$sigma=5.67*10^-8fracWm^2K^4$ (Stefan-Boltzmann constant)
$S = 1367fracWm^2$ (solar constant)
$D = 1.496*10^11 m$ (Earth-Sun median distance)
$R = 6.963*10^8 m$ (radius the the Sun)
$T = (fracPsigma A)^frac14 = (fracS4 pi D^2sigma 4pi R^2)^frac14=(fracSD^2sigma R^2)^frac14 = 5775.8 K$
(Wikipedia provides 5777K because the radius to be rounded come $6.96*10^8m$)
This calculate is perfectly clear.
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But in Gerthsen Kneser Vogel there is an exercise wherein Sherlock Holmes estimated the temperature of the sun only knowing the root of the portion of D and also R. Allows say, he estimated this portion to 225, so the square source is about 15, exactly how does he involved 6000 K ? The value $(fracSsigma)^frac14$ has around the worth 400. It cannot be the approximate mean temperature on earth, i m sorry is about 300K. What perform I miss ?
astrodearteassociazione.org temperature blackbody order-of-magnitude estimate
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edited Nov 14 "14 in ~ 14:10
pentane
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The partnership of temperature in between a planet and also a star based on a radiative power balance is provided by the adhering to equation (from Wikipedia):
$T_p = temperature of the planet$ $T_s = temperature of the star$ $R_s = radius of the star$ $alpha = albedo of the planet$ $epsilon = average emissivity of the planet$ $D = distance between star and planet$
Therefore if Sherlock to know $sqrtfracR_sD = 0.06818$ and can estimate the Earth"s temperature $T_p$ as well as $alpha$ and also $epsilon$ climate he have the right to calculate the temperature top top the surface of the sun which is the unknown variable $T_s$.
Both $alpha$ and $epsilon$ have true values between zero and one. Speak Sherlock assumed $alpha = 0.5$ and $epsilon = 1$ (perfect blackbody). Estimating the temperature of the planet $T_p$ to it is in 270 K and also plugging in every the numbers us have:
Which is really near the true typical temperature that the surface ar of the sun, 5870 K. Case closed!
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edited Jul 14 "15 at 14:15
answered Nov 14 "14 at 6:10
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A stormy estimate of a body"s temperature in the solar mechanism is $$T=frac280KsqrtD_AU$$if us calculate the AU fraction from the Sun"s "edge" come its center, R end D = $4.65x10^-3$, and also substitute this right into the formula, the Sun"s temperature would be around 4100K.Not an extremely close to her 5776 K, however utilizes the square source of the R D fraction.
The formula reflects efficient temperatures. Yet peak, so called sub-solar temperatures, room $sqrt2$ times effective temperatures, which would certainly yield about 5800K.Clever Sherlock!
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edited might 30 "14 in ~ 10:32
answered may 29 "14 in ~ 23:39
Michael LuciukMichael Luciuk
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