In a German Wikipedia page, the following calculation for the temperature on the surface of the Sun is made:

$\sigma=5.67*10^{-8}\frac{W}{m^2K^4}$ (Stefan-Boltzmann constant)

$S = 1367\frac{W}{m^2}$ (solar constant)

$D = 1.496*10^{11} m$ (Earth-Sun average distance)

$R = 6.963*10^8 m$ (radius of the Sun)

$T = (\frac{P}{\sigma A})^\frac{1}{4} = (\frac{S4 \pi D^2}{\sigma 4\pi R^2})^\frac{1}{4}=(\frac{SD^2}{\sigma R^2})^\frac{1}{4} = 5775.8\ K$

(Wikipedia gives 5777K because the radius was rounded to $6.96*10^8m$)

This calculation is perfectly clear.

But in Gerthsen Kneser Vogel there is an exercise where Sherlock Holmes estimated the temperature of the sun only knowing the root of the fraction of D and R. Lets say, he estimated this fraction to 225, so the square root is about 15, how does he come to 6000 K ? The value $(\frac{S}{\sigma})^\frac{1}{4}$ has about the value 400. It cannot be the approximate average temperature on earth, which is about 300K. What do I miss ?


2 Answers 2


The relationship of temperature between a planet and a star based on a radiative energy balance is given by the following equation (from Wikipedia):

energy balance

$T_p = temperature\ of\ the\ planet$
$T_s = temperature\ of\ the\ star$
$R_s = radius\ of\ the\ star$
$\alpha = albedo\ of\ the\ planet$
$\epsilon = average\ emissivity\ of\ the\ planet$
$D = distance\ between\ star\ and\ planet$

Therefore if Sherlock knows $\sqrt{\frac{R_s}{D}} = 0.06818$ and can estimate the Earth's temperature $T_p$ as well as $\alpha$ and $\epsilon$ then he can calculate the temperature on the surface of the sun which is the unknown variable $T_s$.

Both $\alpha$ and $\epsilon$ have true values between zero and one. Say Sherlock assumed $\alpha = 0.5$ and $\epsilon = 1$ (perfect blackbody). Estimating the temperature of the Earth $T_p$ to be 270 K and plugging in all the numbers we have:

enter image description here

Which is very near the true average temperature of the surface of the sun, 5870 K. Case closed!

  • $\begingroup$ $6653K$ is "very near" to $5870K$ ? $\endgroup$
    – Peter
    Commented May 8, 2018 at 15:44
  • 3
    $\begingroup$ @Peter, considering the range of temperatures that exist in the universe (~0 K all the way up to tens of millions K and even higher), 15% accuracy is close. $\endgroup$ Commented May 8, 2018 at 15:49
  • $\begingroup$ @Joshua Sorry, the approximation might be good enough as a rough guess, but it is not "close". $\endgroup$
    – Peter
    Commented May 8, 2018 at 16:00
  • 2
    $\begingroup$ @Peter, how close is close? It's an arbitrary distinction I understand. My close is not your close in this instance. My point is, for Sherlock to know only one number and then make a guess on $\alpha$ and $\epsilon$ and get an answer that is within ~15%, that is quite good. $\endgroup$ Commented May 8, 2018 at 16:27
  • 1
    $\begingroup$ The book says the estimation was 6000 K--only one significant figure-- so you could easily imagine the estimation is good to + or - 1000 K. If Sherlock assumes $\alpha = 0.3$ instead (closer to the real value) then the estimated temperature of the sun is 6123 K, which rounds to 6000 K. $\endgroup$
    – pentane
    Commented May 9, 2018 at 23:39

A rough estimate of a body's temperature in the solar system is $$T=\frac{280K}{\sqrt{D_{AU}}}$$ if we calculate the AU fraction from the Sun's "edge" to its center, R over D = $4.65x10^-3$, and substitute this into the formula, the Sun's temperature would be about 4100K. Not very close to your 5776 K, but utilizes the square root of the R D fraction.

The formula reflects effective temperatures. However peak, so called sub-solar temperatures, are $\sqrt{2}$ times effective temperatures, which would yield about 5800K. Clever Sherlock!

  • $\begingroup$ If I understand the article right, the 5777K is called effective temperature. $\endgroup$
    – Peter
    Commented May 30, 2014 at 13:14
  • $\begingroup$ Peter, effective temperature is basically the average temperature of a body orbiting the Sun. Sub-solar temperatures occur at the Sun's zenith for a body. $\endgroup$ Commented May 31, 2014 at 19:04
  • $\begingroup$ I want you to be aware that my answer was an attempt to answer the question of how Sherlock might have determined the Sun's temperature using your R and D factors. It no way is a valid use of the formula I posted. It was simply a trick using a valid formula. But it answered your question. $\endgroup$ Commented May 31, 2014 at 19:10
  • $\begingroup$ Michael, the units on your equation don't seem to check out. $\endgroup$
    – pentane
    Commented Nov 14, 2014 at 6:11
  • $\begingroup$ pentane, very true. The formula is simply a useful approximation to estimate solar system body temperatures. It ignores factors like albedo, internal energy and atmospheric effects, which explains its simplicity. $\endgroup$ Commented Nov 24, 2014 at 22:11

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