Chapter 54


In this chapter, Kepler approximates the aphelial and perihelial distances of Mars by making use of the distances he calculated in Chapter 51.  The two that he uses in this chapter are the aphelial (green) and perihelial (purple) distances on the left semicircle in the diagram of that chapter, reproduced to the right.  After making corrections to these distances (including accounting for latitude), Kepler determines the two corrected distances to be:

11°52′ mean anomaly
161°30½′ mean anomaly

Accounting for Inclination

Let’s look at how Kepler adjusts the perihelial distance.  The distance of 139,000 is not truly the distance to Mars (red), but rather to the point (black) on the ecliptic from which a perpendicular could be raised to reach Mars.  At this location of 21° Pisces, the planet is 35° from the limit (the location of its maximum inclination), and therefore 55° from the node.  The maximum inclination was determined to be 1°50′ in Chapter 13.  Kepler calculates the inclination (the angle above the ecliptic that the Sun would observe Mars) to be:

1°50′ : Sin(90°) = inclination : Sin(55°)
inclination = 1°31½′.

Kepler then writes: “The excess of the secant will be 35½, which is equivalent to 49 of our units.” From the above diagram it is clear that the actual distance of Mars is the secant, if the distance to the black point is considered as our base unit.  Using a table of sines, we find:

secant 1° 31 ½′ = 100035 ½

Which, when multiplied by 139000, gives 49, meaning the length 139,000 must be increased by 49 units, giving a corrected Sun-Mars distance of 139,049 at this location.

Estimating the Apsidal Distances (Help!)

Now that Kepler has the corrected Sun-Mars distances at locations near the apsides, he must estimate what the length would be were the planet directly at the apsides.  This is difficult! If you can figure out how Kepler got his numbers (or how Donahue got his in footnote 2 on page 539), please send us an email.  Kepler is not assuming a circle when he makes these corrections, but it also does not seem that he is assuming the oval of the chapter 45 hypothesis.  We got 153 for the first example where Kepler got 164, using a circular orbit and the calculated eccentric anomaly of 10°54′ from the spreadsheet below.

If it can be of service, click here for a Microsoft Excel spreadsheet that you can use to get the equated anomalies corresponding to the mean anomalies Kepler gives, using the Vicarious Hypothesis.  Additionally, the spreadsheet gives the eccentric anomalies assuming a circular orbit and an eccentricity of 9265.