For sky wave propagation of a 10, MHz signal, | vedclass

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(A) The critical frequency $(f_c)$ for sky wave propagation is related to the maximum electron density $(N_{\max})$ by the formula: $f_c = 9 \sqrt{N_{

Vedclass · Accepted Answer

$\, 1.2 	imes 10^{12} \, m^{-3}$

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(A) The critical frequency $(f_c)$ for sky wave propagation is related to the maximum electron density $(N_{\max})$ by the formula: $f_c = 9 \sqrt{N_{\max}}$.Given $f_c = 10 \, MHz = 10 	imes 10^6 \, Hz$.Substituting the value: $10 	imes 10^6 = 9 \sqrt{N_{\max}}$.Squaring both sides: $(10^7)^2 = 81 	imes N_{\max}$.$N_{\max} = \frac{10^{14}}{81} \approx 1.23 	imes 10^{12} \, m^{-3}$.Thus, the minimum electron density required is approximately $1.2 	imes 10^{12} \, m^{-3}$.

For sky wave propagation of a $10\, MHz$ signal, what should be the minimum electron density in the ionosphere?

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