<center> <img src = "https://i.imgsafe.org/13/1309f7d6c0.png" /></center>
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# Kepler's Law of Planetary Motion - [Part 2]

## Short but Astonishing

Continuing from the last talk ([<Math & Physics #10>](https://steemit.com/math/@mathsolver/math-and-physics-10-the-beauty-of-ellipse-kepler-s-law-of-planetary-motion-part-1)), in this post I will talk about Kepler's Second Law of Planetary motion - Equal Time Equal Area Law. 

## 0. What is the second law of planetary motion?

It states that
 
___A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.___

For example, look at the following figure.

<center> <img src = "https://i.imgsafe.org/0e/0ee4d1fb49.gif" /></center>
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The planet is rotating around the sun (located at the origin) in elliptical orbit. To satisfy Kepler's second law, at the apoapsis (the point where the distance between the sun and the planet is maximum), the orbital velocity (green vector) should be minimized. On the contrary, at the periapsis (the point where the distance is minimum), the orbital velocity should be maximized. Note that each portion divided by white lines is equal. 

## 1. History Behind the Second law

[Johannes Kepler](https://en.wikipedia.org/wiki/Johannes_Kepler)'s second law of planetary motion first came out in 1609, in his paper [<___Astonomia Nova___> ](https://archive.org/stream/Astronomianovaa00Kepl#page/n3/mode/2up)[3] . At that time Kepler did not present his second law in its modern form. His law only applied to Mars. Kepler restated his second law in 1621, for any planet. In his paper [<___Epitome Astronomiae Copernicanae___>](https://books.google.co.kr/books?id=wa2SE_6ZL7YC&pg=PA668&redir_esc=y#v=onepage&q&f=false) [4], Kepler wrote
> It has been said above that, if the orbit of the planet is divided into the smallest equal parts, the times of the planet in them increase in the ratio of the distances between them and the sun.) That is, a planet's speed along its orbit is inversely proportional to its distance from the Sun.

It is clear that Kepler was referring to what is now called angular velocity and the second law can be easily deduced from it. 

## 2. The actual Verification

Honestly, the actual proof of Kepler's second law is extremely simple. Knowing little bit of vector calculus as well as Newtonian mechanics are enough. For proof I will use the following diagram. 

<center> <img src = "https://i.imgsafe.org/0f/0f8e762d27.png" /> </center>

Define a vector from the sun to planet <img src="http://latex.codecogs.com/gif.latex?P" title="P" /> as <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" />. In infinitesimal time <img src="http://latex.codecogs.com/gif.latex?dt" title="dt" /> , the infinitesimal change of vector <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" /> is <img src="http://latex.codecogs.com/gif.latex?d\mathbf{r}" title="d\mathbf{r}" /> . Now, the infinitesimal area swept by the vector <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" /> during time <img src="http://latex.codecogs.com/gif.latex?dt" title="dt" /> is equal to 

<center> <img src="http://latex.codecogs.com/gif.latex?\lVert&space;d\mathbf{A}&space;\rVert&space;=&space;\frac{1}{2}&space;\lVert&space;\mathbf{r}&space;\times&space;d\mathbf{r}&space;\lVert" title="\lVert d\mathbf{A} \rVert = \frac{1}{2} \lVert \mathbf{r} \times d\mathbf{r} \lVert" /> </center> 

using cross product. (The magnitude of cross product is equal to area of parallelogram so we should take half of it). Now we will investigate the quantity 

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

The rate of sweeping out area due to movement is 

<center> <img src="http://latex.codecogs.com/gif.latex?\begin{align*}&space;\dot{\mathbf{A}}&space;&=&space;\frac{d\mathbf{A}}{dt}&space;\\&space;&=&space;\frac{1}{2}&space;\left(&space;\frac{d\mathbf{r}}{dt}&space;\times&space;d\mathbf{r}&space;&plus;&space;\mathbf{r}&space;\times&space;\frac{d\mathbf{r}}{dt}\right)\&space;(\because&space;d\mathbf{r}&space;\times&space;d\mathbf{r}&space;=&space;0)\\&space;&=&space;\frac{1}{2}&space;\left(&space;\mathbf{r}&space;\times&space;\dot{\mathbf{r}}&space;\right&space;)&space;\end{align*}" title="\begin{align*} \dot{\mathbf{A}} &= \frac{d\mathbf{A}}{dt} \\ &= \frac{1}{2} \left( \frac{d\mathbf{r}}{dt} \times d\mathbf{r} + \mathbf{r} \times \frac{d\mathbf{r}}{dt}\right)\ (\because d\mathbf{r} \times d\mathbf{r} = 0)\\ &= \frac{1}{2} \left( \mathbf{r} \times \dot{\mathbf{r}} \right ) \end{align*}" /> </center>

