A Guide to Kepler’s First Law

Kepler’s first law states an object orbits another object under a gravitational pull in a conic section orbit. If the orbit is closed, it is an ellipse. It also turns out to be really hard to prove. You either have to use calculus and differential equations, or use geometry with lines and stuff and know well the properties of the ellipse. Either way, you have to set up the problems in creative ways.  In this post, I would like to collect all the ways Kepler’s first law can be derived. I have always found it annoying how scattered the proofs were, and I would like to leave this behind for anyone who is itching to find out how Newton’s laws implies elliptical orbit and vice versa.

For a newbie, the best proof out there is in my opinion Feynman’s geometric proof. While it is still complicated, it is not as hard as the other proofs to understand. You don’t have to know anything too complicated except for some of the properties of the ellipse, and get used to the methods of geometry. It is also great for its clarity, unlike Newton’s geometric proof. Newton’s proof is convoluted and use really complicated geometry, but if you would like to know how the master himself did it, there it is.

The most common derivation is the differential equation approach. It is the standard textbook approach, and if you know something about calculus and differential equations, it is easier to swallow. Then there is the more complicated version which takes account of the fact that two move around a center of mass, instead of one around the other.

My favorite version, though, is the one that uses the Laplace-Runge-Lenz vector. Its derivation is elegantly simple, following directly from the approach. In the other differential equation method, you have to find the acceleration in terms of polar coordinates and then do a creative substitution that makes them end up as a simple second order differential equation. This one is somewhat less convoluted than that, and once you get the vector, you are only one step away from Kepler’s first law. In The Mechanical Universe and Beyond, video 22 titled The Kepler Problem uses this derivation.

Finally, I know there is the one that uses the complex function. Unfortunately, I can’t find it online. It is contained in this book, though.

If anyone knows of other alternatives, I can post it here.

Prime Divergence

Did you know that:

is equal infinity? Now, some of you might be thinking, “Oh, I already knew that.” Well, then this post is not for you, move along. Others, though, might think, “hold on, isn’t adding anything up to infinity equals infinity?” Nope because the sum (1/2)+(1/4)+(1/8)+ etc for example is equal to 1. And in fact, all geometric sum (in case you don’t know, it goes like n+n^2+n^3+… etc) with a number less than one converge into a number that is not infinite.

Some others who has learned stuff from second semester calculus class might think “Really? It looks real slow.” Well, it is really slow, but as you shall see, it nevertheless does diverge, and demonstrating this will be the topic of this post.

This post is based on a section in the book “An Imaginary Tale” by Paul Nahin. The book talks about this, and I thought the mathematics was super neat, and so I decided to put it here.

The post is in three parts:

a)I will show a proof originally done by Nicole Oresme (1320-1382) on the divergence of the harmonic series, which is 1+(1/2)+(1/3)+…

b)Then I will change the Riemann zeta function, which is f(x)=1+(1/2^x)+(1/3^x)… into a product of primes, which was done by Leonard Euler (1707-1783).

c)And finally I will use the above results to prove that the sum of all 1/prime does diverge, which was proved by Euler.

Who’s Got More Points?

So, the question is, which of the following has more points: a line segment or an infinite line? You might think the obvious answer is the infinite line. But it is not as simple as you think it is. A simple proof done by the mathematician Georg Cantor will show you the answer:

Imagine two semicircles of the same size and shape. A semicircle is pretty much half a circle. You agree, since both semicircles are the same, they both have the same number of points, right? Beneath the first semicircle, place a line segment that have the same size as the diameter of the semicircle. Beneath the other one, place a line that goes to infinity. Now, your j0b will be to match every single point on the border of the semicircle to every single point on the line segment and every single point in the second semicircle to every single point in the infinite line. The setup is done as shown in the figure below:

In the line segment, I match up every point of it to the semicircle by a straight line connection. By going from left to right (or any direction you want), you realize that you can match every single point on that semicircle to the line segment. For that reason, the line segment has as many points as the semicircle above it.

In the case of the infinite line, I have to do a cleverer maneuvering. From the center of the semicircle, I extend a line straight to the left (or right), and sweep around the semicircle counterclockwise. Doing so will match every point on the semicircle. Yes, even the one point in the infinite line that seems unimaginably distant. After all, all I would have to do is extend a line from the semicircle to that point. As you will see, no matter how far away the point is from the semicircle, it will touch a point on the semicircle. Interesting, it seems like an infinite line has the same amount of points as the semicircle. What is the implication of all of these things I have written?

Well, since the line segment has the same number of points as its corresponding semicircle, and so does the infinite line, and both semicircles have the same number of points, there is only one conclusion I can take. The infinite line has the same number of points as a line segment!

Mind Blown.

-This post has been inspired by the book To Infinite and Beyond: A Cultural History of the Infinite, so I have the book to thank-

Euclid’s The Elements: Preposition One-Really Fun Math Puzzle

This is the nth edition of *drum rolls* Pop Quizz! A quizz in which the answers are not obvious and if it is obvious, and you get it wrong, you are rebuked at!!!

Occasionally, I will occasionally ask a question, and the reader will answer them! For every wrong answer, I will consider one of your comments a spam. Ha! See how you like that suckers! (I am kidding, of course) Sometimes, a wrong answer won’t invite rage, some really obvious one will, and you will be called an idiot if you can’t answer the obvious one. In fact, I may have to make an extra page on my blog for the lists of idiots.

Anyways, once upon a short time ago, I was reading Euclid’s The Elements, which is basically the geometry book to rule them all. First, it defines various terms, and go over postulates. Then it goes through geometric proofs, each different one called prepostitions. Well, the first one I think is a really neat geometry problem. It goes like this:

3. On a given straight line to construct an equilateral triangle.

Meaning, considering you have a line AB, make a triangle ABC with all sides equal. Sounds hard? Well, if you hear the clue, you migh go “OOOOHHHhhhhh!” If not, well… shame on you, mate. ^_^