<p>For a planet with another planet at their <a href="https://en.wikipedia.org/wiki/Lagrange_point#L4_and_L5_points" rel="noreferrer">Lagrange L4 or L5 point</a>, those would be each others' closest neighbours; as you can see from the simulations, Mercury has an average distance a few % more than the average distance between the Sun and the planet (e.g. 1.04 AU for Earth), and the Lagrange points form equilateral triangles with the largest planet and the Sun, so for Earth that would be at a (constant) distance of 1 AU.</p>
<p>Of course, whether the smallest of the two would count as a planet is <a href="https://en.wikipedia.org/wiki/Clearing_the_neighbourhood" rel="noreferrer">debatable</a>. They could be quite large, e.g. Jupiter (317.8 Earth masses) could have a Trojan weighing 317.8 / 24.96 = 12.7 Earth masses.</p>

CC BY-SA 4.0
2021-11-01T13:26:45.723Z
Yep, depends on whether you allow objects at a L4 or L5 to be called planets rather than moons (or even something less "major" )
Carl Witthoft
CC BY-SA 4.0
2021-11-01T16:25:30.717Z
A planet in the L4 or L5 point of a bigger planet has not cleared its neighbourhood and can be a dwarf planet at most.
Pere
CC BY-SA 4.0
2021-11-01T16:36:50.453Z
Do you have a source/explanation for the possible Jupiter Trojan of 12.7 Earth masses? Very interesting (especially for sci-fi ideas), though I don't understand the math you've done
Drake P
CC BY-SA 4.0
2021-11-01T16:46:59.557Z
@DrakeP 24.96 is the minimum mass ratio: https://en.wikipedia.org/wiki/Lagrange_point#cite_ref-exact_stability_threshold_7-0
Glorfindel
CC BY-SA 4.0
2021-11-01T16:49:38.873Z
@Pere exactly, that's why I linked to the Wikipedia article about [Clearing the neighbourhood](https://en.wikipedia.org/wiki/Clearing_the_neighbourhood). Pluto-fanatic might disagree with that. I don't have a particular opinion on this matter.
Glorfindel
CC BY-SA 4.0
2021-11-03T20:34:49.993Z
@Glorfindel I think you're reading it incorrectly.  The M1 is the main body (Sol) and M2 is the secondary body (Jupiter).  The ratio of 24.96 is required for there to be L4 and L5 points at all, it doesn't describe the mass of objects that could be in a stable orbit there. "*indeed, the third body need to have negligible mass*"
Jason Goemaat
CC BY-SA 4.0
2021-11-03T20:49:15.003Z
@JasonGoemaat ... you're right, thanks for pointing out! Any idea where we can find a maximum mass / minimum ratio for M2/M4?
Glorfindel
CC BY-SA 4.0
2021-11-01T10:40:17.913Z
2021-11-01T10:40:17.913Z
Glorfindel
<p>As has been mentioned earlier, the fully general version of this question requires rather tricky analysis to deal with orbits that aren't circular and coplanar. It's even trickier if you want to include the gravitational interactions between the planets, although in most cases those small perturbations won't be large enough to affect which planet has the smallest mean distance to a given planet.</p>
<p>Fortunately, for our Solar System, we can use <a href="https://ssd.jpl.nasa.gov/horizons/" rel="nofollow noreferrer">JPL Horizons</a> to find distances between the planets. I'll show some results derived from Horizons for the mean distance between Earth &amp; Mercury and Earth &amp; Venus later in this answer. But first, I'd like to give a brief analysis of the simple coplanar circular orbits version of this problem.</p>
<p>In our simple flat circular solar system, the planets all orbit the sun S in the same direction. The orbit of our &quot;home&quot; planet H has radius 1 and period 1. The planets obey Kepler's laws, so they move around S with constant orbital speed. A planet P with an orbital radius of <span class="math-container">$r$</span> has a period of <span class="math-container">$T=r^{3/2}$</span>.</p>
<p><a href="https://i.sstatic.net/mEHdJ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/mEHdJ.png" alt="circular orbits" /></a></p>
<p>Relative to H, P has a <a href="https://en.wikipedia.org/wiki/Orbital_period#Synodic_period" rel="nofollow noreferrer">synodic period</a> of
