﻿ 潜艇重力梯度极值和深度的关系研究
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 大地测量与地球动力学  2020, Vol. 40 Issue (12): 1228-1232  DOI: 10.14075/j.jgg.2020.12.004

### 引用本文

ZENG Xianghang, WAN Xiaoyun, SUI Xiaohong, et al. The Relationship between Extreme Values of Gravity Gradients and Depth of Submarine[J]. Journal of Geodesy and Geodynamics, 2020, 40(12): 1228-1232.

### Foundation support

National Natural Science Foundation of China, No. 41674026; Fundamental Research Funds for the Central Universities, No. 2652018027; Fund of China Geological Survey, No.DD20191006;Open Foundation of Key Laboratory of Space Utilization, CAS, No.LSU-KFJJ-201902; National Defense Science and Technology Innovation Special Zone Project and Qian Xuesen Lab-DFH Sat Co Joint Research and Development Fund, No.M-2017-006.

### Corresponding author

WAN Xiaoyun, associate professor, majors in satellite geodesy, E-mail:wanxy@cugb.edu.cn.

### 第一作者简介

ZENG Xianghang, postgraduate, majors in satellite geodesy, E-mail:zengxianghang@qq.com.

### 文章历史

1. 中国地质大学(北京)土地科学技术学院，北京市学院路29号，100083;
2. 资源环境与灾害监测山西省重点实验室，山西省晋中市迎宾西街380号，030600;
3. 钱学森空间技术实验室，北京市友谊路104号，100094

1 理论模型 1.1 潜艇模型的建立

 图 1 潜艇模型示意图 Fig. 1 Diagram of submarine model
1.2 重力梯度的计算

1.2.1 外壳产生的重力梯度

 ${V_1} = G\sigma \iint {\frac{{{\text{d}}s}}{{\sqrt {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} }}}$ (1)

V1xyz求2阶导数就可以得到外壳产生的6个重力梯度分量：

 $\left\{ \begin{gathered} {V_{1xx}} = G\sigma \iint {\frac{{2{{\left( {X - x} \right)}^2} - {{\left( {Y - y} \right)}^2} - {{\left( {Z - z} \right)}^2}}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}s \hfill \\ {V_{1yy}} = G\sigma \iint {\frac{{2{{\left( {Y - y} \right)}^2} - {{\left( {X - x} \right)}^2} - {{\left( {Z - z} \right)}^2}}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}s \hfill \\ {V_{1zz}} = G\sigma \iint {\frac{{2{{\left( {Z - z} \right)}^2} - {{\left( {X - x} \right)}^2} - {{\left( {Y - y} \right)}^2}}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}s \hfill \\ {V_{1xy}} = G\sigma \iint {\frac{{3\left( {X - x} \right)\left( {Y - y} \right)}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}s \hfill \\ {V_{1xz}} = G\sigma \iint {\frac{{3\left( {X - x} \right)\left( {Z - z} \right)}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}s \hfill \\ {V_{1yz}} = G\sigma \iint {\frac{{3\left( {Y - y} \right)\left( {Z - z} \right)}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}s \hfill \\ \end{gathered} \right.$ (2)
1.2.2 潜艇耐压舱产生的质量亏损

 ${V_2} = - G\rho \iiint {\frac{{{\text{d}}v}}{{\sqrt {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} }}}$ (3)

V2xyz求2阶导数就可以得到耐压舱产生的6个重力梯度分量：

 $\left\{ \begin{gathered} {V_{2xx}} = - G\rho \iiint {\frac{{2{{\left( {X - x} \right)}^2} - {{\left( {Y - y} \right)}^2} - {{\left( {Z - z} \right)}^2}}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}x{\text{d}}y{\text{d}}z \hfill \\ {V_{2yy}} = - G\rho \iiint {\frac{{2{{\left( {Y - y} \right)}^2} - {{\left( {X - x} \right)}^2} - {{\left( {Z - z} \right)}^2}}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}x{\text{d}}y{\text{d}}z \hfill \\ {V_{2zz}} = - G\rho \iiint {\frac{{2{{\left( {Z - z} \right)}^2} - {{\left( {X - x} \right)}^2} - {{\left( {Y - y} \right)}^2}}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}x{\text{d}}y{\text{d}}z \hfill \\ {V_{2xy}} = - G\rho \iiint {\frac{{3\left( {X - x} \right)\left( {Y - y} \right)}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}x{\text{d}}y{\text{d}}z \hfill \\ {V_{2xz}} = - G\rho \iiint {\frac{{3\left( {X - x} \right)\left( {Z - z} \right)}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}x{\text{d}}y{\text{d}}z \hfill \\ {V_{2yz}} = - G\rho \iiint {\frac{{3\left( {Y - y} \right)\left( {Z - z} \right)}}{{{{\left[ {{{\left( {X - x} \right)}^2} + {{\left( {Y - y} \right)}^2} + {{\left( {Z - z} \right)}^2}} \right]}^{\frac{5}{2}}}}}}{\text{d}}x{\text{d}}y{\text{d}}z \hfill \\ \end{gathered} \right.$ (4)

