Regarding the calculation of the planetary stationary period, we have already discussed this issue using the method of calculus. Perhaps many friends who do not yet have a foundation in higher mathematics will feel dizzy, so here I provide a derivation discussed from a geometric perspective.
Regarding the "station" (stationary point), many people think it is the moment when the planet’s velocity relative to the Earth is zero. In fact, this statement is slightly inaccurate. Strictly speaking, "velocity" should be changed to "angular velocity" or "tangential velocity" (in astronomy, tangential refers to the direction perpendicular to the line of sight). In actual motion, there is no single moment when the planet’s movement velocity relative to the Earth is zero. Based on this statement, we can create the following diagram (still considering only uniform circular motion):
And list the following: V_p \cos\varphi = V_e \cos(\theta+\varphi) = V_e \cos\theta \cos \varphi - V_e \sin\theta \sin\varphi i.e., V_p = V_e \cos \theta - V_e \tan \varphi \sin\theta At the same time, according to the diagram, we can list: \tan\varphi = \frac{r_e \sin\theta}{r_p - r_e \cos\theta}. Substituting this into the above equation, we get: \begin{aligned} V_p(r_p - r_e \cos\theta) &= V_e r_p \cos\theta - V_e r_e \\ V_p r_p + V_e r_e &= (V_p r_e + V_e r_p) \cos\theta \end{aligned} Using inverse trigonometric functions, we get: \theta = \arccos\left(\frac{V_p r_p + V_e r_e}{V_p r_e + V_e r_p}\right) During the period within an angle \theta before and after opposition, the planet is in a state of retrograde motion; therefore, the total retrograde duration is: T = \frac{2\theta}{\omega_e - \omega_p}
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