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

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Homework Help Overview

The discussion revolves around calculating the rate of solar energy incident on Earth, given its distance from the Sun and the Sun's power output. The subject area involves concepts from astrophysics and energy transfer.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the expression for solar energy incident on Earth, considering the relationship between power, area, and distance. Questions arise regarding the appropriate form of the equation and the interpretation of units.

Discussion Status

Some participants have provided hints about the necessary calculations and units, while others are reflecting on the implications of the total power versus power per unit area. Multiple interpretations of the problem are being explored, but no consensus has been reached.

Contextual Notes

Participants note the absence of specific information regarding the amount of work or time, which complicates the formulation of the problem. The focus remains on understanding the relationship between the variables involved.

student123
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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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