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This question already has an answer here:

Let's calculate the sun's surface temperature using the Stefan-Bolztman law in the form

$$\langle w\rangle=\frac{4\sigma}{c}T^4$$

where $\sigma=5.67\cdot10^{-8}$ W$\cdot$m$^{-2}$/K$^4$ is the Stefan–Boltzmann constant, $\langle w\rangle$ is the timep-average of the energy density of the sun's radiation, and $T$ the temperature we are looking for.

I am given the value $I_0=1.35$ kW/m$^2$ of the solar constant (the power per unit area of the solar radiation on a panel oriented perpendicular to the sun's rays in the high Earth atmosphere).


From electromagnetism we know that the magnitude of the Poynting vector is simply $wc$ and the power per unit area $I_0$ is the time-average of the magntitude Poynting. Hence $I_0=\langle w\rangle c$.

We thus have $T=\left(\dfrac{I_0}{4\sigma}\right)^{1/4}=277$ K

which is obviously false as the real value is $5760$ K.

What is wrong with this temperature calculation?

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marked as duplicate by Kyle Kanos, JMac, sammy gerbil, John Rennie homework-and-exercises Dec 29 '17 at 6:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ σ = 5.670367(13)×10−8 W⋅$m^−2⋅K^−4$ $\endgroup$ – QuIcKmAtHs Dec 28 '17 at 13:06
  • $\begingroup$ it is m^-2, not m^2 $\endgroup$ – QuIcKmAtHs Dec 28 '17 at 13:07
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What is wrong with this temperature calculation?

You have calculated the effective surface temperature of a ball with a radius of one astronomical unit that emits 1.35 kilowatts per square meter at its surface. The Sun is a ball with a radius of 695700 kilometers. At the surface of the Sun, your 1.35 kilowatts per square meter becomes 62.4 megawatts per square meter, resulting in a surface temperature of 5760 kelvins.

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Distance to the sun, or angle subtended by the sun?

The apparent diameter if the sun is half a degree, about 0.01 radian. The solid angle is then about π × 0.01^2/4. This is less than 4π (the whole sky) by a factor of $10^{-4}/16$. So the Sun is hotter than what you calculated by the fourth root of that: a factor 20.

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  • $\begingroup$ Why does it matter. $\endgroup$ – Joshua Benabou Dec 28 '17 at 13:11
  • $\begingroup$ @JoshuaBenabou You calculated the temperature of the upper atmosphere of the Earth. (Because the Earth will radiate as much as it absorbs from the Sun.) $\endgroup$ – Pieter Dec 28 '17 at 15:46

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