deploy: 0f78d6e

_sources/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.

_sources/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:

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

@@ -499,7 +499,7 @@ 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 +509,12 @@ 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.

classical-orbital-elements/orbital-elements-and-the-state-vector.html

@@ -789,7 +789,7 @@ <h2>Orbital Elements → State Vector<a class="headerlink" href="#orbital-elemen
 <p>The first step is rather simple, but the derivation of the second step requires a long digression on the topic of <a class="reference external" href="https://en.wikipedia.org/wiki/Rotation_of_axes">coordinate transformations</a>. So, we’ll skip the derivation and get right to the point.</p>
 <section id="step-1transform-to-perifocal-frame">
 <h3>Step 1—Transform to Perifocal Frame<a class="headerlink" href="#step-1transform-to-perifocal-frame" title="Link to this heading">#</a></h3>
-<p>Remember that the perifocal frame is defined in the orbital plane with the unit vectors <span class="math notranslate nohighlight">\(\uvec{p}\)</span>, <span class="math notranslate nohighlight">\(\uvec{q}\)</span>, and <span class="math notranslate nohighlight">\(\uvec{w}\)</span>, as shown in <a class="reference internal" href="perifocal-frame.html#fig-definition-of-perifocal-frame"><span class="std std-numref">Fig. 41</span></a>. The position and velocity components in the perifocal frame are given by Eq. <a class="reference internal" href="perifocal-frame.html#equation-eq-perifocal-vector-orbit-equation">(167)</a> and Eq. <a class="reference internal" href="perifocal-frame.html#equation-eq-perifocal-simplified-velocity-vector">(170)</a>, respectively.</p>
+<p>Remember that the perifocal frame is defined in the orbital plane with the unit vectors <span class="math notranslate nohighlight">\(\uvec{p}\)</span>, <span class="math notranslate nohighlight">\(\uvec{q}\)</span>, and <span class="math notranslate nohighlight">\(\uvec{w}\)</span>, as shown in <a class="reference internal" href="perifocal-frame.html#fig-definition-of-perifocal-frame"><span class="std std-numref">Fig. 42</span></a>. The position and velocity components in the perifocal frame are given by Eq. <a class="reference internal" href="perifocal-frame.html#equation-eq-perifocal-vector-orbit-equation">(167)</a> and Eq. <a class="reference internal" href="perifocal-frame.html#equation-eq-perifocal-simplified-velocity-vector">(170)</a>, respectively.</p>
 <div class="sd-tab-set docutils">
 <input checked="checked" id="sd-tab-item-14" name="sd-tab-set-7" type="radio">
 </input><label class="sd-tab-label" data-sync-id="tabcode-python" for="sd-tab-item-14">
@@ -813,7 +813,7 @@ <h3>Step 1—Transform to Perifocal Frame<a class="headerlink" href="#step-1tran
 </section>
 <section id="step-2rotate-the-perifocal-frame">
 <h3>Step 2—Rotate the Perifocal Frame<a class="headerlink" href="#step-2rotate-the-perifocal-frame" title="Link to this heading">#</a></h3>
-<p>The second step of this algorithm is to apply the set of coordinate transformations that converts the perifocal frame into the inertial frame. It turns out that a set of three rotations applied sequentially will accomplish this goal. These steps are shown in <a class="reference internal" href="#fig-euler-angle-rotation"><span class="std std-numref">Fig. 46</span></a>.</p>
+<p>The second step of this algorithm is to apply the set of coordinate transformations that converts the perifocal frame into the inertial frame. It turns out that a set of three rotations applied sequentially will accomplish this goal. These steps are shown in <a class="reference internal" href="#fig-euler-angle-rotation"><span class="std std-numref">Fig. 47</span></a>.</p>
 <ol class="arabic simple">
 <li><p>Rotate around the <span class="math notranslate nohighlight">\(\uvec{w}\)</span> axis until the <span class="math notranslate nohighlight">\(\uvec{p}\)</span> axis is aligned with the node line</p></li>
 <li><p>Rotate around the node line until the <span class="math notranslate nohighlight">\(\uvec{w}\)</span> axis is aligned with the <span class="math notranslate nohighlight">\(Z\)</span> axis</p></li>
@@ -822,7 +822,7 @@ <h3>Step 2—Rotate the Perifocal Frame<a class="headerlink" href="#step-2rotate
 <figure class="align-default" id="fig-euler-angle-rotation">
 <img alt="../_images/Euler.gif" src="../_images/Euler.gif" />
 <figcaption>
-<p><span class="caption-number">Fig. 46 </span><span class="caption-text">The sequence of rotations to convert from the perifocal frame to the inertial frame. Adapted from <a class="reference external" href="https://commons.wikimedia.org/wiki/File:Euler2.gif">Juansempere</a>, <a class="reference external" href="https://creativecommons.org/licenses/by-sa/4.0">CC BY-SA 4.0</a>, via Wikimedia Commons.</span><a class="headerlink" href="#fig-euler-angle-rotation" title="Link to this image">#</a></p>
+<p><span class="caption-number">Fig. 47 </span><span class="caption-text">The sequence of rotations to convert from the perifocal frame to the inertial frame. Adapted from <a class="reference external" href="https://commons.wikimedia.org/wiki/File:Euler2.gif">Juansempere</a>, <a class="reference external" href="https://creativecommons.org/licenses/by-sa/4.0">CC BY-SA 4.0</a>, via Wikimedia Commons.</span><a class="headerlink" href="#fig-euler-angle-rotation" title="Link to this image">#</a></p>
 </figcaption>
 </figure>
 <div class="sd-container-fluid sd-sphinx-override sd-mb-4 docutils">