Again differentiating both sides (respect to time) gives 

<center> <img src="http://latex.codecogs.com/gif.latex?\begin{align*}&space;\ddot{\mathbf{A}}&space;&=&space;\frac{d\dot{\mathbf{A}}}{dt}&space;\\&space;&=&space;\frac{1}{2}&space;\left(&space;\dot{\mathbf{r}}&space;\times&space;\dot{\mathbf{r}}&plus;&space;\mathbf{r}&space;\times&space;\ddot{\mathbf{r}}&space;\right)\\&space;&=&space;\frac{1}{2}&space;(\mathbf{r}&space;\times&space;\ddot{\mathbf{r}})\&space;(\because&space;\dot{\mathbf{r}}&space;\times&space;\dot{\mathbf{r}}&space;=&space;0)&space;\end{align*}" title="\begin{align*} \ddot{\mathbf{A}} &= \frac{d\dot{\mathbf{A}}}{dt} \\ &= \frac{1}{2} \left( \dot{\mathbf{r}} \times \dot{\mathbf{r}}+ \mathbf{r} \times \ddot{\mathbf{r}} \right)\\ &= \frac{1}{2} (\mathbf{r} \times \ddot{\mathbf{r}})\ (\because \dot{\mathbf{r}} \times \dot{\mathbf{r}} = 0) \end{align*}" /> </center>

The vector <img src="http://latex.codecogs.com/gif.latex?\ddot{\mathbf{r}}" title="\ddot{\mathbf{r}}" /> is nothing but the accerleration vector of <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" />. 

By [Newton's second law of motion](https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton's_2nd_Law), the gravitational force <img src="http://latex.codecogs.com/gif.latex?\mathbf{F}" title="\mathbf{F}" /> has same direction with the accerleration <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" /> . We all know that the gravitational force between two bodies has direction towards each other, so that 

<center> <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}&space;\parallel&space;\ddot{\mathbf{r}}&space;\implies&space;\mathbf{r}&space;\times&space;\ddot{\mathbf{r}}&space;=&space;0" title="\mathbf{r} \parallel \ddot{\mathbf{r}} \implies \mathbf{r} \times \ddot{\mathbf{r}} = 0" /> </center>

We can therefore say that 

<center> <img src="http://latex.codecogs.com/gif.latex?\ddot{\mathbf{A}}&space;=&space;0&space;\implies&space;\dot{\mathbf{A}}&space;\equiv&space;\mathbf{c}\&space;(\text{constant&space;vector})" title="\ddot{\mathbf{A}} = 0 \implies \dot{\mathbf{A}} \equiv \mathbf{c}\ (\text{constant vector})" /> </center>

Now

<center> <img src="http://latex.codecogs.com/gif.latex?\lVert&space;d\mathbf{A}&space;\rVert&space;=&space;\lVert&space;\mathbf{c}&space;dt&space;\rVert&space;=&space;\lVert&space;\mathbf{c}\rVert&space;dt=&space;cdt" title="\lVert d\mathbf{A} \rVert = \lVert \mathbf{c} dt \rVert = \lVert \mathbf{c}\rVert dt= cdt" /> </center>

which means that the area swept by vector <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" /> is directly proportional to the time duration <img src="http://latex.codecogs.com/gif.latex?dt" title="dt" />. So if two intervals of time have equal length, then the area swept should also be equal, proving the Kepler's second law of planetary motion. 

## 3. Velocity of the Planet 

Remember the diagram we've seen in Section 0? The velocity of the planet changes at every point of the orbit. Especially, at apoapsis it's minimized, and at periapsis it's maximized. Take a loot at the following diagram. 

