= Solution
The <shell theorem> gives enclosed halo <mass> $4\pi\rho r^3/3$. Define $\alpha=4\pi\rho/3$. Then \b[$F_r=-GMm/r^2-\alpha Gmr$]. Interior to the uniform halo its <potential energy> may be written
$$
U(r)=-\frac{GMm}{r}+\frac12\alpha Gmr^2+U_0,
\quad
E=\frac m2(\dot r^2+r^2\dot\phi^2)+U(r).
$$
Outer shells alter the additive constant, not this interior <force>. An actually infinite uniform <mass> distribution has no finite potential referenced to infinity; only local potential differences are needed here.
The <central force> has zero <torque>, so $\mathbf J=\mathbf r\times m\dot{\mathbf r}$ is conserved and the motion is planar. With $J=mr^2\dot\phi$, the effective radial potential and equation are
$$
U_{\rm eff}=\frac{J^2}{2mr^2}-\frac{GMm}{r}+\frac12\alpha Gmr^2,
\qquad \ddot r=\frac{J^2}{m^2r^3}-\frac{GM}{r^2}-\alpha Gr.
$$
For circular radius $r_0$, $J=mr_0^2\Omega$ gives
$$
\boxed{\Omega^2=\frac{GM}{r_0^3}+\alpha G.}
$$
Set $r=r_0+\xi$ at fixed $J$. The <derivative> of the right-hand side at <equilibrium> is $-3\Omega^2+2GM/r_0^3-\alpha G$, so
$$
\boxed{\ddot\xi+\omega^2\xi=0,\qquad
\omega^2=\frac{GM}{r_0^3}+4\alpha G.}
$$
Both <frequencies> are positive, but radial and azimuthal motion no longer close with the Kepler ratio. At first order in <eccentricity> $e$, one may write $r=r_0(1-e\cos\omega t)$ and $\phi=\Omega t+(2\Omega e/\omega)\sin\omega t$. For a weak halo this is a nearly Keplerian ellipse whose apsis rotates at angular rate $\Omega-\omega$.
Put $\Omega_K=\sqrt{GM/r_0^3}$ and $\varepsilon=\alpha r_0^3/M\ll1$. Expansion yields
$$
\boxed{\dot\varpi=\Omega-\omega\simeq-\frac32\Omega_K\varepsilon
=-\frac{3\alpha G}{2\Omega_K}=-\frac{2\pi G\rho}{\Omega_K}.}
$$
The precession is retrograde. In cycles per unit time the precession <frequency> is $\dot\varpi/(2\pi)$; per radial period the apsidal shift is $2\pi(\Omega/\omega-1)\simeq-3\pi\varepsilon$. This weak-halo interpretation avoids confusing a focus-centred Kepler ellipse with the centred ellipse of a purely harmonic potential.
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