<p>Short answer: 92.5 million km or about 0.619 AU.</p>
<p>Long Answer: First we should note that a thin disk containing all the planet's orbits is not necessarily symmetric about any plane passing through the center of the more massive body.  This should be clear by looking at an example of a highly eccentric orbit like the Molniya orbit. <a href="https://i.sstatic.net/eHOTy.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/eHOTy.png" alt="enter image description here" /></a></p>
<p>The thinnest disk aligned with the equatorial plane that contains a Molniya orbit will be mostly above the equator.  The orbits of the planets around the Sun, of course, aren't nearly as eccentric.  Also, a satellite in a Molniya orbit is farthest from the equatorial plane at apogee, which is not generally the case for natural satellites like planets.</p>
<p>On a <a href="https://astronomy.stackexchange.com/questions/39924/what-are-the-correct-positions-or-alignment-of-planets-in-relation-to-each-other/39993#39993">previous answer</a>, I made a diagram of the distribution of the orbits of the planets along the Sun's equatorial plane.  Here is a diagram of the distributions along the ecliptic plane.<a href="https://i.sstatic.net/h7XVG.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/h7XVG.png" alt="enter image description here" /></a></p>
<p>We can only see seven shapes since the Earth's orbital inclination with respect to the ecliptic is zero.  We can see that Neptune's orbit dominates the width of a containing disk.  The maximum y-value is about 140.06 million km, and the minimum y-value is about -137.61 million km.  So the disk width <span class="math-container">$h$</span> is 277.67 million km, or about 1.86 AU.</p>
<p>I already had point sets for all the orbits, so I ran a search through all possible 3d rotations with a granularity of .1 degrees to find the thinnest disk possible.  An azimuth rotation of 151.6 degrees and an elevation of 1.3 degrees yields another plane in which the maximum width is 92.5 million km or about 0.619 AU.  You can see in the orientation along this plane, the maximum distances from Neptune and Uranus to the plane are equalized.</p>
<p><a href="https://i.sstatic.net/NI2NL.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/NI2NL.png" alt="enter image description here" /></a></p>
<p>Here is a figure of the orbits from the positive z-axis:</p>
<p><a href="https://i.sstatic.net/9sDDW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9sDDW.png" alt="enter image description here" /></a></p>
<p>And here is a projection of the 3-d model onto a 2-D surface with the plane horizontal:</p>
<p><a href="https://i.sstatic.net/SpEXa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/SpEXa.png" alt="enter image description here" /></a></p>
<p>The axis units are in km, but please note that the z-axis scale is smaller, so the inclinations appear exaggerated.</p>

