What would be the synodic period of the Moon’s revolution if its movement around the Earth would occur with

What would be the synodic period of the Moon’s revolution if its movement around the Earth would occur with the same speed, but in the opposite direction?

The sidereal period (relative to distant stars) would remain the same T1 = 27.32166 days
The synodic period (relative to the Sun) is now T2 = 29.5306 days. Let’s try to display it. To do this, we denote the angular velocity of the Earth’s revolution around the Sun = w1 = 2pi / 365.25 (in radians per day), and the angular velocity of the Moon around the Earth = w2 = 2pi / 27.32166. Then
w1 * T2 + 2pi = w2 * T2 (overtakes by 2 pi for T2)
T2 = 2pi / (w2-w1) = 1 / (1 / 27.32166 – 1 / 365.25) = 29.5306 days, that is, coincided with the above value. So it remains to twist in the other direction, changing the sign:
1 / (1 / 27.32166 + 1 / 365.25) = 25.4202 days

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