Gravity planetary moments and fluid field models

Spatial integration and vector calculus become particularly effective when symmetry reduces a three-dimensional problem to one radial or one-dimensional equation. We derive a uniform sphere's gravitational potential and inertia, analyze fluid rotation and radial flow, and connect an elastic wave mode to a constant-coefficient ODE. The physical laws are explicitly assumed models, with no invented measurements treated as observations.

1 Uniform-sphere gravitational potential

For a sphere of radius $R>0$, assume Newtonian potential per unit test mass is

$$\Phi(x)=-G\iiint_{|y|\le R}\frac{\rho}{|x-y|}dV_y,$$

for constant mass density $\rho>0$ and positive gravitational constant $G$. Spherical symmetry makes it depend only on $r=|x|$. On a source shell of radius $s$, place the polar axis toward $x$. Then $|x-y|=\sqrt{r^2+s^2-2rs\cos\phi}$ and $dV=s^2\sin\phi\,ds\,d\phi\,d\theta$.

For $r,s>0$, substitute $u=\cos\phi$ to obtain the angular factor

$$2\pi\int_{-1}^1\frac{du}{\sqrt{r^2+s^2-2rsu}} =\frac{2\pi[(r+s)-|r-s|]}{rs}=\frac{4\pi}{\max(r,s)}.$$

At $r=s$, the angular endpoint is improper but integrable, and the same formula follows by direct endpoint evaluation. The proper capped-kernel and small-ball proof below justifies the singular volume integral and the angular-first calculation, with explicit convergence bounds.

For $r\ge R$, this gives $\Phi(r)=-4\pi G\rho R^3/(3r)=-GM/r$, where $M=4\pi\rho R^3/3$. For $0<r<R$, split the source shells at $s=r$:

$$\Phi(r)=-4\pi G\rho\left[\frac1r\int_0^rs^2ds+\int_r^Rs\,ds\right] =-2\pi G\rho\left(R^2-\frac{r^2}{3}\right).$$

The same inside formula extends to $r=0$ by direct integration or continuity. The gravitational field $g=-\nabla\Phi$ is therefore

$$g(r)=\begin{cases}-(4\pi G\rho/3)r\,e_r,&0<r<R,\\-GM r^{-2}e_r,&r>R.\end{cases}$$

The field matches continuously at $R$ and extends as zero at the center. The outside inverse-square law and inside linear law have been derived from the volume integral and uniform density, not assumed for every planetary density profile.

2 A planetary mass moment

The moment of inertia about the $z$ axis in this rigid-body model is

$$I_z=\iiint_{|x|\le R}\rho(x^2+y^2)dV.$$

In the spherical convention $x^2+y^2=r^2\sin^2\phi$ and $dV=r^2\sin\phi\,dr\,d\phi\,d\theta$. Hence

$$I_z=\rho\left(\int_0^Rr^4dr\right) \left(\int_0^\pi\sin^3\phi\,d\phi\right) \left(\int_0^{2\pi}d\theta\right) =\frac{8\pi\rho R^5}{15}=\frac25MR^2.$$

It is not enough to integrate density alone: inertia weights each mass element by squared distance from the rotation axis. A nonuniform density changes the weighting integral, and a nonspherical body changes the geometric bounds.

A curved surface area calculation

For the graph $z=x^2+y^2$ above a disk of radius $A>0$, parameterization $r(\rho_c,\theta)=(\rho_c\cos\theta,\rho_c\sin\theta,\rho_c^2)$ gives cross-product magnitude $\rho_c\sqrt{1+4\rho_c^2}$. Its scalar area is therefore

$$\mathcal A=2\pi\int_0^A\rho_c\sqrt{1+4\rho_c^2}d\rho_c=\frac\pi6[(1+4A^2)^{3/2}-1].$$

This is larger than the projected disk area $\pi A^2$; an upward constant vertical unit field has flux only $\pi A^2$. The distinction between scalar area and oriented flux matters. For a sphere of radius $R>0$, the spherical surface scale is $R^2\sin\phi$, whose integral is $4\pi R^2$. These are surface Jacobians, not the volume factor $r^2\sin\phi\,dr$ reused with a missing differential.