CC BY-SA 4.0
2021-03-14T21:53:58.840Z
Nice work. If you know Python, it's pretty easy to create interactive [3D plots using Sage](https://doc.sagemath.org/html/en/reference/plot3d/). I just posted an example at the end of this old answer: https://astronomy.stackexchange.com/a/28036/16685
PM 2Ring
CC BY-SA 4.0
2021-03-14T22:39:02.257Z
Thanks. I usually use Matlab, but the interactive plots are pretty cool!
Connor Garcia
CC BY-SA 4.0
2021-03-14T20:49:14.393Z
2021-03-18T18:51:02.627Z
2021-03-18T18:51:02.627Z
Connor Garcia
<p>The thickness of the planetary disc is dominated by Neptune, due to its large orbital radius.</p>
<p>We can calculate a planet's maximum distance from the ecliptic <span class="math-container">$h$</span> from the inclination angle of its orbit <span class="math-container">$\theta$</span> and its aphelion distance <span class="math-container">$r$</span>. We get a right triangle, with <span class="math-container">$r$</span> as the hypotenuse, so
<span class="math-container">$$h = r\sin\theta$$</span></p>
<p>The table below was calculated using data from the NASA <a href="https://nssdc.gsfc.nasa.gov/planetary/factsheet/" rel="nofollow noreferrer">Planetary Fact Sheet</a>. Angles are in degrees, distances are in millions of kilometres.</p>
<h3>Planet distance from the ecliptic plane.</h3>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Name</th>
<th>Inclination</th>
<th>Aphelion</th>
<th>Distance</th>
</tr>
</thead>
<tbody>
<tr>
<td>Mercury</td>
<td>7.0</td>
<td>69.8</td>
<td>8.506</td>
</tr>
<tr>
<td>Venus</td>
<td>3.4</td>
<td>108.9</td>
<td>6.458</td>
</tr>
<tr>
<td>Earth</td>
<td>0.0</td>
<td>152.1</td>
<td>0.000</td>
</tr>
<tr>
<td>Mars</td>
<td>1.9</td>
<td>249.2</td>
<td>8.262</td>
</tr>
<tr>
<td>Jupiter</td>
<td>1.3</td>
<td>816.6</td>
<td>18.526</td>
</tr>
<tr>
<td>Saturn</td>
<td>2.5</td>
<td>1514.5</td>
<td>66.062</td>
</tr>
<tr>
<td>Uranus</td>
<td>0.8</td>
<td>3003.6</td>
<td>41.937</td>
</tr>
<tr>
<td>Neptune</td>
<td>1.8</td>
<td>4545.7</td>
<td>142.784</td>
</tr>
</tbody>
</table>
</div>
<p>So the total thickness of the disc is <span class="math-container">$2×142.784 = 285.568$</span> million kilometres, which is almost <span class="math-container">$1.91$</span> au.</p>
<hr />
<p>Here's the Python code I used to create that table:</p>
<pre><code>from math import sin, radians

names = (
 'Mercury', 'Venus', 'Earth',
 'Mars', 'Jupiter', 'Saturn',
 'Uranus', 'Neptune',
)

# Orbit data from https://nssdc.gsfc.nasa.gov/planetary/factsheet/
# Inclination to ecliptic plane
inc = [7.0, 3.4, 0.0, 1.9, 1.3, 2.5, 0.8, 1.8]
# Aphelion in millions of kilometres
aph = [69.8, 108.9, 152.1, 249.2, 816.6, 1514.5, 3003.6, 4545.7]

