= Solution
For Alice's circular geodesic, the conserved quantities or radial geodesic equation give $\Omega=d\phi/dt=\sqrt{r_s/(2r^3)}$. Consequently
$$
\left(\frac{d\tau_A}{dt}\right)^2=f-r^2\Omega^2=1-\frac{3r_s}{2r}.
$$
Bob at rest far away has $d\tau_B/dt\simeq1$. Thus the ratio of ageing rates is
$$
\boxed{k^2=\frac1{1-3r_s/(2r)}=\frac{2r}{2r-3r_s}.}
$$
For Alice's stable orbit $r>3r_s$, the denominator factor exceeds $1/2$, proving \b[$k<\sqrt2$]. Both gravitational time dilation and Alice's orbital motion enter this proper-time ratio.
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