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If a piece of metal were orbiting the Sun at the distance from earth to Mars, always facing the sun, what would it's temperature be.

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    $\begingroup$ Is it shiny and smooth or rough and dark? $\endgroup$ – Organic Marble Feb 10 at 14:44
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    $\begingroup$ Perfectly clean or oxidized? The problem is that an emissivity is needed for visible light and for thermal IR. Also, without any more context, this is really purely a physics question and not specific to Space Exploration. $\endgroup$ – uhoh Feb 10 at 14:58
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    $\begingroup$ Copper or gold or alumin or silver, that would result in a different temperature. $\endgroup$ – Uwe Feb 10 at 16:06
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    $\begingroup$ This is actually a complex problem. As already commented, the precise composition of the metal is very important. In some cases, relatively minor constituents of an alloy can make a big difference. Other factors: the albedo and emissivities of the surfaces (ex: oxidized or not, coated or not) both front and back, and if it's very thick, the sides as well; and the thickness of the metal, since that establishes the net thermal conductivity to the back side, which radiates heat to space. Also I'm unsure what you mean by "...at the distance from earth to Mars..." $\endgroup$ – Tom Spilker Feb 10 at 16:29
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    $\begingroup$ +1 for the answer, it is a very good answer to a very vague question. $\endgroup$ – Magic Octopus Urn Feb 10 at 17:30
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A single number is impossible to give, unless you exactly specify the kind of metal. So I'll answer in the general sense.

The first question to answer is "How much energy does the sheet absorb?". This is given by the Inverse-Square Law:

$$ I={\frac {P}{A}}={\frac {P}{4\pi r^{2}}}$$

For Mars, this works out to $~ 589\frac W {m^2}$ of energy.

The next thing we need to consider is what happens with the energy that the metal sheet absorbs? If it just absorbed the energy, it would get hotter and hotter, melt, then turn into a plasma and at some point leave the realm of physics. What happens is that the metal needs to shed energy/heat. The three principal mechanisms of heat transfer are

  • convection (transfer of heat in gases and fluids)
  • conduction (transfer of heat between touching solid objects)
  • radiation (transfer of heat via thermal radiation)

Obviously, only the third option is available to our sheet of metal. So basically, emission (how much energy is radiated away) and absorption of energy must be equal. Lets call the emitted energy $\Phi_e$ and the absorbed energy $\Phi_a$, then we can state this formally as:

$$\Phi_e = \Phi_a$$

This radiation equilibrium between emitted radiation and absorbed radiation basically applies to every body in thermal equilibrium, including non-ideal black bodies that do not radiate at maximum power. After all, a constant temperature and thus a thermal equilibrium will be reached for every object after a certain time, in which emission and absorption must take place to the same extent. This law is also called Kirchhoff’s law of radiation. (Source).

The unfortunate thing is that we don't know $\Phi_a$ yet. Materials have properties, and three important ones are the spectral absorption component $\alpha$ (Absorption), the spectral reflection component $\rho$ (Reflectance) and the spectral transmission component $\tau$ (Transmittance).

When radiant energy reaches a surface, the energy can be absorbed, transmitted (through), or reflected (or any combination). The sum of these three effects equals the total energy transmitted, and the parameters that describe these three phenomena are given by $$\alpha + \rho + \tau = 1$$ (Source)

So if we know the properties of the material, especially absorption rates, we can calculate how much of the $589 \frac W {m^2}$ our sheet of metal absorbs!

But we still don't know how hot it becomes. But there is a law for that as well:

The radiative heat transfer rate is given by the Stefan-Boltzmann law

$$Q_T = \sigma A T^4$$

where σ is the Boltzmann constant and A is the surface area of the radiating source. The temperature is in an absolute scale (°Kelvin, corresponding to °C, or °Rankin, corresponding to °F)

However, this applies to a perfect black body. Our metal is not. We still need another thing:

In purely formal terms, the emission power of a real body can also be expressed with a factor which then indicates the ratio of the emitted radiation of the real body compared to an ideal black body! This factor is referred to as emissivity ε. (Source)

Plugging this together, we end up with

$$\Phi_{e,real} = \varepsilon \cdot \sigma \cdot A \cdot T^4 $$

And that has to equal $\Phi_a$.

Note, however, that if you consider your metal sheet to be a plane, you absorb only on one side and emit on two sides. Hence you get a 1:2 ratio between the absorbing and emitting area, which you need to account for.

Furthermore, if the thickness of the metal becomes non-negligible, you need to consider all 6 sides. But other than the ratio, the areas cancel out in this formula.

So, if you decide what properties the metal sheet has (especially emissivity and absorption rate), you can plug in the numbers here and find out how hot the metal becomes, everywhere in space, not only in Mars orbit.


Things I deliberately glossed over:

  • out-gassing or loss of mass in general
  • not being in a perfect vacuum (interaction with other particles)
  • other sources of radiation, especially interstellar radiation
  • non-uniformity of the heat distribution
  • and probably a few other things...

Further reading:

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    $\begingroup$ The cube (and other 3D geometries) gets complex. This is a class of problem my thermal-systems colleagues at JPL commonly face. Of course the sun-facing side is the hottest, but though the other five sides are equal in area to the sun-facing one, their temperatures are not uniform — nor is the temperature of the sun-facing side uniform! The temperature of any given spot on the unilluminated surface is a complex interaction of radiation from that spot and conduction from (and to!) the 2pi steradians of metal around that point. Edges are special (1pi), and corners even more so (pi/2). $\endgroup$ – Tom Spilker Feb 11 at 2:49

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