print(&quot;|Name | Inclination | Aphelion | Distance|&quot;)
print(&quot;|-|-|-|-|&quot;)
for n, th, r in zip(names, inc, aph):
    # Perpendicular distance to eciptic
    h = r * sin(radians(th))
    print(f&quot;|{n} | {th} | {r} | {h:.3f}|&quot;)
</code></pre>
<p>Here's a <a href="https://sagecell.sagemath.org/?z=eJxVUsFu2zAMvesriPQQp_BkJ3bSJMAOA7bDBqwdMGyXoQdOlithtixQ9ICu7r9XUoMNkwCJfBQfnkj2NI0wIhuwo5-IIVhXAmFn0QUhHI46wFsoBKw_a1IzPa5LWH_Xbg7J-IDEZl2mKFJGPs3esqZkfkWeyeXoN8JLxq32PDsd0Y0QV3BHPy1Dh4zQJymG2YdzVbkQOiUfQq-kw4DyYfpd-QGdZqTHqkfFwWjNVaT46NRgHbKdHPAEOnqerYL8XFinov4fN7IuoZFtCXWytvKUjqaEndwn7Jjc432ke-eNHhKXdTDaIZkBph5-2WEaNZMOAr1JnIdTTquPmWy_k9tI157kroTj9iAPCdy2ib-p6yb57b7dy5t7ITxZx8VquY31heW_Lyz_FCzw3gZGp_Sy2vzNeXPZEeongtguNrFlSe8f64vcsjJ6qoQodHMWENcVfNHkteusmgck6C7ErxXLBcvv0scIrtMYFJcpKNhsNjn4qqBfLU_uOYp7YpMvyqc5y6Z_jqpeAIVXtGI=&amp;lang=python" rel="nofollow noreferrer">live version of the script</a> running on the SageMathCell server.</p>
<hr />
<p>As John Holtz mentions in the comments, the true <span class="math-container">$h$</span> value for a planet may be smaller than the value shown in my table. The table's <span class="math-container">$h$</span> value only occurs if the planet's <a href="https://en.wikipedia.org/wiki/Argument_of_periapsisis" rel="nofollow noreferrer">argument of periapsis</a> is ±90°. Fortunately, Neptune's argument of periapsis is currently ~272°, so my <span class="math-container">$h$</span> value should be fairly close to the true value.</p>
<hr />
<p>James K has supplied a list of orbit inclinations to the Solar System's <a href="https://en.wikipedia.org/wiki/Invariable_plane" rel="nofollow noreferrer">invariable plane</a>. Here's the table using those values.</p>
<h3>Planet distance from the Solar System invariable plane.</h3>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Name</th>
<th>Inclination</th>
<th>Aphelion</th>
<th>Distance</th>
</tr>
</thead>
<tbody>
<tr>
<td>Mercury</td>
<td>6.34</td>
<td>69.8</td>
<td>7.708</td>
</tr>
<tr>
<td>Venus</td>
<td>2.19</td>
<td>108.9</td>
<td>4.161</td>
</tr>
<tr>
<td>Earth</td>
<td>1.57</td>
<td>152.1</td>
<td>4.167</td>
</tr>
<tr>
<td>Mars</td>
<td>1.67</td>
<td>249.2</td>
<td>7.262</td>
</tr>
<tr>
<td>Jupiter</td>
<td>0.32</td>
<td>816.6</td>
<td>4.561</td>
</tr>
<tr>
<td>Saturn</td>
<td>0.93</td>
<td>1514.5</td>
<td>24.582</td>
</tr>
<tr>
<td>Uranus</td>
<td>1.02</td>
<td>3003.6</td>
<td>53.468</td>
</tr>
<tr>
<td>Neptune</td>
<td>0.72</td>
<td>4545.7</td>
<td>57.121</td>
</tr>
</tbody>
</table>
</div>
<p>That reduces Neptune's <span class="math-container">$h$</span> considerably! Uranus may even be the current &quot;winner&quot;, depending on their arguments of periapsis with respect to the invariable plane.</p>

