Ask HN: Acceleration of a Drop Falling Through Mist?

Long, long ago, in my college days, I worked the problem of a drop falling through a mist (in a vacuum). As it falls, it sweeps out a volume of the mist, and so the mass of the drop keeps increasing. So you have

mg = ma + v dm/dt

and the surprising result is that a = g/7.

I thought this was a fun problem back then, and I've worked it at least once since, just for amusement value. But now when I try, I'm missing a step.

I know that the mass is proportional to r^3, and the volume swept out is proportional to vr^2. Therefore the volume swept out is proportional to vm^(2/3). I think that's part of the answer, but I don't remember the next step. (I think I'm only missing one piece...)

Before anyone asks: No, I am not some college kid trying to get the internet to do my homework. I passed the class where this problem came up back in 1983. Also, I suspect most universities are on break right now.