# Steady State Temperature of an Object Orbiting the Earth

This may be irrelevant or stupid to ask but I couldn't come up with a good answer. At least, we could not agree on with my friend the other day.

I would like an estimate of the temperature of a human body orbiting the earth. I know this may seem strange and a bit funny..

From what I know or read (or maybe 'guess' would be a better word), any body of mass would cool down eventually because it radiates (i.e. most of the objects in space are cool). OK. But imagine that a body is accidentally thrown from a space shuttle orbiting the earth, what would happen? I guess first of all it depends on how it receives the sunlight..

According to @zephyr's answer, the steady state temperature would be around $271 K$, so in which distance to the sun one body should be to have a steady state temperature of (say) $295 K$? Or doesn't the distance make a difference?

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–  Qmechanic Jan 20 '12 at 23:58

Heat is transferred by 3 mechanisms: conduction, convection and radiation. In space, conduction and convection are basically absent, so radiation is the only relevant mechanism.

The radiation of a black body is given by the Stefan-Boltzmann equation, which tells us that the radiated power is proportional to $T^4$. An object in orbit, assuming it is not shadowed by another body, is radiative contact with three thermal reservoirs: the Sun, the earth and the cosmic microwave background. At steady state, the radiation received must balance the radiation emitted, so we can write the equation:

$A_{sun}T_{sun}^4 + A_{earth}T_{earth}^4 + A_{cmb}T_{cmb}^4 = 4\pi T_{object}^4$

Here the $A$s refer to solid angle subtended by the various objects. Plugging in the appropriate values will allow you to solve for the steady-state temperature of the object.

As an example, if we ignore the earth, we have $T_{sun} \approx 5800K$, $A_{sun}\approx 6*10^{-5}$, $A_{cmb}\approx 4\pi$ and $T_{cmb}\approx 0$, yielding $T_{object}\approx 271K$ - quite similar to the mean temperature of the earth.

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Thanks. I added another question after your answer. Doesn't the distance affect the steady state temperature? –  onur güngör Jan 21 '12 at 10:59
@onurgüngör: distance does, but that's taken into account by the solid angle subtended by the Sun, which scales as $1/r^2$. –  genneth Jan 21 '12 at 12:20
@genneth: Can I get the exact $r$ through $4\pi(295)^4 = (1/r^2)6*10^-5(5800)^4$ ? If it is, I get 1.19. What is the unit? According to the formula, $A_{sun}$ has no units? –  onur güngör Jan 22 '12 at 1:13