Practice question
Question
A planet orbits the Sun with a period of 9 years. If Earth’s orbital radius is 1.5×1011m, what is its semi-major axis?
Explanation
Kepler’s third law: Tp2TE2 = ap3aE3. TE = 1year, Tp = 9years, aE = 1.5×1011m. 9212 = ap3(1.5×1011)3. 81 = ap33.375×1033. ap3 = 81×3.375×1033 = 2.73375×1035. ap = (2.73375×1035)1/3≈6.49×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.5 × 10¹¹ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.