At a depth d below the Earth's surface (d ≪ R), the acceleration due to gravity is approximately:
Ag (1 − d/R)
Bg (1 − 2d/R)
Cg (1 + d/R)
Dg (1 + 2d/R)
Answer & Solution
Correct answer: A. g (1 − d/R)
1. For a uniform-density Earth, only the mass enclosed within radius (R − d) attracts an object at depth d.
2. Mass enclosed scales as the cube of the radius: M(R−d) = M · (R − d)³ / R³.
3. g(d) = G M(R−d) / (R − d)² = G M (R − d) / R³.
4. Simplify: g(d) = g · (1 − d/R).
5. So gravity decreases LINEARLY with depth, while it falls off as (1 − 2h/R) with height — different from below.
_Source: NCERT Class 11 Physics Ch 7 "Gravitation", §7.6 g below the surface, p. 133_
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