What is the minimum speed to escape from 7RE from Earth’s center? (g\=9.8m/s2,RE\=6.4×106m)
ve = 2gRE27RE = 2×9.8×6.4×1067. ve = 1.79×107≈4.23×103m/s≈4.2km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.2 km/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.