Question

Radiant energy from the Sun arrives at Earth with an intensity of 1.5kW/m2.1.5 kW/m^2. Making the rough approxima- Sion that all photons are absorbed, find (a) the radiation pressure and (b) the total force experienced by Earth due to this “solar wind.”

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a) The total energy is by definition is the energy of a single photon times the number of photons emitted (nn), equivalently, you can define it as the power (WW) emitted in a given period of time. Starting with this definition and considering the variation in momentum Δp\Delta p in a certain time interval Δt\Delta t, we could derive an expression for the pressure (PP) that relate it to the problem given quantities.

WΔt=nhf=nhcΔλW=IA,Δλ=hΔpIAΔt=nΔpcFtot=nΔpΔt,P=FtotA=nΔpAΔtP=Ic=1.5×103W/m23×108m/sP=5×106Pa\begin{gather*} \therefore W \Delta t = nhf = \frac{nhc}{\Delta \lambda}\\ \because W = IA \:\:,\:\: \Delta \lambda = \frac{h}{\Delta p}\\ \therefore IA\Delta t = n\Delta pc \\ \because F_{\text{tot}} = n \frac{\Delta p }{\Delta t} \:\:,\:\: P = \frac{F_{\text{tot}}}{A} = \frac{n\Delta p }{A\Delta t} \\ \therefore P = \frac{I}{c} = \frac{1.5 \times 10^3\:\:\mathrm{W/m^2}}{3\times 10^8\:\:\mathrm{m/s}} \\ \boxed{\therefore P = \mathbf{5\times 10^{-6} \:\:Pa} } \end{gather*}

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