KEPLER'S LAWS

Kepler's laws are very important

Here is a link to a 5-minute movie about them:LINK TO KEPLER MOVIE

Here is a written summary:

German astronomer Johannes Kepler was born in December 1571, and throughout his 59 years of life, he contributed immensely to science. He is most well-known, however, for his three laws of planetary motion. This work stemmed from a collaboration with Danish astronomer Tycho Brahe. Kepler mathematically analyzed some 20 years of precise planetary observations that Brahe collected. He determined how the solar system’s planets move the way they do and laid the foundation for Isaac Newton’s theory of gravity. What we now call his three laws are theories that are universal, verifiable, and precise — and they don’t just govern the motion of the planets, but also comets, asteroids, and other minor bodies orbiting the Sun.

Kepler’s first law says that the orbit of every planet is an ellipse, with the Sun at one of two focus points. A circle is a special type of ellipse that has just one focus, which is located at its center. All other ellipses look like flattened circles, and have two focus points. If you take any point on the ellipse, the sum of the distances to the focus points is constant. 

Before Kepler’s work, astronomers tried to describe the motion of the planets via interconnected circles, but they struggled to match observations. They also could not predict where planets would appear in the sky. Kepler’s theory changed that and showed how elegantly the planets moved.

Kepler’s second law says that a line joining a planet and the Sun sweeps out equal areas during equal intervals of time. Thus, a planet moves fastest when it’s closest to the Sun (at a point called perihelion) and slowest at its farthest point from the Sun (known as aphelion). You can see this property best with objects that have longer elliptical orbits, like comets. As they near the Sun, they travel much faster than when they are more distant. 

Nine years after Kepler published his first two laws, he determined the last one. Kepler’s third law says that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. The semi-major axis is just a term for half of the longest length of the ellipse. If you measure the period in Earth years and the orbit’s semi-major axis in astronomical units, the equation simplifies to period squared equals semi-major axis cubed. Astronomers use this relation to figure out the orbit of planets around other stars. They directly detect the world’s orbit period and can then figure out how far from the star the planet orbits.

Thanks to all who commented!

A dozen or more students have commented, and will receive Japanese 1-Yen coins after school starts in 2013.

In some cases, it was unclear who was commenting, but we will find you! We will have a practice commenting session early in 2013.

Welcome to 21st century on-line teaching!

Sincerely, Stephen G. "Dr. Steve" Margolis

ABOUT DIVISIBILITY BY 11 AND 13

From a previous post, the rules for divisibility by 11 and 13 are:

11

Alternately add and subtract the digits from left to right. (You can think of the first digit as being 'added' to zero.) If the result (including 0) is divisible by 11, the number is also. Example: to see whether 365167484 is divisible by 11, start by subtracting: [0+]3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11.

13

Delete the last digit from the number, then subtract 9 times the deleted digit from the remaining number. If what is left is divisible by 13, then so is the original number. Here are a couple of examples: 286 Add the first digit to zero, then subtract and add alternately. +2 -8 +6 = 0; zero is divisible by 11, and so is 286 (try it with your calculator). 715 +7 -1 +5 = 11, 11 is divisible by 11 and so is 715 try it with your calculator). 286 again Delete the 6, multiply by 9, subtract from 28. 28 - 54 = -26, -26 is divisible by 13, and so is 286 try it with your calculator). 715 again Delete the 5, multiply by 9. 71-45 = 26, 26 is divisible by 13, so is 715 try it with your calculator).

DIVISIBILITY RULES

The following is borrowed from "Ask Dr. Math" published by Drexel University in Philadelphia PA.