<center> <img src = "https://i.imgsafe.org/10/104377976e.png" /></center>

1. Gravitation constant <img src="http://latex.codecogs.com/gif.latex?G" title="G" />, the mass of sun <img src="http://latex.codecogs.com/gif.latex?M" title="M" /> and the mass of the planet <img src="http://latex.codecogs.com/gif.latex?m" title="m" /> . 

1. Sun is located at the origin. (fixed)

2. A vector from sun to the planet <img src="http://latex.codecogs.com/gif.latex?P" title="P" /> is denoted by <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" />, and the instantaneous velocity vector is denoted by <img src="http://latex.codecogs.com/gif.latex?\mathbf{v}" title="\mathbf{v}" /> . In particular at apoapsis and periapsis, we use subscripts <img src="http://latex.codecogs.com/gif.latex?a" title="a" /> and <img src="http://latex.codecogs.com/gif.latex?p" title="p" /> . 

3. The length of semi-major axis of the elliptical orbit is given by <img src="http://latex.codecogs.com/gif.latex?a" title="a" /> . 

The Goal is to find the speed <img src="http://latex.codecogs.com/gif.latex?v&space;=&space;\lVert&space;\mathbf{v}&space;\rVert" title="v = \lVert \mathbf{v} \rVert" align = "center"/>, resepct to <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"/> . 

---
<h4> Ingredients </h4>

___Conservation of Mechanical Energy___
By the conservation of mechanical energy, we get at any point on the orbit, 

<center> <img src="http://latex.codecogs.com/gif.latex?\frac{1}{2}mv^2&space;-&space;\frac{GmM}{r}&space;\equiv&space;E&space;\text{&space;(constant)}" title="\frac{1}{2}mv^2 - \frac{GmM}{r} \equiv E \text{ (constant)}" /> </center> 

___Kepler's Second law of motion___
From section 2, we obtained the fact <img src="http://latex.codecogs.com/gif.latex?\dot{\mathbf{A}}&space;=&space;\frac{1}{2}&space;(\mathbf{r}&space;\times&space;\dot{\mathbf{r}})&space;=&space;\frac{1}{2}&space;(\mathbf{r}&space;\times&space;\mathbf{v})&space;\equiv&space;\text{constant}" title="\dot{\mathbf{A}} = \frac{1}{2} (\mathbf{r} \times \dot{\mathbf{r}}) = \frac{1}{2} (\mathbf{r} \times \mathbf{v}) \equiv \text{constant}" align = "center"/>. Vector <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" /> and <img src="http://latex.codecogs.com/gif.latex?\mathbf{v}" title="\mathbf{v}" /> are __perpendicular__ to each other  at two points (apoapsis & periapsis), so that 

<center> <img src="http://latex.codecogs.com/gif.latex?\lVert&space;\mathbf{r}_a&space;\times&space;\mathbf{v}_a&space;\rVert&space;=&space;\lVert&space;\mathbf{r}_p&space;\times&space;\mathbf{v}_p&space;\rVert&space;\implies&space;r_av_a&space;=&space;r_pv_p" title="\lVert \mathbf{r}_a \times \mathbf{v}_a \rVert = \lVert \mathbf{r}_p \times \mathbf{v}_p \rVert \implies r_av_a = r_pv_p" /> </center>

___Property of Ellipse___
<center> <img src="http://latex.codecogs.com/gif.latex?r_a&space;&plus;&space;r_p&space;=&space;2a" title="r_a + r_p = 2a" /> </center>

---
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Now let's begin. Put <img src="http://latex.codecogs.com/gif.latex?r_a,&space;v_a" title="r_a, v_a" align = "center"/> and <img src="http://latex.codecogs.com/gif.latex?r_p,&space;v_p" title="r_p, v_p" align = "center"/> into energy equation. 