3 Fluid rotation and radial hydrological flux

For a constant-density fluid, mass conservation $\rho_t+\nabla\cdot(\rho u)=0$ reduces to $\nabla\cdot u=0$ when density is constant in space and time. The velocity field $u=\Omega\rho_c e_\theta$ in cylindrical coordinates (here $\rho_c$ denotes radius, not mass density) has divergence zero and curl $2\Omega e_z$. This is a simple local model of solid rotation. Circulation around radius $a$ is $2\pi\Omega a^2$, equal to curl flux through the disk by Stokes. It illustrates velocity, circulation and vorticity; it is not a full model of oceanic or atmospheric force balance.

For an ideal steady axisymmetric flow through a layer of thickness $L>0$, let total outward flow rate be $Q$. Conservation through cylindrical surfaces requires $q_{\rho_c}2\pi\rho_cL=Q$, so

$$q_{\rho_c}=\frac{Q}{2\pi L\rho_c}.$$

Assume the constitutive law $q=-K\nabla h$ with constant $K>0$ and scalar head $h$. Choose a reference radius $\rho_0>0$. Then

$$h(\rho_c)=h(\rho_0)-\frac{Q}{2\pi KL}\log\frac{\rho_c}{\rho_0},\qquad\rho_c>0.$$

The logarithm needs a dimensionless radius ratio. Positive $Q$ denotes an outward source; inward pumping uses the opposite sign convention. The axis is excluded and must be replaced by a source/well boundary in a realistic model. Conservation and the constitutive law play different roles: one sets flux geometry, the other converts flux into a head gradient.

4 An elastic wave mode and its ODE

In a one-dimensional small-displacement elastic model, assume stress $\sigma=Eu_x$ with constant modulus $E>0$ and mass density $\rho_m>0$. Force balance is $\rho_m u_{tt}=\sigma_x=Eu_{xx}$, so wave speed is $c=\sqrt{E/\rho_m}$.

For a spatial mode with $k>0$, $u(x,t)=\cos(kx)T(t)$, substitution gives $T''+c^2k^2T=0$. The second-order ODE methods therefore produce $T=A\cos(ckt)+B\sin(ckt)$. A displacement field such as $u=A\cos(kx)\cos(ckt)$ is checked by direct differentiation. This connects spatial derivatives and constant-coefficient oscillation without claiming a full seismological wave theory.

For smooth solutions on a fixed interval, define energy per unit cross-sectional area

$$\mathcal E(t)=\frac12\int[\rho_m u_t^2+Eu_x^2]dx.$$

Differentiation and integration by parts give $\mathcal E'=[Eu_tu_x]_{\rm endpoints}$, because the interior terms cancel using the wave equation. Zero boundary power flux conserves energy; nonzero endpoint forcing can add or remove it. Boundary conditions are part of the model, not optional afterthoughts.

A proper-integral truncation proof at the Newtonian singularity

Fix $x\in\mathbb R^3$ and $R>0$. The density in this example is the fixed finite constant $\rho>0$. For an integer $N\ge1$, define $k_N(y)=\min(N,1/|x-y|)$ away from $x$, and $k_N(x)=N$. This function is continuous, including at $x$, and bounded. The proper Jordan theorem and ordinary Fubini therefore apply to each $\rho k_N$ on the source ball, with the coordinate boundary limits proved in the spatial-integration chapter.

The local singularity is integrable with a quantitative bound. On $\varepsilon\le|y-x|\le\delta$, translated spherical coordinates give

$$\int\frac{\rho}{|x-y|}\,dV=4\pi\rho\int_\varepsilon^\delta s\,ds =2\pi\rho(\delta^2-\varepsilon^2).$$

Here the kernel is continuous and bounded on each annulus; take the proper integrals before letting $\varepsilon\downarrow0$. Thus the singular contribution from a ball of radius $\delta$ is at most $2\pi\rho\delta^2$. Integrals on the source ball with a ball of radius $\varepsilon$ about $x$ removed form a Cauchy family: differences lie in one such small annulus and satisfy this bound. Their real limit defines the improper potential integral. Its difference from $\int\rho k_N$ is nonnegative and at most $2\pi\rho/N^2$, since the kernels coincide outside $|y-x|<1/N$. This proves that the capped proper integrals have the same limit, including when $x$ lies on the source boundary.

