Solutions for Chapter 2

Abstract

The power P is given by the product of intensity I, the area A beamwidth corresponds to at distance D and bandwidth B, integrated over the solid angle. Expressing the area as A= ^D2 ∫ dΩ, this is: $P = I{D^2}B{\left( {\int {d\Omega } } \right)^2} = 4{\pi ^2}{D^2}BI{\left( {\int_0^{8'} {\sin \theta d\theta } } \right)^2}.$ Thus, at the distance ofthe Moon, he intensity is $I = \left[ {2 \times {{10}^8}} \right]/\left[ {4{\pi ^2}{{\left( {3.84 \times {{10}^8}} \right)}^2} \times \left( {{{10}^3}} \right) \times {{\left( {2.7 \times {{10}^{ - 6}}} \right)}^2}} \right] = 4.7 \times {10^{ - 3}}W{m^{ - 2}}H{z^{ - 1}}steradia{n^{ - 1}}.$ Since $S = \int {d\Omega }$ , the flux density is $S = 8{\text{ x 1}}{{\text{0}}^{ - 8}}{\text{ W}}{{\text{m}}^{ - 2}}{\text{ }}H{z^{ - 1}}$ At the nearest star, the flux density is $S = P/[2\pi {(3.9{\text{ x 1}}{{\text{0}}^{16}})^2}{\text{ x (1}}{{\text{0}}^3}){\text{ }}x{\text{ (2}}{\text{.7 x 1}}{{\text{0}}^{ - 6}}] = {\text{ 7}}{\text{.7 x 10 - 24 W}}{{\text{m}}^{ - 2}}{\text{ H}}{{\text{z}}^{ - 1}}.$ The peak power density in Wm^-2, the closest safe distance is 343 km. At 10km, the peakpower is 120 kW m^-2.