<p>With copious research and thinking, I managed to answer my own question, but I am going to share the answer seeing that no one has answered it satsifactorily.</p>
<p>Q1: the position vector <span class="math-container">$\vec{r}$</span> of the reduced mass is also the relative distance between the 2 masses, therefore if the reduced mass is undergoing elliptical motion about its centre of mass, it is equivalent to taking the reference frame one of the objects in the binary system and measuring the position vector of the other object since <span class="math-container">$\vec{r}$</span> is relative distance. Therefore what Kepler's 1st Law is really saying, is that in a binary system, the objects will undergo elliptical motion with respect to the other object.</p>
<p>Q2: As the eccentricity tends towards 1, notice that from the formula <span class="math-container">$e=\sqrt{1-\frac{b^2}{a^2}}$</span> that if e were to tend towards 1, <span class="math-container">$\frac{b^2}{a^2}$</span> must tend towards 0, and this would cause the ellipse to approach the shape of a line (but it is important to note that <span class="math-container">$e=1$</span> would result in a parabola, not a straight line). In a line, since the velocity and position vectors are parallel, L would thus be 0 (since <span class="math-container">$L=m\vec{r}\times \vec{v}$</span> and if they are parallel the cross product would be 0)</p>

CC BY-SA 4.0
2020-12-02T02:59:49.770Z
2020-12-02T02:59:49.770Z
Lucas Tan
<p>I must say that I agree with you, this way of derivate the Kepler law is not the most intuitive, it's maybe for this reason they specify: &quot;revisited&quot;.</p>
<p><strong>R1:</strong> The reason is stated at the very begin of the chapter, where you derive the relation between <span class="math-container">$r$</span> and the angle to the perihelion <span class="math-container">$\theta$</span> for the general case of an ellipse (Eq 3 in your book in Cap 2.1 Elliptical orbit, but I can see I have an older version than yours), where:</p>
<p><span class="math-container">$$ r = \frac{a(1-e^2)}{1+e\,cos(\theta)}$$</span></p>
<p>here, the eccentricity of a ellipse is then define as:
<span class="math-container">$$ e = \sqrt{1- \frac{b^2}{a^2}}$$</span></p>
<p>in the case of Kepler's law you can define the eccentricity as <span class="math-container">$e = D/\mu G M$</span> and  <span class="math-container">$a$</span> as:
<span class="math-container">$$ a = \frac{L^2 G\, M}{\mu^2\,G^2\,M^2 - D^2}$$</span></p>
<p>and you can work out in equation (29), obtaining the equation in the same form as for an elliptical orbit.</p>
<p>The reason to derive <span class="math-container">$r$</span> is because it indicates the distance of the object from the focal point (centre of mass of the system), which in the case of the solar system, is the distance of the planet to the Sun (actually the focal point of the solar system, or better to say of the Sun-Jupiter system, is a few kilometre above the solar surface).</p>
<p><strong>R2:</strong> the eccentricity <span class="math-container">$e$</span> quantify the shape of a cone cross-section <a href="https://en.wikipedia.org/wiki/Eccentricity_(mathematics)" rel="nofollow noreferrer">Eccentricity (math) Wikipedia</a>.</p>
<p>For example, in the case of the an ellipse you have <span class="math-container">$0 &lt; e &lt; 1$</span>, and it indicates how much the ellipse is &quot;squashed&quot;. Another example, if you have <span class="math-container">$e=0$</span> than you have a circle.</p>
<p>In the case of <span class="math-container">$e=1$</span> you have a parabola, from the equation of <span class="math-container">$e$</span> above it means that <span class="math-container">$a\gg b$</span>. In a more physics sense, it means that the system <strong>is not in a gravitationally bounded state</strong>, and the less massive object <span class="math-container">$m_1 \ll m_2$</span> is on a parabolic trajectory, deviated by the gravitational pull of <span class="math-container">$m_2$</span>, and will not stay in orbit.</p>

CC BY-SA 4.0
2020-12-02T03:01:45.750Z
Thank you for the insights about eccentricity, however I still don't see how this directly answers the questions. But if you refer to the answer below yours, I managed to come up with solutions to this question using your help, thanks!
Lucas Tan
CC BY-SA 4.0
2020-11-28T15:03:50.783Z
2020-11-28T15:18:29.560Z
2020-11-28T15:18:29.560Z
Michele Bianco
<p>I was going through the derivation of Kepler's 1st Law in the textbook &quot;An Introduction to Modern Astrophysics&quot; by Carroll and Ostlie. From there, I got stuck in a few places in their derivation as I found it confusing. In total, I have 2 questions.</p>
<p><a href="https://i.sstatic.net/wYTGw.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wYTGw.jpg" alt="enter image description here" /></a></p>
<p>Q1: In the second sentence that I highlighted, it is mentioned that when equation 29 is translated into a physical standpoint, it implies that <em>&quot;both objects in a binary orbit move about the centre of mass in ellipses, with the centre of mass occupying one focus of each ellipse&quot;</em> however I don't see how the physical implications of equation 29 will translate into this that they claim. This equation clearly only describes the position vector of the reduced mass, so how can it be extrapolated to talk about the motion of a binary system when there are 2 masses rather than 1 reduced mass. (Clarification: I am aware that reduced mass is a way to model binary systems, but in this case I don't see the link between the motion of the reduced mass and the motion of 2 bodies)</p>
<p>Q2: On the very last line, it is mentioned that <em>&quot;L is at minimum as eccentricity approaches unity, as expected&quot;</em>. However I don't see the qualitative reasoning as to why this would be &quot;expected&quot;. Is there a physical or physics explanation as to why we expect an orbit with a large eccentricity to have a lower angular momentum than one with a low eccentricity?</p>

<orbital-mechanics><celestial-mechanics><johannes-kepler>
How to derive Kepler's 1st Law?
Lucas Tan
# How to derive Kepler's 1st Law?