<span class="math-container">$$T_s = \frac T{|T-1|}$$</span></p>
<p>Let the angle at HSP be <span class="math-container">$\theta$</span>. Then <span class="math-container">$$\theta=\frac{2\pi t}{T_s}$$</span>
where <span class="math-container">$t$</span> is time. So to find the mean distance between H &amp; P we need to average over the synodic period.</p>
<p>Let HP be <span class="math-container">$s$</span>. Then by the cosine rule,
<span class="math-container">$$s^2 = 1 + r^2 - 2r\cos\theta$$</span>
Therefore, the mean distance <span class="math-container">$m$</span> between H &amp; P is
<span class="math-container">$$m = \frac1{2\pi}\int_0^{2\pi} \sqrt{1 + r^2 - 2r\cos\theta}\,d\theta$$</span></p>
<p>That's not an easy integral, but it can be converted to a <a href="https://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_second_kind" rel="nofollow noreferrer">complete elliptical integral of the second kind</a>.</p>
<p>Using Sage, we can plot <span class="math-container">$s$</span> and compute <span class="math-container">$m$</span> for a few values of <span class="math-container">$r$</span>, over two synodic cycles. The blue dotted horizontal line shows the mean distance.</p>
<p><a href="https://i.sstatic.net/Xuu5v.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Xuu5v.gif" alt="animated plot of s" /></a></p>
<p>When <span class="math-container">$r=0$</span>, <span class="math-container">$m$</span> has its minimum value of 1, and <span class="math-container">$m$</span> grows as <span class="math-container">$r$</span> does. For small <span class="math-container">$r$</span>, the graph of <span class="math-container">$s$</span> is <em>almost</em> a sine wave ranging from <span class="math-container">$1-r$</span> to <span class="math-container">$1+r$</span>. But as <span class="math-container">$r$</span> grows, we see that the lower part of the graph is decidedly more &quot;pointy&quot; than the upper part, which shifts the mean value upwards. The pointiness reaches a peak at <span class="math-container">$r=1$</span>, after which it gradually diminishes. (Note that <span class="math-container">$r=1$</span> means that H &amp; P are in the same orbit, thus <span class="math-container">$T_s$</span> is infinite, so the <span class="math-container">$r=1$</span> frame isn't really possible, it's just the limiting case).</p>
<p>You can use <a href="https://sagecell.sagemath.org/?z=eJxVkEFrwzAMhe_9FSKX2qnXJYHuEAgMdh6ENbfSFjdxOoNjZ7Kz0Y3-98ntCqsvtp6ePj0coIJPiWwe5nz2rG1QKNsw61QPe4bVi7NB28lNfm10p5CxTMCKC_BBjV5_qypb5gLILicTqFhxXs6ATk9g_4GB5bAA3BXwAEWKaes8K9JRp4Hzi28gn50GhbqVZh8DHFEa1guIq3LON9n2YhyRmix5k52efJkIQP5ff1XSkjhcxYaouMuXq2vlqWzgEeTBs4aS5HejNS13HQ03d_L6ZF2n26j7a6MmzGhcuKUr6COCDkZV_S0X_OBZQAxDz6FcPvVnAhxRd0Zb5atk0NZhcuMtKog620TcQDhWxHsrLrIPJ2In5Z_dv7svVvNfJLt4Rg==&amp;lang=sage" rel="nofollow noreferrer">this script</a> to generate a static plot for smallish values of <span class="math-container">$r$</span>.</p>
<p>Here's a plot of <span class="math-container">$m$</span> vs <span class="math-container">$r$</span>, computed using elliptic integrals.</p>
<p><a href="https://i.sstatic.net/5gvSW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/5gvSW.png" alt="general mean distance plot" /></a></p>
<p>And here's the <a href="https://sagecell.sagemath.org/?z=eJxNzbEOgjAYBOCdp7h0akkRQV1M-ggurkZJhR_zSymkhUR9etE4eNPlyyXXUItWBrVPsKSDwTYNuQzIUKhL-dVA0xw87DX-HCnIOR4nriuqZdZ9pESOkZNkdMMkWw251tgojWfP3izdPihWzl7JRXMSR9vwHIWGOJD1aDhO1tckznq5_JtWkV9kilWx07gFbhx7ikb09j4Eod5KsDee&amp;lang=sage" rel="nofollow noreferrer">plotting script</a></p>
<hr />
<p>As promised, here's a Horizons plot of the Earth-Mercury distance, over 12 synodic cycles. The starting &amp; ending points are close to local maxima. The time interval is divided into 1390 equal steps (which are slightly less than 1 day).</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th></th>
<th>Julian day number</th>
<th>Date &amp; time</th>
</tr>
</thead>
<tbody>
<tr>
<td>From:</td>
