34: Problem 34.45

INTRO:
The intensity of sunlight reaching the earth is 1360 W/m².

Part A:
What is the power output of the sun?
Express your answer with the appropriate units.

SOLUTION:
givens:
I = 1360 W/m²

By rearranging the Intensity formula we are able to solve, since I = P_source⋅1/(4πr²),
P_source = I⋅4πr²
r is the distance from the sun to earth and that's something we're going to have to google
r = 150,000,000 km = 1.5⋅10⁸km ⋅1⋅10³m/km = 1.5⋅10¹¹m
& r²=2.25⋅10²²m²
→P_source = 1360 W/m²⋅4π⋅2.25⋅10²²m²
=1360⋅4π⋅2.25⋅10²² W
P_source = 3.85⋅10²⁶ W

Part B:
What is the intensity of sunlight on Mars?
Express your answer with the appropriate units.

SOLUTION:
first google the distance between the sun and mars to get r
r = 2.28⋅10¹¹ m & r² = 5.20⋅10²² m²

& plugging into the formula we get
I = 3.85⋅10²⁶ W ⋅1/(4π⋅5.20⋅10²² m²)
≈ 598 W/m²