= Solution
\b[Field of discrete masses.] <Newton's law of universal gravitation> and <superposition> give the <gravitational field> away from the source points:
$$
\boxed{\mathbf g(\mathbf r)=-G\sum_{i=1}^N
M_i\frac{\mathbf r-\mathbf r_i}{|\mathbf r-\mathbf r_i|^3}.}
$$
It is the <force> per unit test <mass>, with units $\mathrm{N\,kg^{-1}}=\mathrm{m\,s^{-2}}$, and points toward each positive source <mass>.
\b[Infinite circular cylindrical rod.] Let $b$ be the radius of the circular cross-section, so $A=\pi b^2$, and write $\lambda=\rho A$ for the <linear mass density>. Choose a coaxial cylindrical <Gaussian surface> with radius $r>b$ and length $L$. Translation and rotation symmetry make $\mathbf g=g_r(r)\widehat{\mathbf r}$. The end caps have zero field <flux> because their normals are axial; on the curved surface the outward-normal component is the constant negative value $g_r$. <Gauss's law for gravity> therefore gives
$$
\oint\mathbf g\cdot d\mathbf S=2\pi rL g_r
=-4\pi G(\rho AL),
\qquad
\boxed{\mathbf g(r)=-\frac{2G\rho A}{r}\widehat{\mathbf r}\quad(r>b).}
$$
The <gravitational field of an infinite uniform cylinder> inside the material instead has enclosed <mass> $\rho\pi r^2L$, giving $g_r=-2\pi G\rho r$. The exterior result assumes a circular cylindrical rod; area by itself would not specify the exact near field of a noncircular cross-section.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018-ia-physics-q14-rod-orbits.png]
{title=Gaussian-cylinder cross-section for an infinite rod and the circular-speed curve of a finite line rod}
In the left panel the dashed circle is a transverse section of the coaxial <Gaussian surface>; the field points inward and the surface normals outward. Its side area is $2\pi rL$.
For an exterior <circular orbit>, <Newton's second law> equates the <force> required for <centripetal acceleration> to gravity:
$$
\frac{mv^2}{r}=m\frac{2G\rho A}{r},\qquad
\boxed{v=\sqrt{2G\rho A}=\sqrt{2G\lambda}.}
$$
The orbital <speed> is independent of radius and of the orbiting object's <mass>, for orbits outside the infinite rod.
\b[Finite rod.] Taking the supplied field as the <gravitational field of a finite uniform line segment>, the corresponding balance at radius $R$ is
$$
\frac{v^2}{R}=\frac{GM}{R\sqrt{R^2+a^2}},\qquad
\boxed{v(R)=\left(\frac{GM}{\sqrt{R^2+a^2}}\right)^{1/2}
=\frac{\sqrt{GM}}{(R^2+a^2)^{1/4}}.}
$$
Since $M=2a\rho A=2a\lambda$, its two limits are
$$
\boxed{R\ll a:\quad v\simeq\sqrt{\frac{GM}{a}}=\sqrt{2G\rho A},
\qquad R\gg a:\quad v\simeq\sqrt{\frac{GM}{R}}.}
$$
The first reproduces the infinite-rod result: a nearby observer sees a very long source. The second is the familiar <circular orbit> speed about a <point mass>, since a distant observer cannot resolve the rod's length. The dimensionless curve is
$$
\frac{v}{\sqrt{GM/a}}=[1+(R/a)^2]^{-1/4}.
$$
It starts at a constant plateau and decreases smoothly, approaching an $R^{-1/2}$ dependence far away. For $R>0$, $dv/dR=-\sqrt{GM}\,R/[2(R^2+a^2)^{5/4}]<0$, consistent with the sketch.
\b[Derivation of the finite-rod field.] Put the source along $-a<z<a$ on the $z$ axis. In its central perpendicular plane, an element $dm=\lambda\,dz$ is at distance $s=\sqrt{r^2+z^2}$. Its field magnitude is $G\lambda\,dz/s^2$, and its inward radial component has the extra factor $r/s$. The axial components cancel between $z$ and $-z$. Thus
$$
g_r=-G\lambda r\int_{-a}^a\frac{dz}{(r^2+z^2)^{3/2}}
=-G\lambda r\left[\frac{z}{r^2\sqrt{r^2+z^2}}\right]_{-a}^{a}
=-\frac{2aG\lambda}{r\sqrt{r^2+a^2}}.
$$
Replacing $2a\lambda$ by $M$ gives
$$
\boxed{\mathbf g(r)=-\frac{GM}{r\sqrt{r^2+a^2}}\widehat{\mathbf r}.}
$$
The integral treats the source as a <line mass>. For a finite cylinder of nonzero cross-section, the displayed finite-rod formula is a thin-rod approximation, not an exact field for all $r$. Its near-rod limit should therefore be read as $b\ll R\ll a$, keeping the orbit outside the material; the apparent line-source singularity at $r=0$ does not describe the interior of a real cylinder.
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