Practice question
Question
A satellite near Earth has a period of 83 minutes. What is its period at h\=8RE? (RE\=6.4×106m)
Explanation
T2∝(RE+h)3. T02 = kRE3, h = 8RE, r = 9RE. T2 = k(9RE)3 = 729kRE3. T = T0729 = 83×27≈2241min. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2250 min. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
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