## TEACHING RESEARCH AND STUDY FOR SPACECRAFT RELATIVE MOTION DYNAMICS

WANG Weilin,1), SONG Xumin, CHAI Hua, LIU Haitao, YUE Zhiyu

Space Engineering University, Beijing 101416, China

Abstract

Spacecraft relative motion dynamics is core content of space service principles and it focuses on two spacecraft relative motion laws based on target orbit coordinate. The relative motion laws in space are contradictory to the relative motion laws on the ground, which pose critical challenge to students' spatial imagination. The teaching team proposed a systematic teaching technique which includes illustration method, interaction method and experimental method. The implicit relative motion model is transferred into explicit expression through formula translation. The statical formula is demonstrated by case application and the abstract relative motion law is represented by vivid pictures and dynamic animation, etc. The practice indicates that the proposed method contributes to student learning enthusiasm and the teaching quality improvement significantly.

Keywords： spacecraft; relative motion; illustration method; interaction method; experimental method

WANG Weilin, SONG Xumin, CHAI Hua, LIU Haitao, YUE Zhiyu. TEACHING RESEARCH AND STUDY FOR SPACECRAFT RELATIVE MOTION DYNAMICS. Mechanics in Engineering, 2022, 44(2): 434-442 DOI:10.6052/1000-0879-21-316

## 2 交互法提供理论依据

### 2.1 施加切向脉冲后的相对运动方程

$\left.\begin{array}{c}x(t)=x_{0}+2 \frac{\dot{z}_{0}}{\omega}-\left(3 \dot{x}_{0}-6 \omega z_{0}\right) t+ \\2\left(2 \frac{\dot{x}_{0}}{\omega}-3 z_{0}\right) \sin (\omega t)-2\left(\frac{\dot{z}_{0}}{\omega}\right) \cos (\omega t) \\y(t)=\frac{\dot{y}_{0}}{\omega} \sin (\omega t)+y_{0} \cos (\omega t) \\z(t)=4 z_{0}-2 \frac{\dot{x}_{0}}{\omega}+\left(2 \frac{\dot{x}_{0}}{\omega}-3 z_{0}\right) \cos (\omega t)+ \\\left(\frac{\dot{z}_{0}}{\omega}\right) \sin (\omega t)\end{array}\right\}$

$\left.\begin{array}{l}x=4 \frac{\dot{x}_{0}}{\omega} \sin (\omega t)-3 \dot{x}_{0} t \\y=0 \\z=2\left(\frac{\dot{x}_{0}}{\omega} \cos (\omega t)-\frac{\dot{x}_{0}}{\omega}\right)\end{array}\right\}$

### 2.2 施加径向脉冲后的相对运动方程

$\left.\begin{array}{l}\dot{a}=\frac{2 a^{2}}{\sqrt{\mu p}}\left[e \sin f \cdot a_{\mathrm{r}}+(1+e \cos f) \cdot a_{\mathrm{t}}\right] \\\dot{e}=\frac{r\left[\sin f(1+e \cos f) \cdot a_{\mathrm{r}}+\left(2 \cos f+e+e \cos ^{2} f\right) \cdot a_{\mathrm{t}}\right]}{\sqrt{\mu p}} \\\dot{i}=\frac{r \cos u}{\sqrt{\mu p}} a_{\mathrm{h}} \\\dot{\Omega}=\frac{r \sin u}{\sqrt{\mu p} \sin i} a_{\mathrm{h}} \\\dot{\omega}=\frac{r\left[-\cos f(1+e \cos f) \cdot a_{\mathrm{r}}+\sin f(2+e \cos f) \cdot a_{\mathrm{t}}\right]}{e \sqrt{\mu p}}-\cos i \cdot \dot{\Omega} \\\dot{M}=\frac{(p \cos f-2 r e) \cdot a_{\mathrm{r}}-(p+r) \sin f \cdot a_{\mathrm{t}}}{e \sqrt{\mu a}}+\sqrt{\frac{\mu}{a^{3}}}\end{array}\right\}$

$\left.\begin{array}{l}x=\frac{2 \dot{z}_{0}}{\omega} \cos (\omega t)-2 \frac{\dot{z}_{0}}{\omega} \\y=0 \\z=-\frac{\dot{z}_{0}}{\omega} \sin (\omega t)\end{array}\right\}$

$\frac{\left(x+2 \frac{\dot{z}_{0}}{\omega}\right)^{2}}{\left(\frac{2 \dot{z}_{0}}{\omega}\right)^{2}}+\frac{(z)^{2}}{\left(-\frac{\dot{z}_{0}}{\omega}\right)^{2}}=1$

