= Solution
For <point masses>, Newton's <force> is attractive along their separation with magnitude $Gm_1m_2/r^2$. 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> gives
$$
\boxed{F_1=Gm\mu\int_{-a}^a\frac{dx}{(b-x)^2}=\frac{GMm}{b^2-a^2},}
$$
directed toward the rod. On the perpendicular bisector, axial components cancel and transverse components add:
$$
\boxed{F_2=Gm\mu r\int_{-a}^a\frac{dx}{(x^2+r^2)^{3/2}}=\frac{GMm}{r\sqrt{r^2+a^2}}.}
$$
At $r=b\gg a$,
$$
F_1=\frac{GMm}{b^2}\left[1+\frac{a^2}{b^2}+O(a^4/b^4)\right],\quad
F_2=\frac{GMm}{b^2}\left[1-\frac{a^2}{2b^2}+O(a^4/b^4)\right].
$$
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 gives
$$
\boxed{v(r)=\sqrt{\frac{GM}{\sqrt{r^2+a^2}}}
=\sqrt{\frac{GM}{a}}[1+(r/a)^2]^{-1/4}.}
$$
For $r\gg a$, it is Keplerian, $v\sim\sqrt{GM/r}$. For rod thickness $\ll r\ll a$, the <force> approximates that of an infinite line and \b[$v\to\sqrt{GM/a}$], a flat <speed> limit rather than a divergence.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1-ia2010-rod-orbit.png]
{title=Circular speed around a uniform rod}
{description=Original orbital speed normalized by sqrt(GM/a), against perpendicular orbit radius divided by the rod half-length. The small-radius plateau and large-radius Kepler asymptote are marked; the zero-thickness rod approximation requires orbit radius larger than the physical rod thickness.}
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