1.5 Shock wave radius of a nuclear explosion Figure 1.7 shows the site and shock wave of Trinityr explosion, the rst test of a full- scale 20 kilotons nuclear bomb at 5:30 a.n1. on 16 July 1945, in New Mexico, USA. The instantaneous energy release from a nuclear explosion causes a shock wave, whose inside pressure is thousands times greater than outside. This pressure difference can push down trees and structures, and tear apart all kinds of objects. The shock wave Figure 1.7 Left: The site of the Trinity nuclear explosion, 16 July 1945. Right: The shock wave at time t = 0.025 [sec]. may be assumed to be spherical and has radius R at it time after the explosion. Given the nuclear energy E, calculate the shock wave radius as a function of time, and hence predict the shock wave's arrival time and prepare for protection. Shock wave occurring in atmosphere due to the supersonic compression of the air from one side so that the air mass from the side accumulates, cannot escape, builds pressure, develops a large pressure difference with respect to the other side, and hence forms a shock Two critical elements here are supersonic push and compressible air. Thus, shock wave radius should be related to density p of a compressible air, total energy E, and time t. Because atmospheric pressure is small compared with the pres- sure caused by the nuclear explosion , gravity can be negligible. Thus, we assume the following R = aEprtC. (1.32) The dimension of the above equation is [R] = [aHEplbltl (1.83) which leads to L = 1 X (MLZTZ)G(ML3)5TC = Ma+bL2a3bT2a+c. (1.84) The exponents of both sides of this equation should be equal: a + b = 0, (1.85) 2a 3b = 1, (1.86) 2a, + c = 0. (1.87) These three equations have a unique solution a: 1/5,b= 1/5,c=2/5. (1.88) You can obtain this solution by a simple R programming code to be discussed in the next chapter, or you can solve it by hand. Observing that the rst two equations have only two variables a. and b, you can thus solve these two equations rst. Multiplying the rst equation by 3 and then adding the results to the second equation, you obtain 50. = 1. (1.89) Hence, a. = 1/5. (1.90) The rst equation implies that b = a. = 1/5. (1.91) The third equation implies c = 20. = 2/5. (1.92) Therefore, R = gal/5p1/5t2/5. (1.93) or 1/5 2 R=a + 211115. (1.96) p Any of the above three formulas can be used to predict the position of the shock wave for a given time, if a is known. Yet, it is not easy to evaluate this a by an experiment since such an experiment is too expensive. By solving another mathematical model, Cambridge University uid dynamicist G. 1 Taylor (1886-1975) estimated that a = 1.0. The shock wave propagation speed is 1/5 11 = \":7? = g (g) 13/5. (197) Because of the negative power, the shock wave propagates very fast initially within the rst second, and slows down after the rst second. Still another way of writing the energy-timeradius equation is _ R59 E _ V\" (1.98) This allows one to estimate the power of an nuclear bomb using news reports on the the shock arrival time at a given location. 1.6 (a) Draw at least two diagrams to illustrate the nuclear shock wave problem in Section 1.5. 1/5 (b) Write down the derivation details for the result equation: R = a (g) 1525. (C) Discuss the problem assumptions and the result