Nuclear detonation X-ray source

The temperature of an X-ray fireball generated by a nuclear detonation can reach 10⁶-10⁷ K. At this temperature, the substances in the area surrounding the detonation are completely ionized, forming high-temperature plasma [14]. The plasma exhibits a large contrast with its surroundings and can be approximated as a blackbody [3, 15]. The normalized energy spectrum can be represented by the Planck blackbody spectrum as follows: $\begin{array}{c}P\left(E,T_{\text{X}}\right)=\frac{15}{{\left(\pi \times T_{\text{X}}\right)}^4}\frac{E^3}{\exp \left(E/T_{\text{X}}\right)-1},\end{array}$ (1) where $E$ is the energy of the X-ray photons and $T_{\text{X}}$ is the equivalent temperature of the X-ray fireball. The energy spectrum represents the probability of the X-ray energy distribution. Assuming that $u=E/T_{\text{X}}$ is a dimensionless energy ratio, the normalized energy spectrum can be written as $\begin{array}{c}P\left(u\right)=\frac{15}{{\pi }^4}\frac{u^3}{\exp \left(u\right)-1}.\end{array}$ (2)

Some studies [16, 17] have shown that the equivalent temperature of X-ray fireballs is related to the design details of the nuclear weapon detonation processes and projectile materials. X-ray sources can be broadly divided into three categories based on the projectile type: fission bombs, ordinary hydrogen bombs, and enhanced X-ray weapons. The X-ray energy spectrum of a fission bomb can be characterized using a single blackbody spectrum with an equivalent temperature of approximately 1–2 keV. The X-ray energy spectra of other types of weapons can be characterized using composite blackbody spectra obtained by combining multiple blackbody spectra in different proportions. Here, the nuclear detonation X-ray source was a fission atomic bomb with an equivalent temperature of 1.4 keV.

Atmospheric transmission model

To calculate the parameters of nuclear detonation X-ray transmission to satellite-borne detectors, it is necessary to establish a mathematical model for nuclear detonation X-ray transmission through the atmosphere. The U.S. Standard Atmosphere 1976 was used in the atmospheric model in this study [18]. The atmosphere above 1000 km is extremely thin and can be considered a vacuum environment with an atmospheric density of 0. Statistically [19], the atmosphere is mostly distributed below 90 km, and the ratios of major elements (C, N, O, etc.) in the atmosphere hardly change in the 40–90 km range. The density of the atmosphere from 90 to 120 km is low, and the proportions of N and O vary, whereas the contents of the other components are very small. Monte Carlo simulations verified that the change in the proportion of each component in the atmosphere with altitude has a negligible effect (with a maximum relative error of less than 0.6% compared with the results for a fixed component) on the X-ray photon energy fluence. Therefore, it was assumed that the atmospheric composition is independent of altitude. Supposing that Earth is a standard sphere, the atmospheric density is approximated to have a spherically symmetric distribution.

The transmission parameters were calculated by establishing the transmission path relationship between detector A (where the altitude of the satellite is h_A), detonation point B(B’) (where the burst height is h_B), and Earth (where the spherical center is O, and the Earth’s radius is R_E). δ is the observation angle, α is the transmission angle, and AB(AB’) is the X-ray transmission path, as shown in Fig. 1.

Fig. 1Full-screen View

(Color online) High-altitude nuclear detonation X-ray transmission

Direct transmission

If a nuclear weapon detonates, X-rays interact with the atmosphere surrounding Earth layer-by-layer, starting at the detonation point, until they reach the satellite-borne X-ray detector. Nuclear detonation X-rays transmitted to satellite detectors consist of two components: directly transmitted [20] and scattered X-rays. First, we calculated the directly transmitted X-rays.

The atmosphere from the detonation point to the atmospheric boundary was assumed to be vertically divided into $N\left(M\right)$ layers according to the 1976 standard atmosphere. The stratification height of the i-th $\left(0\le i\le N\left(M\right)\right)$ atmosphere layer is $\text{Δ}{S}_i$. The layer containing the detonation point is the 0-th layer, with $\text{Δ}{S}_0=0$. Other stratification heights $\text{Δ}{S}_i$ are a piecewise function of altitude $h$: $\begin{array}{c}\text{Δ}{S}_i\left(1\le i\le N\left(M\right)\right)=\{\begin{array}{l}0.1,0\le h\le 30\hfill \\ 2,30<h\le 90\hfill \\ 5,90<h\le 1000\hfill \end{array}\end{array}$ (3)

Calculation of the X-ray transmission attenuation requires the atmospheric column density ${\rho }_{\text{S}}$ along the transmission path [21]. The atmospheric column density, also known as mass absorption distance, is the integral of the atmospheric density per unit length over distance. Two cases were considered for ${\rho }_{\text{S}}$ according to the transmission angle $\alpha$.

