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Ember, issue 30, January 26, 1947

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Stanley's Atomic-Drive Propulsion Unit...continued...you will recall from Ember #29 that Norm Stanley suggests a drive composed of two masses of U-235, both of which are below critical mass, but whose sum is greater than critical. Thus, the distance between the masses is important, and with one mass cylindrically shaped with a concave, focussing face, and the other mass with a doughnut shape; it might be so arranged that the disintegration products would go blasting thru the hole in the doughnut, Norm continues: "Probably the idea's impractical, and objection, aside from the lethal amounts of radiation produced, would be the limitations on the size of the individual propulsion unit, since neither of the two masses could exceed the critical mass, which appears to be somewhere between one and ten pounds. However, let's try making a few rough calculations anyway, and see what we get:
First, how much energy do we get from the fission of U-235? It's reported that the fission of a single U-235 nucleus releases 200 million electron volts (200 Mev) of energy. An atomic weight of U-235 would, of course, be 235 grams and would contain 6.06 x (10^23) x 1.60 x (10^-6) = 1.94 x (10^20) ergs
This is a lot of ergs, but if the U-235 could be completely converted into energy, the annihilation of 235 grams would give by the E=Mc^2 formula:
E = 235 x ((3 x (10^10))^2) = 2.115 x (10^23) ergs
This is a little more than 1000 times the amount liberated by fission. This checks with the statement in the Smyth report that in fission the amount of matter actually converted to energy is 0.1% of the total amount which fissions. The other 99.9% are the fission products, which in our proposed rocket would be shot off to provide the 'kick'.
Now what would the 'exhaust velocity' of these fission products be? If we assume that all energy released appears as kinetic energy of the fission products and neglect the relativistic increase of mass at high velocities, we can get, by the formula for kinetic energy, E = (M x (v^2)) / 2 solved for v
v = square root (2 x E x M) = square root (2 x 1.94 x (10^20) x 235) = 30.2 x (10^10) cm/sec
This is approximately ten times the speed of light! And, of course, meaningless, as we neglected the relativity correction. The calculation does show, however, that the actual velocity attained would be fairly close to the speed of light as the relativistic mass increase only becomes appreciable as that velocity is approached. And, also, not all the energy appears as kinetic energy of the flying particles; furthermore, there is also a wide velocity range involved, with lighter particles moving faster than heavier ones. The electrons emitted would probably be close to light-speed, and the neutrons, too, probably over 100,000 miles per second. What we're interested in, though, are the heavy fission products, of about half the weight of the original nucleus. The Smyth report doesn't state their velocity; Hawley and Leifson, in their book, Atomic Energy in War and Peace, estimate it to be around 150,000 miles per second, which sounds like a wild guess; Campbell, in the Air Trails article, is more conservative, and has them loafing along at a mere 50,000 mps. For my calculations here I'll be still more conservative and give them the same velocity as that of the alpha particles from radium -- 10,000 mps.
So now let us examine the behavior of a rocket whose exhaust has a velocity of 10,000 miles per second... (To be continued next week)
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