 Solution

The far fields obey $E_{\theta }=c{B}_{\varphi }$. The radial component of their Poynting vector is $E_{\theta }{B}_{\varphi }/{\mu }_0$. Averaging ${\ddot{p}}^2=p_0^2{\omega }^4{\cos }^2\omega (t-r/c)$ and integrating $\int {\sin }^2\theta d\Omega =8\pi /3$ givesFor a small isotropically polarizable particle, $p_0=\alpha E_0$ and incident intensity is $I={ϵ}_{0}c{E}_0^2/2$. Thus the total Rayleigh scattering cross-section isEach incident linear polarization gives the same integrated power; averaging unpolarized light does not add another factor of two. The scalar real polarizability approximation neglects absorption and molecular anisotropy.

For Rayleigh attenuation from refractivity in dilute air, polarization density is $\mathbf{P}=N\alpha \mathbf{E}$. Consequently ${ϵ}_r=1+N\alpha /{ϵ}_0$ and, with ${\mu }_r\simeq 1$ and $n-1\ll 1$,The ideal-gas number density at the supplied pressure and temperature is $N=P/(k_{B}T)=2.47\times 10^{25}{\mathrm{m}}^{-3}$, yielding $\alpha =2.15\times 10^{-40}\mathrm{C}{\mathrm{m}}^2{\mathrm{V}}^{-1}$. For an explicit representative visible wavelength $\lambda =550\mathrm{nm}$,Direct-beam intensity follows $I(x)=I(0)e^{-x/L_{\mathrm{att}}}$. Equivalently $L_{\mathrm{att}}=3N{\lambda }^4/[32{\pi }^3(n-1{)}^2]$. Its strong wavelength dependence gives roughly $21$–$200\mathrm{km}$ across $400$–$700\mathrm{nm}$ if refractivity is held fixed. This is a uniform-density dry-air scattering estimate, not absorption or transmission through a vertically stratified atmosphere; aerosols and molecular depolarization corrections are outside the given model.