(I) Case 1: Transmission angle

α≥90°

This case corresponds to detonation-point location $B$ in Fig. 1. The transmission paths are $BC\to BD\to \cdots \to BA$. The layered paths are $BC\to CD\to \cdots \to A$. In this case, the transmission paths are all above the burst height $h_{\text{B}}$ of point $B$. The total number of layers $N$ can be written as $\begin{array}{c}N=N_1+N_2,\end{array}$ (4) where $N_1$ and $N_2$ represent the number of layers within the 30–90 km range and 90–1000 km range, respectively. $\begin{array}{c}{N}_1=\{\begin{array}{cc}\frac{90-h_{\text{B}}}{2},& 30\le h_{\text{B}}\le 90\\ 0,& \text{else}\end{array},\end{array}$ (4a) $\begin{array}{c}{N}_2=\{\begin{array}{ll}182,\hfill & 30\le h_{\text{B}}\le 90\hfill \\ \frac{1000-h_{\text{B}}}{5},\hfill & \text{ else}\hfill \end{array}.\end{array}$ (4b)

For △CBO, according to the cosine theorem, $\begin{array}{c}O{C}^2=B{C}^2+O{B}^2-2BC\times OB\times \cos \alpha ,\end{array}$ (5) where $OC=R_{\text{E}}+h_{\text{B}}+\text{Δ}{S}_1$ and $OB=R_{\text{E}}+h_{\text{B}}$. Suppose that $BC=X_1$, and $\begin{array}{c}{X}_1=\frac{H\times \cos \alpha +\sqrt{H^2\times {\cos }^2\alpha +4\left(\text{Δ}{S}_1{}^2+H\times \text{Δ}{S}_1\right)}}{2},\end{array}$ (6) where $H=2\left(R_{\text{E}}+h_{\text{B}}\right)$, and the length $\text{Δ}{X}_1$ of the first layered path $BC$ is $\begin{array}{c}\text{Δ}{X}_1=X_1.\end{array}$ (7)

By analogy, the recurrence relation $X_n\left(2\le n\le N\right)$ for the length of $BC\to BD\to \cdots \to BA$ is $\begin{array}{c}{X}_n=\frac{H\times \cos \alpha +\sqrt{H^2\times {\cos }^2\alpha +4\left({\left({\displaystyle {\sum }_{i=1}^n\text{Δ}{S}_{\text{i}}}\right)}^2+H\times {\displaystyle {\sum }_{i=1}^n\text{Δ}{S}_i}\right)}}{2}.\end{array}$ (8)

The length $\text{Δ}{X}_n\left(2\le n\le N\right)$ of the $n$-th layered path of $BC\to CD\to \cdots \to A$ is $\begin{array}{c}\text{Δ}{X}_n=X_n-X_{n-1}\end{array}\mathrm{.}$ (9)

The atmospheric column density of the transmission path in Case 1 can be expressed as $\begin{array}{c}{\rho }_{\text{S}}\left(h_{\text{B}},\alpha \ge 90 °\right)={\displaystyle \sum _{n=1}^N\text{Δ}{X}_n\times {\rho }_{\text{V}}}\left(h_{\text{B}}+{\displaystyle \sum _{i=0}^{n-1}\text{Δ}{S}_i}\right),\end{array}$ (10) where ${\rho }_{\text{V}}\left(h\right)$ is the atmospheric density at altitude $h$.

(II) Case 2: Transmission angle

α<90°

This case corresponds to detonation-point location $B^{\prime }$ in Fig. 1. The transmission paths are $B^{\prime }F\to B^{\prime }G\to B^{\prime }{C}^{\prime }\to B^{\prime }{D}^{\prime }\cdots \to B^{\prime }A$. The layered paths are $B^{\prime }F\to FG\to G{C}^{\prime }\to C^{\prime }{D}^{\prime }\cdots \to A$. In this case, some layered paths are partially above the burst height $h_{\text{B}}$, such as $G{C}^{\prime }$; others are below $h_{\text{B}}$, such as $B^{\prime }G$. To distinguish from the layers above the burst height $h_{\text{B}}$, the stratification height in section $B^{\prime }G$ is expressed using $\text{Δ}{s}_i$, whose physical meaning and expression are the same as $\text{Δ}{S}_i$ in Equation (3).

