Our students should look for patterns in mathematics and in science. With this thought in mind we will look briefly at the work of German astronomer Johann Elert Bode (1747-1826). Bode’s project was to try and determine the underlining pattern which “positioned“ the planets in the Sun’s disc. Given his pattern could we tell just where we would expect to find this or that planet?
Simply put, Bode’s “Law” said if you take the number series 0, 3, 6, 12, 24, 48, and so forth, add 4 to each of the numbers and divide by 10, then you will get a nice approximation of the mean distances (in astronomical units) of the planets from the Sun. The problem is that while some of the results are good there is no hard and fast law. Nor has any law been found. I have included a table which you and your students can use to see how close Bodes table matched with reality. Some would like to talk about the asteroid belt which may contain the mix for a “missing planet”. Its been shown that if all of the material in the asteroid belt did combine to form a planet it would be the size of a very small Moon.
Planet Bode Distance Actual Mean Distance
Mercury 0.4 0.39
Venus 0.7 0.72
Earth 1.0 1.00
Mars 1.6 1.52
Asteroid belt 2.8 ——
Jupiter 5.2 5.20
Saturn 10.0 9.53
Uranus 19.6 19.19
Neptune 38.8 30.07
Pluto 77.2 39.5
Let us think about the Sun and how its parallel rays interact with the surface of the Earth. If we cut a circle from hard paper (5 x 8 card) put a brass paper fastener in the middle and attach this to a lined piece of paper. On our paper disk we should see the equator, north and south poles and our real axis with its 23° tilt. Now what do we see? The solar energy which hits the Earth at its Equator is concentrated over a relatively small area. Now let your eyes move to the top of the “sphere” what do you see? Do you expect more intense radiation here? If you were to compare the two locations how would you rate or compare the intensity of the solar radiation? Use a suitable ratio. Try watching the seasons in the hemisphere (north and south). Let the students make observations.
Figure available in printed form
In the classroom I use this hands on model. A dowel marks a point on its circumference. The removable diameter is a strip of formica shaped like an arrow. Holding the arrow a foot or two from the model I ask the class to “questimate” the length of the model’s diameter, two guesses are marked on the arrow which is then hung on the dowel and allowed to swing in the manner described above. Usually by the third time they can call the diameter on the button.
Figure available in printed form
To solve our next problem, that of determining the circumference of a circle, we must rely on the work of Archimedes (pi) if we wish to get any amount of accuracy. If we are looking for simple ideas for the class we can return to the wooden “lollipop” and its flexible (formica strip) which has now been marked off (diameter). We put a random mark on the perimeter of the circle. Next we take the flexible diameter and wrap it around this circular disc and mark how far it went. We continue in this fashion until we return to our first mark. It will be noted that we were able to wrap three diameters around the circle. There is however a problem. The third wrap does not exactly meet the first mark. We have a space which grows proportionally larger as the circles we are measuring grow larger.
Figure available in printed form
If we disregard the space we can formulate a rule for the circumference of a circle. To find the circumference of a circle take the product of its diameter and three and you will have a good estimate (if the circle is not too large)> This system worked for the Chinese 5000 years ago, but it needed to be refined. The time was 300 to 200 B.C. The Greeks (as well as many others) were trying to measure our space ship and the relative distances to the moon and Sun, and they needed accuracy. As we have just seen they could use the results of Archimedes’ computation of pi.
Now to tackle the last of the three questions. How do we determine the area of a circle? I use three approaches. Approach #1. Present each student with a large circle which has been over laid with a square grid. The instructions are simple. Count and record the number of complete squares which appear within the circle. Now count all of the squares which are 90%, 80%, 70%,..., down to 10% complete within the circle. Add the results of each one of these countings and we should have the area of this circle. If you were the “official circle area computing person” in Greece would you look foreword to going to work? If the answer the question is no! Then we should look for a better way.
Figure available in printed form
To prepare for this task we took several large sized wooded disks. The first disk was just a large disk. The next disk was cut through the center to form four pie slices. The next disk was cut through its center 8 times forming 16 equal “pie slices.” These were then taken and bound in a strip of leather, screwed and glued in place. After this two fasteners were installed so that this model could be displayed as a circle or as two semi-circles.
The first circle disk was left on the chalk rail for all to see. The second disk, which was now in four parts was show to the class who were then asked if the resulting shapes reminded them of anything. Hopefully they will see these shapes resemble a series of round bottom triangles.
Figure available in printed form
The remaining part of the demonstration may done in the following way. We take the disk which has cut into 16 parts. We unsnap the leather which binds the disk and open up each semi circle, then push the two parts together to form a rough parallelogram. Since the students should know that the area of a parallelogram equals the product of its base times its height. The height is r (radius of original disk) and the base is one half the circumference of the original disk. The formula for the area of a circle is therefore:
Aparallelogram = Base * Height
Area circle = 1/2 *(2 * pi*r) *r
Area circle = pi * r
2
Figure available in print form