: T goes from 41.5 = 8.00 yr to 81.5 ≈ 22.6 yr.
Check it in the model: Set the Semi-major axis a slider to 4.0 AU, then to 8.0 AU, and compare the Period T readout (8 yr, then 22.6 yr).
star planet and velocity empty focus
v² = GM(2 ÷ r − 1 ÷ a)
T² = a³ (T in years, a in AU)
areal speed = ½ r v⊥ = constant
Try this
- Set e = 0 and note the speed. Then raise e to 0.60 and compare the speeds at closest and furthest approach. What is the ratio vmax ÷ vmin and how does it relate to e?
- With the swept-area shading on, watch the shaded sector at perihelion and at aphelion. Its shape changes but the area readout does not. Why?
- Set a = 4 AU, then a = 8 AU, with e unchanged. By what factor does T change when a doubles?
Exam tip: Kepler's second law is conservation of angular momentum, so at the two ends of the orbit rpvp = rava. Use this to link the speeds at the closest and furthest points before reaching for energy.