Question
(a) Calculate the power per square meter reaching Earth’s upper atmosphere from the Sun. (Take the power output of the Sun to be $4.00 \times 10^{26} \textrm{ W}$). (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of $1.30 \textrm{ kW/m}^2$ reaches Earth’s surface. Calculate the area in $\textrm{km}^2$ of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States’ energy needs $(1.05 \times 10^{20} \textrm{ J})$? Australia’s energy needs $(5.4 \times 10^{18} \textrm{ J})$? China’s energy needs $(6.3 \times 10^{19})$? (These energy consumption values are from 2006.)

a) $1.42 \textrm{ kW/m}^2$

b) $28.8 \textrm{ km}^2$, $A_{US} = 128000 \textrm{ km}^2$, $A_{AU} = 6600 \textrm{ km}^2$, $A_{CN} = 77000 \textrm{ km}^2$

Solution Video

# OpenStax College Physics Solution, Chapter 7, Problem 43 (Problems & Exercises) (6:48)     