What Is Earth's Equilibrium Temperature Assuming Blackbody Radiation?

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SUMMARY

The equilibrium temperature of Earth, assuming it behaves as a blackbody, can be calculated using the solar constant of 1.36x10^3 W/m^2 and the Stefan-Boltzmann law (R=σT^4). The discussion highlights the importance of considering only the circular cross-section of Earth when calculating the energy absorbed from the Sun, rather than the entire surface area. This approach leads to a more accurate determination of Earth's temperature by ensuring the energy balance between incoming solar radiation and outgoing thermal radiation is correctly established.

PREREQUISITES
  • Understanding of the Stefan-Boltzmann law
  • Familiarity with blackbody radiation concepts
  • Knowledge of solar constants and their significance
  • Basic geometry related to Earth's cross-sectional area
NEXT STEPS
  • Calculate Earth's equilibrium temperature using the Stefan-Boltzmann law
  • Explore the implications of Wien's displacement law on blackbody radiation
  • Investigate the effects of albedo on Earth's energy balance
  • Examine real-world applications of blackbody radiation in climate science
USEFUL FOR

Students studying physics, climate scientists, and anyone interested in understanding Earth's energy balance and temperature dynamics.

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Homework Statement


The energy reaching Earth from the Sun at the top of the atmosphere is 1.36x10^3 W/m^2, called the solar constant. Assuming that Earth radiates like a blackbody at uniform temperature, what do you conclude is the equilibrium temperature of Earth?


Homework Equations



Stefan-Boltzmann law: R=\sigmaT^4, Wein's displacement law: (\lambdamax)T=2.898x10^-3 mK

The Attempt at a Solution



I know that the Earth radiates the same amount of energy it takes in if it is acting as a black body, but I am stuck at figuring out how much energy the Earth is actually absorbing from the sun. If I can figure that out I know what to do with the rest of the problem.
 
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If you look up the radius of the Earth and assume it absorbs all of the energy that crosses the circular cross section it presents to the sun, wouldn't that do it?
 
Ah, you're right, I was stupidly using the full surface area of the Earth instead of a cross section.
 

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