Two identical satellites are in circular orbits around Earth. Satellite A is at an altitude of 400 km, while Satellite B is at an altitude of 1600 km. How does the orbital period of Satellite A compare to that of Satellite B?

Core Concept

Using Kepler's third law for circular orbits, T² ∝ r³, where T is the orbital period and r is the orbital radius. The radius of Satellite A's orbit is R + 400 km, and for Satellite B it's R + 1600 km, where R is Earth's radius (approximately 6400 km). So r_A = 6800 km and r_B = 8000 km. The ratio r_A/r_B = 6800/8000 = 0.85. Taking the cube root of this ratio gives (r_A/r_B)^(3/2) = (0.85)^1.5 ≈ 0.78, which is approximately 1/2. Therefore, T_A ≈ (1/2)T_B. Option A would only be correct if the satellites were at the same altitude. Option B would be correct if Satellite A were at a higher altitude. Option D would be correct if the relationship were T ∝ r² rather than T ∝ r^(3/2).