<td>2458159.229166667</td>
<td>A.D. 2018-Feb-09 17:30:00.0000</td>
</tr>
<tr>
<td>To:</td>
<td>2459549.125000000</td>
<td>A.D. 2021-Nov-30 15:00:00.0000</td>
</tr>
</tbody>
</table>
</div>
<p>Mean distance = 154762296 km</p>
<p><a href="https://i.sstatic.net/9GMjg.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9GMjg.png" alt="Earth-Mercury distance" /></a></p>
<p>And here's the Earth-Venus distance, over 12 synodic cycles. The time interval is divided into 879 equal steps (slightly less than four days).</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th></th>
<th>Julian day number</th>
<th>Date &amp; time</th>
</tr>
</thead>
<tbody>
<tr>
<td>From:</td>
<td>2456953.250000000</td>
<td>A.D. 2014-Oct-22 18:00:00.0000</td>
</tr>
<tr>
<td>To:</td>
<td>2460466.604166667</td>
<td>A.D. 2024-Jun-05 02:30:00.0000</td>
</tr>
</tbody>
</table>
</div>
<p>Mean distance = 170272478 km</p>
<p><a href="https://i.sstatic.net/1jlNw.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1jlNw.png" alt="Earth-Venus distance" /></a></p>
<p>Mercury's orbit is rather eccentric, so there's quite a variation in the heights of the maxima &amp; minima, and in the length of the synodic cycle. To get a more accurate value for the mean distance we should really average over a lot more cycles.</p>
<p>The eccentricity of Venus's orbit is quite low, so its graph is much closer to the simple circular orbit plots above.</p>
<p>Although the mean distance of Mercury is certainly smaller than the mean distance of Venus, by 15510182 km, clearly Venus has the lower minimum distance to Earth.</p>
<hr />
<p>Those last two plots were created using this <a href="https://sagecell.sagemath.org/?z=eJx1V3tz2kgS_59PMVFyi5SALGC9t-aKusWOXOuseRSw2cphSjWgAeToddIIQ1z-7tfdIwnJ3qMKpOnX9OPXPYOmaWwmZOKJg2AJD3eCuVxytk2igP0eJd6PKEzZNkoYZ5InOyHZOnJPDQYfHrrwZdE6FcnBC3dsI0IpEqLHfiSZJ03WING_Ek9KEbL1iU1HrDtD6a7V7ZjWldm5amjgRaPhBXGUSJaIFnz_m4lUpg3yY5uFGxlFfspyET_JnA3f7AVoZYnPBkzbSxmn_YuLNHXNx9g3Q55ycxcdLnjsXezzQJyt5wsTKFoDfpyDSFIvClEdPEEX1jwVziZwkaRpjdHwD9uxp7_bIyA0v9nzZmMxvL63ncW3qY2kr_bNYjID8uTPhfPn-G4xR-ofo_bnZmNm3zrT--GYBO2b-7vp4u5Gkeff5gtl80vXsqxmA-04N5PZDGnjydgmknM_vLbv54qmKJ_t-8XQWRSkm_lX53YyGw0XpYOT6y_O5-FiWNUip5HwSzPP9W9lCvWAH1Pvhxh4odR7htFwxZZthdzsHUSCrqreyovbYikQJD6iGH9FbPSpxCprW0zbuw8f5pPbxnORzZfGzWQ0Go4_owvPyuALOG-PFzZF_KyMA22-GM4WzuJuRO4-02ZEnkyr1Cgmoj115nf_KYgCibC3DXurKNEvgBLwC0CZcZRKHTDTIpwPnpuA7YDLZp81pTjK5kuLIUhS4HhhnCFDb0IMzRZGaLwYZPSMnESYqeDJZq8n2nB6x77as_ndZNx_SD_qD_OPhkZgNtG2Ye6SKIv1jrLhbUsz7wasgkeVT_zECRZlq_3FkxBaps9wh0LJS9lz_g5OrzPJ5B5o6SbxYolc7Nrnit0XzchzErxyXP93ajx8eICq2br5kV7tiV11vfA4QMPjKBSvndTGE0adYs_u5prxilu3o8oisyQkU0TzvVCkqlAkaaax70mi6kor3-ch1MzHyAt14i0v-51fVobRqNgMyjwvO_0Vgt1DdPGNJGwHHJQVCAda5-pKK7A90C4t67ceUsAcQW-gda3Or-1bsW5fsc4_-z1LU9BHRrfTHkeHds9incu-pTgCOJ3elUUmAGGCSD2rxaQXAKpuuZ_CfHMjmQ4WSQavmyw5iOI98iNwIxGucgHbstdtsSfPlfvBz7DBYVeY4JmMnCzGPRQp70MoUgizV8ctGCCA7BtvyjUXvthIxmGYCp5KBnVg0ZbVtLQ35VLwoRNi8HZImCkcJLFuFAkt17kZSulZCPNYXYlyVcKNdvq_iIuYiPciEAlIqFMrykL33Ru3q07j4y22cuYyVSy92Woay367s6ImgnMnJJEVCb8nYhI9FeR-gfLoyfhXPVXv2RRZKnT2E8XMvmS-x1H1lAKJUFJF-C2ceX1A00e0vbRWy353VWuBRdRnrBQAL0ki31E6MOKg5fFUXK6K6Ki38iV677WKAESYQQZBQkdrFaC8Z_ZRYtewDfdF6PKEDJV8oOIaezZ6grmKDZf7UFc_h9uCNt_tZRt7oaWuG_mjnVRNP7ot5gAHGiVBv7d-xKWeGudqwJarUrwMGY72GDzVdTRA2kbdpRH_LlQnSh7EackBqHnsH2XDssGAWedM4OcUgMl9gGnMA8-hgkOmt6rLtlgAm4MoaOViWrsCS3IaTbnsE4zUT-y07PbrNsAj8rPuRa74Ca4nMAdBETwqNjABETDyrLodCqmWF5mWSXEBJQT9AqzXmee77Fr88EQ-A1SOZHwAbz3oab1MtgpHwmg7wfdggUAIg1sHWcVS-mfcIeBgoSsVQxF5IrDzrBKZcLeA_B2wyGDpHL4rQUyyNux4ph0xh5JdsN6Zht64J2CASx9BBI3RswZNBGYeJ7pQ8sgfyK9-srAu8NXBImzrngzYp2ugLdyxWwUPhVovFTYFpAPTjiGDIXSFrIJBGHg6BkM0ZR0psDBWdZyoLBYVxBHzeptzht8UpEhmUd-RwEb0AP3hRrC1kE8CbuUSLtJwq_eKYvsRqhZ1xgmUgwouGDUOjp6cFaDlgUrfBUQCkm20VB1cW62-_4A9o1rf7G1f2PegvJ68Z3Mal-jXLuHxHg5AuadlUDOAI5w0pmALF5hqP2qRGCV07-WLVXG60m9L3TfkyRcDra8V0NglnkuMwXKZQWQEyCwf8yIFG1rAH6NEWxUa_ChSx-dr4YOOdp500Iva94CBYFXEoRO9Y3Yu8Za5oxX-GHkQgDu8--iUJ63m_yv38380kvsOh5kaBnDgDjRfbGWZRU-d5mdUkv0Yrk4y1SE8ODPO8ZU1BYfJrcvahuWB_AroZHJNbeTEXO51hda6blnVoetWTrtT4IUAT_j7AcWbmrjEewC-8WNxaTjREAUR6EN6tkkPu9AyrctyauBgg2FbluqVi5RVGuE0ArHZtpA9ilRLA-77WqvSVvk9bO3zzXcoA9yfpbepp1rdcLRqcI8Z3KN4GuO9Co4zLyLOkcI8gu-t1yEfy5CPKuS_SYMaksooTKWjysGRcnBRz4kCEdzqpaMUHPJCt8xfIV25jQt1mSz8TvfRkz4tywv3y0rmzJQfhN4s_j2bwG0a_wP4fqRW&amp;lang=sage" rel="nofollow