Fix a few typos preventing equations display. Add a figure for JWST t…

…rajectory

interplanetary-maneuvers/interplanetary-transfer-phasing.md

@@ -277,10 +277,10 @@ The wait times are shown in {numref}`tab:heliocentric-hohmann-wait-times`. The t
 
 | N | $t_{\text{wait}}$ (years) |
 |---|-----------------------|
-| 0 | {glue:text}`heliocentric-hohmann-t_wait_0` |
-| 1 | {glue:text}`heliocentric-hohmann-t_wait_1` |
-| 2 | {glue:text}`heliocentric-hohmann-t_wait_2` |
-| 3 | {glue:text}`heliocentric-hohmann-t_wait_3` |
+| 0 | {glue:text}`heliocentric-hohmann-t_wait_0:.4f` |
+| 1 | {glue:text}`heliocentric-hohmann-t_wait_1:.4f` |
+| 2 | {glue:text}`heliocentric-hohmann-t_wait_2:.4f` |
+| 3 | {glue:text}`heliocentric-hohmann-t_wait_3:.4f` |
 :::
 
 Clearly, the total mission time is dominated by the transfer time. This is because the synodic period of Venus relative to Neptune is quite small, at only {glue:text}`heliocentric-hohmann-T_syn:.2f` Earth years. Since Venus whips around the Sun, relative to Neptune, the same phase angle occurs relatively often.

the-n-body-problem/circular-restricted-three-body-problem.md

@@ -255,9 +255,9 @@ The characteristic length is the circular orbit radius, $r_{12}$. Using this, we
 :::{math}
 :label: eq:non-dim-r-vectors-cr3bp
 \begin{aligned}
- \vector{\rho} &= \frac{\vector{r}}{r_{12}} = x^_\uvec{\imath} + y^_\uvec{\jmath} + z^_\uvec{k} \\
- \vector{\sigma} &= \frac{\vector{r}_1}{r_{12}} = \left(x^_ + \pi_2\right)\uvec{\imath} + y^_\uvec{\jmath} + z^_\uvec{k} \\
- \vector{\psi} &= \frac{\vector{r}_2}{r_{12}} = \left(x^* - 1 + \pi_2\right)\uvec{\imath} + y^_\uvec{\jmath} + z^_\uvec{k}
+ \vector{\rho} &= \frac{\vector{r}}{r_{12}} = x^*\uvec{\imath} + y^*\uvec{\jmath} + z^*\uvec{k} \\
+ \vector{\sigma} &= \frac{\vector{r}_1}{r_{12}} = \left(x^* + \pi_2\right)\uvec{\imath} + y^*\uvec{\jmath} + z^*\uvec{k} \\
+ \vector{\psi} &= \frac{\vector{r}_2}{r_{12}} = \left(x^* - 1 + \pi_2\right)\uvec{\imath} + y^*\uvec{\jmath} + z^*uvec{k}
 \end{aligned}
 :::
 
@@ -281,7 +281,7 @@ where $\tau = t/t_C$. Making the terms on the right hand side of Eq. {eq}`eq:fiv
 
 :::{math}
 :label: eq:non-dim-five-term-accel-cr3bp
-\ddot{\vector{\rho}} = \left(\ddot{x}^* - 2\dot{y}^* - x^_\right)\uvec{\imath} + \left(\ddot{y}^_ + 2\dot{x}^* - y^_\right)\uvec{\jmath} + \ddot{z}^_\uvec{k}
+\ddot{\vector{\rho}} = \left(\ddot{x}^* - 2\dot{y}^* - x^*\right)\uvec{\imath} + \left(\ddot{y}^* + 2\dot{x}^* - y^_\right)\uvec{\jmath} + \ddot{z}^*\uvec{k}
 :::
 
 Now we have the non-dimensional inertial acceleration, we need to make Eq. {eq}`eq:vector-eom-cr3bp`, the equation of motion, non-dimensional. After a bunch of algebra, not shown here, we end up with:

the-n-body-problem/lagrange-points.md

@@ -207,7 +207,7 @@ ax.plot([0, 1], [0, -1], lw=1.0, color="silver", ls="--", label="$m_2$")
 ax.set_xticks(np.arange(0, 1.1, 0.1))
 ax.set_yticks(np.arange(-1.5, 1.75, 0.25))
 ax.grid()
-ax.set_xlabel("$\pi_2$")
+ax.set_xlabel(r"$\pi_2$")
 ax.set_ylabel("$x^*$")
 ax.annotate("$m_2$", xy=(0.55, 0.5), ha="center", va="bottom")
 ax.annotate("$m_1$", xy=(0.55, -0.5), ha="center", va="bottom")
@@ -228,7 +228,7 @@ On this figure, the $x$ axis is $\pi_2$ and the $y$ axis is $x^*$. For a given v
 The solutions for $x^*$ for the collinear Lagrange points lie on the S-curve shape. For a given value of $\pi_2$, we can see there are 3 solutions of the function, corresponding to the three collinear Lagrange points for that system.
 