$v=\frac{r v^{2}}{\mu}$
$f=\operatorname{atan} 2\left(v \sin \Theta \cos \Theta, v \cos ^{2} \Theta-1\right)$

### 图6

$v_{\mathrm{r}}=v \sin \Theta=\sqrt{\frac{\mu}{p}} e \sin f$
$v_{\mathrm{t}}=v \cos \Theta=\sqrt{\frac{\mu}{p}}(1+e \cos f)$

$\dfrac{v_{\rm r} }{v_{\rm t} }=e$

## 3 实验法提供数据来源

$\left[\begin{array}{c}x_{\mathrm{t}} \\y_{\mathrm{t}} \\z_{\mathrm{t}} \\\dot{x}_{\mathrm{t}} \\\dot{y}_{\mathrm{t}} \\\dot{z}_{\mathrm{t}}\end{array}\right]=\Phi\left[\begin{array}{c}x_{0} \\y_{0} \\z_{0} \\\dot{x}_{0} \\\dot{y}_{0} \\\dot{z}_{0}\end{array}\right]$

${X}_{\rm t} ={\varPhi X}_{0}$

$\Phi=\left[\begin{array}{cccccc}1 & 0 & 6(\omega t-\sin (\omega t)) & \frac{4 \sin (\omega t)-3 \omega t}{\omega} & 0 & \frac{2(1-\cos (\omega t))}{\omega} \\0 & \cos (\omega t) & 0 & 0 & \frac{\sin (\omega t)}{\omega} & 0 \\0 & 0 & 4-3 \cos (\omega t) & \frac{2(\cos (\omega t)-1)}{\omega} & 0 & \frac{\sin (\omega t)}{\omega} \\0 & 0 & 6 \omega(1-\cos (\omega t)) & 4 \cos (\omega t)-3 & 0 & 2 \sin (\omega t) \\0 & -\omega \sin (\omega t) & 0 & 0 & \cos (\omega t) & 0 \\0 & 0 & 3 \omega \sin (\omega t) & -2 \sin (\omega t) & 0 & \cos (\omega t)\end{array}\right]$

$\begin{array}{c}\Phi\left(t, t_{0}\right)=\left[\begin{array}{ll}\Phi_{\mathrm{r}}\left(t, t_{0}\right) & \Phi_{\mathrm{v}}\left(t, t_{0}\right)\end{array}\right]= \\{\left[\begin{array}{ll}\Phi_{\mathrm{rr}}\left(t, t_{0}\right) & \Phi_{\mathrm{rv}}\left(t, t_{0}\right) \\\Phi_{\mathrm{vr}}\left(t, t_{0}\right) & \Phi_{\mathrm{vv}}\left(t, t_{0}\right)\end{array}\right]}\end{array}$

$\left.\begin{array}{l}r(t)=\Phi_{\mathrm{rr}}\left(t, t_{0}\right) r_{0}+\Phi_{\mathrm{rv}}\left(t, t_{0}\right) v_{0} \\v(t)=\Phi_{\mathrm{rr}}\left(t, t_{0}\right) r_{0}+\Phi_{\mathrm{Vv}}\left(t, t_{0}\right) v_{0}\end{array}\right\}$

${v}_{0}^{+} =-\left[ {{\varPhi }_{\rm rv} } \right]^{-1}\left[ {{\varPhi}_{\rm rr} } \right]{r}_{0}$
$\Delta {v}_{1} ={v}^{+}(0)-{v}(0)$

${v}\left( t \right)={\varPhi }_{\rm vr} {r}_{0} +{\varPhi }_{\rm vv} {v}_{0}^{+}$
$\Delta {v}_{2} =-{v}(t)$

$\Delta v=\left\| {\Delta {v}_{1} } \right\|+\left\| {\Delta {v}_{2}} \right\|$

### 图8

MATLAB为遍历计算仿真提供了便利,假设对一个航天器沿各个方向施加等量脉冲,在时间较短时,可达区的形状近似为球形,这一结论同样可以从CW方程中推导得到,式(21)给出了时间较短时三个方向的运动轨迹表达式,在各个方向都近似匀速直线运动。

$\left.\begin{array}{l}x(t) \approx \dot{x}_{0} t \\y(t) \approx \dot{y}_{0} t \\z(t) \approx \dot{z}_{0} t\end{array}\right\}$

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