CC BY-SA 4.0
2021-03-13T20:22:56.420Z
Amazing. Thanks a ton! You went one step ahead by showing python code that I am trying to formulate.
Majoris
CC BY-SA 4.0
2021-03-13T20:33:37.897Z
@Majoris No worries. I figured from your profile that Python code may be useful to you. ;)
PM 2Ring
CC BY-SA 4.0
2021-03-14T10:00:35.700Z
I wonder if we can also get the second part of the question:  the width of a disc that contains all the planets, but not necessarily parallel to the ecliptic.  It is possible to find inclinations relative to the invariant plane ( `inc =[ 6.34, 2.19 ,1.57, 1.67, 0.32 ,0.93 ,1.02 , 0.72]`)  giving '57.1' million km for neptune, but I'm sure there is a narrower disc , but it would require knowing the 3d orientation of each orbit, not just the inclination.
James K
CC BY-SA 4.0
2021-03-14T10:59:40.220Z
@JamesK I suspect that's what planetmaker intends to do. It's a bit tedious, but I suppose you really only need to do the calculations for the 4 giants.
PM 2Ring
CC BY-SA 4.0
2021-03-14T15:43:47.713Z
Since Neptune is the "deciding factor" and its orbit is close to circular, the following does change the answer. You assumed that the aphelion occurs when at the maximum angle above the ecliptic. That is, the argument of perihelion is at 90 degrees. It is possible that the maximum distance above the ecliptic (r*sin theta) occurs when not at aphelion for an eccentric orbit.
JohnHoltz
CC BY-SA 4.0
2021-03-14T16:19:22.327Z
@JohnHoltz That's a very good point! That NASA Planetary Fact Sheet gives 0.011 for Neptune's eccentricity, but https://nssdc.gsfc.nasa.gov/planetary/factsheet/neptunefact.html has 0.0113. However, Wikipedia has 0.008678, which matches what Horizons prints for the Neptune barycenter on J2000.0 Hmmm...
PM 2Ring
CC BY-SA 4.0
2021-03-14T16:49:14.573Z
@JohnHoltz A Horizons Elements query for the Neptune barycenter  also tells me that its argument of perifocus was around 273° on 1 Jan 2000 (it's now around 272.45°), so my tabulated value of $h$ for Neptune should be fairly close to the true value. Phew! :)
PM 2Ring
CC BY-SA 4.0
2021-03-14T18:01:03.977Z
One interesting tidbit is that neptune's inclination (wrt the invariant plane) is decreasing, while uranus's is increasing. so after a few hundred thousand years, Uranus will be the "deciding factor".
James K
CC BY-SA 4.0
2021-03-14T18:11:40.987Z
@JamesK That *is* interesting. I was surprised that the h value for Uranus is less than Saturn's. It seems odd to me that its inclination is so low, considering how tilted its rotation axis is (& the plane of the Uranian satellites). OTOH, the disturbance that created Saturn's rings is relatively recent, and I guess that whatever happened to Uranus is a lot more ancient.
PM 2Ring
CC BY-SA 4.0
2021-03-14T18:59:32.317Z
Calculations in the invariable plane and around the center of mass/barycenter makes much more sense! True measure of the thickness.
Majoris
CC BY-SA 4.0
2021-03-14T19:15:21.487Z
@Majoris That's true, although as AtmosphericPrisonEscape mentions in a comment on your question, treating the planets as a single disc is a somewhat questionable exercise. ;) I guess using the Solar System barycentre is better than using the Sun. OTOH, the Sun's mean position is centred in the invariable plane. A few months ago I used Horizons to create a 3D plot of the Sun's position relative to the barycentre. Its motion up & down the Z axis is around 1 or 2% of its motion in the plane, IIRC.
PM 2Ring
CC BY-SA 4.0
2021-03-13T14:01:08.047Z
2021-03-14T18:28:54.720Z
2021-03-14T18:28:54.720Z
PM 2Ring
<p>What is the thickness in AU of the all planets' orbital plane height combined in the Solar System?
Excluding Pluto.</p>
<p>Looking for <em><code>h</code></em></p>
<p><a href="https://i.sstatic.net/1Vtk6.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1Vtk6.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-03-13T03:15:37.710Z
Are you looking for the width of the thinnest disk that would contain the orbital planes of all the planets?  The thickness of each individual orbital plane, if you assume Keplerian orbits, is zero.
Connor Garcia
CC BY-SA 4.0
2021-03-13T05:02:55.663Z
Looking for Solar system's thin disk's thickness in AU that all major planets orbit it, upper and lower limit from inclinations e.g mercury is ~7 degree. Exclude Pluto
Majoris
CC BY-SA 4.0
2021-03-13T05:27:19.960Z
Does it have to be exactly parallel to the ecliptic, or can it be tilted slightly to be minimized in thickness?
uhoh
CC BY-SA 4.0
2021-03-13T05:57:19.633Z
@uhoh Good point. I would like to know both then, one exactly parallel to ecliptic as you said. And second since earth also has inclination so may be the thickness from orbits above and below of the invariable plane centered around barycenter.
Majoris
CC BY-SA 4.0
2021-03-13T10:19:33.300Z
There is two ways one might define the thickness: as a (more or less) thin disk of equal height irrespective of radial distance. The other way is to assume a radially increasing height like protoplanetary disk show it ([flared disk][1]), so as to define an angle with respect to the [invariable plane][2] (the plane perpendicular to the total angular momentum vector).


