# Temperature of the Sun via Stefan-Boltzmann law [duplicate]

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?

## marked as duplicate by Kyle Kanos, JMac, sammy gerbil, John Rennie homework-and-exercises StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 29 '17 at 6:53

• σ = 5.670367(13)×10−8 W⋅$m^−2⋅K^−4$ – QuIcKmAtHs Dec 28 '17 at 13:06
• it is m^-2, not m^2 – QuIcKmAtHs Dec 28 '17 at 13:07

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.