Circular orbit perpendicular to a uniform rod ($v(r)=\sqrt{GM/a}[1+(r/a{)}^2{]}^{-1/4}$)

For a light test mass in the rod's perpendicular-bisector plane, axial symmetry makes the attraction radial. Equating $m{v}^2/r$ to the transverse force gives the title speed. Far outside the rod it is Keplerian, proportional to $r^{-1/2}$. Well inside the half-length but outside its physical thickness, the rod approximates an infinite line and the circular speed is nearly constant, $\sqrt{GM/a}$. Extrapolation to an orbit inside the material or a zero-radius orbit is outside this thin-rod model.