In this case, the total number of layers $M$ can be written as $M=N+M_{\text{B"G}},$ (11) where $N$ is given by Eq. (4) in Case 1, and M_B'G represents the number of layers in section $B^{\prime }G$. M_B'G is related to the shortest altitude $h_{\text{min}}$ of the transmission path to Earth’s ground and burst height $h_{\text{B}}$, where $\begin{array}{c}OF=\left(R_{\text{E}}+h_{\text{A}}\right)\times \sin \delta =\frac{1}{2}{H}_{\text{A}}\times \sin \delta ,\end{array}$ (12) $\begin{array}{c}{h}_{\text{min}}=OF-R_{\text{E}}.\end{array}$ (13)

Layer M_B'G can be divided into three parts: $\begin{array}{c}{M}_{\text{B"G}}=M_1+M_2+M_3,\end{array}$ (14) where $M_1$, $M_2$, and $M_3$ represent the number of layers below 30 km, between 30 and 90 km, and between 90 and 1000 km, respectively. They can be expressed as $\begin{array}{c}{M}_1=\{\begin{array}{l}\frac{30-h_{\text{min}}}{0.1},h_{\text{min}}\le 30\hfill \\ 0,\text{ else}\hfill \end{array},\end{array}$ (14a) $\begin{array}{c}{M}_2=\{\begin{array}{l}\frac{h_{\text{B}}-30}{2},\left(h_{\text{min}}\le 30\right) \text{and} \left(30\le h_{\text{B}}\le 90\right)\hfill \\ 30,\left(h_{\text{min}}\le 30\right) \text{and} \left(h_{\text{B}}>90\right)\hfill \\ \frac{h_{\text{B}}-h_{\text{min}}}{2},\left(30\le h_{\text{min}}\le 90\right) \text{and} \left(30\le h_{\text{B}}\le 90\right)\hfill \\ \frac{90-h_{\text{min}}}{2},\left(30\le h_{\text{min}}\le 90\right) \text{and} \left(h_{\text{B}}>90\right)\hfill \\ 0,\text{ else}\hfill \end{array}\end{array},$ (14b) $\begin{array}{c}{M}_3=\{\begin{array}{l}\frac{h_{\text{B}}-90}{5},\left(h_{\text{min}}\le 30\right) \text{and} \left(h_{\text{B}}>90\right)\hfill \\ \frac{h_{\text{min}}-90}{5},\left(30\le h_{\text{min}}\le 90\right) \text{and} \left(h_{\text{B}}>90\right)\hfill \\ \frac{h_{\text{B}}-h_{\text{min}}}{5},h_{\text{min}}>90\hfill \\ 0,\text{ else}\hfill \end{array}\end{array}.$ (14c)

In △$B\text{'}C\text{'}O$, where $B\text{'}C\text{'}=X_1$, the length $\text{Δ}{X}_1$ of the first layered path $GC\text{'}$ above the burst height $h_{\text{B}}$ is $\begin{array}{c}\text{Δ}{X}_1=X_1-2F{B}^{\prime }=X_1-2\sqrt{{\left(H/2\right)}^2-{\left(H_{\text{A}}/2\right)}^2\times {\sin }^2\delta }.\end{array}$ (15)

The other recurrence relations for $X_n$ and $\text{Δ}{X}_n$ are the same as those in Equations (8) and (9) in Case 1.

For $B\text{'}G$ below burst height $h_{\text{B}}$, the detonation point can be equated to point $F$, where point $B\text{"}$ is equivalent to the satellite. According to the geometric relationship, $\begin{array}{c}{B}^{\prime }G=2B^{\prime }F,\end{array}$ (16) the transmission paths of section $B^{\prime }F$ have a recurrence relationship similar to that of Case 1. The first-layer transmission path length $x_1$ and layered path length $\text{Δ}{x}_1$ below the burst height $h_{\text{B}}$ have the following relationship: $\begin{array}{c}\text{Δ}{x}_1=x_1=\sqrt{{\left(OF+\text{Δ}{s}_1\right)}^2-O{F}^2}=\sqrt{H_{\text{A}}\times \sin \delta \times \text{Δ}{s}_1+\text{Δ}{s}_1{}^2}.\end{array}$ (17)

