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A circular infinite cylinder of radius $b$ and uniform mass density $\rho$ has linear mass density $\lambda =\pi b^2\rho$. A coaxial Gaussian surface of length $L$ has inward radial gravitational field and no end-cap flux. Gauss's law for gravity gives $2\pi rL{g}_r=-4\pi G\lambda L$ outside, hence the displayed result. Inside, enclosed linear mass density is $\pi r^2\rho$, so $g_r=-2\pi G\rho r$. An exterior circular orbit has speed $\sqrt{2G\lambda }$ independent of radius. The exterior formula presupposes circular axial symmetry; area alone would not specify the near field of a noncircular rod.

The usual Newtonian gravitational potential satisfies $\mathbf{g}=-\nabla \Phi$ and ${\nabla }^2\Phi =4\pi G\rho$. An equally consistent convention sets $\phi =-\Phi$, giving $\mathbf{g}=\nabla \phi$ and ${\nabla }^2\phi =-4\pi G\rho$. With zero potential at infinity, a point mass has $\Phi =-GM/r$ and $\phi =GM/r$. Both conventions describe the same inward gravitational field and the same Gauss's law for gravity. Only the signs of the potential and its differences change. The gravitational potential energy per unit test mass is $\Phi$, or $-\phi$ in the alternative convention. A formula for a potential difference must therefore be read together with its relation to the gravitational field.

For any closed surface enclosing a region $V$, Gauss's law for gravity states that the outward flux of the gravitational field isHere $G$ is the gravitational constant and $M_V$ is the enclosed mass; the negative sign expresses attractive gravity.

Use the gravitational potential sign conventions specified by this paper: $\mathbf{g}=\nabla \phi$. Thus $\phi$ is the negative of the usual Newtonian gravitational potential $\Phi$ defined by $\mathbf{g}=-\nabla \Phi$. For example, setting the value at infinity to zero gives $\phi =GM/r$ for a point mass. The divergence theorem now givesSince $V$ is arbitrary, continuity of the integrands implies the local Poisson's equationConversely, integrating this Poisson's equation over any $V$ and applying the divergence theorem recovers Gauss's law for gravity. At a discontinuity of the mass density, the local equation is interpreted piecewise, or globally in the sense of distributions.

On a sphere of radius $r$, the radial gravitational field has constant magnitude. Thus Gauss's law for gravity becomeswhere $M(r)$ is the enclosed mass. In the hollow core $M(r)=0$. In the material shell, put $u=r/r_0$ and use the spherical coordinates volume element:Outside the shell the enclosed mass is the totalConsequently,At $r=0$ the displayed value refers to the radial derivative of the constant core value. The expressions match at both shell boundaries. There is no surface sheet of mass, so a jump in the volume mass density does not produce a jump in the gravitational field. The core has zero gravitational field, and the exterior gravitational field equals that of a central point mass $M_{\ast }$, as in the shell theorem.

Far from its ends, approximate the body by an infinite uniform cylinder. By cylindrical symmetry, the Newtonian gravitational field is radial. Applying Gauss's law for gravity to a coaxial cylindrical surface of radius $s$ and length $\ell$ outside the material givesAt the surface, $R=50\mathrm{m}$, soThe gravitational acceleration points inward toward the axis. Its small magnitude reflects both the small size and the modest density. The infinite-cylinder approximation uses the specified distance from both ends; a spherical approximation would discard the relevant geometry. No rotational contribution is included, since no rotation rate is supplied.

Field of discrete masses. Newton's law of universal gravitation and superposition give the gravitational field away from the source points:It is the force per unit test mass, with units $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}=\mathrm{m}{\mathrm{s}}^{-2}$, and points toward each positive source mass.

