# What is space temperature around Earth?

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

• 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/… – userLTK Aug 2 '15 at 22:21
• "Black body temperature depends on color and reflective properties." - already said, a black body, so why asking for color? – Anixx Aug 2 '15 at 22:55
• @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. – AtmosphericPrisonEscape Aug 2 '15 at 23:32
• 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. – userLTK Aug 2 '15 at 23:38
• @userLTK tour first link says it is 6 C, not -18 C. – Anixx Aug 3 '15 at 0:58

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 $$T = \left( \frac{f}{4\sigma}\right)^{1/4} = 278.6\ K$$