# What's wrong with this temperature-in-space calculation?

I'm trying to calculate the steady-state temperature of a body in space, but my numbers are coming up much too small. For example, for a 1-meter cube, I'm getting a temperature of 194 K (or -81 C). I'm hoping someone can spot where I'm going wrong.

I'm working from the equations here, which are based on energy intensity (W/m^2) at the surface. The example given assumes that solar radiation the spacecraft is uniform over its entire surface area. I can't make that assumption, so I'm assuming the sun illuminates only one face of the cube, i.e. 1m^2 of area. I'm also using the more accurate figure of 1361 W/m^2 for solar intensity, so this gives 1361 W insolation. As in the example, I'm assuming an absorptivity of 0.3, so we multiply that by 13.61 MW to get 408.3 W absorbed energy.

Now here's where I may be stepping in something: to use the formula given, I need energy input per unit surface area. So I'm just dividing this absorbed solar energy by the total surface area of the cube, 6 m^2, to get Q_in = 68.05 W/m^2. This is quite a lot less than the 300 W/m^2 in the example. But it doesn't seem reasonable to expect the sun to be shining on the craft from all directions, does it?

Anyway, once I have Q_in, I just plug it into the formula (using an epsilon of 0.85), and 194 K pops out, which seems too cold.

So, where have I gone wrong?

• Fun fact: Earth would be frozen over without the greenhouse effect (and if it were perfectly conductive). Note that your factor of 6 for a cube becomes a factor of 4 for a sphere, so a cube will be colder. – user10851 Sep 16 '14 at 17:50
• Joe, many spacecraft have onboard electrical heaters that keep the internal electronics, fuel tanks, transmitters, etc. at a desired temperature (often ~0-20 C, depending on what's being heated). Some instruments, like fluxgate magnetometers, have their own heaters because the instrument's response is temperature dependent. – honeste_vivere Jan 19 '15 at 18:01

## 1 Answer

You are doing fine. You are correct to say the sun only shines on one side of the spacecraft, which the page you link to misses. Increasing the radiating area by a factor of six will decrease the temperature by a factor $\sqrt[4]6 \approx 1.565$ Dividing their $285$ by $\sqrt [4]6$ and multiplying by $\sqrt [4]{1.36}$ (to correct for your more accurate insolation value) gives $196.6$ Why do you think you are wrong? In reality, most spacecraft have solar panels to capture more energy than the body does. The electrical dissipation keeps the spacecraft much warmer.

Incidentally, on that page they consider the energy radiated from deep space at $3K$ to the spacecraft. Though correct, it is truly negligible and complicates the equations.