 By convention, the Lagrange points are numbered such that $L_1$ lies between $m_1$ and $m_2$, $L_2$ lies to the right of $m_2$, and $L_3$ lies to the left of $m_1$. Thus, we can see on the figure that the upper part of the S-curve is the solution for $L_2$. Below $x^{*} = 1.0$, the solution is for $L_1$, since that lies between $m_1$ and $m_2$. Finally, below $x^{*} = -1.0$, the solution is for $L_3$.
-
+<!--
 {numref}`fig:lagrange-points-animation` plots the five Lagrange points in non-dimensional coordinates as a function of the mass ratio $\pi_2$.
 
 ```{code-cell}
@@ -282,20 +282,19 @@ ann = ax.annotate("", xy=(1, 0.85), ha="center", va="center", fontsize=20)
 
 ax.legend(bbox_to_anchor=(0, 1, 1, 0), loc="lower left", mode="expand", ncol=7)
 
-
 def init():
  m2_orb.set_data([], [])
  m1_orb.set_data([], [])
  equil.set_data([], [])
  ann.set_text("")
  return (com, l, m2_orb, m1_orb, equil, ann)
 
-
+# Precompute the arrays, then index by "time step" to return lines, rather than solving on each one.
 def animate(pi_2):
- m_1 = -pi_2
- m_2 = 1 - pi_2
- m1_line.set_data(m_1, 0)
- m2_line.set_data(m_2, 0)
+ m_1 = [-pi_2]
+ m_2 = [1 - pi_2]
+ m1_line.set_data(m_1, [0])
+ m2_line.set_data(m_2, [0])
  x_2 = m_2 * cos
  y_2 = m_2 * sin
  x_1 = m_1 * cos
@@ -306,11 +305,11 @@ def animate(pi_2):
  L_1 = newton(func=collinear_lagrange, x0=0, args=(pi_2,))
  L_3 = newton(func=collinear_lagrange, x0=-1, args=(pi_2,))
  L_4 = L_5 = 0.5 - pi_2
- L1_line.set_data(L_1, 0)
- L2_line.set_data(L_2, 0)
- L3_line.set_data(L_3, 0)
- L4_line.set_data(L_4, np.sqrt(3) / 2)
- L5_line.set_data(L_5, -np.sqrt(3) / 2)
+ L1_line.set_data([L_1], [0])
+ L2_line.set_data([L_2], [0])
+ L3_line.set_data([L_3], [0])
+ L4_line.set_data([L_4], [np.sqrt(3) / 2])
+ L5_line.set_data([L_5], [-np.sqrt(3) / 2])
  equil.set_data([m_1, L_4, m_2, L_5, m_1], [0, np.sqrt(3) / 2, 0, -np.sqrt(3) / 2, 0])
  ann.set_text(fr"$\pi_2$ = {pi_2:.4G}")
 
@@ -329,7 +328,7 @@ glue("lagrange-points-animation", HTML(anim.to_jshtml()), display=False)
 Animation showing the position of the five Lagrange points as the value of $\pi_2$ goes from 0 to 1.
 :::
 
-{numref}`fig:lagrange-points-animation` shows that the solution of the equations of motion for the equilibrium points is symmetrical. We chose $m_1$ to be the larger mass at the start of the problem, but we can interchange $m_1$ and $m_2$ without any problems.
+{numref}`fig:lagrange-points-animation` shows that the solution of the equations of motion for the equilibrium points is symmetrical. We chose $m_1$ to be the larger mass at the start of the problem, but we can interchange $m_1$ and $m_2$ without any problems. -->
 
 For the Earth-Moon system, the value of $\pi_2$ is approximately 0.012.
 
@@ -499,7 +498,7 @@ The Trojan and the Greek asteroids are clusters of asteroids that have collected
 
 The collinear Lagrange points, $L_1$, $L_2$, and $L_3$ are all **saddle points** in {numref}`fig:pseudo-potential-energy-cr3bp`, meaning that the function increases when going in one axis, but decreases going in the other axis. This means that the three collinear Lagrange points are **unstable** and an object placed at one of those points, if perturbed, will diverge from the position.
 
-Nonetheless, these are quite useful points for observation of the solar system. Several satellites have been placed at the $L_1$ point of the Earth-Sun system for solar observation, and the James Webb Space Telescope is planned to launch to the $L_2$ of the Earth-Sun system sometime ~this year~ in 2022.
+Nonetheless, these are quite useful points for observation of the solar system. Several satellites have been placed at the $L_1$ point of the Earth-Sun system for solar observation, and the James Webb Space Telescope (JWST) is located at the $L_2$ point in the Earth-Sun system specifically to avoid sunlight interefering with observations.
 
