<p><a href="https://en.wikipedia.org/wiki/Solar_System#Distances_and_scales" rel="nofollow noreferrer">Wikipedia</a> has a nice planetary distribution picture, but it is in 1-dimension and based off aphelion (furthest position to the sun) and perihelion (closest position to the sun) only.</p>
<p><a href="https://i.sstatic.net/ogQST.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ogQST.png" alt="enter image description here" /></a></p>
<p>Unfortunately, this picture doesn't take into account the planets' orbital inclinations to the Sun's equatorial plane.  Nor does it take into account that the perihelion and aphelion are on opposite sides of the Sun's plane for all 8 planets.</p>
<p>Can a more accurate 2-D representation of this distance picture be produced using Keplerian orbital elements?</p>
<p>We start by calculating the semi-minor axis of the ellipse by <span class="math-container">$b=a\sqrt{1-e^2}$</span> where <span class="math-container">$a$</span> is the semi-major axis and <span class="math-container">$e$</span> is the eccentricity.  Then we use the standard formula for an ellipse <span class="math-container">$x^2/a^2+y^2/b^2 = 1$</span> to generate a point set.  We can put the focus at the origin by translating the ellipse in the negative <span class="math-container">$x$</span> direction by <span class="math-container">$a-p$</span> where <span class="math-container">$p$</span> is the distance at perihelion.  Then we need to rotate the point set counterclockwise by <span class="math-container">$\omega-90$</span> degrees, where <span class="math-container">$\omega$</span> is the argument of perihelion (periapsis), to put the ellipse in the reference plane pictured here.</p>
<p><a href="https://i.sstatic.net/0bQ37.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/0bQ37.png" alt="enter image description here" /></a></p>
<p>We rotate the ellipse up into 3 dimensions using the inclination <span class="math-container">$i$</span> with the coordinate transformation: <span class="math-container">$x'=x\cos(i)$</span>, <span class="math-container">$y'=y$</span>, and <span class="math-container">$z'=x\sin(i)$</span>.</p>
<p>Once we have the points of the orbit ellipse calculated we can plot them like the above wikipedia picture with another coordinate transformation as <span class="math-container">$x''=\sqrt{x'^2+y'^2}$</span> and <span class="math-container">$y''=z'$</span>, with the sun at the origin, and the sun's equator aligned with the x-axis and units in km.  This is the more accurate 2-D representation of the Wikipedia picture at the beginning of this answer.  Each planet always lies somewhere on it's corresponding line.</p>
<p><a href="https://i.sstatic.net/VHtug.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/VHtug.png" alt="enter image description here" /></a></p>
<p>And here is a 'zoom in' on the inner four planets:</p>
<p><a href="https://i.sstatic.net/3HAhI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3HAhI.png" alt="enter image description here" /></a></p>
<p>Is there an equal chance that a randomly sampled planet position on one of these orbit will lie in an equal length region of this distribution (i.e. is this a uniform probabilistic distribution)?  No, since Kepler’s second law says that orbits sweep out equal areas in equal time, it’s more likely to land in a region further to the sun. Since the eccentricities of planets in our solar system are low, this effect is small.</p>
<p>If we want to imagine the planetary position distributions in 3-Dimensions we only need to rotate the above image around the sun's axis.  The result will be concentric <a href="https://en.wikipedia.org/wiki/Toroid" rel="nofollow noreferrer">toroids</a>, like the one pictured here from a square.<a href="https://i.sstatic.net/N9WDx.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/N9WDx.png" alt="enter image description here" /></a></p>
<p>Notes:</p>
<ol>
<li><p>The above is a first order approximation using Keplerian techniques.  If we include barycenters in our calculations for more accuracy, the probability surfaces in 3-D will thicken (on the order of 10^5-10^6 km) along the general direction of the sun.</p>
</li>
<li><p>Some of these orbit shapes are pretty weird looking zoomed in,  Check out Neptune, which kind of looks like a Nike Swoosh.  I've never seen these shapes generated before.</p>
</li>
</ol>
<p><a href="https://i.sstatic.net/eODTO.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/eODTO.png" alt="enter image description here" /></a></p>
<ol start="3">
<li>I can provide my matlab code for generating these plots on request.</li>
</ol>

