Let's see what we have till now:
The Earth's distance from the center of sun: Ro = 1AU (150,000,000 km)
The sun's diameter: 1.39 million km (the sun's radius Rsun = 0.695 million km)
The solar flux (measured) at TOA (also called Solar constant So) So = 1,362 W/m²
σ = 5,67*10⁻⁸ W/m²K⁴, the Stefan-Boltzmann constant
The distance reverse square law So/Ro² = S/R²
The Stefan-Boltzmann law
Jemit = σT⁴ W/m²
By using what we have we estimate sun's Photosphere (effective) temperature: 5,772 K. (from Wikipedia)
At Earth's orbit distance from the sun an imaginary sphere emits: Jemit = So = 1,362 W/m²
When substituting in Stefan-Boltzmann emission law equation
Jemit = σT⁴ W/m² 1362 W/m² = σT⁴ W/m²
and solving for T we shall have
T = 393,68K
it is the temperature of the imaginary sphere with radius R = 1 AU T=394K or almost 110C.
It is the temperature of the "touch" of this sphere to Earth's surface on a cloudless midday in Equatorial zone.
Let's continue about effective temperature:
Sun's Photosphere (effective) temperature: 5,772 K
How Sun's Photosphere (effective) temperature: 5,772 K is calculated?
Here is how: So = 1,362 W/m²
By using the distance reverse square law So/Ro² = Ssun/Rsun² and substituting the values of R = 1AU (150,000,000 km) and Rsun = 0.695 million km = 695,000 km/150,000,000 km = Rsun = 695,000 /150,000,000 = 0.00463 AU
Ssun = So /Rsun² = 1,362 W/m² /(0.00463)² = 1,362 W/m²*46,581 = 63,443,921 W/m²
Sun's photosphere emission intensity
Jsun = 63,443,921 W/m² = σTsun⁴
solving for Tsun we obtain Tsun = 5,783.64 K the Sun's effective temperature at the Photosphere.
"The Sun is the star at the center of the Solar System. It is a nearly perfect ball of hot plasma, heated to incandescence by nuclear fusion reactions in its core, radiating the energy mainly as visible light, ultraviolet light, and infrared radiation. It is by far the most important source of energy for life on Earth. Its diameter is about 1.39 million kilometres (864,000 miles)"
"Temperature Center (modeled): 1.57×107 K Photosphere (effective): 5,772 K". "At the photosphere, the temperature has dropped to 5,700 K and the density to only 0.2 g/m3 (about 1/6,000 the density of air at sea level)."
"At the photosphere, the temperature has dropped to 5,700 K and the density to only 0.2 g/m3 (about 1/6,000 the density of air at sea level)."
An important, very important Notice:
Is it possible to measure with some kind of thermometer the temperature of gas at density 0.2 g/m3 ???
No, it is not possible.
Still we estimate sun's photosphere effective temperature.
Effective temperature of sphere is the S/B radiative temperature of the same radius sphere EM energy emission.
We have not a solid surface here, but we have a uniform for the imaginary surface emission intensity - and thus we calculate the photosphere's effective temperature.
Photosphere's effective temperature is not the gas' of the any measurements escaping density of 0.2 g/m3 measured temperature.
No. A gas of 0.2 g/m3 density cannot emit Sun's photosphere emission intensity of Jsun = 63,443,921 W/m²
The Sun's photosphere emission intensity Jsun = 63,443,921 W/m² at photosphere's radius of Rsun =695,000 km is the weakening with distance but still very powerful EM energy FLUX coming thru from the inner sun's nuclear fusion process.
We have calculated the Jsun = 63,443,921 W/m² by the use of the distance square inverse law... we do not have measured it.
Planets do not meet with this condition. Planets do not have uniform EM outgoing emission energy, so planets do not have uniform surface temperature.
The Stefan-Boltzmann planet blackbody emission equation, and, therefore, the planet effective temperature is only a mathematical abstraction, and that is why the planet effective temperature cannot be compared with the actual planetary mean surface temperature.