Practice question
Question
Minimum speed to escape from 2RE from Earth’s center is? (g\=9.8m/s2,RE\=6.4×106m)
Explanation
ve = 2gRE22RE = gRE. ve = 9.8×6.4×106 = 6.272×107. ve≈7.92×103m/s≈7.9km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.9 km/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
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