Practice question
Question
A planet’s orbital period around the Sun is 8 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 = 8years, aE = 1.5×1011m. 8212 = ap3(1.5×1011)3. 64 = ap33.375×1033. ap3 = 64×3.375×1033 = 2.16×1035. ap = (2.16×1035)1/3≈6.0×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.0 × 10¹¹ m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.