Bump Dependencies and Update Equation (#28)

* Bump dependencies and fix resulting errors * Improve equation formatting and add spelling words * More bumps! * Fix errors
bryanwweber · Jun 16, 2024 · b2b2a74 · b2b2a74
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diff --git a/.vscode/settings.json b/.vscode/settings.json
@@ -1,31 +1,39 @@
 {
- "cSpell.words": [
- "apoapsis",
- "deorbit",
- "Hohmann",
- "Kármán",
- "periapsis",
- "semimajor",
- "Storrs"
- ],
- "cSpell.ignoreWords": ["devcontainer"],
- "cSpell.languageSettings": [
- {
- // use with Markdown files
- "languageId": "markdown",
- "dictionaries": ["latex"],
- // Exclude code
- // "ignoreRegExpList": [
- // "\\$.*\\$",
- // "\\$\\$([\\n\\(\\)]|[^(\\$\\$)])*\\$\\$",
- // "`.*`",
- // "```([\\n\\(\\)]|[^(\\$\\$)])*```",
- // "\\{\\w+\\}`.+`",
- // ":::\\{math\\}([\\n\\(\\)]|[^(\\$\\$)])*:::",
- // "---([\\n\\(\\)]|[^(\\$\\$)])*---"
- // ]
- }
+ "cSpell.words": [
+ "apoapse",
+ "apoapsis",
+ "deorbit",
+ "Hohmann",
+ "Kármán",
+ "Laguerre",
+ "periapsis",
+ "Prussing",
+ "semimajor",
+ "Semiminor",
+ "Storrs",
+ "Stumpff"
+ ],
+ "cSpell.ignoreWords": [
+ "devcontainer"
+ ],
+ "cSpell.languageSettings": [
+ {
+ // use with Markdown files
+ "languageId": "markdown",
+ "dictionaries": [
+ "latex"
  ],
- "editor.formatOnSave": true
-
+ // Exclude code
+ // "ignoreRegExpList": [
+ // "\\$.*\\$",
+ // "\\$\\$([\\n\\(\\)]|[^(\\$\\$)])*\\$\\$",
+ // "`.*`",
+ // "```([\\n\\(\\)]|[^(\\$\\$)])*```",
+ // "\\{\\w+\\}`.+`",
+ // ":::\\{math\\}([\\n\\(\\)]|[^(\\$\\$)])*:::",
+ // "---([\\n\\(\\)]|[^(\\$\\$)])*---"
+ // ]
+ }
+ ],
+ "editor.formatOnSave": true
 }
diff --git a/classical-orbital-elements/scripts/orbital_elements_and_the_state_vector.py b/classical-orbital-elements/scripts/orbital_elements_and_the_state_vector.py
@@ -51,6 +51,7 @@
  warnings.simplefilter("ignore")
  from myst_nb import glue
 
+ np.set_printoptions(legacy="1.25")
  glue("orbital-elements-radius", r, display=False)
  glue("orbital-elements-velocity", v, display=False)
  glue("orbital-elements-v_r", v_r, display=False)

diff --git a/interplanetary-maneuvers/planetary-arrival-flyby.md b/interplanetary-maneuvers/planetary-arrival-flyby.md
@@ -250,6 +250,7 @@ h_1 = np.sqrt(mu_sun * R_Neptune * (1 - e_1))
 from functools import partial
 from myst_nb import glue as myst_glue
 glue = partial(myst_glue, display=False)
+np.set_printoptions(legacy="1.25")
 glue("interplanetary-flyby-e_1", e_1)
 glue("interplanetary-flyby-h_1", h_1)
 ```

diff --git a/orbital-maneuvers/comparison-of-bielliptic-and-hohmann.md b/orbital-maneuvers/comparison-of-bielliptic-and-hohmann.md
@@ -41,6 +41,7 @@ v_3 = np.sqrt(mu / r_3) # km/s
 
 ```{code-cell} ipython3
 :tags: [remove-cell]
+np.set_printoptions(legacy="1.25")
 glue("moon_circle_velocity", v_1)
 glue("leo_circle_velocity", v_3)
 ```

diff --git a/orbital-maneuvers/plane-change-example.md b/orbital-maneuvers/plane-change-example.md
@@ -66,6 +66,7 @@ Delta_v_1 = Delta_v_1_LEO + Delta_v_1_GEO
 
