At what height above Earth’s surface is g reduced to 5.88m/s2? (g0\=9.8m/s2,RE\=6.4×106m)
g(h) = g0(1+h/RE)2. 5.88 = 9.8(1+h/RE)2. (1+h/RE)2 = 1.667. 1+h/RE = 1.667≈1.291. h/RE = 0.291. h = 0.291×6.4×106≈1.86×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.9 × 10⁶ m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.