
NUCLEAR ENERGY IS A POOR CITY DESTROYER. A 1-kiloton conventional explosive has a blast equivalent to a 10-kiloton nuclear explosion, 10 times weaker, or is this a mistake?
The figure shows three maps of three powerful city center explosions, resized to be on the same scale. On the left is the Nagasaki nuclear explosion, on the right is the Hiroshima nuclear explosion (both 10-kiloton). In the center is the 2020 explosion of 3,000 tons of ammonium nitrate in the port of Beirut, with a blast equivalent to 1 kiloton.
Looking at the Beirut explosion chronicle and the image I received, one gets the strong impression that all three explosions are remarkably similar in their blast effect. Yes, we understand that the destruction radius for 1 kt and 10 kt of the same nature will differ by only a factor of (10/1)^(1/3) = 2.15 (the inverse-cube law for any explosion on the bottom of the air ocean). Nevertheless, the feeling remains that an equivalent nuclear explosion is clearly weaker in destructive power than a conventional chemical explosive. In fact, this is noted in many sources. And there's no particular mystery here. This is due precisely to the fact that, with equal explosive energy, the momentum, or destructive impulse, of a chemical explosion is clearly greater due to the greater mass of the substance involved. An explosion with a 3-ton chemical explosive yield will rupture any explosion chamber, but for equivalent 3-ton peaceful mini-thermonuclear explosions, explosion chambers of a quite reasonable size are designed.
Let's do some quick math. The port of Beirut contained 2,750 tons of compacted ammonium nitrate, which exploded with a yield of 1.5 kilotons. The density of ammonium nitrate is 1.72 tons/m³. Therefore, the entire mass that exploded there can be represented as a solid ball of fertilizer with a diameter of 14.5 meters.
In a chemical explosion, 90% of the explosive energy is converted into the kinetic energy of gas expansion (which creates a shock wave in the atmosphere). Therefore, we can calculate the notional total, initial velocity of the explosive ammonium nitrate "piston" using the law of conservation of energy as follows: (2 * 1.5 * 4.18E + 12 * 0.9/ 2750,000) ^ (1/2) = 2026 m/s. Assuming that the entire mass of gases at the moment of explosion acquired this velocity, the momentum of the explosion is the product of the mass of the explosive and the velocity: 2,026 x 2,750,000 = 5.57E + 9 kg m/s, which ultimately translates into the motion of the crushing shock wave.
What about a similar nuclear explosion of 1.5 kt? Such an explosion essentially occurs in a pinpoint device, say, 200 kg in mass. A momentary flash of energy and X-ray radiation occurs, which is intensely absorbed in the air around the explosive device, turning the area into a fireball, and it is in this fireball that the shock wave is generated. The empirical formula for the shock wave's separation radius from the fireball (at the ground's surface) states the following:
R = 47*q^0,324
Here, q is the explosive yield in megatons, and therefore the shock wave's separation radius for a 1.5 kt air explosion is 6.52 m. This means that the diameter of the sphere from which the shock wave is generated is 13 m. The air contained in this sphere essentially acts as the piston that creates the shock wave of a nuclear explosion. With a density of 1.25 kg/m³, we obtain that the mass of the nuclear explosion "piston" (plus the mass of the charge itself) is 1.622 tons. Moreover, based on the fact that only 45% of the explosion energy is converted into shock wave motion (the rest is various types of radiation), from the law of conservation of energy, as in the previous case, we calculate the average speed of the air "piston": (2 * 1.5 * 4.18E + 12 * 0.45/1622) ^ (1/2) ~ 59,000 m/s. Hence, the total momentum (mechanical impulse) generated in the fireball of a nuclear explosion: 59,000 * 1.622 = 8.4E + 7 kg * m/s.
This momentum is 58 times less than that of a chemical explosion of saltpeter of the same energy equivalent.
Of course, my calculation is a rough, "on the napkin" estimate. Nevertheless, the fact that a nuclear airburst was clearly inferior in its destructive effect on the buildings of Hiroshima to conventional explosives was noted back in 1947 by Sir P.M.S. Blackett in his book "Fear, War and the Bomb: Military and Political Consequences of Atomic Energy." A more thorough and dispassionate analysis (the British had good statistics on the effects of conventional bombing on cities) showed that the destruction wrought by the "Little Boy" bomb in Hiroshima, using conventional chemical explosives, could have been achieved using just 2,000 one-ton high-explosive bombs (of which only one-third the mass is explosive) dropped from conventional bombers. Since the yield of the Hiroshima explosion was then officially declared to be 20 kilotons, Mr. Blackett "discovered" a "missing" order of magnitude in the destructive effectiveness of the new American weapons compared to the old ones. Of course, part of the deficiency can be attributed to the fact that a single explosion causes excessive destruction at the epicenter, while the effect drops off sharply with distance to the periphery. However, this is only part of the "missing" factor. The Beirut explosion was also a single-point explosion, and with 10 times less energy, it produced destruction comparable to the atomic bombings. The second factor is that a nuclear explosion is a pulsed release of enormous energy, which is not immediately converted (as in a chemical explosion) into the movement of matter, into a direct high-explosive shock wave impulse. The flash of light must first be absorbed by the air in the fireball, which becomes a "piston" for the shock wave, and due to the low density of air, it is a very poor "piston." Moreover, as the explosive yield increases from the nominal 20 kt (to 200, 2000 kt), the situation with the mechanical efficiency of using the explosive energy to destroy objects around it only worsens. This means that nuclear weapons, especially high-yield ones, are very ineffective weapons of destruction from a physics perspective. One suspects that this is precisely why the "paper" city of Hirashima was chosen for the demonstrative destruction of people and buildings, and its population, like ducks, was long trained not to react to the appearance of a pair of American planes in the air. The goal was to achieve a demonstratively exaggerated effect from the new weapon, one that could never be replicated. The severed head of Medusa can only work once.
Therefore, all this frightening talk about terrifying nuclear megatons hanging over the world like the Sword of Damocles, the eternal appeal to Hiroshima as a model for calculating casualties and destruction (Nagasaki, for example, is used much less frequently and only in passing, because the effect there was much more realistic) is blatant manipulation, essentially a pure lie. The military knows the truth. It was the actual, physical effectiveness of nuclear weapons, after being studied, that caused great disappointment. That's why recently there's even been talk about high-precision weapons being able to replace nuclear weapons! For real warfare, that's (almost) true. But why "disappoint" a flock of self-terrified civilians with this? They (especially the humanist physicists) so desperately wanted to invent a doomsday weapon and thus end wars on Earth once and for all! It so flatters their vanity and their humanist pride!