noreferrer">Horizons Sage / Python script</a>. You can use it to create general distance plots. Any body that Horizons knows about can be given as the target, although only major bodies can be given as the center. There's an earlier version of the script <a href="https://space.stackexchange.com/a/55061/38535">here</a>, along with brief instructions.</p>
<p>The script uses the Horizons range data (&amp; its time derivative) to compute a series of cubic Bézier curves passing through the data points. It determines the mean distance by exact integration of the Bézier curves.</p>




CC BY-SA 4.0
2021-11-02T21:05:56.030Z
This is outstanding analysis, and (at least to me), is the most intuitive explanation and demonstration of why Mercury is (on average) the closest planet to any other planet in our solar system.  However, I don't see how it actually answers the original question.
Connor Garcia
CC BY-SA 4.0
2021-11-02T21:27:04.597Z
Thanks, @Connor. True, I didn't attempt to answer the OP's question about possible complicated stellar systems that don't follow the "innermost is closest on average" rule, I just mentioned that it's quite hard to do (analytically). But in compensation I did provide actual data for Mercury's eccentric orbit. Also, my Horizons script *could* be used on bodies with more eccentric & inclined orbits. (Horizons can provide data for 1,145,682 asteroids and 3,771 comets).
PM 2Ring
CC BY-SA 4.0
2021-11-02T21:27:37.400Z
FWIW, I primarily wrote this answer to complement the existing answers (which do discuss various "oddball" systems), and to add to the explanation from the Physics Today article which the OP linked.
PM 2Ring
CC BY-SA 4.0
2021-11-02T21:39:55.353Z
It is an excellent complement to the existing answers, +1!
Connor Garcia
CC BY-SA 4.0
2021-11-02T18:26:05.670Z
2022-03-31T19:58:35.990Z
2022-03-31T19:58:35.990Z
PM 2Ring
<p>Not for <a href="http://en.wikipedia.org/wiki/Double_planet" rel="noreferrer">double planets</a>, for which each body is always the nearest planet to the other.</p>
<p>Pluto and Charon disputably fit the bill. <a href="http://iau.org/static/public_press/iau_po_report.pdf" rel="noreferrer">This 2006 International Astronomical Union report</a> states</p>
<blockquote>
<p>Q: Is Pluto a planet?</p>
<p>A: Yes. In fact, Pluto’s large companion named
Charon is also large enough and massive enough to satisfy the
definition of “planet”. Because Pluto and Charon are gravitationally
bound together, they are actually now considered to be a “double
planet.”</p>
</blockquote>
<p>Even if one doesn't consider them planets, an Earth-Theia-like collision in another system could conceivably produce a pair with a barycenter outside each body.</p>
<p><a href="http://en.wikipedia.org/wiki/Double_planet" rel="noreferrer">The wiki</a> also states</p>
<blockquote>
<p>...the Moon currently migrates outward from Earth at a rate of
approximately 3.8 cm (1.5 in) per year; in a few billion years, the
Earth–Moon system's center of mass will lie outside Earth, which would
make it a double-planet system.</p>
</blockquote>

CC BY-SA 4.0
2021-11-01T12:17:19.487Z
-1 for cherry-picking the resolution that was described, but not passed.
notovny
CC BY-SA 4.0
2021-11-01T13:07:04.557Z
Actually any 2 planets with the same orbital period around the sun, which double planets also have I guess, potentially have a shorter distance to each other than to the innermost planet, I think
Ivo
CC BY-SA 4.0
2021-11-01T13:58:20.657Z
@IvoBeckers Sharing an orbit requires a very specific spatial configuration. Aside from double planets, you'd need to have a body at the L4 or L5 Lagrange points as described in Glorfindel's answer, but both bodies might not be considered planets.
