= Solution
Put $V_0=200\ \mathrm{km\,s^{-1}}$. A <Taylor expansion> near the centre gives
$$
V_c(r)=V_0\left[\frac r{r_0}-\frac12\left(\frac r{r_0}\right)^2+O\left(\frac{r^3}{r_0^3}\right)\right].
$$
Therefore the circular <angular frequency> is
$$
\boxed{\Omega(r)=\frac{V_c(r)}r=\frac{V_0}{r_0}\left[1-\frac r{2r_0}+O\left(\frac{r^2}{r_0^2}\right)\right]\longrightarrow\frac{V_0}{r_0}.}
$$
Thus the central motion approaches solid-body rotation. Numerically $\Omega_0=1.6204\times10^{-15}\ \mathrm{s^{-1}}$, and the central orbital period tends to
$$
\boxed{P_0=\frac{2\pi}{\Omega_0}=\frac{2\pi r_0}{V_0}=3.8776\times10^{15}\ \mathrm s\simeq1.229\times10^8\ \mathrm{yr}.}
$$
The finite limit of $\Omega$ accompanies $V_c\to0$ at $r\to0$; it does not imply a finite central linear <speed>.
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