 ```{code-cell} ipython3
 :tags: [remove-cell]
+np.set_printoptions(legacy="1.25")
 from functools import partial
 from myst_nb import glue as myst_glue
 glue = partial(myst_glue, display=False)

diff --git a/orbital-maneuvers/single-impulse-example.md b/orbital-maneuvers/single-impulse-example.md
@@ -29,6 +29,8 @@ from matplotlib.patches import Circle, Arc
 import numpy as np
 from myst_nb import glue
 
+np.set_printoptions(legacy="1.25")
+
 R_E = 6378 # km
 orbit_radius = 1000 # km
 mu = 398_600 # km**2/s**3

diff --git a/pdm.lock b/pdm.lock
diff --git a/pyproject.toml b/pyproject.toml
@@ -4,7 +4,7 @@ dependencies = [
  "astropy>=6.0.0",
  "docutils>=0.20.1",
  "ipympl>=0.9.3",
- "jupyter-book>=1.0.0",
+ "jupyter-book>=1.0",
  "matplotlib>=3.8.3",
  "myst-nb-bokeh>=2024.1.0",
  "myst-nb>=1.0.0",
@@ -22,6 +22,8 @@ requires-python = ">=3.10,<3.13"
 name = "orbital-mechanics-notes"
 authors = [{ name = "Bryan Weber", email = "[email protected]" }]
 license = { text = "CC-BY-SA-4.0" }
+description = "A book for Orbital Mechanics"
+version = "0.0.0"
 
 [tool.pdm]
 distribution = false

diff --git a/the-n-body-problem/lagrange-points.md b/the-n-body-problem/lagrange-points.md
@@ -294,8 +294,8 @@ def init():
 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)
+ 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 +306,11 @@ 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}")
 

diff --git a/the-orbit-equation/hyperbolic_trajectory.py b/the-orbit-equation/hyperbolic_trajectory.py
@@ -59,7 +59,7 @@ def init():
 
  def animate(t):
  r = a * (e**2 - 1) / (1 + e * np.cos(t))
- point.set_data(-a - r_p + r * np.cos(t), r * np.sin(t))
+ point.set_data([-a - r_p + r * np.cos(t)], [r * np.sin(t)])
  ann.set_text(rf"$\nu$ = {np.degrees(t):.2F}°")
  return (point, ann)
 

diff --git a/time-since-periapsis-and-keplers-equation/elliptical-orbit-example.md b/time-since-periapsis-and-keplers-equation/elliptical-orbit-example.md
@@ -35,7 +35,6 @@ e = \frac{r_a - r_p}{r_a + r_p}
 :::
 
 ```{code-cell} ipython3
-# %matplotlib notebook
 import numpy as np
 from scipy.optimize import newton
 
@@ -51,6 +50,7 @@ e = (r_a - r_p)/(r_a + r_p)
 :tags: [remove-cell]
 from functools import partial
 from myst_nb import glue as myst_glue
+np.set_printoptions(legacy="1.25")
 glue = partial(myst_glue, display=False)
 glue("ellipse-time-since-periapsis-e", e)
 ```
@@ -149,7 +149,7 @@ def kepler(E, M_e, e):
 
 def d_kepler_d_E(E, M_e, e):
  """The derivative of Kepler's equation, to be used in a Newton solver.
- 
+
  Note that the argument M_e is unused, but must be present so the function
  arguments are consistent with the kepler function.
  """

diff --git a/time-since-periapsis-and-keplers-equation/elliptical-orbit-time-in-shadow.md b/time-since-periapsis-and-keplers-equation/elliptical-orbit-time-in-shadow.md
@@ -29,6 +29,7 @@ from matplotlib.patches import Ellipse, Circle, Arc, Rectangle
 import numpy as np
 
 glue = partial(myst_glue, display=False)
+np.set_printoptions(legacy="1.25")
 
 mu = 3.986004418E5 # km**3/s**2
 R_E = 6378 # km

diff --git a/time-since-periapsis-and-keplers-equation/hyperbolic-trajectory-example.md b/time-since-periapsis-and-keplers-equation/hyperbolic-trajectory-example.md
@@ -41,6 +41,8 @@ e = h**2 / (r_p * mu) - 1
 :tags: [remove-cell]
 from functools import partial
 from myst_nb import glue as myst_glue
+
+np.set_printoptions(legacy="1.25")
 glue = partial(myst_glue, display=False)
 
