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Exercise · Q11

Q.Derive an expression for the acceleration due to gravity at a depth dd below the Earth's surface, assuming the Earth to be a uniform sphere of density ρ\rho. Show that gg becomes zero at the Earth's centre.

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Model the Earth as a uniform sphere of density ρ\rho and consider a point at depth dd below the surface, a distance R−dR-d from the Earth's centre. A standard result for a uniform spherical shell is that it exerts NO net gravitational force on any point strictly inside it (the pulls from every part of the shell cancel out by symmetry) -- so only the solid sphere of radius R−dR-d enclosed WITHIN this point contributes any net pull; the spherical shell of material lying between radius R−dR-d and the full radius RR contributes nothing.\n\nSince the density ρ\rho is uniform, mass scales directly with volume, so the enclosed mass at radius R−dR-d is

M′=M(R−dR)3M' = M\left(\frac{R-d}{R}\right)^3

(the ratio of the two spheres' volumes). The acceleration due to gravity at this depth is then

gd=GM′(R−d)2=GM(R−d)3/R3(R−d)2=GMR3(R−d)=GMR2(1−dR)=g(1−dR)g_d = \frac{GM'}{(R-d)^2} = \frac{GM(R-d)^3/R^3}{(R-d)^2} = \frac{GM}{R^3}(R-d) = \frac{GM}{R^2}\left(1-\frac{d}{R}\right) = g\left(1-\frac{d}{R}\right) …

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