## College Physics (4th Edition)

(a) The average power output of the sun is $3.96 \times 10^{26}~W$ (b) The intensity of sunlight incident on Mercury is $9370~W/m^2$
(a) We can find the power of the source: $I = \frac{P}{A}$ $P = I~A$ $P = I~(4\pi~r^2)$ $P = (1400~W/m^2)(4\pi)~(1.5\times 10^{11}~m)^2$ $P = 3.96 \times 10^{26}~W$ The average power output of the sun is $3.96 \times 10^{26}~W$ (b) We can find the intensity on Mercury: $I = \frac{P}{A}$ $I = \frac{P}{4\pi~r^2}$ $I = \frac{3.96\times 10^{26}~W}{(4\pi)~(5.8\times 10^{10}~m)^2}$ $I = 9370~W/m^2$ The intensity of sunlight incident on Mercury is $9370~W/m^2$