Explain space wave propagation and derive the expression of area covered and maximum distance up to which transmission can be received.

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> [!TLDR]
> Space waves travel by direct line-of-sight; using Earth's curvature geometry, the max range is $d_M=\sqrt{2Rh_T}+\sqrt{2Rh_R}$ and the covered area is $A=2\pi Rh_T$.
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> **Space wave propagation.**
> Radio waves of frequency above about 30 MHz (VHF, UHF, and microwaves) are not reflected by the ionosphere (they pass through it) and are also not diffracted significantly around the Earth's curved surface. Such waves travel from the transmitting antenna to the receiving antenna directly through space, in a (nearly) straight line -- this is called space wave (or line-of-sight, LOS) propagation. It may occur as a direct wave and/or a wave reflected off the ground, the two interfering at the receiver. Because Earth's surface curves away, the range of space-wave communication using ground-based antennas is limited by the line-of-sight distance, set by the heights of the transmitting and receiving antennas.
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> **Derivation of maximum line-of-sight distance.**
> Let the transmitting antenna, of height $h_T$, be at point T on the Earth's surface (radius $R$). The farthest point visible in a straight line from T, grazing the Earth's surface, is at distance $d_T$ (the horizon distance). From the right-angled geometry (tangent line from height $h_T$ to the circle of radius $R$):
> $$(R+h_T)^2 = R^2 + d_T^2$$
> $$R^2 + 2Rh_T + h_T^2 = R^2+d_T^2$$
> Since $h_T \ll R$, neglect $h_T^2$:
> $$d_T^2 \approx 2Rh_T \implies d_T = \sqrt{2Rh_T}$$
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> Similarly for the receiving antenna of height $h_R$: $d_R = \sqrt{2Rh_R}$.
> …

Q.Explain space wave propagation and derive the expression of area covered and maximum distance up to which transmission can be received.

Concept understanding — Space Wave and Line-of-Sight

Q.Explain space wave propagation and derive the expression of area covered and maximum distance up to which transmission can be received.

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