Divisibility by:

2	If the last digit is even, the number is divisible by 2.
3	If the sum of the digits is divisible by 3, the number is also.
4	If the last two digits form a number divisible by 4, the number is also.
5	If the last digit is a 5 or a 0, the number is divisible by 5.
6	If the number is divisible by both 3 and 2, it is also divisible by 6.
7	Take the last digit, double it, and subtract it from the rest of the number; if the answer is divisible by 7 (including 0), then the number is also.
8	If the last three digits form a number divisible by 8, then so is the whole number.
9	If the sum of the digits is divisible by 9, the number is also.
10	If the number ends in 0, it is divisible by 10.
11	Alternately add and subtract the digits from left to right. (You can think of the first digit as being 'added' to zero.) If the result (including 0) is divisible by 11, the number is also. Example: to see whether 365167484 is divisible by 11, start by subtracting: [0+]3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11.
12	If the number is divisible by both 3 and 4, it is also divisible by 12.
13	Delete the last digit from the number, then subtract 9 times the deleted digit from the remaining number. If what is left is divisible by 13, then so is the original number. Consider the number 7560, what is it divisible by? II The last digit is zero, which is even, so it is divisible by 2. III The sum of the digits, 7 + 5 + 6 + 0 = 18, which is divisible by 3, so the number is also divisible by 3. IV The last two digits, 60, are divisible by 4, so the number is divisible by 4. V The last digit is zero, so the number is divisible by 5 VI The number is divisible by 3 and 2; consequently, it is divisible by 6. VII For 7560 to be divisible by 7, 756 must be divisible by 7. Knock off the last digit and consider two numbers, 75 and 6. Double 6 and subtract from 75; 75 -12 = 63.We recognize that 63 is divisible by 7, 63/7 = 9. So 7560 is divisible by 7. VIII The number is divisible by 4 and 2, hence is divisible by 8. Or, looking at the last three digits, 560 divided by 8 is 70, so 7560 is divisible by 8. IX The sum of the digits is 18, which is divisible by 9, so 7560 is divisible by 9. X The last digit is zero, so the number is divisible by 10. The number is not divisible by 11 or 13. XII The number is divisible by 3 and 4, hence is divisible by 12. 7560/12 = 630. In fact, I "cooked up" 7560 as 3 x 5 x 7 x 8 x 9, since 8 x 3 = 24, the number was sure to be divisible by 2, 4, 6 and 12. Methods for divisibility by 11 and 13 will be covered in a subsequent post.

A Word Problem About Primes

There are 2 prime numbers between 100 and 199 such that the tens digit is a prime number, the ones digit is a prime number, and the tens and ones digit taken together are a 2 digit prime number. Find the sum of these 2 prime numbers.

Answer:

The numbers in question obviously start with "1", these are all the prime numbers between 100 and 199:

122
123*
125
127

132
133
135
137*

152
153*
155
157

172
173*
175
177.

There are more, but one or more of their digits are not prime.

Go to Rot13.com, paste the following coded message in the box, and it will instantly decipher.

Erzbir gur cevzrf jubfr graf naq barf qvtvgf ner abg cevzr:

Guvf yrnirf bar-gjragl-frira, bar-guvegl-frira, bar-svsgl-frira naq bar-friragl-guerr.

Erzbir gur cevzrf jubfr graf naq barf qvtvgf gnxra gbtrgure ner abg cevzr:

Guvf yrnirf bayl bar-guvegl-frira naq bar friragl guerr. (Gur bgure gjb-qvtvg ahzoref ner qvivfvoyr ol guerr, 57 = 3 k 19)

Nqq gurz:

Bar-guvegl-frira cyhf bar friragl guerr rdhnyf guerr uhaqerq gra.

Qb lbh frr jung “Gur Fbyire” vf qbvat? Jvgubhg hfvat n yvfg, ur dhvpxyl aneebjrq vg qbja gb 4 pubvprf, gjb bs juvpu jr pna rnfvyl ryvzvangr nf gur fhz bs gurve qvtvgf vf qvivfvoyr ol 3, urapr gubfr ahzoref ner abg cevzr.

