Practice question
Question
What is the minimum speed to escape from 8RE from Earth’s center? (g\=9.8m/s2,RE\=6.4×106m)
Explanation
ve = 2gRE28RE = 2×9.8×6.4×1068. ve = 1.568×107≈3.96×103m/s≈4.0km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.0 km/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
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