Gravitation Inside Earth — Question 5

Section 13-4 · Physics pages 1960–1961

Question 5 An interior gravity profile can be inverted to obtain density For a spherical body, the inward gravitational field magnitude is measured to be g(r) = Ar + Br3 for 0 < r ≤R, with A > 0. Assume no point mass at the center. r R dM = 4πr2ρ dr A thin spherical shell connects the local density to the derivative of enclosed mass. (a) Derive the local density profile. (b) For B < 0, find the restriction on B required for density to remain nonnegative.
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Question 5 - Solution An interior gravity profile can be inverted to obtain density r R dM = 4πr2ρ dr (a) Derivation and calculation Invert the shell-theorem relation: M(r) = r2g(r) G = Ar3 + Br5 G . Differentiate and divide by shell volume per unit radius: 4πr2ρ(r) = dM dr = 3Ar2 + 5Br4 G , ρ(r) = 3A + 5Br2 4πG . (b) Result and interpretation If B < 0, then dρ/dr = 10Br/(4πG) < 0 for r > 0. The minimum occurs at R: ρ(R) ≥0 ⇐⇒ 3A + 5BR2 ≥0. −3A 5R2 ≤B < 0 . At the limiting value, the density vanishes only at the surface. Physical check The condition also ensures g(r) > 0 for 0 < r ≤R; there is no hidden central point mass.

Question 5

An interior gravity profile can be inverted to obtain density

For a spherical body, the inward gravitational field magnitude is measured to be g(r)=Ar+Br3g(r)=Ar+Br^3 for 0<r≤R0<r\le R, with A>0A>0. Assume no point mass at the center.

Diagram shown in the original SVG worksheet.

  1. Derive the local density profile.

  2. For B<0B<0, find the restriction on BB required for density to remain nonnegative.

Question 5 - Solution

An interior gravity profile can be inverted to obtain density

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Invert the shell-theorem relation: M(r)=r2g(r)G=Ar3+Br5G.M(r)=\frac{r^2g(r)}G=\frac{Ar^3+Br^5}G. Differentiate and divide by shell volume per unit radius: 4πr2ρ(r)=dMdr=3Ar2+5Br4G,4\pi r^2\rho(r)=\frac{dM}{dr}=\frac{3Ar^2+5Br^4}G, ρ(r)=3A+5Br24πG.\boxed{\rho(r)=\frac{3A+5Br^2}{4\pi G}}.

(b) Result and interpretation

If B<0B<0, then dρ/dr=10Br/(4πG)<0d\rho/dr=10Br/(4\pi G)<0 for r>0r>0. The minimum occurs at RR: ρ(R)≥0⇔3A+5BR2≥0.\rho(R)\ge0\quad\Longleftrightarrow\quad 3A+5BR^2\ge0. −3A5R2≤B<0.\boxed{-\frac{3A}{5R^2}\le B<0}. At the limiting value, the density vanishes only at the surface.

Physical check

The condition also ensures g(r)>0g(r)>0 for 0<r≤R0<r\le R; there is no hidden central point mass.