PICTURES OF JAPANESE 1-YEN COINS; "FLOATING"

HERE ARE SOME PICTURES OF JAPANESE 1-YEN COINS

I "LIBERATED" SOME TOOLS FROM DELTA AIRLINES

THIS IS THE "OBVERSE" OF A 1-YEN COIN

THIS IS THE REVERSE OF A 1-YEN COIN

PICK UP A COIN WITH A PLASTIC FORK, COURTESY OF DELTA AIRLINES

CAREFULLY PLACE 1 COIN ON THE SURFACE OF THE WATER

CAREFULLY PLACE SECOND COIN ON THE SURFACE OF THE WATER

CAREFULLY PLACE THIRD COIN ON THE SURFACE OF THE WATER

CAREFULLY PLACE FOURTH COIN ON THE SURFACE OF THE WATER; NOTICE THAT THE COINS HAVE DRIFTED TOGETHER

CAREFULLY PLACE FIFTH COIN ON THE SURFACE OF THE WATER

CAREFULLY PLACE SIXTH COIN ON THE SURFACE OF THE WATER; THE PREVIOUS FIVE HAVE DRIFTED TOGETHER AND NUMBER 6 FITS IN, LEAVING ONE SPACE.

CAREFULLY PLACE SEVENTH COIN ON THE SURFACE OF THE WATER; PLACE IT NEAR THE
EMPTY SPOT AND IT DRIFTS IN. NOTICE THAT THERE IS A LITTLE WATER ON TOP OF THE SEVENTH COIN, IT IS LIKELY THAT IF WE ADDED ONE MORE, THE SURFACE TENSION WOULD NOT SUPPORT IT AND THE WHOLE ASSEMBLY WOULD SINK.

SO, FOR NOW, THE WORLD'S RECORD IS SEVEN COINS. CAN YOU BEAT IT?

How to comment

Hello Seward Students:

At the bottom of the blog you will find a pale blue line saying "x comments" where x is some number.

Click on this.

A box will pop up.

In this box you will find a statement like: "jump to comment box."

Do so and type in your comment.

At the bottom you will find an imitation to post your comment,.

Do so. I will immediately get your comment as an email.

FREE GIVEAWAY FOR SEWARD STUDENTS

Seward Students Only: surface tension demonstration.

I have a limited number of Japanese 1-Yen coins. A Japanese Yen is currently worth about 1.2 US cents. Japanese people use these coins to pay sales taxes. I believe it costs the Japanese government more than 1 Yen to make each of these coins, but the Japanese people insist that they be made, so they can pay their sales taxes with them.

The Japanese 1-yen coin has the unique quality that it can "float" on water. It doesn't actually float, it is supported by surface tension.

Amaze your friends and family!

The Japanese 1-Yen coin is deliberately made to have exact metric values. Its diameter is exactly 2 centimeters, so its radius is exactly 1 centimeter. Its mass (weight) is exactly 1 gram. The thickness of the rim is 1.5 millimeters (0.15 centimeters) but it is stamped so the average thickness is 1.18 millimeters (0.118 centimeters)

Compute the volume of the Japanese 1-yen coin: use pi x (r squared) x average thickness: use the centimeter values.

Your result should be in cubic centimeters: please round to three decimal places.

Divide the mass (1 gram) by the just-computed volume: you should get a density in grams per cubic centimeter that exceeds 1.0, so the Yen should sink. (Please round the computed density to two decimal places). Incidentally, the density of a US penny is more than double the density of a Japanese Yen. Pennies won't "float;" they are too heavy for surface tension to support them.

But, if you carefully place the 1 Yen coin on the surface of a glass, cup, or bowl of water, it will appear to float; it is supported by the surface tension of water.

A subsequent blog will include some pictures of Japanese 1-yen coins "floating."

How to get your free Japanese 1-yen coin:

With  permission of your parents, parent, or guardian, go to this blog http://sewardmath1.blogspot.com

The blog accepts comments.

Make a comment: this should include your name, your Seward home-room teacher's name or, for middle-school students, your math teacher's name. Also include your calculations of the volume and density of a Japanese 1-yen coin.

The first thirty students to make a complete comment (name, teacher's name, volume & density) will get a free Japanese 1-yen coin. It will be delivered to the mailbox of your home-room teacher's mailbox or, for middle-school students, your math teacher's mailbox on the first school day following your comment.

