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71 changes: 44 additions & 27 deletions _sources/orbital-mechanics/Lecture2/Lecture2.ipynb
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"The time derivative of $\\bf{h}$ is\n",
"```{math}\n",
":label: L2_9 \n",
"\\frac{d\\bf{h}}{dt} &=\\frac{d}{dt}(\\bf{r}\\times\\bf{\\dot r})\\\\\n",
"\\Rightarrow \\frac{d\\bf{h}}{dt} &= \\bf{\\dot r}\\times\\bf{\\dot r} + \\bf{r}\\times\\bf{\\ddot r}\\\\\n",
"\\Rightarrow \\frac{d\\bf{h}}{dt} &= \\bf{r}\\times\\bf{\\ddot r}\n",
"\\frac{\\mathrm{d}\\bf{h}}{\\mathrm{d}t} &=\\frac{\\mathrm{d}}{\\mathrm{d}t}(\\bf{r}\\times\\bf{\\dot r})\\\\\n",
"\\Rightarrow \\frac{d\\bf{h}}{\\mathrm{d}t} &= \\bf{\\dot r}\\times\\bf{\\dot r} + \\bf{r}\\times\\bf{\\ddot r}\\\\\n",
"\\Rightarrow \\frac{\\mathrm{d}\\bf{h}}{\\mathrm{d}t} &= \\bf{r}\\times\\bf{\\ddot r}\n",
"```\n",
"\n",
"From the two-body dynamics equation {eq}`L1_29`, we can rewrite the RHS of {eq}`L2_9` as:\n",
"```{math}\n",
":label: L2_10\n",
"\\Rightarrow \\frac{d\\bf{h}}{dt} &=\\bf{r}\\times-\\frac{\\mu}{r^3}\\bf{r}\\\\\n",
"\\Rightarrow \\frac{d\\bf{h}}{dt} &=0\n",
"\\Rightarrow \\frac{\\mathrm{d}\\bf{h}}{\\mathrm{d}t} &=\\bf{r}\\times-\\frac{\\mu}{r^3}\\bf{r}\\\\\n",
"\\Rightarrow \\frac{\\mathrm{d}\\bf{h}}{\\mathrm{d}t} &=0\n",
"```\n",
"which tells us that specific angular momentum is a constant.\n",
"\n",
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"$\\bf r$ as\n",
"```{math}\n",
":label: L2_18\n",
"dA = \\frac{1}{2}r\\, v\\, \\mathrm{d}t\\sin(\\phi)\n",
"\\mathrm{d}A = \\frac{1}{2}r\\, v\\, \\mathrm{d}t\\sin(\\phi)\n",
"```\n",
"wihch can be further rewritten\n",
"```{math}\n",
":label: L2_19\n",
"\\frac{dA}{dt} &= \\frac{1}{2}r\\, v\\, \\sin(\\phi)\\\\\n",
"\\Rightarrow \\frac{dA}{dt}&= \\frac{1}{2}r\\, v_\\theta \n",
"\\frac{\\mathrm{d}A}{\\mathrm{d}t} &= \\frac{1}{2}r\\, v\\, \\sin(\\phi)\\\\\n",
"\\Rightarrow \\frac{\\mathrm{d}A}{\\mathrm{d}t}&= \\frac{1}{2}r\\, v_\\theta \n",
"```\n",
"or\n",
"```{math}\n",
":label: L2_20\n",
"\\frac{dA}{dt}=\\frac{1}{2}h = constant\n",
"\\frac{\\mathrm{d}A}{\\mathrm{d}t}=\\frac{1}{2}h = constant\n",
"```\n",
"This is essentially giving us *Kepler's Second Law*, which tells us that areal velocity is\n",
"constant. In other words, it tells us that equal areas are swept in equal intervals of\n",
Expand All @@ -395,54 +395,71 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"If the motion is planar we can use polar coordinates to fully describe it. \n",
"If the motion is planar, we can use polar coordinates to fully describe it.\n",
"Consider a situation where this planar motion lies in the x-y plane shown in\n",
"{numref}`polar-twobody`.\n",
"\n",
"![Figure 3](images/L2_3.png)\n",
"\n",
"- In polar coordinates $\\bf{r} = r\\bf{e_r}$\n",
"```{figure} images/L2_3.png\n",
"---\n",
"height: 350px\n",
"name: polar-twobody\n",
"---\n",
"Polar coordinate system (in blue) to study the two-body problem.\n",
"```\n",
