What is the energy density of sunlight on Earth and near the surface of the sun?

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SUMMARY

The energy density of sunlight at Earth is calculated using the solar constant of 1350 W/m². The Stefan-Boltzmann equation is employed to derive energy density, with the formula L = 4π(D²)S for solar luminosity. Near the surface of the sun, the energy density can be determined by applying the solar luminosity divided by the surface area of the sun. The discussion highlights the importance of using precise equations and constants to achieve accurate results in astrophysical calculations.

PREREQUISITES
  • Understanding of the Stefan-Boltzmann Law
  • Familiarity with solar luminosity calculations
  • Knowledge of the solar constant (1350 W/m²)
  • Basic principles of energy density in physics
NEXT STEPS
  • Study the application of the Stefan-Boltzmann equation in various contexts
  • Learn how to calculate solar luminosity using L = 4π(D²)S
  • Explore Planck's Law and its relevance to energy density calculations
  • Investigate the implications of energy density in astrophysics and solar physics
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Students in physics, astrophysics researchers, and anyone interested in solar energy calculations will benefit from this discussion.

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[SOLVED] Energy Density of Sunlight

Homework Statement


Use the solar constant (1350 W/m^2) to calculate the energy density of sunlight at a)Earth and b) near the surface of the sun.


Homework Equations


Stefan-Boltzmann = (2pi^5)/(15(h^3)(c^2))
h=6.63x10^-34 j.s
L(solar luminosity)=4pi(D^2)S
S=1340 W/m^2


The Attempt at a Solution


I don't know what I'm doing. I'm assuming those are the more important things to use.
I'm getting frustrated with this and really need some help. I'm trying to plug in the power per area into the stefan Boltzmann eq to see if I can find the answer but it isn't working.
 
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Thought about it some...

Could I use Planks formula or Raleight Jeanes law to solve this? Since the wavelength is rather large, I'm thinking I could, but I'm not sure how to apply it.
 
Last edited:
dR/D(lambda)=dU/d(lambda)(4/c)
du=4dR/c
integrate
u=4R/c

where R is 1350(solar constant) and gives the energy density on earth.

for part b

L=4pir^2 S(solar constant)
S=L(luminosity)/4pir(of the sun)^2
then u=4r/c where R=S
 

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