To get a notice of additional posts to this blog, with permission of your parents, parent, or guardian: email me (stephen dot margolis at gmail dot com - replace "dot" and "at" with their usual symbols) your email or address or the the email address of your parents, parent, or guardian (the person or persons who receive your report card). If you do this, I will also verify that your calculations are correct. Use a calculator for these calculations. If you don't have a calculator, send me an email and I will give you one (to keep), if I haven't already. Seward Students Only!

As stated above, a subsequent blog will include some pictures of Japanese 1-yen coins "floating."

Computer security: I gave my email address in the strange form above because there are web crawlers: bots that search the internet for email addresses and then use these for evil purposes. The bots are dumb: they are looking for my.name@someplace.com.
 -

Some Problems With Two Unknowns

All of these problems are easily solved by “textbook” methods. But, the textbook is not a 6th grade textbook. If you can solve them by “textbook” methods, feel free to do so.

6th Grade Math Challenge 2010, Individual Round #1, Problem #1

Two notebooks and one pen cost $3.17.
One notebook and two pens cost $3.97

How much do 5 notebooks and 5 pens cost?

Hints: Let N be the price of a notebook and P be the price of a pen. Confession: I wasted time solving for N and P. Then I thought: what are they really asking for?

An equation always says: left side = right side. Since you can always add or subtract the same thing to each side of an equation, you can add or subtract equations. In this case, if you write two equations and add them, you nearly have what they are asking for.

5th Grade Math Challenge 2007, Tie Breaker, Problem #1. Note: Tie Breakers are usually challenging problems.

1. The owner of a bicycle store had a sale on bicycles (two-wheelers) and tricycles (three-wheelers). When he counted the total number of pedals of the cycles on sale, he got 50. When he counted the total number of wheels of the cycles on sale, he got 64.

How many tricycles were offered in the sale?

(Note: each cycle has two pedals.)

Hints: Let T = the number of tricycles and B = the number of bicycles. Write two equations. Since each cycle has two pedals, you know the number of cycles on sale.  If you solve the second equation for B in terms of T, you know that T is an even number.

Guessing a few even numbers and checking the total number of cycles will lead you to the right answers. (They ask for the number of tricycles.)

5th Grade Math Challenge 2007, Team Round #1, Problem #3. Note: Team Round Problems  are usually challenging.

You took a 20-question exam that was scored in this way: 10 points are awarded for each correct answer and 5 points are deducted for each incorrect answer. You answered all 20 questions and received a score of 125 points.

How many questions did you answer incorrectly?

Hints: You know there are 20 questions. Let C be the number answered correctly and W be the number answered incorrectly, or Wrong. (I used W instead of I because I is easily confused with 1.)

Use C and W to write an equation for score, which you know is 125. Solve for W in terms of C. You will find that C has to be equal to or greater than 13 (otherwise, W would be negative) and must be less than 20 (if you had 20 correct answers you would have no wrong answers and would have scored 200.) There are only 7 possible guesses, and if you start at 13, you will get the correct answer after only a few guesses.

Guess and Check Method for Simultaneous Equations: Shown to Me by Kai Alton.

Here is a typical problem with two unknowns:

The gym has 2 kg and 5 kg disks for weight lifting. There are 14 disks in all.

The total weight of the 2 kg disks is the same as the total weight of the 5 kg disks.

What is the total weight of all the disks?

The first step is to read the problem and turn the English into mathematics. Part of this is: give the unknown quantities one-letter names that are meaningful. Also: identify what they are asking for. In this problem, it is the total weight of all the disks.

The two unknowns are the number of 2 kg disks and the number of  5 kg disks. Let’s call these T for number of  Two kg disks and F for number of Five kg disks.

The first English sentence to turn into mathematics is: “there are 14 disks in all.”

In mathematics, it becomes T + F = 14.

The second English sentence to turn into mathematics is: “the total weight of the 2 kg disks is the same as the total weight of the 5 kg disks.” In algebra, if we write a number next to a letter, it means “multiply.”

So the sentence becomes: 2 T = 5 F.

