<p>Let me use the following image as reference.</p>

<p><a href="https://i.sstatic.net/vyDLV.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vyDLV.png" alt="enter image description here"></a></p>

<p>As the planet moves, its <strong>linear displacement on its orbit relative to Earth</strong> is <span class="math-container">$s = d \cdot \psi$</span>, where <span class="math-container">$d$</span> is its distance to the sun and <span class="math-container">$\psi = \phi - \tau$</span> is its <strong>orbital angular displacement <span class="math-container">$\phi$</span> minus the Earth's orbital angular displacement</strong>, both seen from Sun.</p>

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

CC BY-SA 4.0
2018-10-28T08:51:48.063Z
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)
James K
CC BY-SA 4.0
2018-10-29T13:16:26.680Z
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!
Brasil
CC BY-SA 4.0
2018-10-29T17:55:09.400Z
I don't really follow that!
Fattie
CC BY-SA 4.0
2018-10-27T17:17:33.580Z
2018-10-29T13:31:00.253Z
2018-10-29T13:31:00.253Z
Brasil
<p>There is no simple formula.</p>

<p>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.</p>

<p>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</p>

<p><a href="https://i.sstatic.net/BTIRK.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BTIRK.jpg" alt="enter image description here"></a></p>

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

<p>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.</p>

<p>There is a <a href="http://stjarnhimlen.se/comp/ppcomp.html" rel="nofollow noreferrer">guide online</a> for writing a computer program to do this. </p>

CC BY-SA 4.0
2018-10-27T22:42:16.197Z
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.
Brasil
CC BY-SA 4.0
2018-10-27T20:21:36.567Z
2018-10-27T20:21:36.567Z
James K
<p>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.</p>

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

<positional-astronomy>
Angular speed of a planet across the sky
Brasil
# Angular speed of a planet across the sky

(Asked by: username_0 on 2018-10-27)

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?**

**Tags:** positional-astronomy


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## 2 answers

### Answer by username_0 on 2018-10-27. Score: 2 | Accepted answer

Let me use the following image as reference.

[![enter image description here](https://i.sstatic.net/vyDLV.png)](https://i.sstatic.net/vyDLV.png)

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.

> **Comments:**
> **username_1**: 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)  
> **username_0**: 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!  
> **username_2**: I don't really follow that!  

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### Answer by username_1 on 2018-10-27. Score: 4

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](https://i.sstatic.net/BTIRK.jpg)](https://i.sstatic.net/BTIRK.jpg)

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](http://stjarnhimlen.se/comp/ppcomp.html) for writing a computer program to do this.

> **Comments:**
> **username_0**: 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.  

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