# State Matrix Derivation for Orbital Mechanics

#### dwsmith

##### Well-known member
Has anyone seen the derivation of $$\ddot{\mathbf{r}} = -\frac{\mu}{r^3}\mathbf{r}$$ into state matrix form?

If so, can they provide a link?

Last edited:

#### dwsmith

##### Well-known member
In a book, I have found a derivation I don't fully understand.

For one, it defines $$\ddot{\mathbf{r}} = -\frac{\mu}{r^3}\mathbf{r} = -\nabla V(\mathbf{r})$$. I have never seen this definition. Why/How can Keplerian motion de defined as divergence of the potential of r?

Then
$\mathbf{G} = - \begin{bmatrix} \frac{\partial(\nabla V(\mathbf{r}))}{\partial\mathbf{r}} \end{bmatrix} = - \begin{bmatrix} V_{xx} & V_{xy} & V_{xz}\\ V_{xy} & V_{yy} & V_{yz}\\ V_{xz} & V_{zy} & V_{zz} \end{bmatrix}$
Since $$\mathbf{G} = \mathbf{G}^T$$, the state transition matrix of the Keplerian two body problem is guaranteed to be symplectic.
$\mathbf{x}(t) = \begin{bmatrix} \mathbf{r}(t)\\ \mathbf{v}(t) \end{bmatrix} = \begin{bmatrix} F\cdot\mathbb{I} & G\cdot\mathbb{I}\\ \dot{F}\cdot\mathbb{I} & \dot{G}\cdot\mathbb{I} \end{bmatrix}\mathbf{x}_0$
where $$\mathbb{I}$$ is $$3\times 3$$ and
\begin{align}
F &= 1 - \frac{a}{r_0}(1 - \cos(\Delta E))\\
G &= \Delta t + \sqrt{\frac{a^3}{\mu}}(\sin(\Delta E) - \Delta E)\\
\dot{F} &= -\frac{\sqrt{a\mu}}{rr_0}\sin(\Delta E)\\
\dot{G} &= 1 + \frac{a}{r}(\cos(\Delta E) - 1)
\end{align}
The state transition matrix for this nonliear systme is defined as (Why?)
$\mathbf{\Phi}(t,t_0) = \begin{bmatrix} \mathbf{\Phi}_{11} & \mathbf{\Phi}_{12}\\ \mathbf{\Phi}_{21} & \mathbf{\Phi}_{22} \end{bmatrix} = \begin{bmatrix} \frac{\partial\mathbf{x}(t)}{\partial\mathbf{x}_0} \end{bmatrix}$
Then it says subdividing the $$6\times 6$$ state transition matrix into four $$3\times 3$$ matrices $$\mathbf{\Phi}_{ij}$$, and using $$F$$ and $$G$$ solutions to compute the required partial derivatives, leads to the following results: (Can some one walk we through the first one so I understand how these were derived?)
\begin{align}
\mathbf{\Phi}_{11} &= F\cdot\mathbb{I} + \mathbf{r}_0\frac{\partial F}{\partial\mathbf{r}_0} + \mathbf{v}_0\frac{\partial G}{\partial\mathbf{r}_0}\\
\mathbf{\Phi}_{12} &= G\cdot\mathbb{I} + \mathbf{r}_0\frac{\partial F}{\partial\mathbf{v}_0} + \mathbf{v}_0\frac{\partial G}{\partial\mathbf{v}_0}\\
\mathbf{\Phi}_{21} &= \dot{F}\cdot\mathbb{I} + \mathbf{r}_0\frac{\partial \dot{F}}{\partial\mathbf{r}_0} + \mathbf{v}_0\frac{\partial \dot{G}}{\partial\mathbf{r}_0}\\
\mathbf{\Phi}_{22} &= \dot{G}\cdot\mathbb{I} + \mathbf{r}_0\frac{\partial\dot{F}}{\partial\mathbf{v}_0} + \mathbf{v}_0\frac{\partial\dot{G}}{\partial\mathbf{v}_0}\\
\end{align}
The partial derivatives of the initial orbit radius $$r_0$$ and velocity magnitude $$v_0$$ are given by (Can someone explain this?)
\begin{align}
\frac{\partial r_0}{\partial\mathbf{r}_0} &= \frac{1}{r_0}\mathbf{r}_0^T\\
\frac{\partial r_0}{\partial\mathbf{v}_0} &= \mathbf{0}^T\\
\frac{\partial v_0}{\partial\mathbf{r}_0} &= \mathbf{0}^T\\
\frac{\partial v_0}{\partial\mathbf{v}_0} &= \frac{1}{v_0}\mathbf{v}_0^T
\end{align}
Using the definition of $$\sigma_0\equiv \frac{1}{\sqrt{\mu}}\mathbf{r}_0^T\mathbf{v}_0$$, the partial of $$\sigma_0$$ is
\begin{gather}
\frac{\partial\sigma_0}{\partial\mathbf{r}_0} = \frac{1}{\sqrt{\mu}}\mathbf{v}_0^T\\
\frac{\partial\sigma_0}{\partial\mathbf{v}_0} = \frac{1}{\sqrt{\mu}}\mathbf{r}_0^T
\end{gather}
To find the sensitivities of the semimajor axis $$a$$ with respect to the initial state vectors, we write the energy equation as
$\frac{1}{a} = \frac{2}{r_0} - \frac{v_0^2}{\mu}$
Then let's introduce a place holder vector alpha.