 These satellites orbit around the unstable Lagrange points in a [Lissajous orbit](https://en.wikipedia.org/wiki/Lissajous_orbit). This type of orbit requires a very small amount of propulsion onboard the satellite to keep position, but the orbit can last for a very long time with only a little fuel. One example is the [Wilkinson Microwave Anisotropy Probe](https://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe) (WMAP) which was sent to the $L_2$ point in the Earth-Sun system to study the [Cosmic microwave background](https://en.wikipedia.org/wiki/Cosmic_microwave_background). The trajectory of WMAP is shown in {numref}`fig:wmap-trajectory`.
 
@@ -509,4 +508,12 @@ These satellites orbit around the unstable Lagrange points in a [Lissajous orbit
 The trajectory of the [Wilkinson Microwave Anisotropy Probe](https://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe) (WMAP) as viewed from Earth. Note the distance in the bottom of the animation, showing the satellite as approximately 1.5 million km from the earth. [Phoenix7777](https://commons.wikimedia.org/wiki/File:Animation_of_Wilkinson_Microwave_Anisotropy_Probe_trajectory_-_Viewd_from_Earth.gif), [CC BY-SA 4.0](https://creativecommons.org/licenses/by-sa/4.0), via Wikimedia Commons.
 :::
 
+Another example is the JWST, mentioned previously. JWST has a simpler [halo orbit](https://en.wikipedia.org/wiki/Halo_orbit) around $L_2$. The orbit of JWST is shown in {numref}`fig:jwst-trajectory`.
+
+:::{figure} ../images/jwst-trajectory.gif
+:name: fig:jwst-trajectory
+
+The trajectory of the [James Webb Space Telescope](https://en.wikipedia.org/wiki/James_Webb_Space_Telescope) (JWST) as viewed from above the ecliptic plane with Earth fixed. Note again the distance in the bottom of the animation, showing the satellite as approximately 1.5 million km from the earth. [Phoenix7777](https://commons.wikimedia.org/wiki/File:Animation_of_James_Webb_Space_Telescope_trajectory_-_Polar_view.gif), [CC BY-SA 4.0](https://creativecommons.org/licenses/by-sa/4.0), via Wikimedia Commons.
+:::
+
 $L_1$ and $L_2$ in the Earth-Sun system are about 1.5 million km towards the Sun and away from the Sun, starting at the Earth, respectively. $L_3$ lies on the other side of the Sun, and has long been the predicted location of a hidden planet, since it could not be observed from Earth prior to the advent of satellite observation. Now, of course, we know there is no planet at that location.

the-orbit-equation/hyperbolic-trajectories.md

@@ -26,9 +26,9 @@ From the orbit equation, Eq. {eq}`eq:scalar-orbit-equation`, we see that the den
 
 As the true anomaly approaches $\nu_{\infty}$, $r$ approaches infinity. $\nu_{\infty}$ is restricted to be between 90° and 180°.
 
-For $-\nu_{\infty} < \nu < \nu_{\infty}$, the trajectory of $m_2$ follows the occupied or real trajectory shown on the left in {numref}`fig:hyperbolic-trajectory-animation`. For $\nu_{\infty} < \nu < \left({360}^{\circ} - \nu_{\infty}\right)$, $m_2$ would occupy the virtual trajectory on the figure below. This trajectory would require a repulsive gravitational force for a mass to actually follow it, so it is only a mathematical result.
+For $-\nu_{\infty} < \nu < \nu_{\infty}$, the trajectory of $m_2$ follows the occupied or real trajectory. For $\nu_{\infty} < \nu < \left({360}^{\circ} - \nu_{\infty}\right)$, $m_2$ would occupy the virtual trajectory. This trajectory would require a repulsive gravitational force for a mass to actually follow it, so it is only a mathematical result.
 
-```{code-cell} ipython3
+<!-- ```{code-cell} ipython3
 :tags: [remove-cell]
 
 from IPython.display import HTML
@@ -42,7 +42,7 @@ glue("hyperbolic-trajectory-animation", HTML(anim.to_jshtml()), display=False)
 :name: fig:hyperbolic-trajectory-animation
 
 Animation showing the hyperbolic trajectory and the value of the true anomaly for various positions on the occupied and virtual trajectories.
-:::
+::: -->
 
 Periapsis lies on the apse line, as usual, of the occupied trajectory. Interestingly, apoapsis lies on the virtual trajectory. Halfway between periapsis and apoapsis lies the center of a Cartesian coordinate system.