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Elliptical Orbits in the Solar System

Date: 05/22/2005 at 16:29:42
From: Rick
Subject: Contructing ellipses with specific sizes

I teach middle school math and I want to have my students draw a scale
model of the solar system that shows the orbits of the planets.     

Assuming I have the apogee and perigee of each planet's orbit about
the sun they need to construct 9 ellipses with some degree of accuracy.

I am hoping there is some variation on the tack and string method 
that wouldn't involve nine different tack settings and string 

Date: 05/22/2005 at 22:59:51
From: Doctor Peterson
Subject: Re: Contructing ellipses with specific sizes

Hi, Rick.

Let me start with a quick side comment.  The words "apoGEE" and
"periGEE" refer to the greatest and least distances of a satellite
from the EARTH ("ge" in Greek); for planets, you should use "apHELION"
and "periHELION", which refer to the SUN.  See the following page, for
a term I didn't know! 

If I recall correctly, most of the planets have a small enough
eccentricity that you can get a pretty accurate approximation to the
ellipse by just marking the aphelion and perihelion, then finding the
midpoint and drawing a circle with that center.  You could then check
the minor semiaxis and see whether it is noticeably different from the
radius of the circle you drew.  Here are the basics on ellipses:

The semiaxes are a and b, in the x- and y- directions respectively; 
that is, these are half the "diameters" in each direction.  The 
distance from the center to the focus is c, and

  a^2 = b^2 + c^2

        *      |  \   *
     *         |     \   *
   -a*-c       |       c *a
        *      |      *

If the apoapsis is r1 and the periapsis is r2, then

  r1 = a+c
  r2 = a-c


  a = (r1+r2)/2
  c = (r1-r2)/2
  b = sqrt(a^2-c^2) = sqrt[(a+c)(a-c)] = sqrt(r1*r2)

Note that the major semiaxis is the arithmetic mean of the two 
distances, and the minor semiaxis is the geometric mean!

If your focus (sun) is at the origin, and you mark the perihelion on 
the left and the aphelion on the right, like this:

       -r2         0              r1

then the center of the circle would be at c=(r1-r2)/2, and its 
radius would be a=(r1+r2)/2.  Thus the radius of the circle is the 
correct value of a, the major semiaxis, but the correct value of b 
would be smaller than that.  By how much?

I looked up the data for earth, and find

  aphelion:   r1 = 152.6 million km
  perihelion: r2 = 147.5 million km

  a = 150.05 million km
  c =   2.55 million km
  b = 150.03 million km

You can see that in this case, the minor semiaxis is so close to the 
major semiaxis that there is no need to draw an exact ellipse; it 
looks practically identical to a circle, even though its center is 
offset noticeably from the sun.

At least Pluto, and possibly others, will require you to draw the
ellipse with more care.  I looked up the data for Pluto and find

  aphelion:   r1 = 7,000 million km
  perihelion: r2 = 4,500 million km 

  a = 5750 million km
  c = 1250 million km
  b = 5612.5 million km

In this case the minor semiaxis is significantly smaller than the 
major semiaxis, so you would want to use some method to accurately 
draw the ellipse.  If you don't like pencil-and-string, try this one:

  Accurate Drawing of an Ellipse 

That uses the circle we've already drawn, and another with the same 
center, to draw accurate points on the ellipse, which you can 
connect with a smooth curve.  You can find more on the string method 
at the bottom of this page, and in the links:

  Defining an Ellipse 

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Practical Geometry

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