"Using polar coordinates shown in {numref}`polar-twobody`, we can see that $\\bf{r} = r\\bf{e_r}$. Thus\n",
"\n",
"```{math} \n",
":label: L2_21\n",
"\\frac{d\\bf{r}}{dt} = \\bf{v} = {\\dot r}\\bf{e_r} + {r}\\bf{e_\\theta}\\dot \\theta \n",
"\\frac{d\\bf{r}}{dt} = {\\bf v} = {\\dot r}{ \\bf e}_r + {r}{\\dot \\theta}{\\bf e}_\\theta \n",
"```\n",
"\n",
"which allows us to write\n",
"```{math} \n",
":label: L2_22\n",
"\\bf{h}=\\bf{r}\\times\\bf{v}=r\\bf{e_r}\\times({\\dot r}\\bf{e_r} + {r}\\bf{e_\\theta}\\dot \\theta)\n",
"{\\bf h} &= {\\bf r} \\times {\\bf v}\\\\\n",
"&=r {\\bf e}_r \\times ({\\dot r} {\\bf e}_r + {r} \\dot \\theta {\\bf e}_\\theta)\n",
"```\n",
"\n",
"or\n",
"```{math} \n",
":label: L2_23\n",
"⇒r^2\\dot \\theta{\\bf{k}} = rv_\\theta {\\bf{k}} = constant\n",
" {\\bf h} &= r^2\\dot \\theta \\hat{\\bf{k}}\\\\ &= rv_\\theta \\hat{\\bf{k}} = constant\n",
"```\n",
"\n",
">\n",
"where $\\hat{\\bf{k}}$ is a unit vector orthogonal to ${\\bf e}_r$ and ${\\bf e}_\\theta$.\n",
"Further, we can substitute for $h$ from {eq}`L2_23` into to {eq}`L2_20`, which gives\n",
"```{math} \n",
":label: L2_24\n",
"\\frac{dA}{dt}=\\frac{1}{2}h=\\frac{1}{2}r^2\\dot \\theta\n",
"\\frac{\\mathrm{d}A}{\\mathrm{d}t} &= \\frac{1}{2}h\\\\\n",
"\\frac{\\mathrm{d}A}{\\mathrm{d}t} &= \\frac{1}{2}r^2\\dot \\theta\n",
"```\n",
"\n",
"Taking the second derivatibe of $\\bf r$ using {eq}`L2_21`, we get\n",
"```{math} \n",
":label: L2_25\n",
"{\\frac{d^2\\bf{r}}{dt}} = {\\bf{a}} = ({\\ddot r} - r\\dot\\theta^2) {\\bf{e_r}} + (r\\ddot \\theta+2\\dot r\\dot \\theta) {\\bf{e_\\theta}}\n",
"{\\frac{d^2\\bf{r}}{\\mathrm{d}t^2}} = \\ddot{\\bf r}\n",
"= ({\\ddot r} - r\\dot\\theta^2) {\\bf{e_r}} + (r\\ddot \\theta+2\\dot r\\dot \\theta) {\\bf{e_\\theta}}.\n",
"```\n",
"\n",
"We can also write the RHS of {eq}`L1_30` in polar coordinates as\n",
"```{math} \n",
":label: L2_26\n",
"\\frac{-\\mu}{r^2}\\bf{e_r}\n",
"```\n",
"\n",
"which upon comparing to {eq}`L2_25` gives the following two scalar ODEs along the ${\\bf e}_r$ and ${\\bf e}_\\theta$ directions, respectively:\n",
"```{math}\n",
":label: L2_27\n",
"{\\ddot r} - r\\dot \\theta^2 = \\frac{-\\mu}{r^2}\n",
"{\\ddot r} - r\\dot \\theta^2 &= \\frac{-\\mu}{r^2}\\\\\n",
"r\\ddot \\theta + 2{\\dot r}\\dot \\theta &= 0\n",
"```\n",
"\n",
"Note that the second of Equations {eq}`L2_27` can also obtained from differentiating\n",
"the specific angular momentum\n",
"```{math}\n",
":label: L2_28\n",
"r\\ddot \\theta + 2{\\dot r}\\dot \\theta = 0 ↔ \\frac{d}{dt}(r^2\\dot \\theta)=2r{\\dot r}\\dot \\theta+r^2\\ddot \\theta=r(r\\ddot \\theta+2{\\dot r}\\dot \\theta) =0\n",
"\\frac{d}{dt}(h) &= \\frac{d}{dt}(r^2\\dot \\theta)\\\\\n",
"\\Rightarrow &=2r{\\dot r}\\dot \\theta+r^2\\ddot \\theta\\\\\n",
"\\Rightarrow &=r(r\\ddot \\theta+2{\\dot r}\\dot \\theta)\\\\\n",
"\\Rightarrow &=0\n",
"```\n",
"\n",
"- {eq}`L2_28` is equivalent to $\\bf{h}$ = constant which is a scalar"
"In other words, the second of Equations {eq}`L2_27` is equivalent to $h = constant$."
