The Sun's Power Output: Solving for P

AI Thread Summary
The Sun emits electromagnetic waves uniformly, with an intensity of 1.4 kW/m² at Earth's upper atmosphere. To calculate the Sun's power output, the formula I = P/A is used, where I is intensity and A is the surface area of a sphere with a radius equal to the distance from the Earth to the Sun. The correct area formula is A = 4πR², not 4πR³, as the latter refers to volume. The discussion also highlights confusion over the meaning of E² in the notes, which remains unresolved. Understanding these calculations is essential for determining the Sun's total energy output.
mikefitz
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The Sun emits electromagnetic waves (including light) equally in all directions. The intensity of the waves at the Earth's upper atmosphere is 1.4 kW/m2. At what rate does the Sun emit electromagnetic waves? (In other words, what is the power output?)

I = P/A

I know I need to convert 1.4kW/m^2 ==> 1400W/m^2
Then I set that value = to P/A

What I'm unclear about is, what is the Area I am supposed to insert in the equation?
 
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*also, looking over my notes, I have A= 4piR^2 - E^2

I assume R = distance from Earth to sun, but what would E^2 represent?
 
any ideas? This is one of two problems I have left for the weekend - thanks!
 
I'm not sure what your E2 term represents here, but your area should be the surface area of a sphere A = 4\pi r^3.
 
Hoot...r^3?
 
No surface "area" is proportional to r^2. Volume, which is not relevant here, is proportional to r^3.

mikefitz said:
*also, looking over my notes, I have A= 4piR^2 - E^2

I assume R = distance from Earth to sun, but what would E^2 represent?
Darned if I know! You didn't even tell us what A represents. It's certainly not a good idea to copy a formula to your notes without writing down what each of the parameters means!

You are told that " The intensity of the waves at the Earth's upper atmosphere is 1.4 kW/m2." Find the surface area of a sphere with radius equal to the distance from the sun to the earth. Find the area that a disk with radius equal to the distance from the center of the Earth to the "upper atmosphere" (not the surface area of that sphere- you want the "cross section area" that intercepts the suns rays). The total energy output of the sun, divided by that first surface area, is equal to the sun's total energy output (in kilowatts) divided by the 1.4 kw/m2 that you are given.
 
neutrino said:
Hoot...r^3?
Damn typo :blushing:
 
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