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For a cold, unmagnetized, collisionless plasma, take an equilibrium Electron density $n_e$, Electrons of mass $m_e$ and charge $-e$, and approximately stationary ions. Use fields proportional to $e^{-i\omega t}$. The Electron displacement satisfiesIts electric polarization is $\mathbf{P}=-n_{e}e\mathbf{x}$, hence the cold collisionless plasma dielectric responseThe negative susceptibility arises from the inertial Electron response, rather than a restoring force binding an Electron to a particular atom.

A transverse electromagnetic wave then obeysFor $\omega >{\omega }_p$, the plasma supports propagation, with refractive index $n_r=\sqrt{1-{\omega }_p^2/{\omega }^2}<1$. The phase velocity is greater than $c$, while the group velocity is $c\sqrt{1-{\omega }_p^2/{\omega }^2}<c$; their product is $c^2$, and no superluminal signal is implied. Below the plasma frequency, the bulk wavenumber is imaginary and the field is evanescent. A sufficiently thick plasma largely reflects such radiation in the lossless model.

Including an Electron momentum-relaxation rate ${\nu }_c$ replaces the denominator by $\omega (\omega +i{\nu }_c)$:Its imaginary part describes absorption with the stated time convention. Finite Electron temperature also produces plasma screening of slowly varying electrostatic fields, with ${\lambda }_D=\sqrt{{ϵ}_0{k}_B{T}_e/(n_e{e}^2)}$. That spatial, longitudinal screening problem is distinct from simply taking $\omega =0$ in the cold transverse-wave formula. Magnetic bias, ion motion and kinetic resonances require more general dielectric responses.

Two interface effects. For $n_2<n_1$, Snell's law gives $\sin {\theta }_t=(n_1/n_2)\sin {\theta }_i$. If $\sin {\theta }_i>n_2/n_1$, no propagating transmitted ray exists: there is total internal reflection, with an evanescent wave on the far side. An optical fibre uses this to guide light.

For ordinary isotropic nonmagnetic media, $\tan {\theta }_i=n_2/n_1$ is the Brewster angle. The reflected p-polarized component, whose electric field lies in the incidence plane, vanishes; the reflected s-polarized component generally remains. Reflection polarizers and polarized glare reduction exploit this selectivity.

The plasma response. Neglect collisions, thermal-pressure corrections and background magnetic field. Let the wave vary as $e^{-i\omega t}$. An electron displacement obeysIts dipole moment is $-e{\mathbf{x}}_e$, so the electron electric polarization is ${\mathbf{P}}_e=-n{e}^2\mathbf{E}/(m_e{\omega }^2)$. Ions of charge $Ze$ give a further contribution ${\mathbf{P}}_i=-nZ{e}^2\mathbf{E}/(m_i{\omega }^2)$, smaller by $Z{m}_e/m_i$. Thus heavy ions move negligibly at these electron-wave frequencies. Using $\mathbf{D}={\epsilon }_0\mathbf{E}+\mathbf{P}$ gives the cold collisionless plasma dielectric response,For this model ${\mu }_r\simeq 1$, so $n_{\mathrm{refr}}^2={\epsilon }_r$ and $c^2{k}^2={\omega }^2-{\omega }_p^2$ for the transverse electromagnetic wave.

The vertical cutoff and density. Below the plasma frequency, the squared refractive index is negative: the wave is an evanescent wave instead of propagating through a layer hundreds of kilometres thick. The observed $5\mathrm{MHz}$ cutoff therefore identifies $f_p=5.0\times 10^6\mathrm{Hz}$, with ${\omega }_p=2\pi f_p$. ConsequentlyThis uses ${\epsilon }_0=8.854\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, $m_e=9.109\times 10^{-31}\mathrm{kg}$ and $e=1.602\times 10^{-19}\mathrm{C}$. The density is also $3.10\times 10^5\mathrm{c}{\mathrm{m}}^{-3}$.

The transmission elevation. At $f=8\mathrm{MHz}$, the ionosphere has refractive indexIf $\alpha$ is elevation above horizontal, incidence at the layer is ${\theta }_i=\pi /2-\alpha$. Avoiding total internal reflection requires $\sin {\theta }_i=\cos \alpha <n_i$. The limiting elevation is thereforeStrictly, it must point above this limiting angle to carry upward power through the finite layer: at equality the transmitted wave is grazing. With a suitable satellite position, the stated orbital altitude imposes no additional angular constraint in this flat-layer approximation.

Polarization-selective ground reflection. The ray's incidence plane is vertical. Horizontal electric polarization is s polarization; the other transverse polarization lies in the vertical incidence plane and is p polarization. Reflection of s but not p identifies the Brewster angle, not total internal reflection, which would reflect both in this lossless isotropic model. ThusFor a reflecting height $h=200\mathrm{km}$, the horizontal range consists of two equal legs. The Brewster reflection from an ionospheric interface givesAt this angle ${\theta }_B$ is below the critical incidence angle, so a transmitted ray also exists; the detected horizontal component is an ordinary partial reflection. The s-polarized reflected power fraction is approximately $[(1-n_i^2)/(1+n_i^2){]}^2\simeq 0.059$, whereas the ideal p-polarized reflected fraction is zero. These results use the question's sharp planar interface model and exclude upper-interface reflections.

Use a representative conduction-electron density $n\sim 10^{29}{\mathrm{m}}^{-3}$, roughly one mobile Electron per atom in a dense metal, and the bare Electron mass for this estimate. The plasma frequency is thenThis is an order-of-magnitude answer because no particular metal or carrier density is specified. It corresponds to an ordinary frequency $f_p\simeq 2.8\times 10^{15}\mathrm{Hz}$, excitation energy $\hslash {\omega }_p\simeq 12\mathrm{eV}$, and free-space wavelength near $1.1\times 10^{-7}\mathrm{m}$.

The physical oscillation is collective motion of the electron gas against the positive ionic background. Displacing the Electrons relative to the ions produces a restoring electric field. In the simplest uniform-displacement picture, $\ddot{x}+{\omega }_p^{2}x=0$: the frequency increases with the density of mobile electric charge and decreases with carrier inertia. Quantizing this longitudinal collective oscillation gives a plasmon.

The same scale controls the ideal collisionless Drude model optical response. Its cold collisionless plasma dielectric response is ${ϵ}_r(\omega )=1-{\omega }_p^2/{\omega }^2$. Below ${\omega }_p$, the negative relative permittivity makes a transverse electromagnetic wave evanescent, producing strong reflection; above it, the ideal model admits propagation. Real-metal collisions, interband photon absorption and background polarization modify this simple optical edge. Thus ${\omega }_p$ is a collective charge-response scale, not a single-electron scattering rate or a sharp transparency threshold for every real metal.