]
},
{
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2 changes: 1 addition & 1 deletion attitude-dynamics/Lecture14/dynamics-1.html
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Expand Up @@ -674,7 +674,7 @@ <h2>Free body diagrams to determine forces<a class="headerlink" href="#free-body
<figure class="align-default" id="spring-forces">
<img alt="../../_images/spring-forces.png" src="../../_images/spring-forces.png" />
<figcaption>
<p><span class="caption-number">Fig. 19 </span><span class="caption-text">Spring forces</span><a class="headerlink" href="#spring-forces" title="Permalink to this image">#</a></p>
<p><span class="caption-number">Fig. 20 </span><span class="caption-text">Spring forces</span><a class="headerlink" href="#spring-forces" title="Permalink to this image">#</a></p>
</figcaption>
</figure>
</section>
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67 changes: 43 additions & 24 deletions orbital-mechanics/Lecture2/Lecture2.html
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Expand Up @@ -707,13 +707,13 @@ <h3>Gravitational Force<a class="headerlink" href="#gravitational-force" title="
{\bf{h}} &amp;= \bf{r} \times \bf{\dot r}\end{split}\]</div>
<p>The time derivative of <span class="math notranslate nohighlight">\(\bf{h}\)</span> is</p>
<div class="math notranslate nohighlight" id="equation-l2-9">
<span class="eqno">(41)<a class="headerlink" href="#equation-l2-9" title="Permalink to this equation">#</a></span>\[\begin{split}\frac{d\bf{h}}{dt} &amp;=\frac{d}{dt}(\bf{r}\times\bf{\dot r})\\
\Rightarrow \frac{d\bf{h}}{dt} &amp;= \bf{\dot r}\times\bf{\dot r} + \bf{r}\times\bf{\ddot r}\\
\Rightarrow \frac{d\bf{h}}{dt} &amp;= \bf{r}\times\bf{\ddot r}\end{split}\]</div>
<span class="eqno">(41)<a class="headerlink" href="#equation-l2-9" title="Permalink to this equation">#</a></span>\[\begin{split}\frac{\mathrm{d}\bf{h}}{\mathrm{d}t} &amp;=\frac{\mathrm{d}}{\mathrm{d}t}(\bf{r}\times\bf{\dot r})\\
\Rightarrow \frac{d\bf{h}}{\mathrm{d}t} &amp;= \bf{\dot r}\times\bf{\dot r} + \bf{r}\times\bf{\ddot r}\\
\Rightarrow \frac{\mathrm{d}\bf{h}}{\mathrm{d}t} &amp;= \bf{r}\times\bf{\ddot r}\end{split}\]</div>
<p>From the two-body dynamics equation <a class="reference internal" href="#equation-l1-29">(30)</a>, we can rewrite the RHS of <a class="reference internal" href="#equation-l2-9">(41)</a> as:</p>
<div class="math notranslate nohighlight" id="equation-l2-10">
<span class="eqno">(42)<a class="headerlink" href="#equation-l2-10" title="Permalink to this equation">#</a></span>\[\begin{split}\Rightarrow \frac{d\bf{h}}{dt} &amp;=\bf{r}\times-\frac{\mu}{r^3}\bf{r}\\
\Rightarrow \frac{d\bf{h}}{dt} &amp;=0\end{split}\]</div>
<span class="eqno">(42)<a class="headerlink" href="#equation-l2-10" title="Permalink to this equation">#</a></span>\[\begin{split}\Rightarrow \frac{\mathrm{d}\bf{h}}{\mathrm{d}t} &amp;=\bf{r}\times-\frac{\mu}{r^3}\bf{r}\\
\Rightarrow \frac{\mathrm{d}\bf{h}}{\mathrm{d}t} &amp;=0\end{split}\]</div>
<p>which tells us that specific angular momentum is a constant.</p>
<div class="math notranslate nohighlight" id="equation-l2-11">
