Assuming circular orbit, the evaluation of the angular speed of a planet around the Sun, just requires its orbital period. Similarly, to calculate its angular speed relative to the Earth (also seen from Sun and considering the approximate circular orbit), its synodic period is needed.

The question is: what is the angular speed of an interior planet across the sky in the conjunction with Sun and how can I deduce its formula?


2 Answers 2


There is no simple formula.

First, it's not true that the angular speed of a planet relative to the sun can be found from its orbital period. This is because planets don't move in circles, but in eclipses and the speed of the planet changes, as described by Kepler's second Law.

When describing the motion of a planet in the sky you need to consider the elliptical motion of both the Earth and the planet, the superposition of these motions causes the planets to make looping motions in the sky, sometimes slowing down and moving backwards

enter image description here

The image shows how Mars would have appeared to loop and move backwards in 2003 and 2005.

The motion in the sky can be calculated, but not by a simple formula. You start with the orbital elements of the planets, solve Kepler's equation to determine the position on their elliptical orbit. Solve some trigonometry to determine the position of each planet relative to the sun, then some vector calculations and spherical geometry to find the position of the planet relative to Earth.

There is a guide online for writing a computer program to do this.

  • $\begingroup$ Hi, @JamesK, thank you very much for clarifying everything. I will edit my question so that the answer I proposed will be acceptable. My original Idea was to show the relations between relative and wathced angular speed of planets against the Sun in situations of conjunction specially for interior planets. $\endgroup$
    – Brasil
    Oct 27, 2018 at 22:42

Let me use the following image as reference.

enter image description here

As the planet moves, its linear displacement on its orbit relative to Earth is $s = d \cdot \psi$, where $d$ is its distance to the sun and $\psi = \phi - \tau$ is its orbital angular displacement $\phi$ minus the Earth's orbital angular displacement, both seen from Sun.

From Earth, we see the same linear displacement, but given by $s = D \cdot \theta$, where $D$ is our distance to the Sun and $\theta$ is the angular distance we watched the planet moving. Therefore $D \cdot \theta = d \cdot \psi$, and therefore, the angular speed $\dot{\theta}$ across the sky is simply $\dot{\theta} = d \cdot \dot{\psi}/D$, where $\dot{\psi}$ is the angular speed of the planet relative to Earth, i.e., $\dot{\psi} = 2\pi/T_\mathrm{s}$, where $T_\mathrm{s}$ is the synodic period of the planet.

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    $\begingroup$ This appears to assume that the Earth is stationary. Are you doing the calculations in a rotating frame of reference? Should D be the Earth-planet distance? Are you assuming circular orbits ( a good approximation for Venus, less good for Mercury) $\endgroup$
    – James K
    Oct 28, 2018 at 8:51
  • $\begingroup$ Hi, @JamesK. Yes, I am doing simplified calculations assuming circular orbits just for the sake of visualization of the configurations. More precise calculus would require Kepler's laws analysis. Here, the Earth is not stationary, but the frame of reference to calculate the speed $\dot{\theta}$. I will edit the answer to make it clear that the speed $\dot{\psi}$ is actually the one associated to the synodic period of the planet. Thank you! $\endgroup$
    – Brasil
    Oct 29, 2018 at 13:16
  • $\begingroup$ I don't really follow that! $\endgroup$
    – Fattie
    Oct 29, 2018 at 17:55

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