Infinite circular cylindrical rod. Let $b$ be the radius of the circular cross-section, so $A=\pi b^2$, and write $\lambda =\rho A$ for the linear mass density. Choose a coaxial cylindrical Gaussian surface with radius $r>b$ and length $L$. Translation and rotation symmetry make $\mathbf{g}=g_r(r)\hat{\mathbf{r}}$. The end caps have zero field flux because their normals are axial; on the curved surface the outward-normal component is the constant negative value $g_r$. Gauss's law for gravity therefore givesThe gravitational field of an infinite uniform cylinder inside the material instead has enclosed mass $\rho \pi r^{2}L$, giving $g_r=-2\pi G\rho r$. The exterior result assumes a circular cylindrical rod; area by itself would not specify the exact near field of a noncircular cross-section.

In the left panel the dashed circle is a transverse section of the coaxial Gaussian surface; the field points inward and the surface normals outward. Its side area is $2\pi rL$.

For an exterior circular orbit, Newton's second law equates the force required for centripetal acceleration to gravity:The orbital speed is independent of radius and of the orbiting object's mass, for orbits outside the infinite rod.

Finite rod. Taking the supplied field as the gravitational field of a finite uniform line segment, the corresponding balance at radius $R$ isSince $M=2a\rho A=2a\lambda$, its two limits areThe first reproduces the infinite-rod result: a nearby observer sees a very long source. The second is the familiar circular orbit speed about a point mass, since a distant observer cannot resolve the rod's length. The dimensionless curve isIt starts at a constant plateau and decreases smoothly, approaching an $R^{-1/2}$ dependence far away. For $R>0$, $dv/dR=-\sqrt{GM}R/[2(R^2+a^2{)}^{5/4}]<0$, consistent with the sketch.

Derivation of the finite-rod field. Put the source along $-a<z<a$ on the $z$ axis. In its central perpendicular plane, an element $dm=\lambda dz$ is at distance $s=\sqrt{r^2+z^2}$. Its field magnitude is $G\lambda dz/s^2$, and its inward radial component has the extra factor $r/s$. The axial components cancel between $z$ and $-z$. ThusReplacing $2a\lambda$ by $M$ givesThe integral treats the source as a line mass. For a finite cylinder of nonzero cross-section, the displayed finite-rod formula is a thin-rod approximation, not an exact field for all $r$. Its near-rod limit should therefore be read as $b\ll R\ll a$, keeping the orbit outside the material; the apparent line-source singularity at $r=0$ does not describe the interior of a real cylinder.

Newton's law of universal gravitation gives the force $-GmM\hat{\mathbf{r}}/r^2$ on a test mass a distance $r$ from a point source. Gauss's law for gravity statesThe inverse-square point field has the stated flux through a surrounding sphere. Superposition extends it to distributed sources; the divergence theorem relates the integral and local forms. Symmetry is still needed to turn a flux into a pointwise field value.

For the gravitational field of an infinite uniform cylinder, a coaxial cylindrical Gaussian surface has no end-cap flux and constant inward field on its curved surface. The enclosed mass per unit length is $\pi \rho r^2$ inside and $\pi \rho R^2$ outside. HenceIts magnitude rises linearly from zero, reaches $2\pi G\rho R$ at the surface, then decreases as $1/r$. The field is continuous at $R$, while its radial derivative changes because the density has a sharp edge.

Use signed distance $y$ along the diameter hole. A rod displaced by $x$ occupies $y\in [x-R,x+R]$, so its force is $F(x)=\lambda {\int }_{x-R}^{x+R}{g}_y(y)dy$. At equilibrium symmetry makes $F(0)=0$. Differentiating the moving endpoints givesThus the leading restoring force and oscillation frequency areFor an explicit positive-displacement check, put $C=2\pi G\rho$ and integrate the inside and outside pieces for $0<x<2R$:Its first-order term is $-2C\lambda Rx$, agreeing with the endpoint argument. The result is the diameter-rod oscillation in cylindrical gravity; it assumes a sufficiently long source cylinder and a narrow hole whose removed mass can be neglected.