From what I can tell from reading various articles about space, a metal bar on the side of the earth exposed to the sun will heat up to about +250 °F (120 °C) and on the shadowed side will cool down to about -250 °F (-160 °C). Well, that is a lot warmer than absolute zero which is about -460 °F. Why would the metal bar stay at -250 °F instead of cooling down to -450 °F or so?

  • $\begingroup$ Unfortunately, that's not how it works. The ISS wouldn't need such big radiators if it could cool down that much in Earth's shadow. $\endgroup$
    – MSalters
    Commented Feb 19, 2018 at 8:13
  • $\begingroup$ Are there no logs of micro-satellite temperatures that could give a clue. They will generate comparatively little internal heat from consumed solar energy. There has been at least one successful crowd funded micro-satellite on KickStarter. $\endgroup$
    – KalleMP
    Commented Feb 19, 2018 at 11:37
  • $\begingroup$ Do you have a good understanding of how thermal radiation works? Every object "seen" by another object is exchanging thermal energy with each other, so an object in the shadow of the Earth is being warmed by the Earth and vice versa, depending on which is cooler. $\endgroup$
    – endolith
    Commented Feb 19, 2018 at 14:56
  • $\begingroup$ The real question is how the metal bar goes down to even -160C. I did a quick calculation with Stephan-Boltzmann's law and found that it takes an hour for a 1mm-radius lead ball to drop from 300K to 297K (see this question too). $\endgroup$
    – LLlAMnYP
    Commented Feb 19, 2018 at 15:45

1 Answer 1


Even on the shadowed face there are other sources of radiation that will control the temperature of the bar. For example, on the example that you read, the earth itself will radiate!

You may think that if you remove the Earth and other bodies floating around, including the Sun then the temperature will drop down to absolute zero (0 K). But even then the bar will stay warmer, at $T_{\rm CMB} = 2.72 ~{\rm K}$ (or $-454.76$ F), the reason being that the universe itself is permeated from a rather uniform bath of radiation coming from the Big Bang: the Cosmic Microwave Background (CMB)

  • 16
    $\begingroup$ @AmbroseSwasey The Earth still takes up a large part of the rods field of view if it is orbiting Earth. This means that it a good amount of radiation from it, it has good visibility. Now consider how warm the Earth is. The coldest the surface drops is −128.6°F (on record), so the radiation source should be pretty well above -250°F, and an equilibrium temperature like that seems pretty reasonable. $\endgroup$
    – JMac
    Commented Feb 18, 2018 at 17:31
  • 1
    $\begingroup$ @AmbroseSwasey: Also, to be in Earth's shadow, it needs to be pretty close. $\endgroup$ Commented Feb 18, 2018 at 18:54
  • 6
    $\begingroup$ It also doesn't stay in the shadow that long and radiation is a slow way to transfer heat, so you should consider the heat capacity of the unspecified bar. $\endgroup$ Commented Feb 18, 2018 at 19:36
  • 1
    $\begingroup$ @JMac Surface temperatures may not be so relevant in this context since most heat radiation escaping Earth is emitted by the atmosphere. But the atmosphere also rarely goes below -128.6°F, so your point still holds. $\endgroup$
    – jkej
    Commented Feb 19, 2018 at 11:08
  • 1
    $\begingroup$ The issue is more heat flux out. Radiating is atrociously ineffective at low temperatures, it is proportional to ${T^4}$. The energy absorbed from the earth, the moon, micro impacts, etc will be enough to keep it far away from $T_{\rm CMB}$ $\endgroup$
    – Stian
    Commented Feb 19, 2018 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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