A problem about the motion of celestial bodies

Note: The shortest time is the time from the west distance to the next east distance.

As can be seen from the figure, the observed planet is an inner planet, and its period of revolution t must be smaller than the period of revolution t of the earth.

How to ask? Find r first, and when r is known, you can get r = r=R*sinθ, t =(sinθ is multiplied by the third power under the root at t). I don't need to say this? )

Planetary angular velocity: 2π/t

Earth angular velocity: 2π/T

Angular velocity difference: (2π/t-2π/T)

Turning angle from east distance to west distance: subtract the sum of two angles from 2π, and the subtracted angle is the complementary angle of θ in the triangle. Just make the following picture yourself.

How long will it take to open such a big angle: divide this angle by the angular velocity difference and it will come out?

I hope you can distinguish right from wrong.