Nuclear Hoagie
CC BY-SA 4.0
2021-11-01T14:20:21.633Z
@notovny: That's why I wrote "disputably". It depends on the definition of "planet" and is subject to change at any time.
Gnubie
CC BY-SA 4.0
2021-11-01T10:32:59.937Z
2021-11-01T11:38:36.113Z
2021-11-01T11:38:36.113Z
Gnubie
<p>The reason Mercury is the closest planet to all of the planets on average is in part because of the fact that, because of Kepler's laws, period scales with the semimajor axis of an orbit, and so big orbits are going to be slow and spend lots of time far away from any given object because the orbits are so slow. At opposition, for any two objects, they will be (given a near-circular orbit) the sum of the two semi-major axes away from each other. So naturally, if a planet has a small semimajor axis it's going to move fast around and be a lot closer a lot more of the time than another object with a larger semimajor axis and slower orbit.</p>
<p>This principle is easily expandable to other systems, as Kepler's laws apply to any planetary system. As for the binary system, dynamics close to a binary star system can get pretty complicated so I'm not sure what would happen exactly, but inasmuch as the binary system could be approximated as one mass because of distances, then this would still hold.</p>
<p>Let me know if this explanation is confusing; I can edit and reword this to make it clearer if necessary.</p>
<p>Edit: Upon further consideration and with the help of the comments, one needs to consider what kind of average we're looking at here. In our solar system, a lot of the orbits are near-circular, and in situations like these, circular approximations can yield the possibility of purely spatial averages because of uniform angular velocity. However, as seen by Kepler's laws, orbits are ellipsis, and technically except in the case of a perfectly circular orbit, have varying angular velocities (and by extension tangential velocities). For an orbit with an eccentricity that can't be approximated by a circular orbit, a temporal average must be considered as opposed to just a spatial one.</p>
<p>A good example of this is given by d_e in the comment; he says</p>
<blockquote>
<p>Let's imagine 3 planets - one is very close to the Sun. the other 2 are very far (in aphelion) from the Sun and have about the same semimajor axis - and the same direction of the apsides line. Now, the trick is to make them have very very high eccentricity - so the time they spend near the sun is very short; therefore most of the time the planets will spend in their aphelion (which their aphelion, as we said is near to each other)</p>
</blockquote>
<p>For circular orbits (or near thereunto) a simple thought experiment can make the reasoning for the mercury conclusion a bit clearer. Let's take mercury, Saturn and Uranus for example. The semi-major axes for each of these is (roughly) .4, 10 and 20 AU (again there's appreciable rounding going on here, but this is a heuristic example). Assuming circular orbits, Saturn will only ever be 9.6 or 10.4 AU away from mercury at all points in its orbit,which means the average would have to fall in there. For Saturn and Uranus, they will be about 10 AU at there closest, and 30 at their farthest. Given the range here, it's not difficult to imagine that the average will likely be a decent amount larger than 10.4, the absolute maximum possible distance average between Saturn and Mercury.</p>
<p>While this example is almost completely qualitative, I hope it's useful for believing the math and quantitative analysis presented by the link posted in the question.</p>
<p>For non-circular orbits, like d_e illustrated quite well, this reasoning does not necessarily extend, and so the answer would be <strong>no</strong>, this does not extend generally. Each system would have to be analyzed individually. But for a group of near-circular orbits, this seems to be the case.</p>
<p>I have left my previous, faulty answer here as context for the comments below, which are quite illustrative of some important principles.</p>

CC BY-SA 4.0
2021-11-01T08:19:40.860Z
I can't see how it is easily expandable to other systems. Let's imagine 3 planets - one is very close to the Sun. the other 2 are *very* far (in aphelion) from the Sun and have about the same semimajor axis - and the same direction of the apsides line. Now, the trick is to make them have very very high eccentricity - so the time they spend near the sun is very short; therefore most of the time the planets will spend in their aphelion (which their aphelion, as we said is near to each other). I believe in this case it will not generalize.
d_e
CC BY-SA 4.0
2021-11-01T08:50:41.820Z
This orbital period argument doesn't make sense. A planet having a fast orbital period doesn't mean it's going to be "a lot closer a lot more of the time" to other planets. The orbital periods determine how *frequently* the orbits will line up, but not what overall proportion of time the bodies will spend close to or far from each other.
user2357112
CC BY-SA 4.0
2021-11-01T08:57:14.040Z
If you could magically speed up or slow down Mercury's orbit without changing the orbit's shape, it would *still* be the planet with the shortest time-averaged distance to Earth, or to other planets.
user2357112
CC BY-SA 4.0
2021-11-01T08:59:50.997Z
If you read the article linked in the question, you'll see that the math doesn't involve orbital periods at all.
user2357112
CC BY-SA 4.0
2021-11-01T10:10:26.613Z
There was a CGP Grey video about this: [Which Planet is the Closest?](https://www.youtube.com/watch?v=SumDHcnCRuU)
Rebecca J. Stones
CC BY-SA 4.0
2021-11-01T15:37:01.060Z
Perhaps I emphasized the wrong thing here; Its not the orbital period of mercury that matters as much as how long farther planet's orbital periods are; the fact that they are so long and also different from other planets means that they're going to be spending a lot of time away from each other, which will, in turn, affect the average. Sorry guys I don't think this is as clear as I wanted it to be, will edit in a bit when I have more time.
