= Index-one plane-parallel polytrope
{title2=$\rho(z)=\rho_c\cos(kz),\quad k^2=2\pi G/K$}
For a symmetric self-gravitating slab with <polytropic equation of state> $p=K\rho^2$ and constant $K>0$, <hydrostatic equilibrium> gives $2K\rho'=-\psi'$. The plane-parallel <Poisson equation for Newtonian gravity> then yields $\rho''+(2\pi G/K)\rho=0$. Reflection symmetry at the midplane fixes the title profile. For a vacuum surface, retain the nonnegative profile only on $|z|\le z_s=\pi/(2k)$ and put $\rho=0$ outside. The half-thickness is independent of central density in this ideal index-one model, while the surface mass density is $\Sigma=2\rho_c/k$.
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