<span class="eqno">(43)<a class="headerlink" href="#equation-l2-11" title="Permalink to this equation">#</a></span>\[{\bf h} = \textbf{constant}\]</div>
Expand Down Expand Up @@ -787,14 +787,14 @@ <h3>Areal velocity and Kepler’s Second Law<a class="headerlink" href="#areal-v
<p>From <a class="reference internal" href="#direction-angmom"><span class="std std-numref">Fig. 4</span></a>, we can write the infinitesimal area <span class="math notranslate nohighlight">\(\mathrm{d}A\)</span> swept by
<span class="math notranslate nohighlight">\(\bf r\)</span> as</p>
<div class="math notranslate nohighlight" id="equation-l2-18">
<span class="eqno">(49)<a class="headerlink" href="#equation-l2-18" title="Permalink to this equation">#</a></span>\[dA = \frac{1}{2}r\, v\, \mathrm{d}t\sin(\phi)\]</div>
<span class="eqno">(49)<a class="headerlink" href="#equation-l2-18" title="Permalink to this equation">#</a></span>\[\mathrm{d}A = \frac{1}{2}r\, v\, \mathrm{d}t\sin(\phi)\]</div>
<p>wihch can be further rewritten</p>
<div class="math notranslate nohighlight" id="equation-l2-19">
<span class="eqno">(50)<a class="headerlink" href="#equation-l2-19" title="Permalink to this equation">#</a></span>\[\begin{split}\frac{dA}{dt} &amp;= \frac{1}{2}r\, v\, \sin(\phi)\\
\Rightarrow \frac{dA}{dt}&amp;= \frac{1}{2}r\, v_\theta \end{split}\]</div>
<span class="eqno">(50)<a class="headerlink" href="#equation-l2-19" title="Permalink to this equation">#</a></span>\[\begin{split}\frac{\mathrm{d}A}{\mathrm{d}t} &amp;= \frac{1}{2}r\, v\, \sin(\phi)\\
\Rightarrow \frac{\mathrm{d}A}{\mathrm{d}t}&amp;= \frac{1}{2}r\, v_\theta \end{split}\]</div>
<p>or</p>
<div class="math notranslate nohighlight" id="equation-l2-20">
<span class="eqno">(51)<a class="headerlink" href="#equation-l2-20" title="Permalink to this equation">#</a></span>\[\frac{dA}{dt}=\frac{1}{2}h = constant\]</div>
<span class="eqno">(51)<a class="headerlink" href="#equation-l2-20" title="Permalink to this equation">#</a></span>\[\frac{\mathrm{d}A}{\mathrm{d}t}=\frac{1}{2}h = constant\]</div>
<p>This is essentially giving us <em>Kepler’s Second Law</em>, which tells us that areal velocity is
constant. In other words, it tells us that equal areas are swept in equal intervals of
time along the orbit in a body whose dynamics are governed by the simple two-body system described
Expand All @@ -803,30 +803,49 @@ <h3>Areal velocity and Kepler’s Second Law<a class="headerlink" href="#areal-v
</section>
<section id="two-body-dynamics-in-polar-coordinates">
<span id="content-polar-coordinates"></span><h2>Two-body Dynamics in Polar Coordinates<a class="headerlink" href="#two-body-dynamics-in-polar-coordinates" title="Permalink to this heading">#</a></h2>
<p>If the motion is planar we can use polar coordinates to fully describe it.</p>
<p><img alt="Figure 3" src="../../_images/L2_3.png" /></p>
<ul class="simple">
<li><p>In polar coordinates <span class="math notranslate nohighlight">\(\bf{r} = r\bf{e_r}\)</span></p></li>
</ul>
<p>If the motion is planar, we can use polar coordinates to fully describe it.
