Practice question
Question
A satellite’s period near Earth’s surface is 88 minutes. What is its period at h\=4RE? (RE\=6.4×106m)
Explanation
T2∝(RE+h)3. T02 = kRE3, h = 4RE, r = 5RE. T2 = k(5RE)3 = 125kRE3. T = T0125 = 88×11.18≈983min. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 990 min. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
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