Similarly, the transmission path length $x_m$ and layered path length $\text{Δ}{x}_m\left(2\le m\le M_{\text{B"G}}\right)$ below burst height $h_{\text{B}}$ have a recursive relationship: $\begin{array}{c}{x}_m=\sqrt{H_{\text{A}}\times \sin \delta \times {\displaystyle \sum _{i=1}^m\text{Δ}{s}_i}+{\left({\displaystyle \sum _{i=1}^m\text{Δ}{s}_i}\right)}^2},\end{array}$ (18) $\begin{array}{c}\text{Δ}{x}_m=x_m-x_{m-1}.\end{array}$ (19)

The atmospheric column density of $B^{\prime }F$ is defined as extra ${\rho }_{{\text{S}}_{\text{e}}}$ and is expressed as $\begin{array}{c}{\rho }_{{\text{S}}_{\text{e}}}={\displaystyle \sum _{m=1}^M\text{Δ}{x}_m\times {\rho }_{\text{V}}}\left(h_{\text{min}}+{\displaystyle \sum _{i=0}^{m-1}\text{Δ}{s}_i}{\displaystyle \sum }_{i=0}^{m-1}\right).\end{array}$ (20)

The atmospheric column density of the transmission path in Case 2 can be written as $\begin{array}{c}{\rho }_{\text{S}}\left(h_{\text{B}},\alpha <90 °\right)={\displaystyle \sum _{n=1}^N\text{Δ}{X}_n\times {\rho }_{\text{V}}}\left(h_{\text{B}}+{\displaystyle \sum _{i=0}^{n-1}\text{Δ}{S}_i}\right)+2{\rho }_{{\text{S}}_{\text{e}}}.\end{array}$ (21)

After calculating the atmospheric column density ${\rho }_{\text{S}}$ for Cases 1 and 2, the X-ray transmission attenuation by the atmosphere can be calculated based on ${\rho }_{\text{S}}$. The mean free path of photons is a general measure of the opacity of a substance and the ability of a medium to absorb radiation [22]. The mean free path $\lambda$ of X-rays with energy $E$ between the detector and detonation point is defined as $\begin{array}{c}\lambda \left(E\right)={\mu }_{\text{m}}\left(E\right)\times {\rho }_{\text{S}} ,\end{array}$ (22) where ${\mu }_{\text{m}}\left(E\right)$ is the mass absorption coefficient of X-rays with energy $E$, which was interpolated based on discrete data provided on the official NIST website [23].

After detonation, the radiation generated by the X-ray source spreads spherically to the surrounding areas. The radiant intensity of nuclear detonation pulsed X-rays with energy $E$ at the detector can be expressed as $\begin{array}{c}{I}^{\left(0\right)}\left(E\right)=\underset{E}{\overset{E+\text{Δ}E}{{\displaystyle \int }}}{K}_{\text{Q}}\times \frac{\eta \times Q\times f\left(t\right)\times P\left(E,T_{\text{X}}\right)}{4\pi \times R_0{}^2}\times \text{exp}\left(-\lambda \left(E\right)\right)\text{d}E,\end{array}$ (23) where $I^{\left(0\right)}\left(E\right)$ is the intensity of direct-transmission X-rays with energy $E$; $K_{\text{Q}}$ is the equivalent-energy conversion factor ($K_{\text{Q}}\approx 2.6114\times {10}^{25}\text{keV}/\text{kt TNT}$), with the equivalent expressed in kt TNT; $f\left(t\right)$ is the X-ray-normalized time spectrum; $R_0$ is the distance from the detonation point to the satellite-borne X-ray detector; and $\eta$ is the proportion of X-ray equivalent $Q_{\text{X}}$ in the total detonation equivalent $Q$. $\eta$ is related to the nuclear weapon design and its value is generally taken from 0.7 to 0.85 [3]. The expression for the relationship is as follows: $\begin{array}{c}\eta =\frac{Q_{\text{X}}}{Q}.\end{array}$ (24)