  [1]: https://www.aanda.org/articles/aa/full/2002/45/aa2934/aa2934.html
  [2]: https://www.aanda.org/articles/aa/full_html/2012/07/aa19011-12/aa19011-12.html
I will not have to transform it in decent answer before Monday.
planetmaker
CC BY-SA 4.0
2021-03-13T10:25:58.133Z
@planetmaker Please Try, I will be looking forward.
Majoris
CC BY-SA 4.0
2021-03-13T22:05:09.963Z
@planetmaker That paper is gold. Big thank you!!
Majoris
CC BY-SA 4.0
2021-03-14T13:12:54.160Z
It doesn't make sense to formulate the thickness of a low number of massive objects as a single thickness. The solar system is not a gas disk. Instead, for the solar system and exosystems one often uses an appropriate average over the angles by which the orbital planets of the planets deviate from the systems' midplane.
AtmosphericPrisonEscape
<orbit><solar-system><orbital-mechanics>
What is the thickness of the Solar System disk?
Majoris
# What is the thickness of the Solar System disk?

(Asked by: username_0 on 2021-03-13)

What is the thickness in AU of the all planets' orbital plane height combined in the Solar System?
Excluding Pluto.

Looking for *`h`*

[![enter image description here](https://i.sstatic.net/1Vtk6.png)](https://i.sstatic.net/1Vtk6.png)

**Tags:** orbit, solar-system, orbital-mechanics
> **Comments:**
> **username_1**: Are you looking for the width of the thinnest disk that would contain the orbital planes of all the planets?  The thickness of each individual orbital plane, if you assume Keplerian orbits, is zero.  
> **username_0**: Looking for Solar system's thin disk's thickness in AU that all major planets orbit it, upper and lower limit from inclinations e.g mercury is ~7 degree. Exclude Pluto  
> **username_2**: Does it have to be exactly parallel to the ecliptic, or can it be tilted slightly to be minimized in thickness?  
> **username_0**: @uhoh Good point. I would like to know both then, one exactly parallel to ecliptic as you said. And second since earth also has inclination so may be the thickness from orbits above and below of the invariable plane centered around barycenter.  
> **username_3**: There is two ways one might define the thickness: as a (more or less) thin disk of equal height irrespective of radial distance. The other way is to assume a radially increasing height like protoplanetary disk show it ([flared disk][1]), so as to define an angle with respect to the [invariable plane][2] (the plane perpendicular to the total angular momentum vector).     [1]: https://www.aanda.org/articles/aa/full/2002/45/aa2934/aa2934.html   [2]: https://www.aanda.org/articles/aa/full_html/2012/07/aa19011-12/aa19011-12.html I will not have to transform it in decent answer before Monday.  
> **username_0**: @planetmaker Please Try, I will be looking forward.  
> **username_0**: @planetmaker That paper is gold. Big thank you!!  
> **username_4**: It doesn't make sense to formulate the thickness of a low number of massive objects as a single thickness. The solar system is not a gas disk. Instead, for the solar system and exosystems one often uses an appropriate average over the angles by which the orbital planets of the planets deviate from the systems' midplane.  

---

## 2 answers

### Answer by username_5 on 2021-03-13. Score: 6 | Accepted answer

The thickness of the planetary disc is dominated by Neptune, due to its large orbital radius.
We can calculate a planet's maximum distance from the ecliptic $h$ from the inclination angle of its orbit $\theta$ and its aphelion distance $r$. We get a right triangle, with $r$ as the hypotenuse, so
$$h = r\sin\theta$$
The table below was calculated using data from the NASA Planetary Fact Sheet. Angles are in degrees, distances are in millions of kilometres.
Planet distance from the ecliptic plane.




Name
Inclination
Aphelion
Distance




Mercury
7.0
69.8
8.506


Venus
3.4
108.9
6.458


Earth
0.0
152.1
0.000


Mars
1.9
249.2
8.262


Jupiter
1.3
816.6
18.526


Saturn
2.5
1514.5
66.062


Uranus
0.8
3003.6
41.937


Neptune
1.8
4545.7
142.784




So the total thickness of the disc is $2×142.784 = 285.568$ million kilometres, which is almost $1.91$ au.

