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..

Anyways, thanks for your answers!


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?


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.

  • $\begingroup$ Thanks. I added another question after your answer. Doesn't the distance affect the steady state temperature? $\endgroup$ Jan 21 '12 at 10:59
  • 1
    $\begingroup$ @onurgüngör: distance does, but that's taken into account by the solid angle subtended by the Sun, which scales as $1/r^2$. $\endgroup$
    – genneth
    Jan 21 '12 at 12:20
  • $\begingroup$ @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? $\endgroup$ Jan 22 '12 at 1:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.