Gravitation Inside Earth — Question 2

Question 2

Equal gravity reductions above and below the surface occur at different distances

A nonrotating uniform-density planet has radius $R$ and surface gravity $g_s$. Use the shell theorem: at interior radius $r$, only enclosed mass contributes to $g(r)$.

1. Derive the interior field and the gravity at depth $d$.

2. Find the exterior altitude giving the same gravity as depth $d=R/4$.

Question 2 - Solution

Equal gravity reductions above and below the surface occur at different distances

(a) Derivation and calculation

For uniform density, $$M(r)=\frac{4\pi}3\rho r^3=M(r/R)^3.$$ By the shell theorem, $$\boxed{g(r)=\frac{GM(r)}{r^2}=g_s\frac rR},\qquad \boxed{g(R-d)=g_s(1-d/R)}.$$

(b) Result and interpretation

At $d=R/4$, $g=3g_s/4$. Matching the exterior field, $$\frac{g(h)}{g_s}=\frac1{(1+h/R)^2}=\frac34.$$ $$\boxed{h/R=\frac2{\sqrt3}-1=0.154701}.$$ The interior field varies linearly with radius, while the exterior field varies as its inverse square.

Physical check

Both laws equal $g_s$ at the surface, and the uniform-sphere field vanishes at the center.