classical-orbital-elements/perifocal-frame.html

@@ -515,10 +515,10 @@ <h1>Perifocal Frame<a class="headerlink" href="#perifocal-frame" title="Link to
 <figure class="align-default" id="fig-definition-of-perifocal-frame">
 <a class="reference internal image-reference" href="../_images/definition-of-perifocal-frame.svg"><img alt="../_images/definition-of-perifocal-frame.svg" src="../_images/definition-of-perifocal-frame.svg" width="75%" /></a>
 <figcaption>
-<p><span class="caption-number">Fig. 41 </span><span class="caption-text">The definition of the perifocal frame. The <span class="math notranslate nohighlight">\(\uvec{w}\)</span> direction is pointing direction towards the viewer.</span><a class="headerlink" href="#fig-definition-of-perifocal-frame" title="Link to this image">#</a></p>
+<p><span class="caption-number">Fig. 42 </span><span class="caption-text">The definition of the perifocal frame. The <span class="math notranslate nohighlight">\(\uvec{w}\)</span> direction is pointing direction towards the viewer.</span><a class="headerlink" href="#fig-definition-of-perifocal-frame" title="Link to this image">#</a></p>
 </figcaption>
 </figure>
-<p>As shown in <a class="reference internal" href="#fig-definition-of-perifocal-frame"><span class="std std-numref">Fig. 41</span></a>, <span class="math notranslate nohighlight">\(\uvec{p}\)</span> points along the apse line to the right of the focus, towards periapsis. <span class="math notranslate nohighlight">\(\uvec{q}\)</span> points 90° true anomaly from <span class="math notranslate nohighlight">\(\uvec{p}\)</span>. Finally, <span class="math notranslate nohighlight">\(\uvec{w}\)</span> points in the same direction as the angular momentum, and can be defined by:</p>
+<p>As shown in <a class="reference internal" href="#fig-definition-of-perifocal-frame"><span class="std std-numref">Fig. 42</span></a>, <span class="math notranslate nohighlight">\(\uvec{p}\)</span> points along the apse line to the right of the focus, towards periapsis. <span class="math notranslate nohighlight">\(\uvec{q}\)</span> points 90° true anomaly from <span class="math notranslate nohighlight">\(\uvec{p}\)</span>. Finally, <span class="math notranslate nohighlight">\(\uvec{w}\)</span> points in the same direction as the angular momentum, and can be defined by:</p>
 <div class="math notranslate nohighlight" id="equation-eq-perifocal-w-unit-vector">
 <span class="eqno">(164)<a class="headerlink" href="#equation-eq-perifocal-w-unit-vector" title="Link to this equation">#</a></span>\[\uvec{w} = \frac{\vector{h}}{h}\]</div>
 <section id="position-vector">

