Variation Of G With Depth

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A mining engineer descends to a depth equal to one-fourth of the Earth's radius, assuming the Earth has uniform density throughout its interior. What is the local value of gravity there?

RankGuru · Accepted Answer

As described in NCERT Class 11, Chapter 8 (Gravitation), for a uniform-density Earth only the mass enclosed within the sphere below an object contributes to the gravity it feels, because the shell of matter above exerts zero net force by the shell theorem. This leads to the linear relation $g_d = g\left(1 - \dfrac{d}{R}\right)$, which decreases steadily with depth and contrasts sharply with the squared decrease seen for altitude above the surface. At the centre, where $d = R$, gravity becomes exactly zero, consistent with matter pulling equally in all directions there. Substituting $d = \dfrac{R}{4}$ gives $g_d = g\left(1 - \dfrac{1}{4}\right) = \dfrac{3g}{4}$. The option $g/4$ confuses the fraction of gravity removed with the fraction remaining. The option $4g/3$ inverts the factor and is unphysically larger than surface gravity, which is impossible as one moves inward. The option $g/2$ would correspond to a depth of $R/2$, not $R/4$. As a plausibility check, gravity must fall as one descends because the outer shells contribute no net force, and $\dfrac{3g}{4}$ correctly lies between the full surface value and zero, exactly matching the expected linear trend with depth.

Variation Of G With Depth

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