Consider a situation where this planar motion lies in the x-y plane shown in
<a class="reference internal" href="#polar-twobody"><span class="std std-numref">Fig. 5</span></a>.</p>
<figure class="align-default" id="polar-twobody">
<a class="reference internal image-reference" href="../../_images/L2_3.png"><img alt="../../_images/L2_3.png" src="../../_images/L2_3.png" style="height: 350px;" /></a>
<figcaption>
<p><span class="caption-number">Fig. 5 </span><span class="caption-text">Polar coordinate system (in blue) to study the two-body problem.</span><a class="headerlink" href="#polar-twobody" title="Permalink to this image">#</a></p>
</figcaption>
</figure>
<p>Using polar coordinates shown in <a class="reference internal" href="#polar-twobody"><span class="std std-numref">Fig. 5</span></a>, we can see that <span class="math notranslate nohighlight">\(\bf{r} = r\bf{e_r}\)</span>. Thus</p>
<div class="math notranslate nohighlight" id="equation-l2-21">
<span class="eqno">(52)<a class="headerlink" href="#equation-l2-21" title="Permalink to this equation">#</a></span>\[\frac{d\bf{r}}{dt} = \bf{v} = {\dot r}\bf{e_r} + {r}\bf{e_\theta}\dot \theta \]</div>
<span class="eqno">(52)<a class="headerlink" href="#equation-l2-21" title="Permalink to this equation">#</a></span>\[\frac{d\bf{r}}{dt} = {\bf v} = {\dot r}{ \bf e}_r + {r}{\dot \theta}{\bf e}_\theta \]</div>
<p>which allows us to write</p>
<div class="math notranslate nohighlight" id="equation-l2-22">
<span class="eqno">(53)<a class="headerlink" href="#equation-l2-22" title="Permalink to this equation">#</a></span>\[\bf{h}=\bf{r}\times\bf{v}=r\bf{e_r}\times({\dot r}\bf{e_r} + {r}\bf{e_\theta}\dot \theta)\]</div>
<span class="eqno">(53)<a class="headerlink" href="#equation-l2-22" title="Permalink to this equation">#</a></span>\[\begin{split}{\bf h} &amp;= {\bf r} \times {\bf v}\\
&amp;=r {\bf e}_r \times ({\dot r} {\bf e}_r + {r} \dot \theta {\bf e}_\theta)\end{split}\]</div>
<p>or</p>
<div class="math notranslate nohighlight" id="equation-l2-23">
<span class="eqno">(54)<a class="headerlink" href="#equation-l2-23" title="Permalink to this equation">#</a></span>\[⇒r^2\dot \theta{\bf{k}} = rv_\theta {\bf{k}} = constant\]</div>
<span class="eqno">(54)<a class="headerlink" href="#equation-l2-23" title="Permalink to this equation">#</a></span>\[\begin{split} {\bf h} &amp;= r^2\dot \theta \hat{\bf{k}}\\ &amp;= rv_\theta \hat{\bf{k}} = constant\end{split}\]</div>
<p>where <span class="math notranslate nohighlight">\(\hat{\bf{k}}\)</span> is a unit vector orthogonal to <span class="math notranslate nohighlight">\({\bf e}_r\)</span> and <span class="math notranslate nohighlight">\({\bf e}_\theta\)</span>.
Further, we can substitute for <span class="math notranslate nohighlight">\(h\)</span> from <a class="reference internal" href="#equation-l2-23">(54)</a> into to <a class="reference internal" href="#equation-l2-20">(51)</a>, which gives</p>
<div class="math notranslate nohighlight" id="equation-l2-24">
<span class="eqno">(55)<a class="headerlink" href="#equation-l2-24" title="Permalink to this equation">#</a></span>\[\frac{dA}{dt}=\frac{1}{2}h=\frac{1}{2}r^2\dot \theta\]</div>
<span class="eqno">(55)<a class="headerlink" href="#equation-l2-24" title="Permalink to this equation">#</a></span>\[\begin{split}\frac{\mathrm{d}A}{\mathrm{d}t} &amp;= \frac{1}{2}h\\
\frac{\mathrm{d}A}{\mathrm{d}t} &amp;= \frac{1}{2}r^2\dot \theta\end{split}\]</div>
<p>Taking the second derivatibe of <span class="math notranslate nohighlight">\(\bf r\)</span> using <a class="reference internal" href="#equation-l2-21">(52)</a>, we get</p>
<div class="math notranslate nohighlight" id="equation-l2-25">
<span class="eqno">(56)<a class="headerlink" href="#equation-l2-25" title="Permalink to this equation">#</a></span>\[{\frac{d^2\bf{r}}{dt}} = {\bf{a}} = ({\ddot r} - r\dot\theta^2) {\bf{e_r}} + (r\ddot \theta+2\dot r\dot \theta) {\bf{e_\theta}}\]</div>
<span class="eqno">(56)<a class="headerlink" href="#equation-l2-25" title="Permalink to this equation">#</a></span>\[{\frac{d^2\bf{r}}{\mathrm{d}t^2}} = \ddot{\bf r}
