What is the equilibrium temperature that a black body will reach at the Earth's distance from Sun?

  • $\begingroup$ Black body temperature depends on color and reflective properties. Also, a rotating object or non rotating? Permanently facing the sun it would be fairly hotish, but a rotating object like the earth, the average black body temp of the Earth is somewhere around -18 degrees (our atmosphere traps about 27 degrees and the internal heat from the earth adds the other 5, bringing the average temp up to about 15 on the surface. Google black body temperature of earth or moon. en.wikipedia.org/wiki/… $\endgroup$ – userLTK Aug 2 '15 at 22:21
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    $\begingroup$ "Black body temperature depends on color and reflective properties." - already said, a black body, so why asking for color? $\endgroup$ – Anixx Aug 2 '15 at 22:55
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    $\begingroup$ @Anixx: Because any physical (non-ideal) black-body will have a finite reflectance that plays a decisive role in how much energy it finally recieves and can absorb. $\endgroup$ – AtmosphericPrisonEscape Aug 2 '15 at 23:32
  • $\begingroup$ I've seen the term used both ways. docs.kde.org/trunk5/en/kdeedu/kstars/ai-blackbody.html Didn't mean to not answer the question. $\endgroup$ – userLTK Aug 2 '15 at 23:38
  • $\begingroup$ @userLTK tour first link says it is 6 C, not -18 C. $\endgroup$ – Anixx Aug 3 '15 at 0:58

Assume you have a spherical blackbody.

The solar flux at the radius of the Earth is given to a good approximation by $L/4\pi d^2$, where $d = 1$ au. This is $f=1367.5$ W/m$^2$ (though note the distance between the Earth and the Sun has an average of 1 au).

If it is a blackbody sphere it absorbs all radiation incident upon it. Assuming this is just the radiation from the Sun (starlight being negligible), then an easy bit of integration in spherical polar coordinates tells us that the body absorbs $\pi r^2 f$ W, where $r$ is its radius.

If it is then able to reach thermal equilibrium and it entire surface is at the same temperature, then it will re-radiate all this absorbed power. Hence $$\pi r^2 f = 4 \pi r^2 \sigma T^4,$$ where $T$ is the "blackbody equilibrium temperature". Hence $$ T = \left( \frac{f}{4\sigma}\right)^{1/4} = 278.6\ K$$


This is not a simple question to answer. Outside the influence of the earth's magnetic field then "space" is actually filled with the solar wind, which has a high nominal temperature but low energy density - in other words the particles in the solar wind are moving very fast but there are not that many of them.

If you mean what temperature a black body in constant sunlight would reach then probably not that dis-similar from the temperature of the Earth, but you are not measuring the temperature of space then, but rather the intensity of solar radiation.


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