What is the minimum speed to escape from 10RE from Earth’s center? (g\=9.8m/s2,RE\=6.4×106m)
ve = 2gRE210RE = 2×9.8×6.4×10610. ve = 1.254×107≈3.54×103m/s≈3.5km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.5 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.