What is the thickness of the Solar System disk?

Question

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

Looking for h

Answer 1

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.

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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.

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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.

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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.

Answer 2

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.

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, 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.

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.

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

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

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