Gravitation Inside Earth — Question 3

Section 13-4 · Physics pages 1956–1957

Question 3 A thick spherical shell has three distinct field regions A spherical body has uniform density ρ only for a < r < b, with an empty concentric cavity at r < a. Use the shell theorem. a b Cross-section of a spherically symmetric mass distribution. (a) Find the inward field magnitude in all three radial regions. (b) Check continuity at a and b, and determine whether the field increases through the material.
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Question 3 - Solution A thick spherical shell has three distinct field regions a b (a) Derivation and calculation Only enclosed mass contributes. With C = 4πGρ/3, M(r) = 4πρ 3        0, r < a, r3 −a3, a < r < b, b3 −a3, r > b. Thus the inward magnitude is g(r) = C        0, r < a, r −a3/r2, a < r < b, (b3 −a3)/r2, r > b. (b) Result and interpretation At the boundaries, g(a−) = g(a+) = 0, g(b−) = g(b+) = C b3 −a3 b2 . Within the material, g′(r) = C(1 + 2a3/r3) > 0, while outside g′ = −2g/r < 0. Hence the global maximum is at r = b. Physical check Finite volume density changes field slope at a boundary without creating a jump in the field.

Question 3

A thick spherical shell has three distinct field regions

A spherical body has uniform density ρ\rho only for a<r<ba<r<b, with an empty concentric cavity at r<ar<a. Use the shell theorem.

Diagram shown in the original SVG worksheet.

  1. Find the inward field magnitude in all three radial regions.

  2. Check continuity at aa and bb, and determine whether the field increases through the material.

Question 3 - Solution

A thick spherical shell has three distinct field regions

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Only enclosed mass contributes. With C=4πGρ/3C=4\pi G\rho/3, M(r)=4πρ3{0,r<a,r3−a3,a<r<b,b3−a3,r>b.M(r)=\frac{4\pi\rho}{3} \begin{cases}0,&r<a,\\r^3-a^3,&a<r<b,\\b^3-a^3,&r>b.\end{cases} Thus the inward magnitude is g(r)=C{0,r<a,r−a3/r2,a<r<b,(b3−a3)/r2,r>b.\boxed{g(r)=C\begin{cases}0,&r<a,\\r-a^3/r^2,&a<r<b,\\(b^3-a^3)/r^2,&r>b.\end{cases}}

(b) Result and interpretation

At the boundaries, g(a−)=g(a+)=0,g(b−)=g(b+)=Cb3−a3b2.g(a^-)=g(a^+)=0,\qquad g(b^-)=g(b^+)=C\frac{b^3-a^3}{b^2}. Within the material, g′(r)=C(1+2a3/r3)>0,g'(r)=C(1+2a^3/r^3)>0, while outside g′=−2g/r<0g'=-2g/r<0. Hence the global maximum is at r=br=b.

Physical check

Finite volume density changes field slope at a boundary without creating a jump in the field.