classical-orbital-elements/right-ascension-declination.html

@@ -514,16 +514,16 @@ <h2>Celestial Coordinate System<a class="headerlink" href="#celestial-coordinate
 <figure class="align-default" id="fig-ra-and-dec-on-celestial-sphere">
 <a class="reference internal image-reference" href="../_images/Ra_and_dec_on_celestial_sphere.png"><img alt="../_images/Ra_and_dec_on_celestial_sphere.png" src="../_images/Ra_and_dec_on_celestial_sphere.png" style="width: 50%;" /></a>
 <figcaption>
-<p><span class="caption-number">Fig. 42 </span><span class="caption-text">The coordinate system on the celestial sphere. <a class="reference external" href="https://commons.wikimedia.org/wiki/File:Ra_and_dec_on_celestial_sphere.png">Tfr000 (talk) 15:34, 15 June 2012 (UTC)</a></a>, <a class="reference external" href="https://creativecommons.org/licenses/by-sa/3.0">CC BY-SA 3.0</a>, via Wikimedia Commons.</span><a class="headerlink" href="#fig-ra-and-dec-on-celestial-sphere" title="Link to this image">#</a></p>
+<p><span class="caption-number">Fig. 43 </span><span class="caption-text">The coordinate system on the celestial sphere. <a class="reference external" href="https://commons.wikimedia.org/wiki/File:Ra_and_dec_on_celestial_sphere.png">Tfr000 (talk) 15:34, 15 June 2012 (UTC)</a></a>, <a class="reference external" href="https://creativecommons.org/licenses/by-sa/3.0">CC BY-SA 3.0</a>, via Wikimedia Commons.</span><a class="headerlink" href="#fig-ra-and-dec-on-celestial-sphere" title="Link to this image">#</a></p>
 </figcaption>
 </figure>
-<p>The celestial sphere is shown in <a class="reference internal" href="#fig-ra-and-dec-on-celestial-sphere"><span class="std std-numref">Fig. 42</span></a>. The fundamental plane in this coordinate system is the earth’s equatorial plane. The intersection of the earth’s equatorial plane and the celestial sphere is the <strong>celestial equator</strong>. The coordinates are similar to latitude and longitude on the earth, but projected out onto the celestial sphere.</p>
+<p>The celestial sphere is shown in <a class="reference internal" href="#fig-ra-and-dec-on-celestial-sphere"><span class="std std-numref">Fig. 43</span></a>. The fundamental plane in this coordinate system is the earth’s equatorial plane. The intersection of the earth’s equatorial plane and the celestial sphere is the <strong>celestial equator</strong>. The coordinates are similar to latitude and longitude on the earth, but projected out onto the celestial sphere.</p>
 <p>The celestial north and south poles are aligned with the earth’s north and south poles. The angle from the equator to the poles is called the <a class="reference external" href="https://en.wikipedia.org/wiki/Declination"><strong>declination</strong></a>. Like latitude, declination is 0° at the celestial equator and ±90° at the poles. Declination is usually abbreviated DEC.</p>
 <p>The other coordinate is called <a class="reference external" href="https://en.wikipedia.org/wiki/Right_ascension"><strong>right ascension</strong></a>, abbreviated RA. The zero point of right ascension is at the March, or vernal, equinox and increases going eastward. It takes the earth about 1 hour to rotate through 15° of right ascension, so lines of right ascension are also called hour lines.</p>
 </section>
 <section id="ecliptic">
 <h2>Ecliptic<a class="headerlink" href="#ecliptic" title="Link to this heading">#</a></h2>
-<p>Also shown on <a class="reference internal" href="#fig-ra-and-dec-on-celestial-sphere"><span class="std std-numref">Fig. 42</span></a> is a line representing the ecliptic. This is the apparent path that the sun follows through the sky if we measured the right ascension and declination of the sun at the same time every day. The ecliptic is also the plane of the earth’s orbit around the sun. The reason that the sun does not follow the celestial equator is because the earth’s rotation axis is tilted by 23.4°.</p>
+<p>Also shown on <a class="reference internal" href="#fig-ra-and-dec-on-celestial-sphere"><span class="std std-numref">Fig. 43</span></a> is a line representing the ecliptic. This is the apparent path that the sun follows through the sky if we measured the right ascension and declination of the sun at the same time every day. The ecliptic is also the plane of the earth’s orbit around the sun. The reason that the sun does not follow the celestial equator is because the earth’s rotation axis is tilted by 23.4°.</p>
 <p>Another way to view the equinox is as the line generated by the intersection of the equatorial plane and the ecliptic plane.</p>
 </section>
 <section id="stars">