Practice question
Question
A satellite near Earth has a period of 86 minutes. What is its period at h\=5RE? (RE\=6.4×106m)
Explanation
T2∝(RE+h)3. T02 = kRE3, h = 5RE, r = 6RE. T2 = k(6RE)3 = 216kRE3. T = T0216 = 86×14.7≈1264min. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1270 min. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
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