 glue("hyperbolic-time-since-perigee-h", h)
@@ -117,7 +119,7 @@ def kepler(F, M_h, e):
 
 def d_kepler_d_F(F, M_h, e):
  """The derivative of Kepler's equation, to be used in a Newton solver.
- 
+
  Note that the argument M_h is unused, but must be present so the function
  arguments are consistent with the kepler function.
  """
@@ -129,7 +131,7 @@ F_2 = newton(func=kepler, fprime=d_kepler_d_F, x0=np.pi, args=(M_h2, e))
 With this value for $F$, we can calculate the value for $\nu$. To avoid quadrant ambiguity problems, we will use Eq. {eq}`eq:eccentric-anomaly-true-anomaly-hyperbola`.
 
 ```{code-cell} ipython3
-sqrt_e = np.sqrt((e + 1) / (e - 1)) 
+sqrt_e = np.sqrt((e + 1) / (e - 1))
 nu_2 = (2 * np.arctan(sqrt_e * np.tanh(F_2 / 2))) % (2 * np.pi)
 ```
 
@@ -181,12 +183,12 @@ function kepler
  v_p = 15; % km/s
  h = r_p * v_p;
  e = h^2 / (mu * r_p) - 1;
- 
+
  nu_1 = deg2rad(100);
  F_1 = 2 * atanh(sqrt((e - 1) / (e + 1)) * tan(nu_1 / 2));
  M_h1 = e * sinh(F_1) - F_1;
  t_1 = h^3 / mu^2 * 1 / (e^2 - 1)^(3 / 2) * M_h1;
- 
+
  t_2 = t_1 + 3 * 3600;
  M_h2 = mu^2 / h^3 * (e^2 - 1)^(3 / 2) * t_2;
 
@@ -198,7 +200,7 @@ function kepler
  t2 = 2 * atan(sqrt((e + 1) / (e - 1)) * tanh(F_2 / 2));
  nu_2 = mod(t2, 2 * pi);
  disp(rad2deg(nu_2))
- 
+
  r_2 = h^2 / mu * 1 / (1 + e * cos(nu_2));
  v_perp = h / r_2;
  v_r = mu / h * e * sin(nu_2);

diff --git a/time-since-periapsis-and-keplers-equation/universal-variables.md b/time-since-periapsis-and-keplers-equation/universal-variables.md
@@ -306,13 +306,13 @@ Alternatively, a more refined estimate can be determined using a secant estimate
 where $f(\chi^{+})$ is the solution of Kepler's equation with the $\chi^{+}$ value.
 
 ::::{note}
-Prussing and Conway {cite}`Prussing2013`, citing Conway {cite}`Conway1986` suggest that faster convergence in the solution of Kepler's equation can be achieved by using the **Laguerre algorithm**, rather than Newton's algorithm. Another advantage of the Laguerre algorithm is that it is relatively insensitive to the value of the initial guess.
+Prussing and Conway {cite}`Prussing2013`, citing Conway {cite}`Conway1986` suggest that faster convergence in the solution of Kepler's equation can be achieved by using the [**Laguerre algorithm**](https://en.wikipedia.org/wiki/Laguerre%27s_method), rather than Newton's algorithm. Another advantage of the Laguerre algorithm is that it is relatively insensitive to the value of the initial guess.
 
 The Laguerre algorithm can be implemented as:
 
 :::{math}
 :label:
-\chi_{i + 1} = \chi_{i} - \frac{n f(\chi_i)}{f'(\chi_i) \pm \left[\left(n - 1\right)^2 \left(f'(\chi_i)\right)^2 - n\left(n - 1\right) f(\chi_i)f''(\chi_i)\right]^{1/2}}
+\chi_{i + 1} = \chi_{i} - \frac{n f(\chi_i)}{f'(\chi_i) \pm \sqrt{\left(n - 1\right)^2 \left[f'(\chi_i)\right]^2 - n\left(n - 1\right) f(\chi_i)f''(\chi_i)}}
 :::
 
 The sign ambiguity in the denominator is determined by taking the sign of the numerical value of $f'(\chi_i)$. In addition, the solution is relatively insensitive to the choice of the value of $n$, which is an integer constant. It seems as though $n = 5$ is a reasonable value. Choosing $n = 1$ gives the standard Newton's algorithm.