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# Kepler's Law of Planetary Motion [Part 3]

This will be the last talk about Kepler's Law of Planetary Motion. Last two posts,[<Math and Physics #10>](https://steemit.com/math/@mathsolver/math-and-physics-10-the-beauty-of-ellipse-kepler-s-law-of-planetary-motion-part-1) and [<Math and Physics #11>](https://steemit.com/math/@mathsolver/math-and-physics-11-kepler-s-law-of-planetary-motion-simple-but-astonishing-part-2) were about Kepler's first and second law. Today, I'm gonna talk about the third law, which states about the relationship between orbital period and semi-major axis of orbit. 

## 0. What is Kepler's Third Law?

It states that 

___The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.___

Formally, a planet revolving around the sun with semi-major axis <img src="http://latex.codecogs.com/gif.latex?a" title="a" /> and period <img src="http://latex.codecogs.com/gif.latex?T" title="T" /> obeys the following law. 

<center> <img src="http://latex.codecogs.com/gif.latex?T^2&space;\propto&space;a^3" title="T^2 \propto a^3" /> </center>

That is, as the orbit gets larger, the length of 1 year in that planet, gets longer. The following diagram shows the orbit of 5 planets - Mercury, Venus, Earth, Mars and Jupiter. As you can see, planets that are close to the sun revolves extremely fast. On the other hand, Jupiter; far away from the sun, revolves in slow pace. 

<center> <img src = "https://i.imgsafe.org/26/260a6dc228.gif" /></center>
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## 1. Why is it Important?

Kepler's third law was first published in 1619, on his paper [<___Harmonices Mundi___>](https://books.google.co.kr/books?id=ZLlCAAAAcAAJ&pg=PA189&redir_esc=y#v=onepage&q&f=false). It was about 10 years after his discovery of first and second laws. During that period, Kepler worked on finding the ultimate __law of harmony__ of the sky using his first and second law of planetary motion. Although Kepler was not able to prove it mathematically, he was confident on his discovery of the relation 

<center> <img src="http://latex.codecogs.com/gif.latex?T^2&space;\propto&space;a^3" title="T^2 \propto a^3" /> </center> 

. Later, Newton - the giant - not only proved it but also found the actual proportionality constant hidden under the relation, which was 

<center> <img src="http://latex.codecogs.com/gif.latex?T^2&space;=&space;\frac{4\pi^2}{GM}a^3" title="T^2 = \frac{4\pi^2}{GM}a^3" /></center>

where <img src="http://latex.codecogs.com/gif.latex?G,&space;M" title="G, M" align = "center"/> are gravitational constants and mass of the sun respectively. Thus, if we neglect the gravitational force __between the planets__ , the proportionality constant is independent of the variables related to specific planets. So, all the planets - Mercury to Neptune, will have __same__ period to axis ratio, <img src="http://latex.codecogs.com/gif.latex?T^2/a^3" title="T^2/a^3" align = "center"/> . 

## 2. The Law of Harmony - Verification

In order to verify Kepler's third law of motion, we need previous facts about first and second law as well as basic geometric facts about ellipse. First let's look at the diagram below.

<center> <img src = "https://i.imgsafe.org/27/2762097698.png" /></center>

The sun is fixed at the origin, and the planets is on elliptical orbit with semi-major axis <img src="http://latex.codecogs.com/gif.latex?a" title="a" /> and eccentricity  <img src="http://latex.codecogs.com/gif.latex?e" title="e" /> . 

---

[___Kepler's first law___](https://steemit.com/math/@mathsolver/math-and-physics-10-the-beauty-of-ellipse-kepler-s-law-of-planetary-motion-part-1) states that the equation of the orbit in polar form is 

<center> <img src="http://latex.codecogs.com/gif.latex?r(\theta)&space;=&space;\frac{a(1-e^2)}{1&space;&plus;&space;e\cos&space;\theta}" title="r = \frac{a(1-e^2)}{1 + e\cos \theta}" /> </center> 

[___Kepler's second law___](https://steemit.com/math/@mathsolver/math-and-physics-11-kepler-s-law-of-planetary-motion-simple-but-astonishing-part-2) states that the rate of change of area swept by the vector <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" /> respect to time is constant, so that 

