$sigma=5.67*10^-8fracWm^2K^4$ (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.

You are watching: How do you measure the temperature of the sun

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 ?

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edited Nov 14 "14 in ~ 14:10

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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

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