Peter sent me images of the ideal crescent moon for a range of Earth albedos — 0.0, 0.29, 0.30 and 0.31. I made a map of the scattered light to see how much brighter the ES (Earthshine) is than the scattered light from the crescent. Over most of the dark side, ES does indeed dominate over the scattered component — which augers well for measuring its ratio relative to the BS (bright side).
Top left: ideal moon outside the atmosphere with NO earthshine at all (Earth albedo = 0) i.e. we see the BS (brightside) only.
Top right: scattered light from the BS only.
Bottom left: scattered light of the same ideal moon, but with the ES for albedo=0.30 included.
Bottom right: the difference between the top right and bottom left images.
Interestingly, one can subtract the modeled BS to obtain the ES over the entire lunar disc. The isophotes of the remaining light are then nicely round around the full, ES-only illuminated disc, as seen in the lower right panel. This might be a way to check that the solutions are coming out the way they should — i.e. fit the BS and scattered light from it only, remove it — and the rest should be the ES. Errors in modeling the BS will show up as asymmetries in the ES that remains. Perhaps we could think about solving for the ES directly, by fitting a lunar model with a BS only, and recovering the ES on the lunar face directly?