a) Intensity of solar radiation at Sun's surface: [tex]69,700 kW/m^2[/tex]
b) Temperature at the surface of the Sun: 5900 K
Explanation:
a)
The intensity of the Sun's radiation decreases following an inverse square law:
[tex]I\propto \frac{1}{r^2}[/tex]
where r is the distance from the source of radiation (the centre of the Sun). This can be rewritten as
[tex]Ir^2 = const.[/tex] (1)
The intensity of the solar radiation when it reaches the upper atmosphere of the Earth is
[tex]I_1 = 1.50 kW/m^2[/tex]
And the distance of the upper atmosphere from the Sun's centre is
[tex]r_1=1.50\cdot 10^{11} m[/tex]
Here we want to find the intensity of solar radiation at the Sun's surface, so at a distance of
[tex]r_0 = 6.96\cdot 10^8 m[/tex]
from the centre of the Sun.
Using eq.(1), we can find the intensity of solar radiation at the Sun's surface:
[tex]I_0 r_0^2 = I_1 r_1^2\\I_0 = \frac{I_1 r_1^2}{r_0^2}=\frac{(1.50)(1.50\cdot 10^{11})^2}{(6.96\cdot 10^8)^2}=69,700 kW/m^2[/tex]
b)
We can solve this problem by using Stephan-Boltzmann law:
[tex]J=\epsilon \sigma T^4[/tex]
where
[tex]J[/tex] is emittance, the power emitted per unit area
[tex]\epsilon[/tex] is the emissivity
[tex]\sigma = 5.67 \cdot 10^{-8} W/m^2 K^4[/tex] is Stephan-Boltzmann constant
T is the absolute temperature at the surface
Here we have:
[tex]J=I_0 = 69,700 kW/m^2=69.7\cdot 10^6 W/m^2[/tex]
[tex]\epsilon=1[/tex] for an ideal black body
Therefore, solving for T, we find the temperature at the surface of the Sun:
[tex]T=\sqrt[4]{\frac{I_0}{\epsilon \sigma}}=\sqrt[4]{\frac{69.7\cdot 10^6}{(1)(5.67\cdot 10^{-8})}}=5900 K[/tex]
Learn more about temperature and power:
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