In my last post about the Russian space program I said that cosmonauts regularly throw stuff in space so it will burn up and not result in permanent space junk. A reader asks whether you can actually do this.. Man, way to ask a hard question.
Orbital mechanics says "if a space vehicle comes within 120 to 160 km of the Earth's surface, atmospheric drag will bring it down in a few days, with final disintegration occurring at an altitude of about 80 km", and we can work out how much delta-v a cosmonaut has to impart to get the semi-major axis of the orbit of the debris below 160 km.
dVA = sqrt(GM*(2.0/rA - 1.0/((rA + rB) / 2.0))) - sqrt(GM/rA)
where rB = 160km + roE
where rA = 278km + roE to 460km + roE
where roE = 6378.1km
where GM = 6.67300 * 5.9742 * 10^24 * 10^-11
with the ISS at 278km the delta-v retrograde is 34.684365m/s or 77.5867148 mph, which is major league baseball.
with the ISS at 428km the delta-v retrograde is 77.243278m/s or 172.788292 mph, with is space cannon territory.
Ok, let's work out how long it will take to degrade from 250km as a throw to that altitude from the ISS is pretty easy most the time.
We need to know how big the thing we're throwing is, let's say 1m x 1m and 100kg, with a drag coefficient of 2.67.
a = 250km + roE = 6628100
darev = (-2 * pi * Cd * A * p * a^2) / m
darev = (-2 * pi * 2.67 * 1 * 2.62 * 10^-12 * 6628100^2) / 100
darev = -19.3094776
L ~ -H / darev
L ~ -58200 / -19.3094776
L ~ 3014.06393 orbits
P = sqrt(4 * pi^2 * a^3 / GM) = 5369.860522 seconds
L ~ 3014.06393 * 5369.860522 seconds
L ~ 187.32758 days
So the answer is: it's unlikely. It depends how high up the ISS is and how long you're willing to wait for the debris to fall below the 160km altitude. Just letting the part go without a throw will cause it to reenter eventually because no low-Earth orbit is stable, but if you want it to come down in just months you've gotta throw it, and note that you don't throw it "down", you throw it retrograde, and preferably at periapsis, but that's any point on a circular orbit.
Thanks for the question Ian.