Two planets orbit the Sun in circular orbits, with their radius of orbit as R₁ = R and R₂ = 4R. Ratio of their periods (T₁/T₂) around the Sun will be
- (a)1/16
- (b)1/8
- (c)1/4
- (d)1/2
Correct — B, 1/8. Kepler's third law (the law of periods) states that the square of a planet's orbital period is proportional to the cube of its orbital radius: T² ∝ R³. Taking the ratio, (T₁/T₂)² = (R₁/R₂)³ = (R/4R)³ = (1/4)³ = 1/64. Taking the square root, T₁/T₂ = 1/8. The inner planet (smaller radius) has the shorter period, consistent with the answer.
- (a)1/16 — This is (1/4)², which would follow from wrongly assuming T ∝ R² instead of the T² ∝ R³ law of periods.
- (c)1/4 — This assumes the period scales linearly with radius (T ∝ R), ignoring the 3/2 power that Kepler's third law requires.
- (d)1/2 — This is (1/4)^(1/2), which comes from mistakenly using T² ∝ R; the correct exponent gives (1/4)^(3/2) = 1/8.
Kepler's third law relates orbital period to orbital size: T² = k·R³ for bodies orbiting the same central mass. Equivalently, T ∝ R^(3/2). It follows from Newton's law of gravitation providing the centripetal force for a circular orbit, and it explains why outer planets take far longer to circle the Sun than inner ones.
The examiner tests whether you apply the 3/2 power correctly. Because the radius grows by a factor of 4, the period grows by 4^(3/2) = 8, so the ratio of the smaller to the larger period is 1/8. The tempting wrong answers all come from using the wrong exponent.
- Kepler's third law: T² ∝ R³ (the 'law of periods').
- Equivalently, T ∝ R^(3/2).
- Here R₂/R₁ = 4, so T₂/T₁ = 4^(3/2) = 8, giving T₁/T₂ = 1/8.
- The law follows from gravity supplying the centripetal force in a circular orbit.
Radius up by 4, so period up by 4^(3/2) = 8 — the inner planet's period is 1/8 of the outer's.
- Forgetting the 3/2 power and scaling the period linearly with radius.
- Squaring instead of taking the square root when solving for the ratio of periods.
Asked as a direct ratio problem — set T² ∝ R³, plug in the radius ratio and take the 3/2 power.
No directly related past PYQ was found.
- practice — not a real PYQ
A planet's orbital radius is increased by a factor of 9. By what factor does its orbital period change?
- (a)9
- (b)27
- (c)81
- (d)3
Answer(b) 27 — period scales as R^(3/2), and 9^(3/2) = 27.
- practice — not a real PYQ
Kepler's third law states that the square of the orbital period is proportional to the
- (a)orbital radius
- (b)square of the orbital radius
- (c)cube of the orbital radius
- (d)square root of the orbital radius
Answer(c) cube of the orbital radius — T² ∝ R³.