Answer
(a) The average power output of the sun is $3.96 \times 10^{26}~W$
(b) The intensity of sunlight incident on Mercury is $9370~W/m^2$
Work Step by Step
(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 $9370~W/m^2$