<center> <img src="http://latex.codecogs.com/gif.latex?\frac{d\mathbf{A}}{dt}&space;=&space;\frac{1}{2}&space;(\mathbf{r}&space;\times&space;\mathbf{v})&space;\equiv&space;\text{constant}" title="\frac{d\mathbf{A}}{dt} = \frac{1}{2} (\mathbf{r} \times \mathbf{v}) \equiv \text{constant}" /> </center>

and we also obtained formula for speed respect to the distance <img src="http://latex.codecogs.com/gif.latex?r&space;=&space;\lVert&space;\mathbf{r}&space;\rVert" title="r = \lVert \mathbf{r} \rVert" align = "center"/> by 

<center> <img src="http://latex.codecogs.com/gif.latex?v=\sqrt{GM\left(\frac{2}{r}&space;-&space;\frac{1}{a}&space;\right&space;)}" title="v=\sqrt{GM\left(\frac{2}{r} - \frac{1}{a} \right )}" /> </center>

[___Total Area of an Ellipse___](https://en.wikipedia.org/wiki/Ellipse#Area) is given by 

<center><img src="http://latex.codecogs.com/gif.latex?\pi&space;a^2&space;\sqrt{1-e^2}" title="\pi a^2 \sqrt{1-e^2}" /></center>

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Now it's time for derivation. Distance between the sun and the planet when the planet is located at <img src="http://latex.codecogs.com/gif.latex?y" title="y" align = "center"/> axis is given by 

<center> <img src="http://latex.codecogs.com/gif.latex?r_0&space;=&space;r(\pi/2)=&space;a(1-e^2)" title="r_0 = r(\pi/2)= a(1-e^2)" /> </center> 

Also, <img src="http://latex.codecogs.com/gif.latex?r_a&space;&plus;&space;r_p&space;=&space;2a" title="r_a + r_p = 2a" align = "center"/> so that 

<center> <img src="http://latex.codecogs.com/gif.latex?\begin{align*}&space;\frac{dA}{dt}&space;&=&space;\frac{1}{2}&space;\lVert&space;\mathbf{r}&space;\times&space;\mathbf{v}&space;\lVert&space;\\&space;&=&space;\frac{r_a&space;v_a}{2}&space;=&space;\frac{r_a}{2}&space;\sqrt{GM&space;\left(&space;\frac{2}{r_a}&space;-&space;\frac{1}{a}&space;\right&space;)}&space;\\&space;&=\frac{a(1&plus;e)}{2}\sqrt{GM&space;\left(\frac{2a-r_a}{ar_a}&space;\right&space;)}&space;\\&space;&=\frac{a(1&plus;e)}{2}&space;\sqrt{GM&space;\left(&space;\frac{1-e}{a(1&plus;e)}&space;\right&space;)}&space;\\&space;&=&space;\sqrt{\frac{GM}{4}a(1-e^2)}=&space;\sqrt{\frac{GMr_0}{a}}&space;\end{align*}" title="\begin{align*} \frac{dA}{dt} &= \frac{1}{2} \lVert \mathbf{r} \times \mathbf{v} \lVert \\ &= \frac{r_a v_a}{2} = \frac{r_a}{2} \sqrt{GM \left( \frac{2}{r_a} - \frac{1}{a} \right )} \\ &=\frac{a(1+e)}{2}\sqrt{GM \left(\frac{2a-r_a}{ar_a} \right )} \\ &=\frac{a(1+e)}{2} \sqrt{GM \left( \frac{1-e}{a(1+e)} \right )} \\ &= \sqrt{\frac{GM}{4}a(1-e^2)}= \sqrt{\frac{GMr_0}{a}} \end{align*}" /> </center>

Now the area swept by the vector <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" /> during one period <img src="http://latex.codecogs.com/gif.latex?T" title="T" /> is equal to the total area of ellipse <img src="http://latex.codecogs.com/gif.latex?\pi&space;a^2&space;\sqrt{1-e^2}" title="\pi a^2 \sqrt{1-e^2}" align = "center"/> . This gives 