There are formal mathematical methods for solving the two equations simultaneously, but I would like to show you a method shown to me by a Seward student, Kai Alton. It works best where the total of the two unknowns is given, as in T + F = 14.

First:  subtraction <= undoes => addition (read it forwards or backwards).
Also: division <= undoes => multiplication (read it forwards or backwards).

A summary of the method is:

1. Make a smart guess for one unknown.
2. Solve for the other unknown in terms of your smart guess, using the “undoing” rules.
3. Check to see if the two (now known) unknowns add up to the given total.
4. If not, make another smart guess until the total is right.
5. Use the (now known) unknowns to give them what they are asking for.

Using the “undo” on 2 T = 5 F, divide both sides of the equation by 5, undoing one multiplication.

The result is F = (2/5) T . Both F and T are whole numbers, so a smart guess for T is a whole number divisible by 5, for example, guess T = 5. Then, F = 2 and the total number of weights is 7. The total should be 14, so a better guess is T = 10, which gives F = 4 and a total of 14. Checking,  5 F= 5 x 4 = 20 kg  and 2 T = 2 x 10 = 20 kg (the total weight of the 5 kg weights equals the total weight of the 2 kg weights) and what they asked for, the total weight of all the disks, is 20 + 20 = 40 kg.

The problem is solved by smart guessing, checking, and arithmetic.

Test Post

This is a test post.

The arithmetic of the election

The 2012 Presidential election will take place on Tuesday, November 6.

It is then that you will hear about electoral votes.

There are 538 electoral votes. Minnesota has 10.

The electoral system was created by the United States Constitution, Article II, modified by the 12th and 23rd amendments. The 23rd amendment was added in 1961 and gives 3 electoral votes to The District of Columbia; this leads to the present total of 538 electoral votes.

The number 538 is divisible by 2 (its last digit is the even number 8). Half of 538 is 269, hence, if either candidate gets 270 votes or more, he wins. Because 538 is divisible by 2, a tie is possible; this happened twice, in the elections of Thomas Jefferson and John Quincy Adams. In those days, the number of electoral votes was smaller and a tie more likely.

In the case of a tie, the President is chosen by the House of Representatives and the Vice President by the Senate.

There are many arguments for and against the use of electoral votes but one is surely true; it preserves the influence of states with small populations. You may have been in some of those states; North Dakota, South Dakota, Wyoming and Montana have small populations but each has 3 electoral votes.

You could add up the populations of the 4 states above and find that their combined population is less that of Minnesota, but they have 12 electoral votes to Minnesota’s 10.

To read more about the electoral vote system, go to:

http://en.wikipedia.org/wiki/U.S._Electoral_College

This is a long article, but if you stay up until the election is decided, you may have time to read it.

The electoral votes will not be officially counted until December, 2012.

Solution to An Old Problem About Shopping and Taxes

I am going to use rot13 to give answers:

Copy the encrypted text below: (use the Copy command) which is Ctrl-C on Windows computers and Command-C on Apple computers.

Zl Erprvcg sebz Cnaren Oernq

SBBQ: fvk qbyynef naq 59 pragf
GNK: sbegl guerr pragf
GBGNY: frira qbyynef naq gjb pragf
PHFGBZRE CNVQ: frira qbyynef naq svir pragf
PUNATR QHR: guerr pragf

Go to:  http://www.rot13.com

Paste the encrypted text into the box, using (Ctrl–V or Command-V) Click on “Cypher.” You should see the decoded answer.

Answer to Units Conversion As A Joke

I am going to use rot13 to give answers:

Copy the encrypted text below: (use the Copy command) – it is Ctrl-C on Windows computers and Command-C on Apple computers.

Wbxr nafjre: ur pbhyq svaq bhg ol purpxvat uvf fcrrqbzrgre.

Havgf pbairefvba:

60 srrg   3600 frpbaq   1 zvyr
------- k      ------ k ------ =   sbegl cbvag avar zvyrf cre ubhe
frpbaq         ubhe    5280 srrg

Ur vf fcrrqvat ohg whfg n yvggyr ovg.