<center> <img src="http://latex.codecogs.com/gif.latex?E&space;=&space;\frac{1}{2}m&space;v_a^2&space;-&space;\frac{GmM}{r_a}&space;=&space;\frac{1}{2}mv_p^2&space;-&space;\frac{GmM}{r_p^2}&space;\\&space;\implies&space;\frac{v_a^2}{2}&space;-&space;\frac{v_b^2}{2}&space;=&space;\frac{GM}{r_a}&space;-&space;\frac{GM}{r_p}" title="E = \frac{1}{2}m v_a^2 - \frac{GmM}{r_a} = \frac{1}{2}mv_p^2 - \frac{GmM}{r_p^2} \\ \implies \frac{v_a^2}{2} - \frac{v_b^2}{2} = \frac{GM}{r_a} - \frac{GM}{r_p}" /> </center>

<img src="http://latex.codecogs.com/gif.latex?v_p&space;=&space;\frac{r_a}{r_p}&space;v_a" title="v_p = \frac{r_a}{r_p} v_a" align = "center"/> via Kepler's law, so that 

<center> <img src="http://latex.codecogs.com/gif.latex?\frac{1}{2}&space;\left(&space;1&space;-&space;\frac{r_a^2}{r_p^2}&space;\right&space;)&space;v_a^2&space;=&space;GM&space;\left(&space;\frac{1}{r_a}&space;-&space;\frac{1}{r_p}&space;\right&space;)&space;\\\\\\&space;\iff&space;\frac{1}{2}&space;\left(&space;\frac{r_p^2&space;-&space;r_a^2}{r_p^2}&space;\right&space;)&space;v_a^2&space;=&space;GM&space;\left(&space;\frac{r_p&space;-&space;r_a}{r_ar_p}&space;\right&space;)\\\\\\&space;\iff&space;\frac{1}{2}&space;\left(&space;\frac{r_p&space;&plus;&space;r_a}{&space;r_p^2}&space;\right&space;)&space;v_a^2&space;=&space;\frac{GM}{r_ar_p}\&space;(\because&space;r_p^2&space;-&space;r_a^2&space;=&space;(r_p&space;&plus;&space;r_a)(r_p-r_a))\\\\\\&space;\iff&space;av_a^2&space;=&space;GM&space;\frac{r_p}{r_a}&space;=&space;\frac{GM(2a&space;-&space;r_a)}{r_a}\&space;(\because&space;r_a&space;&plus;&space;r_p&space;=&space;2a)&space;\\\\\\&space;\iff&space;\frac{1}{2}v_a^2&space;-&space;\frac{GM}{r_a}&space;=&space;-\frac{GM}{2a}\&space;(\text{divide&space;both&space;sides&space;by&space;}&space;2a)" title="\frac{1}{2} \left( 1 - \frac{r_a^2}{r_p^2} \right ) v_a^2 = GM \left( \frac{1}{r_a} - \frac{1}{r_p} \right ) \\\\\\ \iff \frac{1}{2} \left( \frac{r_p^2 - r_a^2}{r_p^2} \right ) v_a^2 = GM \left( \frac{r_p - r_a}{r_ar_p} \right )\\\\\\ \iff \frac{1}{2} \left( \frac{r_p + r_a}{ r_p^2} \right ) v_a^2 = \frac{GM}{r_ar_p}\ (\because r_p^2 - r_a^2 = (r_p + r_a)(r_p-r_a))\\\\\\ \iff av_a^2 = GM \frac{r_p}{r_a} = \frac{GM(2a - r_a)}{r_a}\ (\because r_a + r_p = 2a) \\\\\\ \iff \frac{1}{2}v_a^2 - \frac{GM}{r_a} = -\frac{GM}{2a}\ (\text{divide both side by } 2a)" /> </center>

So for any <img src="http://latex.codecogs.com/gif.latex?r" title="r" /> , we have 

<center> <img src="http://latex.codecogs.com/gif.latex?E&space;=&space;\frac{1}{2}mv^2&space;-&space;\frac{GmM}{r}&space;=&space;\frac{1}{2}mv_a^2&space;-&space;\frac{GmM}{r_a}&space;=&space;-&space;\frac{GmM}{2a}" title="E = \frac{1}{2}mv^2 - \frac{GmM}{r} = \frac{1}{2}mv_a^2 - \frac{GmM}{r_a} = - \frac{GmM}{2a}" /> </center>

thus 

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

a simple result!

