zlacker

Space is a vacuum and yet here we are on a rock floating in space warmed by the sun and the temperate is actually pretty comfortable. Indeed, without the greenhouse effect it would be positively chilly. An important part of a thermos is that you have to use high albedo materials in the vacuum chamber or else it would lose heat too quickly to radiation.

A satellite as a whole will come to thermal equilibrium with space at a fairly reasonable temperature, the problematic part is that the properties of electricity make it easy to concentrate a good part of the incoming energy in a small area where the GPU is. Heat is harder to move than electricity and getting that heat back out to the solar panels or radiators requires either heavy heat pipes or complex coolant pumps.

>>Symmet+(OP)
a data center in space doesn't have a gigantic rock taking up most of its area, a data center in space is 100% data center 0% rock.

If it had the same data center to rock ratio as earth, it would just end up being earth in the end, and earth doesn't seem to be wanting to stick to its equilibrium either right now

>>its-su+g2
Well, they do have silicon, with some more additives they can make rocks in space! And throw them at earth, that will show em

>>its-su+g2
The rock in this case acts as extra thermal mass that makes it take longer to reach thermal equilibrium, but doesn't change what the ultimate thermal equilibrium is. Only the configuration of the parts of the surface that can absorb or radiate electromagnetic radiation do that. And because rock is a fairly good insulator we only really benefit from the top layer and if the sun went out we would all freeze in a week or so.

>>Symmet+54
it changes the amount of exposed area to release heat back into the universe. if you have a non-negligible amount of compute compared to earth, you are going to be approaching a non-negligible amount of space required to radiate that away, along with all the other costs and maintainability issues

>>its-su+f4
The formula for the equilibrium temperature for a sphere in sunlight is

    2 * pi * r^2 * L / (4 * pi * d) * (1 -a) = 4 * pi * r^2 * sigma * T^4

As you can see there are pi*r^2 on both sides of the equation, the surface area to cross section ratio of a sphere doesn't change as it gets bigger and so the equilibrium temperature doesn't change no matter how big the sphere is. (d is the distance to the Sun, nothing to do with the sphere itself).