Blackbody radiation is characteristic of every object in thermodynamic equilibrium and black bodies at constant uniform temperature.
At any temperature objects emit thermal radiation. EM radiation is emitted because inside the object, due to thermal motion of particles charged particles/dipoles start to oscillate, electromagnetic radiation is emitted because of these vibrations. If the object is a black body at constant uniform temperature, the radiation is called blackbody radiation. The energy emitted by any object is always finite with certain distribution over the frequencies with peak at some frequency. We cannot naively expect the energy emitted with all the frequencies carrying equal weight. This is a phenomenon which happens and is observed. This is explained quantum mechanically, infact this led to the development of quantum mechanics.
So a cavity with a small hole with EM radiation inside it is appropriate to study mathematically and is a near perfect blackbody because the hole allows negligible radiation to enter the cavity so that it affects negligibly the thermal equilibrium condition and we can have a very near thermal equilibrium and observe blackbody radiation from it. Rayleigh and jeans couldn't explain blackbody spectrum at higher frequencies, their law predicted infinte spectral radiance at infinite frequencies. Planck gave the solution to the ultraviolet catastrophe(infinite spectral radiance at infinite frequencies) and explained the spectrum of blackbody radiation by assuming the energy of the oscillators inside the cavity to be series of discrete values but not continuous which eventually results in spectral radiance going to zero at higher and infinite frequencies with peak at some frequency.
Radiations emitted by ordinary objects can be approximated as blackbody radiation, they are nearly in thermal equilibrium.
One of the importance is that to know the temperature of a star, the relation between the temperature and wavelength of the peak, called wien's displacement law, evaluated from planck's radiation formula, is used approximating the radiation to be blackbody radiation.
Interesting and complicated question. The things to consider:
"Black body radiation" assumes perfect absorption / radiation at all wavelengths. The greenhouse effect comes about from having absorption in the IR: the hot (short wavelength) radiation from the sun can penetrate the atmosphere, but the cooler earth radiates at a lower temperature - longer wavelength. And that longer wavelength light is reflected by the atmosphere (water, carbon dioxide, methane, etc). Think of the story of Winnie the Pooh visiting Rabbit's burrow. He goes in through the hole, has a "little smackerel" of honey (read: the whole pot), and then is too fat to get out again - commemorated on a postage stamp:
That's your photon. It had no difficulty penetrating the atmosphere as a short wavelength photon - but as a long wavelength photon it gets stuck as it tries to leave earth...
If you worry about your spaceship getting too cold (how big is it?) you should probably consider lowering its reflectivity - this directly scales with the heat loss. Notice how space ships are often "shiny metal". This isn't just because paint is expensive to lift into orbit (it is), but also to keep the emitted power down - keep the people inside protected from too much heat loss when not in the sun, and too much heat gain when they are. If you want to simulate the "Venus" effect you would want to create your own greenhouse effect - add a film that is transparent in the visible and opaque in the near IR.
Either way, your black body model needs to take account of the reflectivity as a function of wavelength - and instead of using the simple Stefan-Boltzmann law (which deals with total power per unit area), use the wavelength formulation (Planck law):
$$S_\lambda=\frac{2\pi hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}$$
But yes - the amount of heat that a large object loses through radiation is substantial, even when it is at room temperature. There's a handy calculation on wolframalpha.com - it shows that the heat loss for an emissivity of 0.1 is still over $40 W/m^2$ at 298 K. The best thing you can do to insulate yourself is not let the outer shell get so hot in the first place - if you used a double shell, with the outer being thermally insulated from the inner, then you can see how this will lower the power emitted at the outer shell comes to equilibrium at some temperature $T_o$.
Assuming that the outer shell reflects half its power back to the inner shell, and half to the universe (which is so close to absolute zero that we ignore the difference), you can write
$$\epsilon \sigma T_i^4 = 2 \epsilon \sigma T_o^4$$
since the outer shell loses heat from both surfaces; thus if the inner shell is at $298 K$ the outer shell temperature will be at 250 K, but the inner shield is now losing heat at
$$\epsilon \sigma (T_i^4-T_o^4) = \frac12 \epsilon \sigma T_i^2$$
In other words - you halved it. If you add additional skins, the heat loss will be further reduced.
I must admit that I did that last bit of analysis "by the seat of my pants". It makes sense, intuitively, that the heat loss is reduced by a radiation shield; I have never attempted to come up with a number before, nor do I remember seeing this analysis. There could be a blooper in here - in which case I would be happy to have someone point it out.
I did find an online book that seemed to follow a similar approach but had a cylindrical geometry and uses different reflectivity on the inner and outer faces, which complicated matter further. But they show that multiple layers of shielding can significantly reduce these heat loads - which was really what I tried to say.
Best Answer
Both in practice and in theory, the best way to get a black body spectrum is to make a small hole in the wall of a hot cavity. Any light entering the hole will probably bounce around many times, and eventually be absorbed, before it could leave the cavity.
The invariance of the spectrum follows from the Second Law of Thermodynamics. (This argument doesn't give any insight to the mechanisms, but is nevertheless enlightening.) Two cavities, made of different materials but at the same temperature, face each other. All the radiation from one enters the other and vice versa. If their total emitted radiation was different, there would be a net flow of energy from one to the other - which is forbidden by the Second Law (and the Zeroth law too, for that matter). If the total radiation was the same, but the spectra were different, you could come up with a simple scheme such as placing a bandpass filter between them, to make sure that if body A emits more than body B at a specific wavelength, only that wavelength is allowed through, still causing energy to pass from A to B.
Your questions, @CuriousOne:
It is applicable to any body, but only if it is actually black at all wavelengths. A microscopic object can also be a black body. For example, a plasma consisting of one free proton and one electron inside a perfectly reflecting box with a small hole is a black body. (See later answer for details.)
As long as there is some mechanism of generating photons of any required wavelength, thermodynamics will see to it that such photons are generated in the required numbers. If there is no such mechanism, the body will not be black. For example, if you found a material that was completely transparent to green light because it had no atoms capable of absorbing green light, then this body would not generate the green part of the black body spectrum - because it is not black.
The one-electron plasma I mentioned would be OK because the free electron can have any energy, and can therefore emit and absorb at any wavelength.
See previous answer.
Whenever free electrons are present, they will contribute to the emitted radiation through their translational motion. In a plasma, this mechanism is called bremsstrahlung. In a metal, it is called reflection. In an atom, emission can only occur if an electron is excited. An atom in the ground state cannot emit. A molecule can emit even if there is no electronic excitation, because it also has vibrational and rotational levels.