How to calculate the period of parade of planets?

Question

Astronomy

Is there any formula? In which direction to dig, if the periods of revolution of the planets are given?

Answer 1

The visible parade of planets is called a planetary configuration, when the five bright planets of the solar system (Mercury, Venus, Mars, Jupiter and Saturn) in their movement across the sky come close to each other and become visible at the same time in a small sector (10 - 40 degrees ) sky.

Let's say we know the date when the sun and 5 planets were exactly on the same line, let's figure out when they will be in the same sector again, say 30°:

from itertools import count

n = 12  # число секторов, мы же взяли сектор в 30°
year = 365.256363004
periods = (
    87.969,  # Меркурий
    224.698,  # Венера
    686.98,  # Марс
    4332.589,  # Юпитер
    10759.22  # Сатурн
)
speeds = [n / p for p in periods]  # угловые скорости планет, сектор/день
for t in count(40):  # стартанём не сразу, а погодя, чтобы планеты разошлись,
    # за 40 дней меркурий убежит почти на 180°
    mercury, *others = (int(t * speed) % n for speed in speeds)
    if all(planet == mercury for planet in others):
        y = int(t / year)
        print(f'{y} лет {round(t - y * year)} дней')
        break

Exhaust - 377 years 212 days.
This is a very rough estimate, because I considered the position of the planets as a parade in one sector with the center in the sun, but it should have been in one sector with the center at the observation point, those on Earth. But an honest calculation is much more difficult.
UPDATE
Yep, that's better.
Here is a neater code, it is easy to see that there is no period, and it is not difficult to understand why: the periods of the planets are not a multiple of any extended time.
Well, there is no gap in (2161 - 1982) years - after all, I started from an imaginary moment in the past when the planets were strictly on the same line - and this probably never happened.

n = 360.  # просто 360°
occupied = 45.  # размер сектора, в который должны впихнуться планеты
year = 365.256363004
periods = (
    87.969,  # Меркурий
    224.698,  # Венера
    year,  # Земля
    686.98,  # Марс
    4332.589,  # Юпитер
    10759.22,  # Сатурн
    60190.03,  # Нептун
    # 90553.02  # Плутон # всё равно он не планета
)
speeds = [n / p for p in periods]  # угловые скорости планет
t = t0 = 0  # текущая дата, дата 'нулевого' парада
for _ in range(32):  # число выводимых парадов
    while True:
        t += 1
        if t - t0 < 43:
            continue  # пусть планеты разойдутся, за 43 дня меркурий убежит на 180°


        def width():
            planets = sorted((t * speed) % n for speed in speeds)
            a, max_free_angle = planets[-1] - n, 0.
            for b in planets:
                if max_free_angle < b - a:
                    max_free_angle = b - a
                a = b
            return 360. - max_free_angle


        if width() < occupied:  # подкараулим день, когда сектор будет минимальным
            u = v = width()
            while u >= v:
                t += 1
                u, v = v, width()
            print(f'через {round((t - t0 - 1) / year):>4} лет '
                  f'планеты соберутся в секторе {round(u)}°')
            t0 = t - 1
            break