Here's the Python code I used to create that table:
from math import sin, radians

names = (
 'Mercury', 'Venus', 'Earth',
 'Mars', 'Jupiter', 'Saturn',
 'Uranus', 'Neptune',
)

# Orbit data from https://nssdc.gsfc.nasa.gov/planetary/factsheet/
# Inclination to ecliptic plane
inc = [7.0, 3.4, 0.0, 1.9, 1.3, 2.5, 0.8, 1.8]
# Aphelion in millions of kilometres
aph = [69.8, 108.9, 152.1, 249.2, 816.6, 1514.5, 3003.6, 4545.7]

print("|Name | Inclination | Aphelion | Distance|")
print("|-|-|-|-|")
for n, th, r in zip(names, inc, aph):
    # Perpendicular distance to eciptic
    h = r * sin(radians(th))
    print(f"|{n} | {th} | {r} | {h:.3f}|")

Here's a live version of the script running on the SageMathCell server.

As John Holtz mentions in the comments, the true $h$ value for a planet may be smaller than the value shown in my table. The table's $h$ value only occurs if the planet's argument of periapsis is ±90°. Fortunately, Neptune's argument of periapsis is currently ~272°, so my $h$ value should be fairly close to the true value.

James K has supplied a list of orbit inclinations to the Solar System's invariable plane. Here's the table using those values.
Planet distance from the Solar System invariable plane.




Name
Inclination
Aphelion
Distance




Mercury
6.34
69.8
7.708


Venus
2.19
108.9
4.161


Earth
1.57
152.1
4.167


Mars
1.67
249.2
7.262


Jupiter
0.32
816.6
4.561


Saturn
0.93
1514.5
24.582


Uranus
1.02
3003.6
53.468


Neptune
0.72
4545.7
57.121




That reduces Neptune's $h$ considerably! Uranus may even be the current "winner", depending on their arguments of periapsis with respect to the invariable plane.

> **Comments:**
> **username_0**: Amazing. Thanks a ton! You went one step ahead by showing python code that I am trying to formulate.  
> **username_5**: @Majoris No worries. I figured from your profile that Python code may be useful to you. ;)  
> **username_6**: I wonder if we can also get the second part of the question:  the width of a disc that contains all the planets, but not necessarily parallel to the ecliptic.  It is possible to find inclinations relative to the invariant plane ( `inc =[ 6.34, 2.19 ,1.57, 1.67, 0.32 ,0.93 ,1.02 , 0.72]`)  giving '57.1' million km for neptune, but I'm sure there is a narrower disc , but it would require knowing the 3d orientation of each orbit, not just the inclination.  
> **username_5**: @JamesK I suspect that's what planetmaker intends to do. It's a bit tedious, but I suppose you really only need to do the calculations for the 4 giants.  
> **username_7**: Since Neptune is the "deciding factor" and its orbit is close to circular, the following does change the answer. You assumed that the aphelion occurs when at the maximum angle above the ecliptic. That is, the argument of perihelion is at 90 degrees. It is possible that the maximum distance above the ecliptic (r*sin theta) occurs when not at aphelion for an eccentric orbit.  
> **username_5**: @JohnHoltz That's a very good point! That NASA Planetary Fact Sheet gives 0.011 for Neptune's eccentricity, but https://nssdc.gsfc.nasa.gov/planetary/factsheet/neptunefact.html has 0.0113. However, Wikipedia has 0.008678, which matches what Horizons prints for the Neptune barycenter on J2000.0 Hmmm...  
> **username_5**: @JohnHoltz A Horizons Elements query for the Neptune barycenter  also tells me that its argument of perifocus was around 273° on 1 Jan 2000 (it's now around 272.45°), so my tabulated value of $h$ for Neptune should be fairly close to the true value. Phew! :)  
> **username_6**: One interesting tidbit is that neptune's inclination (wrt the invariant plane) is decreasing, while uranus's is increasing. so after a few hundred thousand years, Uranus will be the "deciding factor".  
> **username_5**: @JamesK That *is* interesting. I was surprised that the h value for Uranus is less than Saturn's. It seems odd to me that its inclination is so low, considering how tilted its rotation axis is (& the plane of the Uranian satellites). OTOH, the disturbance that created Saturn's rings is relatively recent, and I guess that whatever happened to Uranus is a lot more ancient.  
> **username_0**: Calculations in the invariable plane and around the center of mass/barycenter makes much more sense! True measure of the thickness.  
> **username_5**: @Majoris That's true, although as AtmosphericPrisonEscape mentions in a comment on your question, treating the planets as a single disc is a somewhat questionable exercise. ;) I guess using the Solar System barycentre is better than using the Sun. OTOH, the Sun's mean position is centred in the invariable plane. A few months ago I used Horizons to create a 3D plot of the Sun's position relative to the barycentre. Its motion up & down the Z axis is around 1 or 2% of its motion in the plane, IIRC.  