(Asked by: username_0 on 2020-11-25)

I was going through the derivation of Kepler's 1st Law in the textbook "An Introduction to Modern Astrophysics" by Carroll and Ostlie. From there, I got stuck in a few places in their derivation as I found it confusing. In total, I have 2 questions.

[![enter image description here](https://i.sstatic.net/wYTGw.jpg)](https://i.sstatic.net/wYTGw.jpg)

Q1: In the second sentence that I highlighted, it is mentioned that when equation 29 is translated into a physical standpoint, it implies that *"both objects in a binary orbit move about the centre of mass in ellipses, with the centre of mass occupying one focus of each ellipse"* however I don't see how the physical implications of equation 29 will translate into this that they claim. This equation clearly only describes the position vector of the reduced mass, so how can it be extrapolated to talk about the motion of a binary system when there are 2 masses rather than 1 reduced mass. (Clarification: I am aware that reduced mass is a way to model binary systems, but in this case I don't see the link between the motion of the reduced mass and the motion of 2 bodies)

Q2: On the very last line, it is mentioned that *"L is at minimum as eccentricity approaches unity, as expected"*. However I don't see the qualitative reasoning as to why this would be "expected". Is there a physical or physics explanation as to why we expect an orbit with a large eccentricity to have a lower angular momentum than one with a low eccentricity?

**Tags:** orbital-mechanics, celestial-mechanics, johannes-kepler

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## 2 answers

### Answer by username_0 on 2020-12-02. Score: 2 | Accepted answer

With copious research and thinking, I managed to answer my own question, but I am going to share the answer seeing that no one has answered it satsifactorily.

Q1: the position vector $\vec{r}$ of the reduced mass is also the relative distance between the 2 masses, therefore if the reduced mass is undergoing elliptical motion about its centre of mass, it is equivalent to taking the reference frame one of the objects in the binary system and measuring the position vector of the other object since $\vec{r}$ is relative distance. Therefore what Kepler's 1st Law is really saying, is that in a binary system, the objects will undergo elliptical motion with respect to the other object.

Q2: As the eccentricity tends towards 1, notice that from the formula $e=\sqrt{1-\frac{b^2}{a^2}}$ that if e were to tend towards 1, $\frac{b^2}{a^2}$ must tend towards 0, and this would cause the ellipse to approach the shape of a line (but it is important to note that $e=1$ would result in a parabola, not a straight line). In a line, since the velocity and position vectors are parallel, L would thus be 0 (since $L=m\vec{r}\times \vec{v}$ and if they are parallel the cross product would be 0)


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### Answer by username_1 on 2020-11-28. Score: 2

I must say that I agree with you, this way of derivate the Kepler law is not the most intuitive, it's maybe for this reason they specify: "revisited".

**R1:** The reason is stated at the very begin of the chapter, where you derive the relation between $r$ and the angle to the perihelion $\theta$ for the general case of an ellipse (Eq 3 in your book in Cap 2.1 Elliptical orbit, but I can see I have an older version than yours), where:

$$ r = \frac{a(1-e^2)}{1+e\,cos(\theta)}$$

here, the eccentricity of a ellipse is then define as:
$$ e = \sqrt{1- \frac{b^2}{a^2}}$$

in the case of Kepler's law you can define the eccentricity as $e = D/\mu G M$ and $a$ as:
$$ a = \frac{L^2 G\, M}{\mu^2\,G^2\,M^2 - D^2}$$

and you can work out in equation (29), obtaining the equation in the same form as for an elliptical orbit.

The reason to derive $r$ is because it indicates the distance of the object from the focal point (centre of mass of the system), which in the case of the solar system, is the distance of the planet to the Sun (actually the focal point of the solar system, or better to say of the Sun-Jupiter system, is a few kilometre above the solar surface).

**R2:** the eccentricity $e$ quantify the shape of a cone cross-section [Eccentricity (math) Wikipedia](https://en.wikipedia.org/wiki/Eccentricity_(mathematics)).

For example, in the case of the an ellipse you have $0 < e < 1$, and it indicates how much the ellipse is "squashed". Another example, if you have $e=0$ than you have a circle.

In the case of $e=1$ you have a parabola, from the equation of $e$ above it means that $a\gg b$. In a more physics sense, it means that the system **is not in a gravitationally bounded state**, and the less massive object $m\_1 \ll m\_2$ is on a parabolic trajectory, deviated by the gravitational pull of $m\_2$, and will not stay in orbit.

> **Comments:**
> **username_0**: Thank you for the insights about eccentricity, however I still don't see how this directly answers the questions. But if you refer to the answer below yours, I managed to come up with solutions to this question using your help, thanks!  

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