 Solution

The oscillatory numerator $\cos (\nu \theta )+\nu \sin (\nu \theta )$ has amplitude $\sqrt{1+{\nu }^2}$. Therefore the reciprocal radius stays bounded away from zero exactly whenIn this case the radial bounds areFor completeness, the potential energy per unit mass is $\Phi =-1/(ar)+1/(2r^2)$, and conservation of mechanical energy givesThis effective potential diverges positively at zero, has one minimum at $r=a(1+L^2)$, and tends to zero from below at infinity. Since the initial radial velocity is negative, the particle first reaches an inner turning point. If $L^2\ge 1/2$, the subsequent outward motion has no finite outer turning point and escapes. At equality the escape speed tends to zero, but the radius is still unbounded; infinity is approached only as time tends to infinity. HenceThe minimum of $U$ permits a circular orbit for other initial data, but the nonzero initial radial velocity here excludes that exceptional solution.