Go to: http://www.rot13.com

Paste  using (Ctrl–V or Command-V) the encrypted text into the box. Click on “Encrypt.” You should see the decoded answer.

Units conversion as a joke, corrected

Dr. Steve is driving his car. He is traveling at 60 feet per second and the speed limit is 40 miles per hour. Is Dr. Steve speeding?

Answer in next blog.

Units Conversion As A Joke

Dr. Steve is driving his car. He is traveling at 60 feet per second and the speed limit is 40 miles per hour. I Dr. Steve speeding?

Answer in next blog.

An Old Problem About Shopping and Taxes

On February 10, 2007 “Steve” went to Panera Bread in Eagan MN and bought a “U Pick 2” (soup, sandwich and an apple) for $6.59.

Sales tax in Eagan was 6.5% back then. In Minnesota, tax is rounded to the nearest cent.

Three questions:

What was the tax?

What was the total, including tax?

How much change would Steve get if he presented a $5 bill, a $2 bill, and a nickel (five cents)?

-------------------------------------------

Note: November 2012

The sales tax in Eagan has gone up to 7 1/8 %  (7.125%). The actual ancient receipt will be posted, somewhere in Seward.

Crypt Arithmetic Problems That Are Easy To Solve, Revised

Students in 5th grade math challenge at Seward.

Please perform both checks and write on paper.
Write your name and your E2 teacher's name on
the paper and hand in to Karen Utter during your normal Monday or Wednesday class, but as soon as possible.

We want to verify that you are reading this blog.
Please tell your friends about it.

The assignment may seem trivial but we just want to verify that you are reading this blog.

2012 6th Grade Math Challenge Problem #1
162
+XD
-----
2D7
D must equal 5 so we have:
162
+X5
-----
257
When we add X to 6, we must get 5 and a carry of 1.
So, 6 + X = 15, or X = 9
162
+95
-----
257

Check: Do the above with a calculator and verify that it is correct.

2012 6th Grade Math Challenge Problem #5
ON + ON + ON + ON = GO
Adding ON 4 times is the same as multiplying by 4.
ON
x4
-----
GO
Try various values for N.
N can't be zero because then, N and O would be the same.
If N is 1, O = 4 and 4 x O would be 16, generating a carry and giving a three-digit answer.
If N is 2, O = 8 and 4 x O would be 32, again generating a carry and giving a three-digit answer.
If N is 3, 4 x 3 = 12. That makes O = 2 with a carry of 1.
G = 4 x 2 + 1, or 9.
This gives:
23
x4
----
92
Check: 23 + 23 + 23 + 23 = 92
Verify this with a calculator.

Crypt Arithmetic Problems That Are Simple to Explain

Crypt Arithmetic Problems That Are Simple to Explain


Students in 5th grade math challenge at Seward.
Please perform both checks and write on paper.
Write your name and your E2 teacher's name on
the paper and hand in to Karen Utter on Monday
Nov 5 or Wed Nov 7.

2012 6th Grade Math Challenge Problem #1
162
+XD
-----
2D7

D must equal 5 so we have:

162
+X5
-----
257

When we add X to 6, we must get 5 and a carry of 1.
So, 6 + X = 15, or X = 9

162
+95
-----
257

Check: Do the above with a calculator and verify that it is correct.



2012 6th Grade Math Challenge Problem #5

ON + ON + ON + ON = GO

Adding ON 4 times is the same as multiplying by 4.

ON
x4
-----
GO

Try various values for N.

N can't be zero because then, N and O would be the same.

If N is 1, O = 4 and 4 x O would be 16, generating a carry
and giving a three-digit answer.

If N is 2, O = 8 and 4 x O would be 32, again generating a
carry and giving a three-digit answer.

If N is 3, 4 x 3 = 12. That makes O = 2 with a carry of 1.
G = 4 x 2 + 1, or 9.

This gives:

23
x4
----
92

Check: 23 + 23 + 23 + 23 = 92

Verify this with a calculator.