CC BY-SA 4.0
2020-11-22T23:50:21.480Z
2020-11-23T02:35:15.173Z
2020-11-23T02:35:15.173Z
Connor Garcia
<p>It is impossible to have the planets line up like in your image. This is done for illustration purposes only. If such an image was to scale, either the bodies would be extremely small and impossible to see, or the image would be extremely large and impractical to use.</p>
<p>For example, the Sun is about 1,392,000 km in diameter. The Earth is about 12,756 km in diameter (so roughly 110x smaller), and orbits at an average distance of roughly 149,600,000 km. If the Sun was a 40 cm ball, the Earth would be about 3.6 mm in diameter and located roughly 43 m away from the Sun-ball. At that same scale, Neptune is about 1,3 km away. (Before COVID, I regularly hosted “Solar System Walks” for the city of Montréal, and that was the size of my scale model.)</p>
<p>Now to get back to “lining up” planets… Let’s define a reference plane: the ECLIPTIC is the plane of Earth’s orbit around the Sun. All the other planets have orbits tilted with respect to that plane. For example, Mercury’s orbit is tilted 7° to the ecliptic, and Venus’ orbit 3.4°.</p>
<p>So there IS a point on a planet’s orbit where it lies on the same plane as the Earth—it’s called a NODE, and there’s an ASCENDING NODE, where the planet crosses from “below” (south of) the ecliptic to “above” (north of) it as well as a DESCENDING NODE where the planet goes from above to below.</p>
<p>HOWEVER…</p>
<p>1-The Earth is not necessarily in line with the Sun and the planet at that moment (explaining, for example, how the passage of Venus in front of the Sun is such a rare event [last ones were in 2004 and 2012; next ones will be in 2117 and 2125]);</p>
<p>2-Planets don’t [necessarily] cross the ecliptic plane as the same time as other planets; and</p>
<p>3-The imaginary line between a planet’s two nodes is not [necessarily] coincident with that of another planet. For example, Mercury’s ascending node is at an ecliptic longitude of 48.33° while Venus’ ascending node is at 76.68°.</p>
<p>4-Finally, depending on other specifics of each planet’s orbits (its so-called “orbital elements”), a planet’s orbit is not necessarily divided exactly between “above” the ecliptic and “below” it.</p>
<p>So… To answer your question specifically… The planets are indeed in “weird” positions and not in line, but their specific positions “above” or “below” the Earth–Sun line change with time, and it’s basically impossible for them to be in the plane of the ecliptic at the same time, even roughly speaking.</p>
<p>(Addendum: It applies to planetary alignments “seen from above,” but Jean Meeus has an excellent explanation of the impossibility of aligning more than 2 planets at the same time in his book <em>Mathematical Astronomy Morsels</em> [chapter 31].)</p>

CC BY-SA 4.0
2020-11-20T04:23:32.583Z
+1 vote! Welcome to the astronomy stack exchange! Your addendum to this answer really intrigues me, do you think you could use it to answer this question: https://astronomy.stackexchange.com/questions/39754/occultation-of-planets-by-other-planets ?
Connor Garcia
CC BY-SA 4.0
2020-11-20T13:15:51.570Z
Nice answer! 3.4° doesn't sound like much, but the sine of that angle is roughly 1/17, so at aphelion Venus is more than 6.4 million km from the ecliptic. That's around 530× its diameter.
PM 2Ring
CC BY-SA 4.0
2020-11-20T03:40:44.527Z
2020-11-20T03:40:44.527Z
Pierre Paquette
<p>From a lot a illustrations and videos, the planets seem to align in vertical center to each other like this:</p>
<p><img src="https://i.pinimg.com/600x315/4c/fc/01/4cfc01f1515122c6132fc13fabde6d76.jpg" alt="illustration" /></p>
<p>Is this really true? I like to imagine that the planets are actually in weird alignment to each other, for example, maybe Venus is 50,000Km below earths bottom etc</p>

