
Index 8b. Parallax 8c. Moon dist. (1) 8d. Moon dist. (2) 9a. Earth orbits Sun? 9b. The Planets 9c. Copernicus to Galileo 10. Kepler's Laws 10a1. Kepler's Third Law (optional) Kepler's Laws (for teachers) 10a. Scale of Solar Sys. 11. Graphs & Ellipses 11a. Ellipses and First Law 12. Second Law 12a. More on 2nd Law 12b. Orbital Motion 
Like many laws in nature, Kepler's laws are easily stated, but their discovery was not quite simple. Consider the 3rd law relating the orbital periods (T_{1}, T_{2}) to their mean distances from the Sun (R_{1}, R_{2}), by asserting that
How did Kepler deduce this law, using Tycho's pretelescope observations? All he had were Tycho's measurements, to the limits of the precision of the human eye, of the positions of planets, moving points of light lined up more or less along the ecliptic. How did he deduce periods and distances (or at least their ratios) from such information? Actually, I do not know, and have no text to refer to. Below however is an attempt at a simplified reconstruction of arguments and calculations which Kepler might have used. Copernicus assumed that each major planet moved with a constant speed in a circle around the Sun. Kepler's first two laws (based on a study of the motion of Mars) modified this, but the Copernican model is still a fairly good approximationorbits are approximately circular and planets move at approximately constant speeds. Orbital Period of Planet more distant than Earth(corresponding to 366.24 rotations of Earth around its axis). Kepler also knew (from older observations) the apparent motion of the Sun around the ecliptic, from one constellation of the Zodicac to the next, and could calculate, for any given day and hour, the position in the sky exactly opposite the Sun, the exact position of midnight. (It is just possible that such calculations were originally started by astrologers.) 
Suppose that at some initial time t=t_{1} Jupiter had occupied that midnight position, and that it returned to midnight at t=t_{2}. The period of Jupiter is NOT the elapsed time t_{2}–t_{1}, because all observations are from our own moving Earth, whose orbital period is a big factor here.
You may visualize the motion of both planets around the Sun using the face of a giant clock: the big hand points at Earth, and the small hand, which moves more slowly, points at Jupiter. At t=t_{1} the clock's time is noon since both hands point in the same direction. If T_{E} and T_{J} are the orbital periods of Earth and Jupiter, when a time t has elapsed from the starting position, Earth (large pointer) has rotated in its orbit an angle (t/T_{E})360° , and Jupiter (small pointer) has rotated (t/T_{J})360°. The Earth rotates faster, so the angle between the lines from the Sun to the planets (the clock's "pointers") is At some time t the Earth overtakes Jupiter again, making that angle is 360° and placing Jupiter again at midnight (a different constellation is behind it, however). At that time DistanceNow to the distances R_{E} and R_{J}. From observations such as those available to Kepler and Copernicus, one could only derive relative distances, ratios such asStart again with Jupiter in its midnight position J_{1}, while Earth is at E_{1}. Let us call here the SunEarthJupiter line "The Main Axis" for short. The drawing at the far right assumes a view from high up on the northern side of the plane of the ecliptic, so both planets orbit counterclockwise (CCW). Earth orbits the Sun faster than Jupiter and overtakes it, so after (say) 2 months Earth has advanced to E_{2} while Jupiter has only advanced to J_{2}, a smaller distance. The line of sight EarthJupiter has therefore rotated clockwise and Jupiter is in fact undergoing retrograde motion. After 3 months (a quarter orbit) Earth has reached its greatest distance from the main axis and begins to move closer to it again, while Jupiter continues its slow advance as before. Therefore a time T_{0} will come when Earth at E_{3} and Jupiter at J_{3} will be at the same distance from the "main axis", and observers at Earth will see Jupiter in front of the same distant stars as were behind it when it passed midnight (though, because the Sun is now in a different direction, they are no longer at midnight). At this time, Earth has gone around its orbit a total of The two distances are equal, because the EarthJupiter line is parallel to the main axis. Therefore We already have T_{E} and T_{J}, and if the value T_{0} is observed, we can deduce R_{J}/R_{E} which is about 5.2. Kepler may have done the calculation more accurately, but this simple version gives the idea. The Inner PlanetsThe planets Venus and Mercury orbit is inside the orbit of Earth, but a similar method may be used for them, too. However, the "reference axis" most convenient here is the line from Earth to the planet when its position in the sky is furthest from the Sun or at "maximum elongation." It happens in two positions of the orbitin one, the planet is visible on the night sky before sunrise, and the other, after sunset. At maximum elongation the planet is at about as bright as it gets, and of course, it never passes midnight.Assume we try to derive the orbital period T_{V}Venus, and the time which has passed since maximum elongation (say, on the evening side) is t. Venus, moving faster than Earth, has covered an angle (t/T_{V})360° and Earth an angle (t/T_{E})360°. The angular separation between the lines connecting them to the Sun is At time T_{sV} (synodic period of Venus) the separation has reached 360°, a full circle, which means Venus has again reached maximum elongation on the evening side (following sunset). As before, we find This gives the distance of Venus as 0.731 AU (assuming circular orbits, constant speed and all orbits in the same plane). ================================ Kepler thus knew the orbital periods and relative distances from the Sun for all the planets visible to him. Why then did he wait until 1620 to announce his third law, about 10 years after the other two? I do not know and can only guess. It could be that he stuck to his older mystical interpretation of planetary distances, connected to the 5 Platonic solidsthe only 5 solid figures whose faces are identical equilateral polygons and which fit snugly inside (or outside) a sphere, the only perfectly symmetric polyhedrons in 3dimensional space. These were the tetrahedron (4 triangles, a 3sided pyramid), cube (6 squares), octahedron (8 triangles), dodecahedron (12 pentagons) and icosahedron (20 triangles). If each planetary orbit defined a sphere, the space between the spheres of Mercury and Venus enclosed a tetrahedron, between Venus and Earth a cube, and so on. A closer examination showed that such a "law" gave a rather poor fit. By 1620, however, logarithms had been discovered and Kepler embraced them with enthusiasm. By plotting log T against log R, the 3/2 power dependence was clearly seen. 
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Author and Curator: Dr. David P. Stern
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