# 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