As an introduction to the problem, you will construct the orbit of Mars in much the same fashion as Kepler did. Like Kepler, we can then compare this orbit with what we would expect if the orbits of the planets were circles centered on the Sun.

In order to determine the shape of the orbit of Mars, we will use a triangulation technique. Kepler was aware of the sidereal period of Mars, that is the period for Mars to return to the same position in its orbit, relative to the Sun. This period is 687 days. In the time that Mars revolves once around the Sun, the Earth will have made about 1.8 revolutions about the Sun because the orbital period of the Earth is 365 days.

Suppose we observe Mars at a given time and note its position in the sky relative to the background stars. Then we wait 687 days and observe Mars’ position again. Although Mars is in the same position in its orbit around the Sun as when we first observed it, Mars now appears at a different position in the sky because the Earth has moved. We can use this information to triangulate the position of Mars.

Kepler had available five pairs of Brahe’s observations made about 687 days apart. The observations consisted of the heliocentric (relative to the Sun) longitude of the Earth and the geocentric (relative to the Earth) longitude of Mars. As a starting hypothesis, he assumed that the orbit of the Earth is a circle centered on the Sun. Now we’ll see whether that makes sense.

The following are the 5 pairs of Brahe's observations Kepler used:

The applet below shows the location of the Sun (at the center) and the Earth’s orbital circle.  Enter the data from Brahe's observations in order to locate Mars at various points along its orbit.  The straight edges will move to the proper angle indicating the line of sight direction, allowing the position of Mars to be triangulated. The zero (0^o) longitude direction is horizontal and toward the right.  For your convenience, the data from the 5 pairs of observations above is repeated on the last page of your worksheet.

This is JAVA. You can't do it.

The first two pairs of observations (designated 1 and 2) determine Mars’ largest distance from the sun (aphelion) and its smallest distance (perihelion) respectively.  In the applet above, click Find Center.  A line will be drawn between aphelion and perihelion. The midpoint of this line is indicated by a small cross.. A circle is also drawn centered on the midpoint and passing through the aphelion (1) and perihelion (2) points.