 Solution

For point masses, Newton's force is attractive along their separation with magnitude $G{m}_1{m}_2/r^2$. Treat the narrow rod as uniform, with linear mass density $\mu =M/(2a)$; a mass profile would otherwise be needed beyond total mass and length.

For axial and transverse gravity of a uniform rod, on the axis at $b>a$, integrating the element force givesdirected toward the rod. On the perpendicular bisector, axial components cancel and transverse components add:At $r=b\gg a$,Both approach the point-mass force, but $F_1/F_2=1+3a^2/(2b^2)+O(a^4/b^4)$, so the axial attraction is slightly larger.

For a circular orbit perpendicular to a uniform rod, rotational symmetry about the rod makes a centred circular orbit possible in its perpendicular-bisector plane. With $m\ll M$, centrifugal balance givesFor $r\gg a$, it is Keplerian, $v\sim \sqrt{GM/r}$. For rod thickness $\ll r\ll a$, the force approximates that of an infinite line and $v\to \sqrt{GM/a}$, a flat speed limit rather than a divergence.