Justin T
CC BY-SA 4.0
2021-11-01T18:30:48.567Z
@user2357112supportsMonica The assumptions in the articles linked are assuming circular orbits, which is a good approximation for the bodies in our solar system, but not for even moderately eccentric orbits, and like is illustrated by d_e, time dependence comes up because orbital velocities aren’t constant in elliptical orbits. Although I would revise my statement and say perhaps the period itself is less important, and perhaps what’s happening during the orbit time-wise. For a circular orbit the average can be purely spatial, but for these elliptical orbits, it would have to be a time average
Justin T
CC BY-SA 4.0
2021-11-01T18:59:29.480Z
Ok edited for clarity and to reflect good points mentioned in comments
Justin T
CC BY-SA 4.0
2021-11-02T01:18:54.763Z
This answer is embarrassingly wrong, from start to finish. How on earth does it have four upvotes?
TonyK
CC BY-SA 4.0
2021-11-02T18:23:02.027Z
@TonyK Care to actually add some useful insight? Maybe elaborate a bit more on points you think should be clarified or corrected
Justin T
CC BY-SA 4.0
2021-11-03T03:25:35.053Z
Kepler's laws don't have too much to do with this.  The result would be true even if all the planets moved at constant speed.  I'm finding it difficult to spot anything in this answer that's either useful or correct.  Downvoted.  Sorry.
Dawood ibn Kareem
CC BY-SA 4.0
2021-11-03T03:52:13.633Z
@DawoodibnKareem First Kepler law states all orbits are elliptical, which leads to the second law saying that the orbital speed is not constant. You can look at d_e’s example for why this matters; in a highly eccentric orbit, the vast majority of the time it will be extremely far away from the object, making the average distance different than if it spent the same amount of time at all points of the orbit. Does this make sense? This of course is for temporal averages, since purely spatial averages are useless in this kind of scenario; it wouldn't mean anything.
Justin T
CC BY-SA 4.0
2021-11-03T03:55:11.287Z
This further leads to our conclusion, which is that the results in the paper linked by OP cannot be generalized, since they assume constant orbital speeds which allow spatial averages. Carl Witthoft also mentions this as a warning in the comments of the OP's post
Justin T
CC BY-SA 4.0
2021-11-03T03:55:56.697Z
No, a circular orbit complies with Kepler's laws, and has constant orbital speed.  The law that you've cited in your opening paragraph - that period scales with the semimajor axis - does not in any way affect the result about which planet is closest to any other.
Dawood ibn Kareem
CC BY-SA 4.0
2021-11-03T03:59:32.063Z
@DawoodibnKareem If you read the end of the answer, you’ll see that "I have left my previous, faulty answer here as context for the comments below, which are quite illustrative of some important principles." The stuff post the ‘edit’ are the adjusted answer.
Justin T
CC BY-SA 4.0
2021-11-03T04:24:54.120Z
You have asserted without proof (or any kind of supporting argument) that the result for circular orbits does not extend to non-circular orbits; even though your first attempt at answering implied that it did.  In fact, your original answer was closer to the truth than the edit, even though its reasoning was faulty.  There are only a few edge cases where the result doesn't hold, such as double planets.  In a solar system where no pair of planets are gravitationally bound to each other, and no pair of planets are the same distance from the sun, the result DOES apply.
Dawood ibn Kareem
CC BY-SA 4.0
2021-11-03T05:42:58.060Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/131098/discussion-between-justin-tackett-and-dawood-ibn-kareem).
