At a height h above the Earth's surface (h ≪ R), the acceleration due to gravity g(h) is approximately:
Ag (1 + h/R)
Bg (1 − h/R)
Cg (1 − 2h/R)
Dg (1 + 2h/R)
Answer & Solution
Correct answer: C. g (1 − 2h/R)
1. At height h: g(h) = G M / (R + h)².
2. Factor R: g(h) = (G M / R²) · (1 + h/R)⁻² = g · (1 + h/R)⁻².
3. For h ≪ R, use the binomial approximation (1 + x)⁻² ≈ 1 − 2 x.
4. So g(h) ≈ g (1 − 2 h / R).
5. Distractor (B) drops the factor of 2 — a classic mistake.
_Source: NCERT Class 11 Physics Ch 7 "Gravitation", §7.6 g above the surface, p. 133_
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