Fix Hamiltonian form in Kepler problem example

The Hamiltonian at any point should not depend on the generalized velocity, but instead the momentum. For this problem it doesn't make as much difference, but users can get confused about the equations of motion if the Hamiltonian already includes the generalized velocities.

Author: abhro

docs/src/examples/kepler_problem.md

@@ -1,13 +1,22 @@
 # The Kepler Problem
 
-The Hamiltonian $\mathcal {H}$ and the angular momentum $L$ for the Kepler problem are
+The (non-dimensional) Hamiltonian $\mathcal {H}$ and the angular momentum $L$ for the Kepler problem are
 
-$$\mathcal {H} = \frac{1}{2}(\dot{q}^2_1+\dot{q}^2_2)-\frac{1}{\sqrt{q^2_1+q^2_2}},\quad
-L = q_1\dot{q_2} - \dot{q_1}q_2$$
+```math
+\begin{align*}
+\mathcal{H}(q_1, p_1, q_2, p_2) &= \frac{1}{2}(p^2_1+p^2_2)-\frac{1}{\sqrt{q^2_1+q^2_2}}, \\
+L &= q_1 p_2 - p_1 q_2
+\end{align*}
+```
 
 Also, we know that
 
-$${\displaystyle {\frac {\mathrm {d} {\boldsymbol {p}}}{\mathrm {d} t}}=-{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {q}}}}\quad ,\quad {\frac {\mathrm {d} {\boldsymbol {q}}}{\mathrm {d} t}}=+{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {p}}}}}$$
+```math
+\begin{align*}
+\frac{\mathrm{d} \boldsymbol{p}}{\mathrm{d} t} &= - \frac {\partial \mathcal{H}}{\partial \boldsymbol{q}} , \\
+\frac{\mathrm{d} \boldsymbol{q}}{\mathrm{d} t} &= + \frac {\partial \mathcal{H}}{\partial \boldsymbol{p}}
+\end{align*}
+```
 
 ```@example kepler
 import OrdinaryDiffEq as ODE, LinearAlgebra, ForwardDiff, NonlinearSolve as NLS, Plots