= ({\ddot r} - r\dot\theta^2) {\bf{e_r}} + (r\ddot \theta+2\dot r\dot \theta) {\bf{e_\theta}}.\]</div>
<p>We can also write the RHS of <a class="reference internal" href="#equation-l1-30">(31)</a> in polar coordinates as</p>
<div class="math notranslate nohighlight" id="equation-l2-26">
<span class="eqno">(57)<a class="headerlink" href="#equation-l2-26" title="Permalink to this equation">#</a></span>\[\frac{-\mu}{r^2}\bf{e_r}\]</div>
<p>which upon comparing to <a class="reference internal" href="#equation-l2-25">(56)</a> gives the following two scalar ODEs along the <span class="math notranslate nohighlight">\({\bf e}_r\)</span> and <span class="math notranslate nohighlight">\({\bf e}_\theta\)</span> directions, respectively:</p>
<div class="math notranslate nohighlight" id="equation-l2-27">
<span class="eqno">(58)<a class="headerlink" href="#equation-l2-27" title="Permalink to this equation">#</a></span>\[{\ddot r} - r\dot \theta^2 = \frac{-\mu}{r^2}\]</div>
<span class="eqno">(58)<a class="headerlink" href="#equation-l2-27" title="Permalink to this equation">#</a></span>\[\begin{split}{\ddot r} - r\dot \theta^2 &amp;= \frac{-\mu}{r^2}\\
r\ddot \theta + 2{\dot r}\dot \theta &amp;= 0\end{split}\]</div>
<p>Note that the second of Equations <a class="reference internal" href="#equation-l2-27">(58)</a> can also obtained from differentiating
the specific angular momentum</p>
<div class="math notranslate nohighlight" id="equation-l2-28">
<span class="eqno">(59)<a class="headerlink" href="#equation-l2-28" title="Permalink to this equation">#</a></span>\[r\ddot \theta + 2{\dot r}\dot \theta = 0 ↔ \frac{d}{dt}(r^2\dot \theta)=2r{\dot r}\dot \theta+r^2\ddot \theta=r(r\ddot \theta+2{\dot r}\dot \theta) =0\]</div>
<ul class="simple">
<li><p><a class="reference internal" href="#equation-l2-28">(59)</a> is equivalent to <span class="math notranslate nohighlight">\(\bf{h}\)</span> = constant which is a scalar</p></li>
</ul>
<span class="eqno">(59)<a class="headerlink" href="#equation-l2-28" title="Permalink to this equation">#</a></span>\[\begin{split}\frac{d}{dt}(h) &amp;= \frac{d}{dt}(r^2\dot \theta)\\
\Rightarrow &amp;=2r{\dot r}\dot \theta+r^2\ddot \theta\\
\Rightarrow &amp;=r(r\ddot \theta+2{\dot r}\dot \theta)\\
\Rightarrow &amp;=0\end{split}\]</div>
<p>In other words, the second of Equations <a class="reference internal" href="#equation-l2-27">(58)</a> is equivalent to <span class="math notranslate nohighlight">\(h = constant\)</span>.</p>
</section>
<section id="conservation-of-specific-orbital-energy-e">
<span id="content-conservation-of-specific-orbital-energy-e"></span><h2>Conservation of Specific Orbital Energy: <span class="math notranslate nohighlight">\(E\)</span><a class="headerlink" href="#conservation-of-specific-orbital-energy-e" title="Permalink to this heading">#</a></h2>
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2 changes: 1 addition & 1 deletion orbital-mechanics/Lecture3/Lecture3.html
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Expand Up @@ -637,7 +637,7 @@ <h3>Properties<a class="headerlink" href="#properties" title="Permalink to this
<figure class="align-center" id="fig1">
<a class="reference internal image-reference" href="../../_images/L3_ellipse_drawing.png"><img alt="../../_images/L3_ellipse_drawing.png" src="../../_images/L3_ellipse_drawing.png" style="width: 75%;" /></a>
<figcaption>
<p><span class="caption-number">Fig. 5 </span><span class="caption-text">An ellipse demostrating the different properties of an elliptical orbit, all shown in different colours.</span><a class="headerlink" href="#fig1" title="Permalink to this image">#</a></p>
<p><span class="caption-number">Fig. 6 </span><span class="caption-text">An ellipse demostrating the different properties of an elliptical orbit, all shown in different colours.</span><a class="headerlink" href="#fig1" title="Permalink to this image">#</a></p>
</figcaption>
</figure>
<p><span style="color:#6A73D9">a = semi-major axis</span> <br>
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