Does colour/reflectivity affect equilibrium temperature in a vacuum? I take two objects, identical save for their surface colour, and place them in orbit around the sun (at equal radius of orbit). One has a very shiny/reflective surface, while one has a very matt/dark surface.
Can I say that these two objects will end up at the same temperature? My gut says yes, but I can't quite work out how to extend the reasoning from the situation of two objects in thermal equilibrium with each other (where colour/reflectivity cannot matter as they must reach the same temperature by 2nd law) to this scenario where they are emitting and absorbing at different wavelengths. 
Thanks in advance.
 A: The objects can equilibrate at different temperatures because the environment that you describe is not a uniform temperature.
While orbiting the sun, the object can "see" in one direction something that is ~3K and in another direction can see something that is ~3000K.
The object will exchange radiation with each, coming to an equilibrium temperature that is somewhere between the two.  By changing the efficiency of the exchange with one or the other, it can shift where the equilibrium point falls.
If you instead place it in an oven with a uniform temperature, then it will equilibrate at that temperature, regardless of the materials.
A: For any surface, the absorption and emission coefficient is the same at a given wavelength. You didn't specify whether your two surfaces were "equally grey" (that is, the reflectivity was constant for all wavelengths, just a different value) or "colored" (that is, one might be blue, and have low emissivity in the IR, while the other is white).
I believe that difference matters. In atmospheric science, we recognize albedo (the fraction of incident light reflected) and emissivity (the fraction of black body radiation emitted). When you have zero albedo and 100% emissivity, the equilibrium temperature of the object will be that of a black body in space. But if the albedo is high the object will be cooler than that; and if the emissivity is low, the object may be hotter. The equation (see this answer to see the derivation) is
$$T_o = T_S \sqrt{ \frac{ R_S \sqrt{\frac{1-\alpha}{\overline{\epsilon}}} }{2 D_{o-s} } }$$
Where I use the subscript $_o$ for the object and $_s$ for the sun.
 $R_s$ is the radius of the sun, $D$ is the distance from the (center of the) sun, $\alpha$ is the albedo (properly averaged over the wavelengths of incident light),  $\epsilon$ the emissivity (again, properly averaged over the wavelengths of the emitted light - by the Wien displacement law we know the emitted wavelength will be longer because the object is cooler than the sun; so in general $(1-\alpha) \ne \epsilon$ and the object's equilibrium temperature depends on both of these.
Now if the object is "uniformly gray" (whether $(1-\alpha)=\epsilon=0.9$ or $(1-\alpha)=\epsilon=0.1$), the result will be the same. So a black body, a grey body, and a white body, will all reach the same equilibrium temperature. But a "blue body" might not.
