Home exercise 6.2.2: Sounding the F layer

$f_o{F}_2=8.98\sqrt{1.5\times 10^{12}}=11.0\mathrm{MHz}$.

$f_p=8\mathrm{MHz}$ where $N=(8\times 10^6/8.98{)}^2=7.9\times 10^{11}{\mathrm{m}}^{-3}$, about half of $N_{\max }$.

$f_{\mathrm{MUF}}=f_o{F}_2\sec {\theta }_i=11.0/\cos 70^{\circ }=32.2\mathrm{MHz}$.

At $N_{\max }$, $\sqrt{1-(11.0/20{)}^2}=0.835$, so

$v_p{v}_g=3.59\times 2.51\times 10^{16}=9.0\times 10^{16}=c^2$. The $20\mathrm{MHz}$ wave crosses the peak, as it must, since $20>11.0\mathrm{MHz}$. It escapes at vertical incidence but is returned at ${\theta }_i=70^{\circ }$, since $20<32.2\mathrm{MHz}$: oblique rays reach far stations with frequencies that would pierce the layer overhead. This is the origin of the skip zone of Application 6.2.2.