Justin T
CC BY-SA 4.0
2021-11-01T03:55:45.413Z
2021-11-01T18:56:29.203Z
2021-11-01T18:56:29.203Z
Justin T
<h2>Not necessarily</h2>
<p>The authors of the article linked above assume that the planets involved are</p>
<blockquote>
<p>in roughly circular, concentric, and coplanar orbits</p>
</blockquote>
<p>Suppose Mercury and Earth had circular co-planar orbits with Mercury at 0.4 AU and Earth at 1 AU from the Sun.  With a quick program and Monte Carlo analysis, we get the same result as the above referenced article author's that the average distance from the Earth to Mercury is 1.04 AU.</p>
<p>Alter the above scenario slightly so that Mercury now has a stellar orbital inclination offset from Earth by 90 degrees, i.e. their orbital planes are perpendicular.  Now a Monte Carlo analysis reveals an average distance of 1.0584 AU.</p>
<p>So, clearly (notwithstanding orbital instabilities) we could have an innermost planet with an inclination perpendicular to the ecliptic, which on average is further away from Earth than the next planet out with a co-planar inclination.</p>
<p>Add in the possibility for a highly eccentric orbiting innermost planet (where semi-major axis is our measure of stellar proximity) and we can get an even more glaring discrepancy in average distance since the vast amount of time for such a planet is spent far away from the star at apoapsis.</p>
<p><strong>Note:</strong>
A specific exoplanet counterexample of this kind could be exceedingly difficult to find as:</p>
<ol>
<li>planets tend to evolve in near co-planar orbits.</li>
<li>Such a system would almost certainly be unstable.</li>
<li>Perpendicular orbital planes would make detection of both planets via transit methods extremely rare.</li>
</ol>
<p><strong>Code co-planar:</strong></p>
<pre><code>n=1e7;
theta = rand(1,n)*2*pi;
points =[0.4*sin(theta)-1;0.4*cos(theta)];
aveDist = mean(sqrt(points(1,:).^2+points(2,:).^2));
</code></pre>
<p><strong>Code perpendicular:</strong></p>
<pre><code>n=1e7;
theta = rand(1,n)*2*pi;
phi = rand(1,n)*2*pi;
points =[0.4*cos(phi).*sin(theta)-1;0.4*sin(phi).*cos(theta);0.4*cos(theta)];
aveDist = mean(sqrt(points(1,:).^2+points(2,:).^2+points(3,:).^2));
</code></pre>

CC BY-SA 4.0
2021-11-03T22:39:18.240Z
"Such a system would almost certainly be unstable." - I think that's putting it mildly.  Say Mercury's orbit were tilted at 90 degrees to get the value 1.056.  For a planet *further out* to be closer on average it would still have to have an orbit within  0.47 au (Mercury is 0.39, Venus is 0.73).  I think that is close enough to Mercury to make instability a certainty, even if Mercury's orbit weren't tipped 90 degrees.
Jason Goemaat
CC BY-SA 4.0
2021-11-04T15:32:09.243Z
@JasonGoemaat Agreed.  Though I suppose all n-body systems are unstable, it is just a matter of the time-scale.  The point I make in this post is that even if we are unlikely to ever see a counter-example, we can't rule it out based purely on the orbital geometry.
Connor Garcia
CC BY-SA 4.0
2021-11-01T20:22:28.180Z
2021-11-01T20:22:28.180Z
Connor Garcia
<p>The page you link to and the paper refer to PCM (the Point Circle Method), and the PCM treats the orbits of two objects as circular, concentric, and coplanar.  If planets in a system had odd (and probably unstable) orbits, this might not be the case.</p>
<p>The PCM method treats the planet's position at any given time as a uniform probabilistic distribution.  That means that if you look at a random point in time over the last billion years, the each planet has about the same probability of being at any point in it's orbit.  What that gives you is the ability to treat one planet as a point (P), and calculate an average for all points in the other planet's circular orbit as if they are equally likely (C).</p>
<p>In general then for any two planets you look at, the average distance between them will get smaller as the inner planet's orbit gets smaller.  If you made a list of the closest planets to any other, they would always be in order of closest distance to the sun (smallest orbit).</p>
<p>In the picture below there is the test planet (outer light blue), and two inner planets that are green and blue where we want to find which one is closest on average.</p>
<p><a href="https://i.sstatic.net/XKrxG.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/XKrxG.png" alt="enter image description here" /></a></p>
<p>PCM uses a static position for the test planet (outer light blue) and calculates the average distance to each point on the circles representing the other planets' orbits.</p>
<p>You don't need to do any calculations to tell that the average distance will be smaller for the inner planet though.  If we take any angle around the circle and plot a vertical line (on the right), we can see that the outer planet will be further to the side at that point in it's orbit.  It will be closer 1/2 the time and further 1/2 the time vertically than the outer planet, but in all cases except for when they're aligned with our test planet, the outer planet will have a larger horizontal component to the distance.</p>

CC BY-SA 4.0
2021-11-08T21:00:56.167Z
2021-11-08T21:00:56.167Z
Jason Goemaat
<p>[This is complementary answer to Justin Tackett answer. I felt somewhat compelled in light of the comments (even after edits), and to illustrate my example (referenced in the answer) with a diagram.]</p>
<p>I say there are 2 naïve situations where the innermost planet is not the closet. In the attached illustration there are only 2 planets and I consider the distance to the Sun (As we might assume the innermost planet is so close to the Sun - so it won't make any difference to take the distance to the Sun instead).</p>
<ol>
<li><strong>In the first example the period (and therefore the semi-major axis) and the pehilion angle of the outer planets are the same. They differ only in eccentricity.</strong> We can see clearly see that the distance between the planets is always shorter than the distance to the Sun. (I used to same orbit phase - but we can allow some small difference - also in the pehilion location)<a href="https://i.sstatic.net/B6hk1.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/B6hk1.gif" alt="enter image description here" /></a></li>
</ol>
<p>One can justly claim that this setting is not possible in real-life - and I'll agree with him. The value in my answer is primarily to demonstrate that there is nothing universal here that can be deduced from the Keplerian system; rather, that every system has its own characteristics.</p>
