Gravitational field of an infinite uniform cylinder ($g_r=-2G\lambda /r(r>b)$)

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.

A pendulum of length $L$ in vertical gravity $g$ is weakly attracted horizontally by an added long circular cylinder of line density $\lambda$. Outside it, $g_h=2G\lambda /r$ at axis distance $r$. Small-angle equilibrium gives horizontal deflection $d\simeq L{g}_h/g=2GL\lambda /(rg)$. Removing the cylinder's mass reverses the change. It is a before/after differential estimate; finite length, a nonhorizontal radial direction and mechanical motion require corrections.

A thin rod of length $2R$ and linear mass density $\lambda$ slides along a diameter of an infinite uniform source cylinder of radius $R$. Neglect the narrow hole's perturbation of the source field. If the rod centre is displaced by $x$, $F(x)=\lambda {\int }_{x-R}^{x+R}{g}_y(y)dy$. The endpoint derivative gives $F^{\prime }(0)=\lambda [g_y(R)-g_y(-R)]=-4\pi G\rho \lambda R$. Dividing the restoring coefficient by rod mass $2R\lambda$ gives the displayed small-oscillation frequency.