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Practice question

MCQ

Question

At what height above Earth’s surface is g reduced to 6.12m/s2? (g0\=9.8m/s2,RE\=6.4×106m)

Options

Choose one · Correct answer highlighted

Explanation

g(h) = g0(1+h/RE)2. 6.12 = 9.8(1+h/RE)2. (1+h/RE)2 = 1.601. 1+h/RE = 1.601≈1.265. h/RE = 0.265. h = 0.265×6.4×106≈1.7×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.7 × 10⁶ m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

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