Data comes from NASA planetary fact sheets (amazing resource btw).
Overall equation structure:
Circular area that can receive sunlight: pi x radius ^ 2
Total incoming power : solar irradiance x circular area
Spherical area : 4 x pi x radius ^ 2
Black body radiation : stefan-boltzmann constant x Area x Temperature ^ 4
You know have incoming power and outgoing power.
Percentage : 100 x (Outgoing - incoming) / outgoing
I don’t think so.
Even out at Mars you already have significantly diminished solar incidence.
I think that past Saturn you probably start to have so little incoming solar energy that it’s almost impossible to retain it.
EDIT:
Saturn receives around 1% of the solar irradiance of earth.
Pluto receives 0.064%. less than 1W/m2.
With a radius of 1188km, the absolute maximum incident solar energy is 3.8E12 W. (Assuming no efficiency loss as the angle of incidence decreases due to curvature)
The radiating surface is the full sphere, and using Earth’s black body temperature of 254K.
Therefore, Pluto would be radiating approximately 5.67E-8 x 254^4 x 4 x pi x 1188000 ^2 = 7.38 E14.
In other words, you would need to retain at least 99.5% of all energy emitted by pluto. Mirrors reflect around 95% of visible light, and infrared is even more difficult to reflect.
Another Earth
50/50
Grave of the fireflies
5 centimetres per second
The greenhouse effect still has a limit to how much it can trap.
At the end of the day infrared radiation is still basically light.
Even on the cloudiest day, or when there is super dense smoke or ash, it is still not pitch black out. Some light gets through. If you are looking into a mirror, it might seem like it reflects 100% of light. But they only reflect around 95%.
You would require something which can let through 100% of all sunlight, but still trap 99.5% from leaving.
You could have a look at how one-way mirrors work, to understand the percentages of light passed through and reflected.