<center> <img src="http://latex.codecogs.com/gif.latex?\sqrt{\frac{GMa(1-e^2)}{4}}&space;=&space;\frac{\pi&space;a^2&space;\sqrt{1-e^2}}{T}&space;\\\\&space;\iff&space;\sqrt{\frac{GMa}{4}}&space;=&space;\frac{\pi&space;a^2}{T}&space;\\\\&space;\iff&space;\frac{GMa}{4}&space;=&space;\frac{\pi^2&space;a^4}{T^2}&space;\\\\&space;\iff&space;T^2&space;=&space;\frac{4\pi^2}{GM}&space;a^3" title="\sqrt{\frac{GMa(1-e^2)}{4}} = \frac{\pi a^2 \sqrt{1-e^2}}{T} \\\\ \iff \sqrt{\frac{GMa}{4}} = \frac{\pi a^2}{T} \\\\ \iff \frac{GMa}{4} = \frac{\pi^2 a^4}{T^2} \\\\ \iff T^2 = \frac{4\pi^2}{GM} a^3" /> </center> 

Using Newtonian mechanics, we get the proportionality constant, soley dependent on gravitational constant and mass of the sun, which is 

<center> <img src="http://latex.codecogs.com/gif.latex?\frac{4\pi^2}{GM}" title="\frac{4\pi^2}{GM}" /> </center>

## 3. Application to our Solar System 

Orbital Characteristics of planets in our solar system are as follows. - [3]

|   Name  | Semi-major axis (AU) | Orbital Period (years) | Orbital eccentricity |
|:-------:|:---------------------:|:----------------------:|:--------------------:|
| Mercury |          0.39         |          0.24          |         0.206        |
| Venus   |          0.72         |          0.62          |         0.007        |
| Earth   |           1           |            1           |         0.017        |
| Mars    |          1.52         |          1.88          |         0.093        |
| Jupiter |          5.2          |          11.86         |         0.048        |
| Saturn  |          9.54         |          29.46         |         0.054        |
| Uranus  |         19.22         |          84.01         |         0.047        |
| Neptune |         30.06         |          164.8         |         0.009        |

Theoretically, the proportional constant should be 

<center> <img src="http://latex.codecogs.com/gif.latex?\begin{align*}&space;\frac{4\pi^2}{GM}&space;&=&space;9.4664&space;\times&space;10^{-20}&space;\text{&space;s}^2/\text{m}^3&space;\\&space;&=&space;1.001&space;\text{&space;year}^2/\text{AU}^3&space;\end{align*}" title="\begin{align*} \frac{4\pi^2}{GM} &= 9.4664 \times 10^{-20} \text{ s}^2/\text{m}^3 \\ &= 0.31867 \text{ year}^2/\text{AU}^3 \end{align*}" /> </center>

The real values are 
|   Name  | Proportionality constant (year^2 /AU^3) | Error (= differnece with 1.001) |
|:-------:|:--------------------------------------:|:-------------------------------:|
| Mercury |                0.971021                |             -0.02998            |
|  Venus  |                1.029878                |             0.028878            |
|  Earth  |                    1                   |              -0.001             |
|   Mars  |                1.006433                |             0.005433            |
| Jupiter |                1.000367                |             -0.00063            |
|  Saturn |                0.999586                |             -0.00141            |
|  Uranus |                0.994035                |             -0.00696            |
| Neptune |                0.999879                |             -0.00112            |

Are these errors small enough to neglect? Well, at first you can think so because the error terms look small. It can be a good approximation if you are dealing with small periods of time. But if you are considering large periods of time (such as month, year,...) actual positions of planets differ greatly with the hypothesis <img src="http://latex.codecogs.com/gif.latex?T^2&space;=&space;\frac{4\pi^2}{GM}&space;a^3" title="T^2 = \frac{4\pi^2}{GM} a^3" align  = "center"/>. You should take __interplanetary gravitational forces__ into account, which makes the problem extremely hard (actually there is no closed form expression for velocity and orbit.) See [three body problem](https://en.wikipedia.org/wiki/Three-body_problem) - [4] for more information. 

## 4. Conclusion

Kepler's third law is astonishing in the sense that it describes the relationship between the orbital period and semi-major axis without any specific considerations to the planet itself. However, it only applies to planetary system with single planet. If someone is dealing with planetary system having multiple planets, they must consider perturbations due to interplanetary gravitational forces. 


## 5. Citations

[1] https://gfycat.com/gifs/detail/SpitefulWiltedEasternglasslizard

[2] https://superstarfloraluk.com/2756447-Planets-Solar-System-Orbits-Animation.html

[3] https://en.wikipedia.org/wiki/Planet#Planetary_attributes

[4] https://en.wikipedia.org/wiki/Three-body_problem

All other derivations and geogmetric figures are made by myself using LATEX and GeoGebra.


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