# Angle-Only Cooperative Orbit Determination Considering Attitude Uncertainty
## 3 Attitude and Orbit Determination Method with Angle-Only Measurements
### Case II: The Orbit of \(S_{1}\) Is Known, and the Attitude of \(S_{1}\) Is Unchanged  
In this case, the orbit of \(S_{1}\) is known, and the attitude of \(S_{1}\) is unknown but unchanged; hence, the state to be estimated is given by:  
\[\mathbf{\dot{X}}=\mathbf{F}(\mathbf{X})=\left[\mathbf{\dot{X}}_{2}^{\mathrm{T}},\mathbf{\dot{q}}^ {\mathrm{T}}\right]^{\mathrm{T}}=\left[\mathbf{f}(\mathbf{X}_{2})^{\mathrm{T}},\mathbf{g}_ {1}(\mathbf{q})^{\mathrm{T}}\right]^{\  
Hence, the STM \(\mathbf{\Phi}(t_{k},t_{0})\) has the following format:  
\[\mathbf{\Phi}(t_{k},t_{0})= \begin{bmatrix}\mathbf{\Phi}_{rr}(t_{k},t_{0})&\mathbf{\Phi}_{rv}( t_{k},t_{0})&\mathbf{0}_{3\times 4}\\ \mathbf{\Phi}_{vr}(t_{k},t_{0})&\mathbf{\Phi}_{vv}(t_{k},t_{0})&\mathbf{0}_{3\times 4}\\ \mathbf{0}_{4\times 3}&\mathbf{0}_{4\times 3}&\mathbf{I}_{4\times 4}\end{bmatrix}  \tag{42}\]  
Case III: Both the Orbits of \(S_{1}\) and \(S_{2}\) Are Unknown, and the Attitude of \(S_{1}\) Is Unchanged  
In this case, the orbits of \(S_{1}\) and \(S_{2}\) as well as the attitude of \(S_{1}\) are estimated. The only information for the navigation system is that \(\mathbf{\omega}=\left[0\ 0\ 0\right]^{\mathrm{T}}\). Therefore, the state to be estimated is:  
\[\mathbf{X}=\left[\mathbf{X}_{1}^{\mathrm{T}},\mathbf{X}_{2}^{\mathrm{T}},\mathbf{q}^{\mathrm{ T}}\right]^{\mathrm{T}}=\left[x_{1},y_{1},z_{1},\dot{x}_{1},\dot{y}_{1},\dot{z}_{1},x_{2},y_{2},z _{2},\dot{x}_{2},y_{2},\dot{z}_{2},q_{0},q_{1},q_{2},q_{3}\right]^{\mathrm{T}} \tag{43}\]  
The state model is then given by:  
\[\dot{\mathbf{X}}=\mathbf{F}(\mathbf{X})=\left[\dot{\mathbf{X}}_{1}^{\mathrm{T}},\dot{\mathbf{X}}_{ 2}^{\mathrm{T}},\mathbf{q}^{\mathrm{T}}\right]^{\mathrm{T}}=\left[\mathbf{f}(\mathbf{X}_{ 1})^{\mathrm{T}},\mathbf{f}(\mathbf{X}_{1})^{\mathrm{T}},\mathbf{g}_{1}(\mathbf{q})^{\mathrm{T }}\right]^{\mathrm{T}}=\left[\mathbf{f}(\mathbf{X}_{1})^{\mathrm{T}},\mathbf{f}(\mathbf{X}_{1} )^{\mathrm{T}},[0,0,0,0]\right]^{\mathrm{T}} \tag{44}\]  
Similar to Equation (40), the partial differential matrix of measurement is given as:  
\[ \begin{split}&\mathbf{H}_{k}=\left.\frac{\partial\mathbf{y}}{ \partial\mathbf{X}}\right|_{t=t_{k}}=\left[\left.\frac{\partial\mathbf{y}}{\partial \mathbf{X}_{1}}\right|_{t=t_{k}}\left.\frac{\partial\mathbf{y}}{\partial\mathbf{X}_{2}} \right|_{t=t_{k}}\left.\frac{\partial\mathbf{y}}{\partial\mathbf{x}_{2}}\right|_{t=t_{ k}}\right]=\left[\left.\frac{\partial\mathbf{y}}{\partial\mathbf{r}_{1}}\right|_{t=t_{k}} \left.\frac{\partial\mathbf{y}}{\partial\mathbf{r}_{1}}\right|_{t=t_{k}}\left.\frac{ \partial\mathbf{y}}{\partial\mathbf{r}_{2}}\right|_{t=t_{k}}\left.\frac{\partial\mathbf{y }}{\partial\mathbf{y}_{2}}\right|_{t=t_{k}}\left.\frac{\partial\mathbf{y}}{\partial\bm {y}}\right|_{t=t_{k}}\right]\\ &=\left.\left[\mathbf{R}_{E}^{B}\frac{\partial}{\partial\mathbf{r}_{1} }\right(\frac{\mathbf{r}_{2}-\mathbf{r}_{1}}{\left\|\mathbf{r}_{2}-\mathbf{r}_{1}\right\|}), \ \mathbf{R}_{E}^{B}\frac{\partial}{\partial\mathbf{r}_{1}}\right(\frac{\mathbf{r}_{2}-\mathbf{ r}_{1}}{\left\|\mathbf{r}_{2}-\mathbf{r}_{1}\right\|}),\ \mathbf{R}_{E}^{B}\frac{\partial}{\partial\mathbf{r}_{2}}\right(\frac{\mathbf{r}_{2}- \mathbf{r}_{1}}{\left\|\mathbf{r}_{2}-\mathbf{r}_{1}\right\|}),\ \mathbf{R}_{E}^{B}\frac{\partial}{\partial\mathbf{r}_{2}} \right(\frac{\mathbf{r}_{2}-\mathbf{r}_{1}}{\left\|\mathbf{r}_{2}-\mathbf{r}_{1}\right\|}), \ \mathbf{R}_{E}^{B}\frac{\partial}{\partial\mathbf{r}_{2}}\right(\frac{\mathbf{r}_{2}-\mathbf{ r}_{1}}{\left\|\mathbf{r}_{2}-\mathbf{r}_{1}\right\|}),\ \mathbf{R}_{E}^{B}\frac{ \partial}{\partial\mathbf{r}_{2}}\right(\frac{\mathbf{r}_{2}-\mathbf{r}_{1}}{\left\|\mathbf{r}_{ 2}-\mathbf{r}_{1}\right\|}),\ \frac{\partial\mathbf{R}_{E}^{B}}{\partial\mathbf{q}}\right(\frac{\mathbf{r}_{2}-\mathbf{ r}_{1}}{\left\|\mathbf{r}_{2}-\mathbf{r}_{1}\right\|})\Bigg{|}_{t=t_{k}}\end{split}