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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

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  • $\begingroup$ 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. $\endgroup$ – Connor Garcia Mar 13 at 3:15
  • $\begingroup$ 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 $\endgroup$ – Majoris Mar 13 at 5:02
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    $\begingroup$ Does it have to be exactly parallel to the ecliptic, or can it be tilted slightly to be minimized in thickness? $\endgroup$ – uhoh Mar 13 at 5:27
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    $\begingroup$ @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. $\endgroup$ – Majoris Mar 13 at 5:57
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    $\begingroup$ 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]: aanda.org/articles/aa/full/2002/45/aa2934/aa2934.html [2]: aanda.org/articles/aa/full_html/2012/07/aa19011-12/… I will not have to transform it in decent answer before Monday. $\endgroup$ – planetmaker Mar 13 at 10:19
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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.

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  • $\begingroup$ Amazing. Thanks a ton! You went one step ahead by showing python code that I am trying to formulate. $\endgroup$ – Majoris Mar 13 at 20:22
  • $\begingroup$ @Majoris No worries. I figured from your profile that Python code may be useful to you. ;) $\endgroup$ – PM 2Ring Mar 13 at 20:33
  • $\begingroup$ 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. $\endgroup$ – James K Mar 14 at 10:00
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    $\begingroup$ @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! :) $\endgroup$ – PM 2Ring Mar 14 at 16:49
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    $\begingroup$ Calculations in the invariable plane and around the center of mass/barycenter makes much more sense! True measure of the thickness. $\endgroup$ – Majoris Mar 14 at 18:59
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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

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.enter image description here

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

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

enter image description here

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

enter image description here

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

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