A satellite orbits Earth at a height equal to RE/2. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)
r = RE+RE2 = 1.5RE. v = GMEr = gRE21.5RE = gRE1.5. v = 9.8×6.4×1061.5 = 4.181×107. v≈6.47×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.5 × 10³ m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.