Build-up factor

In addition to being attenuated by collisions with atmospheric particles, X-rays undergo scattering. In this study, the correction for direct-transmission orientation scattering was made using the build-up factor. Bridgman defined the build-up factor as “direct and unreacted photons plus the scattered contribution.” [24] The build-up of X-ray intensity $I$ for direct and scattered transmissions in the direct-transmission orientation comprises three components: intensity of direct transmission $I^{\left(0\right)}$, intensity $$ after being scattered once in the direct-transmission orientation $I^{\left(1\right)}$, and intensity after $m\left(m\ge 2\right)$ scattering $I^{\left(m\right)}$. $\begin{array}{c}I=I^{\left(0\right)}+I^{\left(1\right)}+I^{\left(m\right)}.\end{array}$ (25)

The single-scattering intensity was derived by Bigelow and Winfield [25], and the results of multiple scattering were derived by Renken using the Boltzmann transfer equation [26]. The relationship between the build-up factor $F_{\text{B}}$ and radiation fluence in an infinite homogeneous atmosphere for X-rays with energy $E$ is given by $\begin{array}{c}I\left(E\right)=I^{\left(0\right)}\left(E\right)\times F_{\text{B}}\left(E\right).\end{array}$ (26)

Tayler fitted the build-up factor as a function [27] based on Monte Carlo simulation results: $\begin{array}{c}{F}_{\text{B}}\left(E\right)=A_1\times exp\left(c_1\times \lambda \left(E\right)\right)+A_2\times exp\left(c_2\times \lambda \left(E\right)\right),\end{array}$ (27) where A₁, A₂, c₁ and c₂ are coefficients related to the X-ray photon energy $E$.

The X-ray build-up factor is related to Compton scattering. Kalansky concluded that, based on the Compton scattering of photons, the value of the build-up factor can be divided into three segments depending on the X-ray energy $E$ [28]:

(I) For $E<12\text{ keV}$, the Compton scattering is negligible, with $F_{\text{B}}=1$.

(II) For $12\text{ keV}\le E\le 750\text{ keV}$, Talyer's fitting equation can be used and the Compton scattering cross-section of X-rays with energy greater than 100 keV starts to decrease.

(III) For E>750 keV, the electron pair effect is dominant, and this fitting formula is no longer applicable.

For the nuclear detonation X-ray source adopted in this study (fission atomic bomb), only approximately 1% of the X-ray energy was greater than 10T_X, which corresponded to an energy of 14 keV. The proportion of X-rays with $E>750\text{ keV}$ was low and could be ignored [3].

The energy fluence represents the total energy of photons entering a spherical body per unit cross-sectional area during X-ray emission [29]. The initial energy fluence ${\text{Φ}}_0$ is defined as the sum of the X-ray energy fluences transmitted and scattered in the direct-transmission orientation [30]. This is the integral of the X-ray intensity over time after correction for the build-up factor. $\begin{array}{c}{\text{Φ}}_0=\int I\text{d}t=K_{\text{Q}}\times \frac{\eta \times Q}{4\pi \times R_0{}^2}\underset{0}{\overset{t}{{\displaystyle \int }}}f\left(t\right)\text{d}t\underset{0}{\overset{E}{{\displaystyle \int }}}{F}_{\text{B}}\left(E\right)\times P\left(E,T_{\text{X}}\right)\times \text{exp}\left(-\lambda \left(E\right)\right)\text{d}E.\end{array}$ (28)

When a nuclear weapon is detonated at a high altitude, the spectrum peak and duration remain in the order of nanoseconds to hundreds of nanoseconds [17]. Thus, it can be assumed that the X-ray energy spectrum does not vary with time during the observation [16]; therefore, the effect of time $t$ was ignored.

In the direct-transmission orientation, the transmission coefficient [31] $\tau \left(E\right)$ represents the probability that an X-ray with energy $E$ is transmitted through the atmosphere to the satellite-borne detector without considering geometric attenuation. Owing to the build-up factor, the sum of the transmission coefficients may be greater than 1. The sum of the transmission coefficients at all energies is expressed as $\begin{array}{c}\sum \tau \left(E\right)={\displaystyle \sum _{E=0}^{\infty }{F}_{\text{B}}\left(E\right)}\times P\left(E,T_{\text{X}}\right)\times \text{exp}\left(-\lambda \left(E\right)\right).\end{array}$ (29)

By ignoring the influence of time $t$, the initial energy fluence can be written as $\begin{array}{c}{\text{Φ}}_0=K_{\text{Q}}\times \frac{\eta \times Q}{4\pi \times R_0{}^2}\times \sum \tau \left(E\right).\end{array}$ (30)

Monte Carlo-based scattering coefficient correction

Because the nuclear detonation source can be approximated as a point source, the X-rays emitted after detonation radiate spherically toward the surrounding area. Therefore, in numerical simulations, in addition to considering the correction of direct-transmission scattering using build-up factors, it is necessary to consider the scattering effect of X-rays from other orientations. This study established a correction factor table to correct the energy fluence based on Monte Carlo simulation results. X-ray transport simulations were performed using the SuperMC Monte Carlo simulation software.

