A certain imaginary stellar system had once a water world. But instead of evaporating away under the stellar wind, this lucky (or unlucky?) water world (made of only water and a gaseous atmosphere) interacted with a huge gas giant planet and got expelled from its parent stellar system. Now this planet flies away, solitary, without any nearby star.

My question is:

Will the temperature of this planet reach the cosmic background radiation temperature if left alone for long enough? Will it freeze solid? Will its atmosphere turn liquid? Will its atmosphere turn solid?

  1. Assume a planet the size of Earth.
  2. It is made of almost pure water.
  3. Its atmosphere is made out of mostly hydrogen and helium.
  4. It has no rocky core.
  • $\begingroup$ I beg to differ, if there is a layer of atmosphere to do enough greenhouse effect remember all objects even planet emits infrared (heat) this is just my baseless opinion anyway drink on me lol. $\endgroup$
    – user6760
    Apr 24, 2015 at 9:00

3 Answers 3


Eventually the planet will cool down, that's unavoidable.

The question is how long it will take and there are three things slowing that down.

  1. Space is a good insulator - heat loss by radiation is actually quite slow
  2. The planet contains a lot of mass already heated
  3. Planets generate a certain amount of heat themselves (mostly through nuclear processes). Not having a solid/rocky core will limit that process but you are still going to have some interesting water states near the center of the planet that might allow some heat generation.

So the planet will gradually cool but it will take thousands of years to do so. If you were to use the world as a setting for something you could choose any point in the freezing process and get the conditions you desire. Isolated pockets of the planet would stay warmer than the rest for longer too.


$$ t_{cooling} = \frac{-3Nk}{2\epsilon\sigma{}A} \int^{T_{final}}_{T_{hot}}\frac{1}{T^4}dT = \frac{Nk}{2\epsilon\sigma{}A}\left[ \frac{1}{T^3_{final}} - \frac{1}{T^3_{hot}} \right] $$

where $T$ is absolute temperature, $k$ is the constant of proportionality, $\sigma$ is the Boltzmann constant, $\epsilon$ is the emissivity of the surface, and $A$ is surface area.

This formula gives the time of cooling. In calculating, I made the assumption that it starts out at 70 degrees Fahrenheit. Also, I solved without taking in account the atmosphere, but I will explain why this won't matter much. Doing the calculation, the water freezes in 8434 years. Now, back to why atmosphere doesn't matter. The formula is an approximation for your planet because of the nature of the planet. In the end, the atmosphere won't affect the estimation so much that it needs to be taken into account. The amount of years I gave is the the shortest case, and due to the nature of the plan and the formula will be longer. 8434 years is the shortest time for freezing to occur

  • 5
    $\begingroup$ This answer would be better if the symbols used were defined. $\endgroup$
    – user
    Oct 26, 2015 at 9:47
  • 1
    $\begingroup$ I think the atmosphere does matter--the greenhouse effect applies to energy loss and thus will work just as well in deep space as around a star. I also think your math is wrong--I see no energy of fusion, that looks like the time to cool to 32F, not the time to freeze. Do we even know the heat of fusion for ice XI, anyway? $\endgroup$ Oct 27, 2015 at 1:16
  • $\begingroup$ What is $N$? You should take into account internal heating. Earth is still getting terawatts of heat from radioactive decay. $\endgroup$
    – kingledion
    Oct 21, 2016 at 15:18
  • $\begingroup$ @kingledion I clearly stated that my calculation was an approximation. I also stated that 8434 years was the shortest time for freezing to occur. Of course other factors such as radioactive decay would make it longer. $\endgroup$
    – Jimmy360
    Oct 21, 2016 at 20:51

Not necessarily (will it cool down). Apparently a rogue planet with a high pressure hydrogen atmosphere would be so well 'insulated' that the background nuclear decay in the planet would keep it warm. There could also be liquid water under an ice surface (WARNING: pdf), called a Steppenwolf Planet.

  • $\begingroup$ ...anyone want to lay odds that the authors chose that name for the band, and invented the bit about the rogue planet being like a lone wolf on the galactic steppe as an excuse? $\endgroup$
    – KRyan
    Oct 28, 2015 at 19:53

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