CC BY-SA 4.0
2020-11-18T16:31:40.120Z
See https://en.wikipedia.org/wiki/Ecliptic#Plane_of_the_Solar_System & https://astronomy.stackexchange.com/questions/tagged/ecliptic
PM 2Ring
<planet><solar-system><positional-astronomy><planetary-systems><alignment>
What are the correct positions or alignment of planets in relation to each other and the Sun, is it vertical center, baseline or completely random
ifeoluwa king
# What are the correct positions or alignment of planets in relation to each other and the Sun, is it vertical center, baseline or completely random

(Asked by: username_0 on 2020-11-18)

From a lot a illustrations and videos, the planets seem to align in vertical center to each other like this:

![illustration](https://i.pinimg.com/600x315/4c/fc/01/4cfc01f1515122c6132fc13fabde6d76.jpg)

Is this really true? I like to imagine that the planets are actually in weird alignment to each other, for example, maybe Venus is 50,000Km below earths bottom etc

**Tags:** planet, solar-system, positional-astronomy, planetary-systems, alignment

> **Comments:**
> **username_1**: See https://en.wikipedia.org/wiki/Ecliptic#Plane_of_the_Solar_System & https://astronomy.stackexchange.com/questions/tagged/ecliptic  

---

## 2 answers

### Answer by username_3 on 2020-11-20. Score: 8 | Accepted answer

It is impossible to have the planets line up like in your image. This is done for illustration purposes only. If such an image was to scale, either the bodies would be extremely small and impossible to see, or the image would be extremely large and impractical to use.

For example, the Sun is about 1,392,000 km in diameter. The Earth is about 12,756 km in diameter (so roughly 110x smaller), and orbits at an average distance of roughly 149,600,000 km. If the Sun was a 40 cm ball, the Earth would be about 3.6 mm in diameter and located roughly 43 m away from the Sun-ball. At that same scale, Neptune is about 1,3 km away. (Before COVID, I regularly hosted “Solar System Walks” for the city of Montréal, and that was the size of my scale model.)

Now to get back to “lining up” planets… Let’s define a reference plane: the ECLIPTIC is the plane of Earth’s orbit around the Sun. All the other planets have orbits tilted with respect to that plane. For example, Mercury’s orbit is tilted 7° to the ecliptic, and Venus’ orbit 3.4°.

So there IS a point on a planet’s orbit where it lies on the same plane as the Earth—it’s called a NODE, and there’s an ASCENDING NODE, where the planet crosses from “below” (south of) the ecliptic to “above” (north of) it as well as a DESCENDING NODE where the planet goes from above to below.

HOWEVER…

1-The Earth is not necessarily in line with the Sun and the planet at that moment (explaining, for example, how the passage of Venus in front of the Sun is such a rare event [last ones were in 2004 and 2012; next ones will be in 2117 and 2125]);

2-Planets don’t [necessarily] cross the ecliptic plane as the same time as other planets; and

3-The imaginary line between a planet’s two nodes is not [necessarily] coincident with that of another planet. For example, Mercury’s ascending node is at an ecliptic longitude of 48.33° while Venus’ ascending node is at 76.68°.

4-Finally, depending on other specifics of each planet’s orbits (its so-called “orbital elements”), a planet’s orbit is not necessarily divided exactly between “above” the ecliptic and “below” it.

So… To answer your question specifically… The planets are indeed in “weird” positions and not in line, but their specific positions “above” or “below” the Earth–Sun line change with time, and it’s basically impossible for them to be in the plane of the ecliptic at the same time, even roughly speaking.

