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The figure shows a planet P orbiting the Sun S in an ellipse, where AB is the minor axis and CD is the major axis. Let t1 be the time the planet takes along the path ACB and t2 the time along the path BDA. Compare t1 and t2 (assume C is the end of the major axis nearer the Sun, i.e. perihelion side).

1. t1 < t2
2. t1 = t2
3. t1 > t2
4. nothing can be concluded

Correct answer: t1 < t2

Solution

By Kepler's second law, equal areas are swept in equal times, so time spent on a path is proportional to the area swept by the radius vector from the Sun (at a focus). The path ACB passes the perihelion C (Sun side), enclosing a smaller area between the orbit and the focus than the path BDA which passes the aphelion D. Smaller swept area means less time. Hence t1 < t2.

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