It remains to justify the angular-first evaluation of that limit. For $r=|x|>0$ and source radius $s>0$, put $a=|r-s|$, $b=r+s$. Substituting the distance $d=\sqrt{r^2+s^2-2rsu}$ into the proper capped angular integral gives

$$I_N(r,s)=\frac{2\pi}{rs}\int_a^b\min(Nd,1)\,dd.$$

The uncapped expression is $I(r,s)=2\pi(b-a)/(rs)=4\pi/\max(r,s)$. Their difference is nonnegative, and

$$0\le I-I_N\le\frac{2\pi}{rs}\int_0^{1/N}(1-Nd)\,dd=\frac{\pi}{rsN}.$$

Multiplying by $\rho s^2$ and integrating $0<s<R$ bounds the radial error by $\pi\rho R^2/(2rN)\to0$. Each capped integral was a proper integral, so this explicit error proves that the displayed angular-first radial calculation equals the improper volume integral. It does not appeal to a general interchange theorem for singular kernels. For $r=0$, the angular factor is $4\pi\min(N,1/s)$ and the radial integral tends directly to $4\pi\rho\int_0^R s\,ds=2\pi\rho R^2$; the missing radial part is controlled by the same small-ball estimate. Differentiating the resulting explicit radial potential, rather than differentiating the singular integrand, gives the gravitational field stated above.

For the rotation model, explicitly extend the cylindrical notation to $u=(-\Omega y,\Omega x,0)$ on the entire disk neighborhood. This is a smooth Cartesian field with curl $(0,0,2\Omega)$, including the axis, so the disk version of Stokes applies. The radial flow $Q/(2\pi L\rho_c)$ has no such extension at the axis when $Q\ne0$ and retains its excluded-axis domain.

5 Graded practice

E1 Gravity. Find gravitational acceleration magnitude at $r=R/2$ inside the uniform sphere and compare it with the surface magnitude.

E2 Planetary moment. If a uniform sphere's radius doubles while density remains fixed, by what factors do mass and moment of inertia change?

E3 Hydrology. For positive outward $Q$, find $h(2\rho_0)-h(\rho_0)$ and state its sign under the specified constitutive law.

E4 Circulation. For $u=\Omega\rho_c e_\theta$, calculate circulation around a circle of radius $a$ directly from the line integral.

E5 Wave mode. Verify $u=\sin(kx)\cos(ckt)$ solves $u_{tt}=c^2u_{xx}$ and identify a zero-power-flux interval when $k>0$.

E6 Surface area. Find the scalar area of $z=x^2+y^2$ over the unit disk and compare it with upward flux of $(0,0,1)$.

6 Full solutions

S1. Inside magnitude is $(4\pi G\rho/3)r$. At $R/2$ it is $2\pi G\rho R/3$, exactly half the surface magnitude $4\pi G\rho R/3$. This linear comparison relies on uniform density.

S2. Mass scales as $R^3$, so it multiplies by 8. Inertia scales as $\rho R^5$, so it multiplies by 32, not merely 8. Its extra squared-distance weight explains the difference.

S3. The difference is $-[Q/(2\pi KL)]\log2<0$. Under outward positive flow, head decreases with increasing radius. Reversing the definition of positive flow reverses the sign.

S4. Along the circle, $dr=a e_\theta d\theta$ and velocity is $\Omega a e_\theta$. Thus the integral is $\int_0^{2\pi}\Omega a^2d\theta=2\pi\Omega a^2$, consistent with vorticity flux.

S5. Both $u_{tt}$ and $c^2u_{xx}$ equal $-c^2k^2\sin(kx)\cos(ckt)$. On $[0,\pi/k]$, $u$ and hence $u_t$ vanish at both endpoints, so $Eu_tu_x$ vanishes there and energy is conserved. The boundary statement follows from the selected mode, not from the PDE alone.

S6. Put $A=1$ in the derived formula: area is $\pi(5\sqrt5-1)/6$. The upward vertical flux is the projected area $\pi$, smaller than the scalar surface area. The graph area factor $\sqrt{1+4\rho_c^2}$, not a flux sign, creates the difference.

Summary

Spherical integration derives gravitational potential and weighted planetary moments. Divergence and Stokes connect fluid flux and circulation to local operators; a constitutive law supplies the additional physical relation. Elastic normal modes reduce a spatial model to ODEs, and an energy identity exposes the role of boundary input. Each conclusion remains tied to the stated symmetry, material and boundary assumptions.