## 4. Analysis of the Results obtained in 2 & 3

Going back to  verification steps in sections 2, you'll notice that the key concept we used to prove Kepler's Law was the fact that acceleration vector <img src="http://latex.codecogs.com/gif.latex?\ddot{\mathbf{r}}" title="\ddot{\mathbf{r}}" /> is parallel to <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" />. As long as the force between the planet and the sun is parallel to <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" />, the proof is valid.  Only the direction of the force matters, NOT the magnitude. So Kepler's second law not only applies to planets in gravitational field, but also to objects rotating around fixed body in any field satisfying

<center> <img src="http://latex.codecogs.com/gif.latex?\mathbf{F}(\mathbf{r})&space;\parallel&space;\mathbf{r}" title="\mathbf{F}(\mathbf{r}) \parallel \mathbf{r}" /> </center>

;)

Suppose we want to launch a spacecraft exploring deep space; far beyond the solar system. The spacecraft should overcome sun's gravitaional field with great escape velocity. In order to be successful, the semi-major axis of orbit should diverge to <img src="http://latex.codecogs.com/gif.latex?\infty" title="\infty" /> , in other words, the escape velocity is equal to 

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


## 5. Real Applications

Using information from last talk [<Math & Physics #10>](https://steemit.com/math/@mathsolver/math-and-physics-10-the-beauty-of-ellipse-kepler-s-law-of-planetary-motion-part-1), <img src="http://latex.codecogs.com/gif.latex?a&space;=&space;149.6&space;\times&space;10^6&space;\text{&space;km}" title="a = 149.6 \times 10^6 \text{ km}" />. The speed of revolution at apoapsis (minimum) is about 

<center> <img src="http://latex.codecogs.com/gif.latex?v_a&space;=&space;29,295&space;\text{&space;m/s},\&space;v_p&space;=&space;30,290&space;\text{&space;m/s}" title="v_a = 29295 \text{ m/s},\ v_p = 30290 \text{ m/s}" /> </center>

hmm, not much difference considering the magnitude of speed isn't it? Also at the surface of the earth, the escape velocity __respect to the sun's gravitational field__ is about 

<center> <img src="http://latex.codecogs.com/gif.latex?v_{\infty,&space;sun}&space;=&space;41.78&space;\text{&space;km/s}" title="v_{\infty, sun} = 41.78 \text{ km/s}" /> </center>

slightly differs from real value <img src="http://latex.codecogs.com/gif.latex?42.1&space;\text{&space;km/s}" title="42.1 \text{ km/s}" align = "center"/> . The escape velocity __resepct to the earth's gravitational field__ is theoretically, 

<center> <img src="http://latex.codecogs.com/gif.latex?v_{\infty,&space;earth}&space;=&space;11.186&space;\text{&space;km/s}" title="v_{\infty, earth} = 11.186 \text{ km/s}" /> </center>

a perfect match with real value. 

As you can see, due to mass difference, escaping from the sun is way more difficult than the earth. 

## 6. Conclusion

1. Keppler's second law states that the line joining a planet and the Sun sweeps out equal areas during equal intervals of time.

2. At any point, the cross product between position vector <img src="http://latex.codecogs.com/gif.latex?\mathbf{r}" title="\mathbf{r}" /> and velocity vector <img src="http://latex.codecogs.com/gif.latex?\mathbf{v}" title="\mathbf{v}" /> is constant. 

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

I will finish this post with simulation of earth's revolution. 

<center> <img src = "https://i.imgsafe.org/13/1392b1667c.gif" /> </center>

Can you see the difference? Well...not quite :(
## 7. Citations

[1] https://www.ck12.org/c/physics/keplers-laws-of-planetary-motion/lesson/Kepler%E2%80%99s-Laws-of-Planetary-Motion-PHYS/ (only image is used)

[2] https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Second_law_of_Kepler (gif)

[3] https://archive.org/stream/Astronomianovaa00Kepl#page/166/mode/2up

[4] https://books.google.co.kr/books?id=wa2SE_6ZL7YC&pg=PA668&redir_esc=y#v=onepage&q&f=false

[5] https://en.wikipedia.org/wiki/Vis-viva_equation

<center> <img src = "https://steemitimages.com/DQmeJgsbM5K3pUC8kPBToDKRxE2gijUXvgvX6oUiBgaaiyk/atom.gif" /></center>