MHB State Matrix Derivation for Orbital Mechanics

[Dustinsfl]

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?

 

[Dustinsfl]

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.

To be continued... but if you know how to answer anything already asked, please do.