1.2.3 潜艇产生的重力梯度

 ${V_{ij}} = {V_{1ij}} + {V_{2ij}}, i = x, y;j = x, y$ (5)
2 数值分析 2.1 梯度极值点与潜艇深度关系的数值分析

 图 2 潜艇上方600 m处产生的重力梯度 Fig. 2 The gravity gradients generated by submarine while z=600m

 图 3 潜艇极值点坐标与潜艇到观测点的垂直距离的关系 Fig. 3 The relationship between the submarine extreme point coordinates and the vertical distance from the submarine to the observation point

 $\left\{ \begin{array}{l} {X_{A1}} = \frac{{ - 172.9297}}{{{d^{0.0085}}}} + 166.2401 \hfill \\ {Y_{A1}} = 0.0603 \hfill \\ {X_{B1}} = - 1.2639d + 53.5005 \hfill \\ {Y_{B1}} = 0.0020 \hfill \\ {X_{C1}} = 1.2664d - 47.7261 \hfill \\ {Y_{C1}} = 0.0020 \hfill \\ \end{array} \right.$ (6)
 $\left\{ \begin{array}{l} {X_{A2}} = - 0.5816d + 19.7724 \hfill \\ {Y_{A2}} = - 0.5830d + 8.3924 \hfill \\ {X_{B2}} = - 0.5815d + 19.7240 \hfill \\ {Y_{B2}} = 0.5831d - 8.4583 \hfill \\ {X_{C2}} = 0.5869d - 17.3083 \hfill \\ {Y_{C2}} = - 0.5829d + 8.0917 \hfill \\ {X_{D2}} = 0.5870d - 17.3614 \hfill \\ {Y_{D2}} = 0.5828 - 8.0516 \hfill \\ \end{array} \right.$ (7)
 $\left\{ \begin{array}{l} {X_{A3}} = - 0.4978d + 15.2091 \hfill \\ {Y_{A3}} = - 1.140{\text{ }}5 \times {10^{ - 7}}d - 7 \times {10^{ - 4}} \hfill \\ {X_{B3}} = 0.5038d - 13.4695 \hfill \\ {Y_{B3}} = - 1.9008 \times {10^{ - 7}}d - 0.0024 \hfill \\ \end{array} \right.$ (8)
 $\left\{ \begin{array}{l} {X_{A4}} = \frac{{ - 23.1460}}{{{d^{0.3052}}}} + 6.0236 \hfill \\ {Y_{A4}} = 0.0603 \hfill \\ {X_{B4}} = \frac{{ - 55.9930}}{{{d^{0.5720}}}} + 4.3604 \hfill \\ {Y_{B4}} = - 1.2343d + 12.3085 \hfill \\ {X_{C4}} = \frac{{- 55.9548}}{{{d^{0.5718}}}} + 4.3616 \hfill \\ {Y_{C4}} = 1.2343d - 12.3100 \hfill \\ \end{array} \right.$ (9)
 $\left\{ \begin{array}{l} {X_{A5}} = \frac{{ - 22.0907}}{{{d^{0.2530}}}} + 7.0582 \hfill \\ {Y_{A5}} = - 0.5019d + 3.4077 \hfill \\ {X_{B5}} = \frac{{ - 21.8708}}{{{d^{0.2476}}}} + 7.1683 \hfill \\ {Y_{B5}} = 0.5018d - 3.3806 \hfill \\ \end{array} \right.$ (10)
 ${X_{A6}} = \frac{{- 24.6826}}{{{d^{0.0933}}}} + 16.1134, {Y_{A6}} = 0.0603$ (11)

 ${X_i} = {a_i}d + {b_i}$ (12)
 $\sigma _d^2 = \frac{1}{{a_i^2}}\sigma _{{X_i}}^2$ (13)
 ${P_i} = \frac{1}{{\sigma _{{d_i}}^2}}$ (14)
2.2 潜艇深度反演结果分析

3 结语