---

### Answer by username_1 on 2021-03-14. Score: 1

Short answer: 92.5 million km or about 0.619 AU.

Long Answer: First we should note that a thin disk containing all the planet's orbits is not necessarily symmetric about any plane passing through the center of the more massive body. This should be clear by looking at an example of a highly eccentric orbit like the Molniya orbit. [![enter image description here](https://i.sstatic.net/eHOTy.png)](https://i.sstatic.net/eHOTy.png)

The thinnest disk aligned with the equatorial plane that contains a Molniya orbit will be mostly above the equator. The orbits of the planets around the Sun, of course, aren't nearly as eccentric. Also, a satellite in a Molniya orbit is farthest from the equatorial plane at apogee, which is not generally the case for natural satellites like planets.

On a [previous answer](https://astronomy.stackexchange.com/questions/39924/what-are-the-correct-positions-or-alignment-of-planets-in-relation-to-each-other/39993#39993), I made a diagram of the distribution of the orbits of the planets along the Sun's equatorial plane. Here is a diagram of the distributions along the ecliptic plane.[![enter image description here](https://i.sstatic.net/h7XVG.png)](https://i.sstatic.net/h7XVG.png)

We can only see seven shapes since the Earth's orbital inclination with respect to the ecliptic is zero. We can see that Neptune's orbit dominates the width of a containing disk. The maximum y-value is about 140.06 million km, and the minimum y-value is about -137.61 million km. So the disk width $h$ is 277.67 million km, or about 1.86 AU.

I already had point sets for all the orbits, so I ran a search through all possible 3d rotations with a granularity of .1 degrees to find the thinnest disk possible. An azimuth rotation of 151.6 degrees and an elevation of 1.3 degrees yields another plane in which the maximum width is 92.5 million km or about 0.619 AU. You can see in the orientation along this plane, the maximum distances from Neptune and Uranus to the plane are equalized.

[![enter image description here](https://i.sstatic.net/NI2NL.png)](https://i.sstatic.net/NI2NL.png)

Here is a figure of the orbits from the positive z-axis:

[![enter image description here](https://i.sstatic.net/9sDDW.png)](https://i.sstatic.net/9sDDW.png)

And here is a projection of the 3-d model onto a 2-D surface with the plane horizontal:

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

The axis units are in km, but please note that the z-axis scale is smaller, so the inclinations appear exaggerated.

> **Comments:**
> **username_5**: Nice work. If you know Python, it's pretty easy to create interactive [3D plots using Sage](https://doc.sagemath.org/html/en/reference/plot3d/). I just posted an example at the end of this old answer: https://astronomy.stackexchange.com/a/28036/16685  
> **username_1**: Thanks. I usually use Matlab, but the interactive plots are pretty cool!  

---

