 Solution

By reflection symmetry the vertical gravitational field of the self-gravitating incompressible slab vanishes at its mid-plane. Inside the slab, the gravitational Poisson equation therefore integrates to $d\psi /dz=4\pi G\rho z$. Treat the star as a test particle, so its own mass does not alter that field. Its vertical stellar orbit in a uniform slab obeysHence $z(t)=a\cos (\omega t)$ with $\omega =\sqrt{4\pi G\rho }$. The first mid-plane crossing occurs after one quarter of this simple harmonic motion period:Its independence of $a$ follows from the linear interior restoring force. The calculation neglects drag and changes to the slab, and it uses the infinite planar approximation rather than the field of a finite Galactic disk.