A planet orbits the Sun with a period of 12 years. If Earth’s orbital radius is 1.5×1011m, what is its semi-major axis?
Kepler’s third law: Tp2TE2 = ap3aE3. TE = 1year, Tp = 12years, aE = 1.5×1011m. 12212 = ap3(1.5×1011)3. 144 = ap33.375×1033. ap3 = 144×3.375×1033 = 4.86×1035. ap = (4.86×1035)1/3≈7.86×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.9 × 10¹¹ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.