(Addendum: It applies to planetary alignments “seen from above,” but Jean Meeus has an excellent explanation of the impossibility of aligning more than 2 planets at the same time in his book *Mathematical Astronomy Morsels* [chapter 31].)

> **Comments:**
> **username_2**: +1 vote! Welcome to the astronomy stack exchange! Your addendum to this answer really intrigues me, do you think you could use it to answer this question: https://astronomy.stackexchange.com/questions/39754/occultation-of-planets-by-other-planets ?  
> **username_1**: Nice answer! 3.4° doesn't sound like much, but the sine of that angle is roughly 1/17, so at aphelion Venus is more than 6.4 million km from the ecliptic. That's around 530× its diameter.  

---

### Answer by username_2 on 2020-11-22. Score: 5

[Wikipedia](https://en.wikipedia.org/wiki/Solar_System#Distances_and_scales) has a nice planetary distribution picture, but it is in 1-dimension and based off aphelion (furthest position to the sun) and perihelion (closest position to the sun) only.

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

Unfortunately, this picture doesn't take into account the planets' orbital inclinations to the Sun's equatorial plane. Nor does it take into account that the perihelion and aphelion are on opposite sides of the Sun's plane for all 8 planets.

Can a more accurate 2-D representation of this distance picture be produced using Keplerian orbital elements?

We start by calculating the semi-minor axis of the ellipse by $b=a\sqrt{1-e^2}$ where $a$ is the semi-major axis and $e$ is the eccentricity. Then we use the standard formula for an ellipse $x^2/a^2+y^2/b^2 = 1$ to generate a point set. We can put the focus at the origin by translating the ellipse in the negative $x$ direction by $a-p$ where $p$ is the distance at perihelion. Then we need to rotate the point set counterclockwise by $\omega-90$ degrees, where $\omega$ is the argument of perihelion (periapsis), to put the ellipse in the reference plane pictured here.

[![enter image description here](https://i.sstatic.net/0bQ37.png)](https://i.sstatic.net/0bQ37.png)

We rotate the ellipse up into 3 dimensions using the inclination $i$ with the coordinate transformation: $x'=x\cos(i)$, $y'=y$, and $z'=x\sin(i)$.

Once we have the points of the orbit ellipse calculated we can plot them like the above wikipedia picture with another coordinate transformation as $x''=\sqrt{x'^2+y'^2}$ and $y''=z'$, with the sun at the origin, and the sun's equator aligned with the x-axis and units in km. This is the more accurate 2-D representation of the Wikipedia picture at the beginning of this answer. Each planet always lies somewhere on it's corresponding line.

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

And here is a 'zoom in' on the inner four planets:

[![enter image description here](https://i.sstatic.net/3HAhI.png)](https://i.sstatic.net/3HAhI.png)

Is there an equal chance that a randomly sampled planet position on one of these orbit will lie in an equal length region of this distribution (i.e. is this a uniform probabilistic distribution)? No, since Kepler’s second law says that orbits sweep out equal areas in equal time, it’s more likely to land in a region further to the sun. Since the eccentricities of planets in our solar system are low, this effect is small.

If we want to imagine the planetary position distributions in 3-Dimensions we only need to rotate the above image around the sun's axis. The result will be concentric [toroids](https://en.wikipedia.org/wiki/Toroid), like the one pictured here from a square.[![enter image description here](https://i.sstatic.net/N9WDx.png)](https://i.sstatic.net/N9WDx.png)

Notes:

1. The above is a first order approximation using Keplerian techniques. If we include barycenters in our calculations for more accuracy, the probability surfaces in 3-D will thicken (on the order of 10^5-10^6 km) along the general direction of the sun.
2. Some of these orbit shapes are pretty weird looking zoomed in, Check out Neptune, which kind of looks like a Nike Swoosh. I've never seen these shapes generated before.

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

3. I can provide my matlab code for generating these plots on request.


---