The average energy fluence $\overline{\epsilon }$ of a single photon at the satellite altitude was calculated using Monte Carlo simulation [32]. Energy fluence ${\text{Φ}}_{\text{MC}}$ can be calculated by the average energy fluence of individual photons $\overline{\epsilon }$ amplifying the number of X-rays $n_{\text{X}}$ when the detonation equivalent is $Q$: $\begin{array}{c}{\text{Φ}}_{\text{MC}}=n_{\text{X}}\times \overline{\epsilon } ,\end{array}$ (31) where $n_{\text{X}}$ can be expressed as $\begin{array}{c}{n}_{\text{X}}={\displaystyle \sum _{i=1}^W{K}_{\text{Q}}}\times \frac{\eta \times Q}{{\epsilon }_i}\times \underset{E_{i-1}}{\overset{E_i}{{\displaystyle \int }}}P\left(E,T_{\text{X}}\right)\text{d}E.\end{array}$ (32)

Here, $W$ is the number of energy intervals, and the X-ray energy spectrum was divided into $2\times {10}^5$ energy intervals. ${\epsilon }_i\left(1\le i\le W\right)$ is the average photon energy of the i-th energy interval.

The detector counting results simulated using the Monte Carlo method were considered valid when the statistical error was within 5% [33]. The maximum permissible number of particles for each simulation condition was $2\times {10}^9$. Under these limitations, the Monte Carlo simulation results demonstrated that the lowest credible burst height was approximately 42 km.

By comparing the data of Monte Carlo simulated energy fluence ${\text{Φ}}_{\text{MC}}$ with the data of initial energy fluence ${\text{Φ}}_0$, we found a strong linear relationship between the two. Therefore, X-ray energy fluence $\text{Φ}$ at satellite altitudes can be further corrected and expressed as $\begin{array}{c}\text{Φ}=K\left(h_{\text{B}}\right)\times {\text{Φ}}_0,\end{array}$ (33) where $K$ is the scattering correction coefficient related to burst height $h_{\text{B}}$ (listed in Table 1), and $R^2$ is the coefficient of determination. After testing in areas with burst heights $h_{\text{B}}>$ 100 km, the simulation results show that $K\approx 1$ is applicable for 100 km < $h_{\text{B}}<$ 1500 km.

Comparison with Monte Carlo simulation results

The calculation conditions for the analytical simulation of the atmospheric transmission of pulsed X-rays from high-altitude nuclear detonations are listed in Table 2.

For each burst height, we randomly selected no less than six transmission angles from 89, 90, 95, 100, 110, 120, 130, 140, 150, and 160° and calculated the corresponding initial energy fluence ${\text{Φ}}_0$ corrected by the build-up factor and ${\text{Φ}}_{\text{MC}}$ using the Monte Carlo method. Figure 2 shows that ${\text{Φ}}_0$ and ${\text{Φ}}_{\text{MC}}$ exhibit a good linear relationship. The slope of the fitting line in Fig. 2 represents the scattering correction coefficient.

Fig. 2Full-screen View

(Color online) Fitting diagram of initial energy fluence and Monte Carlo results

For each burst height, we randomly selected no less than six transmission angles from 89, 90, 95, 100, 110, 120, 130, 140, 150, and 160° and calculated the corresponding initial energy fluence ${\text{Φ}}_0$ corrected by the build-up factor and ${\text{Φ}}_{\text{MC}}$ using the Monte Carlo method. Figure 2 shows that ${\text{Φ}}_0$ and ${\text{Φ}}_{\text{MC}}$ exhibit a good linear relationship. The slope of the fitting line in Fig. 2 represents the scattering correction coefficient.

