 Gravitational field of a finite uniform line segment

Spread mass $M$ uniformly along $-a<z<a$ on an axis, with linear mass density $\lambda =M/(2a)$. In its perpendicular bisector plane, axial field components cancel, and the radial component isThis is an exact line mass model and a thin-rod approximation for a physical cylinder. A centred circular orbit in this symmetry plane has $v^2=GM/\sqrt{R^2+a^2}$. Thus $v\to \sqrt{GM/a}=\sqrt{2G\lambda }$ for $R/a\to 0$, and $v\sim \sqrt{GM/R}$ for $R/a\to \infty$. For a finite-width rod, the first limit is useful only while its transverse size remains much smaller than $R$.