 Solution

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.