Radioactive decay as direct rocket propulsion

When an atom emits a particle say an alpha, according to Mr I. Newton the atom should recoil backwards, however slightly.
So, if you took radioactive atoms with the most energetic and heavy particle ejections and cooled them down as much as possible so that thermal effects are reduced: would you be able to observe the atoms bouncing around like fleas?
And if you made a very tiny rocket with shielding on one side so that particles are directed in a certain direction only would it (possibly very slowly) start to put-put its way across the universe?

Yes, but the amount of “Thrust” is so weak as to be non-existant.

An isotope’s “burn time” is dictated by its half-life. Suppose you use plutonium or enriched uranium…the thrust would continue for thousands of years, but with the force of less than the weight of a piece of paper lying on the palm of yor hand.

I suppose a Doper has the math to describe the variables in the equation; mass of the vehicle, half-life of the isotope, mass of isotope at beginning of event, mass of isotope at end of event, etc.

I’m thinking of a more “vicious” element than plutonium/uranium that, although of short half life, would spew out lots of heavy and energetic particles.
It would only have to “fly” for a day or so; maybe even less.

The initial thrust is given by (to first order):

F = ln(2) × M[sub]Sample[/sub] × (2 m Q)[sup]1/2[/sup] / (M[sub]Atom[/sub] T)

Where M[sub]Sample[/sub] is how much stuff you’ve got, m is the mass of the decay particle, Q is the energy released per atomic decay, M[sub]Atom[/sub] is the mass of one atom before decay, and T is the half-life. In units I can grasp, that’s:

F = 982 M[sub]Sample[/sub] × (m Q)[sup]1/2[/sup] / (M[sub]Atom[/sub] T)

Where F is in metric tons (times g), M[sub]Sample[/sub] is in kilograms, m and M[sub]Atom[/sub] are in amu, Q is in MeV, and T is in seconds.

For U-238, M[sub]Atom[/sub] = 238, m = 4, Q = 4.27, and T = 1.4×10[sup]17[/sup]. If we let M[sub]Sample[/sub] = 1, then F = 1.2×10[sup]-16[/sup]. Not a lot. Clearly, it’s T that’s holding it back. There are two problems with getting a more active substance, though. First, where are you going to keep it until the launch. And second, notice that I called this the initial thrust. After you get started, the thrust will die off exponentially, with a characteristic timescale of T. So, in order to fly for a day, it would have to have a half-life of, say, a day. If we could find something with the same properties as U-238, except with T = 86400, that would boost our thrust up to a whopping 0.0002 tons per kilogram of sample. It couldn’t even lift itself, much less a rocket.

Your problem is that any element that has a long enough half-life that enough would still be around after a day doesn’t “spew out lots of heavy and energetic particles”. There is, very generally speaking, a reverse correlation between breakdown energies and half lives. Anything with a half-life as long as, say an hour, is going to spew out particles at a very low ratio of particle to total mass of the element. (Of course, by the end you will have only 1/2[sup]24[/sup] of the original material left, so the total thrust will be greatly reduced. This reduction gets much worse with shorter half-lives. [But then, you might find a path-way of radioactive by-products that would give off their own particles. The calculations would be interesting.])

I’m also having problems with you’re being able to direct your thrust. Doesn’t shielding normally absorb particles rather than having them bounce off backwards? If so, then your thrusts will be even fore and aft and your craft won’t move at all.

And you have to be able to have enough thrust to move your cooling system as well. (If your cooling is extremely effective your atoms will actually be moving relatively slowly, not like fleas.

I see on preview that Achernar has put what I said into real math. Just let my post reinforce his. Maybe this’ll work as alien high school science project, but not in reality.

Yeah, this is another good point I didn’t mention. In my formula, I assumed you have a perfect parabolic reflector. In real life, I imagine, it would be much worse.