Chapter 2-1. The Cycle of Lunar Eclipses
Recommended Reading : 【Earth Science】 Chapter 2. The Solar System
Q.
Find the formula for calculating the cycle of lunar eclipses occurring on Earth in 2015 on the internet, and explain how this formula was derived mechanically and astronomically.
A.
The orbital period of the moon orbiting the Earth is approximately 27.212 days. During this time, the moon roughly draws a circle centered on the Earth. Thus, with respect to the celestial sphere centered on the Earth, the moon draws a great circle (i.e., the ecliptic) with a period of about 27.212 days, centered on the Earth.
Additionally, it takes about 346.6 days for the sun to return to the same node after crossing it. Therefore, with respect to the celestial sphere centered on the Earth, the sun moves along a great circle (i.e., the ecliptic) with a period of about 346.6 days, centered on the Earth.
Now, let’s find the conditions for a lunar eclipse to occur.
Figure. 1. Model for calculating the cycle of lunar eclipses
The distance from the common tangent point of the Earth and the sun to the Earth is,
Figure. 2. Model for calculating the cycle of lunar eclipses
Therefore, for a lunar eclipse to occur, the angle between the extended line connecting the sun and the Earth and the line connecting the moon and the Earth must be less than or equal to 0.692° + 0.253° = 0.945°. For a total lunar eclipse to occur, the angle must be less than or equal to 0.692° - 0.253° = 0.439°. (It is self-evident that the moon must be near the opposite side of the sun because a lunar eclipse is a phenomenon where the Earth’s shadow cast by the sun covers the moon.)
A total lunar eclipse occurred on April 4, 2015. Therefore, to simplify the discussion, let’s assume that on April 4, 2015, the moon was on the line connecting the sun and the Earth. However, since the ecliptic is a great circle passing through the center of the celestial sphere, the opposite side of the sun is also on the ecliptic. Thus, based on the above assumption, the moon is at the node of the ecliptic and the great circle, so the sun is also at the node of the ecliptic and the great circle.
Now, let’s consider the cycle of the next lunar eclipse. Assuming the celestial sphere is a unit sphere, the coordinates of the sun on the celestial sphere can be represented as (cosωSt, sinωSt cosρ, sinωSt sinρ) and the coordinates of the moon can be represented as (-cosωMt, -sinωMt, 0). This can also be expressed in terms of the dot product of the vector corresponding to the point opposite the sun (-cosωSt, -sinωSt cosρ, -sinωSt sinρ) and the vector corresponding to the coordinates of the moon with respect to the origin (Earth). This is left as a suggestion.
To find the cycle more simply, we use the idea of the least common multiple. Given the conditions for a lunar eclipse, the sun, Earth, and moon must be almost in a straight line. This means that the sun and moon must be at opposite nodes of the ecliptic and the great circle. Therefore, the condition for another lunar eclipse to occur after some time T is that the sun has rotated by a half-integer multiple of the ecliptic and the moon has rotated by a half-integer multiple of the great circle. Moreover, the sun and moon must be positioned opposite each other.
According to this logic, 346.6 × 19 = 6585.4 (days) and 27.212 × 242 = 6585.3 (days), which comparatively satisfy the cycle of lunar eclipses. (The term ‘comparatively’ can be proven as suggested.)
This is called the Saros cycle.
Entered : 2015.09.18 20:08