Calculate Solar Energy Incident on Earth at 1.5x10^11 m from Sun

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To calculate the solar energy incident on Earth from a distance of 1.5x10^11 m, one must consider the Sun's power output of approximately 4x10^26 watts. The discussion emphasizes the need to determine the total power incident on Earth rather than the power per square meter. The concept of a glass sphere surrounding the Sun is introduced to visualize how the Earth's radius affects the fraction of solar energy it intercepts. The equation P = W/T is deemed unhelpful in this context since the focus is on total power rather than work over time. Ultimately, the goal is to express the incident solar energy in watts.
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Hi, so I have been asked to find an expression that governs the rate of solar energy incident upon the Earth, if the Earth lies at a distance, D from the Sun of 1.5x10^11 m, and the Earth has a radius, RE. I also know the Sun has a power output of about 4x10^26 watts.

At this point I have considered looking at P = W/T, I think the main issue is I have no clue what the rate of solar energy incident upon the Earth is supposed to look like written out in equation form. I know the question is asking the rate at which the Earth absorbs energy from the sun, but I'm not sure what to do or where to go from here. Please help me! Thank you!
 
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The answer will have units "Watts per square meter".

Imagine a glass sphere of radius D surrounding the sun. All the light from the sun must pass through that sphere. It has area = ??. The Earth blocks what fraction of that area?
 
CWatters said:
The answer will have units "Watts per square meter".
My reading is that it's the total power incident on the Earth that's required, so that will be in Watts.
I have considered looking at P = W/T
Not useful here since there is no mention of an amount of work or a period of time. Just follow CWatters' hint.
 
haruspex said:
My reading is that it's the total power incident on the Earth that's required, so that will be in Watts.
I've had another read of the question and think haruspex is right. They probably want the total power in watts rather than the power per square meter.
 
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