Axial and transverse gravity of a uniform rod ($F_{\parallel }=GMm/(b^2-a^2),F_{\perp }=GMm/[r\sqrt{r^2+a^2}]$)

Integrate Newtonian point-force elements along a uniform thin rod of total mass $M$ and half-length $a$. The first title formula applies on the axis outside the endpoint, $b>a$; the second applies on its perpendicular bisector. Both approach the point-mass result at large distance, but at equal distance their ratio is $1+3a^2/(2b^2)+O(a^4/b^4)$. A nonuniform linear mass profile cannot be replaced by these formulas using only total mass and length.

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.