# Temperature of a black-body in LEO on the dark side of the Earth

Questions about the temperature of something in space are often very hard to pin down (example), since there is radiative transfer to/from many different regions in the field of view at dramatically different temperatures - leading to different answers depending on an object's position and radiative properties. I want to ask about the most simple case of an object in Low Earth Orbit (LEO) I can think of.

Let's ignore the fact that an object in orbit on the night side of Earth will eventually move to the sunny side. Let's say there is a black-body at an orbital altitude above Earth on the opposite side as the sun. No internal heat production. Completely steady state. What would the temperature of the object be?

Some initial thinking:

For simplicity, we'll make it a plate. From this point, I don't know where to go exactly. The Earth occupies an even $2 \pi$ solid angle of the field of view. We know $T$ of the Earth and space, but do we need more information? The answer would probably be a function of the emissivity of the Earth.

Alternatively, maybe we would find $\dot{Q}$ from the Earth and assume that all outgoing radiation from the satellite is lost. This last argument is certainly true, but is a source of my confusion.

Why I ask, my confusion (not needed to answer the question):

What does it mean that the greenhouse effect insulates the earth radiatively? We have the surface at a given $T$, then some of the outgoing thermal radiation is blocked. What does that mean? Is the outgoing radiation then the $T$ of the colder upper atmosphere? Is that how it reduces the thermal radiation? Or is the thermal radiation the same average $T$, but at a lower intensity? This last explanation would seem to violate the 2nd law. If we consider the plate above the night side of Earth, and the far side of the plate is perfectly insulated from the rest of space... then it must reach equilibrium at the $T$ of Earth (be it the surface or upper atmosphere $T$, either is possible). Or did I interpret the law wrong? This is a common contradiction I run into with thinking about radiative heat transfer, and any correct solution to this problem should clear it up.

The concept of emissivity $<1$ makes sense to me. But that's only allowed because the surface will also proportionally reflect more light. Thus, a neighboring object absorbs less light from it, but it will then absorb more of its own reflection, leading to thermal equilibrium again, with the two objects at the same $T$. But that approach doesn't work for the Earth. As far as a satellite is concerned, any radiation it emits is gone, because the Earth is very very big. Would insulation via the greenhouse effect then just cause greater coupling between the satellite and the vacuum of space? You can imagine a $T$ of Earth and a temperature of space, with the object falling in-between them. Is the correct conclusion then that the object's temperature then lies closer to the temperature of space?

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–  Qmechanic Apr 11 '13 at 0:43

## 2 Answers

The Earth receives a certain amount of radiation from the Sun, and also generates a certain amount of heat from radioactivity in the core. It must radiate this amount of energy because otherwise it would heat up until the thermal emission matched the rate of energy input. It's this heat flux that you'd use when calculating how much your plate was heated by the Earth. The sums seem simple enough and I'm sure Google would be able to retrieve the values of heat received from the Sun and generated internally.

Given the heat flux from the Earth you can calculate a temperature using the Stefan Boltzmann law. This temperature wouldn't match the temperature of the Earth's surface, but then why should it? As you say, between the surface and space is a layer of insulating gas. The Stefan Boltzmann temperature would be some kind of average for the various parts of the Earth from which radiation is received. Increasing CO$_2$ or other greenhouse gases would change the temperature profiles between the surface and space, but not the overall amount of energy being emitted.

You can define an emissivity for the Earth, but I'm not sure how helpful this is. If you compare the temperature at the surface with the Stefan Boltzmann temperature you'd get an emissivity less than one, but then this isn't an especially useful comparison.

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Total geothermal heat is estimated at 40--44 TW of which 20--24 can be accounted for from uranium and thorium chain decays (as measured by large, low background LOS neutrino dectors such as Borexino and KamLAND). Heat from potassium 40 decays is not measured by these experiments due to thresholds. –  dmckee Dec 13 '12 at 20:00

I'm going to try to do the basic calculations as simply as possible. You can find the outgoing radiation intensity from the Earth:

http://en.wikipedia.org/wiki/File:AIRS_OLR.png

It seems to be on the order of $300 MW/m^2$, although it changes dramatically depending on what specific spot you're talking about.

The argument that the albedo of the Earth is functionally zero seems to be fairly strong. It would matter but only if the object was of a similar size as the Earth. To do the calculations, start with the Stephen-Boltzman Law:

$$P = 300 W/m^2 \times A = A_{total} \sigma \epsilon (T^4-T_0^4)$$

The plate has two sides so $A_{total}=2 A$. Of course, $\epsilon=1$. Once we've accounted for the heat input from Earth's outgoing longwave radiation, then we need to consider no other heat source and all radiation the object emits is lost. So the other temperature it exchanges with is the $2.7 K$ of space itself. We'll call that close enough to zero to complete the above equation.

$$300 W/m^2 = 2 \sigma T^4$$

With this, I obtain $-46^{\circ} C$, which is similar to the $-30^{\circ} C$ figure I've heard in other places.

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Cool! (pun intended :-) –  John Rennie Oct 25 '12 at 17:33