Kepler's Third Law
An astronomer observes a planet orbiting a Sun-like star at a mean distance of 4 astronomical units, where 1 astronomical unit corresponds to an orbital period of 1 year. According to Kepler's third law, what is this planet's orbital period?
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About This Question
- Subject
- physics
- Chapter
- gravitation
- Topic
- kepler's third law
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
8 years
Kepler's third law states that the square of a planet's orbital period is proportional to the cube of the semi-major axis of its orbit, written as T2∝a3. Using the Earth's orbit as the reference where a=1 AU gives T=1 year, the constant of proportionality becomes one in these units, so T2=a3. For a=4 AU, T2=43=64, giving T=64=8 years. The option of 16 years wrongly squares the distance instead of taking the three-halves power. The option of 64 years mistakes the period for a3 itself, forgetting the square root. The option of 4 years assumes a linear relation between period and distance. This is exactly the NCERT and historical Kepler relationship governing planetary motion, which Newton later derived from his universal law of gravitation combined with the centripetal requirement of circular orbits. The proportionality constant GM4π2 depends only on the mass of the central star, so the identical numerical relationship holds for every planet orbiting that star, which is precisely why expressing distances in astronomical units and periods in years makes the constant equal to one. This universality is what allows astronomers to deduce orbital periods purely from measured distances. A plausibility check confirms that a more distant planet must take longer to orbit, and the T=a3/2 scaling gives a sensible eight-year period for a fourfold-larger orbit.
This medium difficulty physics question is from the chapter gravitation, covering the topic of kepler's third law. It appeared in the 2025 exam.
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