Chapter 41: Problem 36

Question: (a) Show that the average intensity of the solar radiation that falls normally on a surface just outside the Earths atmosphere is $I=1.3~{\rm kW/m^2}$. (b) What radiation pressure is exerted on this surface, assuming complete absorption? (c) Hows does this pressure compare with the Earths sea-level atmospheric pressure of $101.3~{\rm kPa}$.

Solution: The power output of the sun is $P~=~ 3.9\times 10^{26}~{\rm W}$ (page A4), and the mean distance to the earth is $r=1.5\times 10^{11}~{\rm m}$, which gives an intensity at the earth of

$$I~=~{P\over 4 \pi r^2}~=~1.38\times 10^3~{\rm W/m^2} \ \ \ .$$ (16)

Radiation pressure for complete absorption, can be found by computing the rate of momentum absorption per unit area. We see that the average pressure is the average power divided by the speed of light,

$${\rm Pressure}~=~{ {\rm Force}\over {\rm Area}}~=~{1\over {\rm Ar... ...{dE\over dt}~=~{{\rm Intensity}\over c}~=~4.6\times 10^{-6}~{\rm N/m^2} \ \ \ .$$ (17)

Since $1~{\rm Pa}=1~{\rm N/m^2}$, we see that the radiation pressure we have computed is tiny in comparison to atmospheric pressure.