# Why does a metal bar in space in the shadow of the earth not go to absolute zero?

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?

• 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. – MSalters Feb 19 '18 at 8:13
• 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. – KalleMP Feb 19 '18 at 11:37
• 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. – endolith Feb 19 '18 at 14:56
• 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). – LLlAMnYP Feb 19 '18 at 15:45

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)
• 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 $T_{\rm CMB}$ – Stian Yttervik Feb 19 '18 at 13:14