The energy fluence $\text{Φ}$ was corrected by the build-up factor and scattering correction coefficient. For cases where the transmission angle was below 90°, the relative error between $\text{Φ}$ and ${\text{Φ}}_{\text{MC}}$ exceeded 15%, when the burst height was less than 120 km. The main reason for the large relative error in these cases is that the interaction process between the X-rays and atmosphere becomes more complex after adding the extra atmospheric column density, and the scattering process becomes more difficult to correct. If the transmission angle is appropriately selected based on different burst height conditions, the relative error is within a reasonable range (within 15% is permissible for detection), as listed in Table 3.

To validate the algorithm, we compared the proposed analytical method and another method without scattering correction with the Monte Carlo method. The results are presented in Table 4, where the uncorrected energy fluence $\phi$ is the energy fluence at the detector for direct transmission of attenuated X-rays.

Based on the data in Table 4, the results obtained by the uncorrected analytical method are significantly different from those obtained by the Monte Carlo simulation. Because the uncorrected analytical method does not consider the scattering effect, all its energy fluence values are lower than those of the Monte Carlo simulation. Under a burst height of 42 km and transmission angle of 120°, the relative error between the two reaches 67.79%. An improvement in accuracy is evident after scattering correction, with an increase of more than 60% in this case.

To further verify the accuracy, the results under different satellite altitudes, burst heights, and equivalent conditions were compared, as listed in Table 5.

The data in Table 5 indicate that the relative error between and can be controlled within 10% by interpolating the scattering correction coefficient for burst heights that are not listed in Table 1. Moreover, the value applies to satellite altitudes greater than 10000 km. Good agreement between the proposed analytical method and Monte Carlo simulation can be observed for the selected available transmission angles.

The computation time [34] of the proposed analytical algorithm is reduced to 1/48000 of the time taken by the Monte Carlo method. Table 6 presents a comparison of the time required by the two methods to calculate six different burst heights, with each height corresponding to 10 transmission angles in parallel.

Numerical simulation calculation results

First, the variation in atmospheric column density corresponding to different transmission angles for different burst heights was calculated, as depicted in Fig. 3. As the transmission angle decreases, the transmission path and atmospheric column density increase. The atmospheric column density changes considerably when the transmission angle is approximately 90°. This is because when the transmission angle is less than 90°, the X-rays pass through the low-altitude and high-density atmosphere below the detonation point, resulting in a significant increase ($2{\rho }_{{\text{S}}_{\text{e}}}$ is added) in the atmospheric column density. At different burst heights, even at the same transmission angle, the atmospheric column density differs by one order of magnitude. However, for transmission angles greater than 100°, the decrease in atmospheric column density exhibits a similar pattern of variation.

Fig. 3Full-screen View

(Color online) Variation of atmospheric column density

Subsequently, the X-ray build-up factors were calculated for different burst heights and transmission angles, as displayed in Fig. 4. The build-up factor increases from 0 to 75 keV before gradually decreasing. The burst height and transmission angle significantly affect the build-up factor. The build-up factor varies substantially for transmission angles above and below 90°. The atmospheric density is relatively high when the burst height is less than 80 km and the build-up factor is relatively large. It can be concluded that there are many scattered X-rays in the direct-transmission orientation. Moreover, X-rays in the energy band of 50–150 keV have scattering intensities exceeding direct transmission by a factor of 100 in the case of a 50 km burst height and 89° transmission angle. As the burst height increases, the atmosphere gradually becomes thinner and the scattering effect decreases. The build-up factor is close to 1 above a burst height of 100 km, and the influence of direct-transmission orientation scattering can be ignored.

Fig. 4Full-screen View

(Color online) Build-up factors for different burst heights and transmission angles

Finally, as illustrated in Fig. 5, under an initial explosive equivalent $Q=100\text{ kt}$ TNT and X-ray equivalent share of 0.7, the burst height of 80 km is considered as the variation boundary. This is because the X-ray mean free path is approximately equal to the atmospheric elevation at that altitude [16]. Therefore, for detonations above 80 km, X-rays emitted upward escape from the atmosphere, whereas X-rays with burst heights below 80 km are strongly absorbed during transmission, and the entire energy emitted downward is deposited in the atmosphere. The energy fluence spans four orders of magnitude. At a burst height of 42 km and transmission angle of 120°, the energy fluence is 2.2744 × 10⁶ keV/cm². However, it can reach 5.9504 × 10¹⁰ keV/cm² at a burst height of 1500 km and transmission angle of 180°. These results provide a solid basis for the design of satellite-borne X-ray detectors.

Fig. 5Full-screen View

(Color online) Energy fluence for different burst heights and transmission angles