Chapter 15



Understanding Mean Anomaly

Let's take the first two years of oppositions in Kepler's table, 1580 and 1582. The date, the longitude, and the latitude seem clear enough: they are when the observation was made, and where Mars was seen. But what about the Mean Longitude on the right?

Working it through

The mean anomaly is a measure of time. You can think of it as a clock. Although we do not always have the sun at its limit at 12 noon, we still use our watches without reseting them every day to solar noon. (The difference between solar time and clock time is called the equation of time, which is one of the causes of the analemma phenomenon.) For the planet Mars, the mean longitude is a measure of how much angle the planet would have traversed, if it always moved at the same rate in its orbit. So let's recreate Kepler's creation of the right column.

First, let's see how many days have passed between Nov 18, 1580 and Dec 28, 1582. We have two years (2×365.25), one month (30 days has November), and ten more days (from the 18th to the 28th). Together, that gives:

2×365.25 + 30 + 10 = 770.5 days.

(We won't factor in the hours and minutes of time for this rough example.) Now, how many Mars orbits do 770.5 days make, and how many degrees would a uniformly moving Mars move? One Mars year is about 686.6 days, so 770.5/686.6 tells us how many Mars years we have, and multiplying by 360° will give us the degrees. When we do the math, we get:

404°, which, removing 360°, is 44°.

Compare that to the difference of Mean Longitude in the right column. The S stands for "Sign", which is 30°. We have (ignoring minutes and seconds),

3S 9° - 1S 25°
99° - 55° = 44°,
the same as we had found above.