<ol start="2">
<li>The second example that was given in Justin Tackett answer, includes <strong>the two planets to have very high eccentricity and about the same semi-major axis and pehilion angle.</strong> I admit I thought - in the numbers below given - the effect would be stronger; nevertheless even here it is quite easy to see that the average distance between the planets is shorter than the the distance to the Sun. Note that the diagram is not on scale: the X-axis is longer - had it on scale the orbits would seem much more like strong ellipses. At most of the time the planets are at X&gt;3 while the sun is at 0. this enough to establish I believe my claim here. <a href="https://i.sstatic.net/UHpHM.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UHpHM.gif" alt="enter image description here" /></a></li>
</ol>

CC BY-SA 4.0
2021-11-06T22:02:20.253Z
FWIW, the actual numbers in case of someone doubts  (using different periods of time, and gap between measure to measure) is: Green planet to the Sun: ~3.51. Green planet to blue planet: ~2.08
d_e
CC BY-SA 4.0
2021-11-06T10:57:03.987Z
2021-11-06T16:39:08.860Z
2021-11-06T16:39:08.860Z
d_e
<p>Lets take a more general look at the situation in systems with one star:</p>
<p>What is the average distance to a planet?  Break this down into two components, the distance along a line from the planet you are measuring from (which I will call O), through the barycenter of the planetary system which I will call X, and the distance the planet is offset from this line which I will call Y.</p>
<p>Now, what is the average X?  While the planet will spend more time near it's apoapsis than it's periapsis this is independent of the orbit of the planet you are measuring from and thus will average to zero.  Thus the average X is always the distance from O to the barycenter.  This is the same for <strong>all</strong> planets.</p>
<p>This leaves the Y.  Once again, the shifting angles between O and the target average things out--it's going to vary from zero to the average orbital radius of the target.  My calculus is far too rusty to sum this but it doesn't matter--it's clearly dependent on the orbital radius and nothing else.</p>
<p>This we have X fixed and Y varying at the orbital radius.  Thus sqrt(x^2 + y^2) will likewise be ordered by orbital radius although the relationship is no longer linear.  Thus the average distance to a planet will be ordered by the orbital radius of the planet.  The inner planet is always the closest.</p>
<p>Now, there are some special cases that have been presented in other answers:  double planets, L4 &amp; L5 objects.  However, one of the requirements to be a planet is to have cleared one's orbit--and all such cases mean you have two objects in the orbit.  Thus they can't be planets to be a closer planet.</p>

CC BY-SA 4.0
2021-11-04T00:50:53.520Z
2021-11-04T00:50:53.520Z
Loren Pechtel
<p>In a binary star system, there are two types of orbits which planets can have.</p>
<p>A planet in an S-type or non circumbinary orbit will orbit one of the two stars.  Thus the distance between the two stars must be at least several times as great as the semi-major axis of the planet's orbit, if the planet is to have a stable orbit.</p>
<p>A planet in an p-type or circumbinary orbit will orbit both of the two stars. Thus the semi-major axis of the planet's orbit must be at least a few times as great as the distance between the two stars for the planet to have a stable orbit.</p>
<p>It is possible for a binary star system to have planets in S-type orbits around one of the stars, or both of the stars, and to have planets in P-type orbits around both of the stars.</p>
<p>Each planet around a star will have a forbidden zone around its orbit where no other planet can orbit because of gravitational interactions.  The closer two planets are to their star, the smaller their forbidden zones will be, and the closer the two orbits can be.</p>
<p>If a binary system has two stars, A and B, and planets in S-type orbits around each of them, it seems obvious to me that the innermost planet around A will be - on average, as the question asks,-closest to each of the other planets orbiting around A.  And similarly the innermost planet orbiting around B will be, on average, as the question asks, closest to each of the other planets orbiting around B.</p>
<p>If the separation between A and B must be at least, for example, at 5 times the distance of the orbit of the outermost planet around either star, we can make a picture of the system.</p>
<p>Assume that the outermost planet around each of the stars orbits at 10 units, and the separation of the two stars in their orbits is 50 units.  Obviously the separation between the outermost planets of each star will vary between 40 to 60 units, with an average separation of about 50.99 units.</p>
<p>But an inner planet orbiting star A at 5 units, for example, planet A I, will always be between 5 and 15 units from the outermost planet orbiting Star A at 10 units, planet A II, and the average Separation between the two planets will be about 11.180 units according to the Pythagorean theorm.</p>
<p>So any planet orbiting star A will always be closer to any planet orbiting star A than any planet orbiting star B can ever get, even at its closest.</p>
<p>Assume that there are also planets in P-type or circumbinary orbits around the center of mass of the two stars A &amp; B.   Imagine that planet AB I orbits at 250 units and planet AB II orbits at 500 Units and planet AB III orbits at 1,000 units.</p>
<p>The distance bween AB I and AB II will always be between 250 and 750 units with an average of 559.01 units.</p>
<p>The distance between AB II and AB III will always be between 500 and 1,500 units with an average of 1118.03 units.</p>
<p>The distance between AB I and AB III Will always be between 750 and 1,250 units, with an average of 1,030.77 units.</p>
<p>So that indicates that the innermost planet orbiting in a P-type orbit around the center of gravity of the two stars A and B, namely planet AB I, will be the closest - on the average, as the question asks - to the other planets orbiting. in P-type orbits around both of the stars.</p>
<p>So that leaves the question of whether any of the planets in S-type orbits around only one of the stars, A or B, will be closer - on the average, as the question asks - to the planets in P-type orbits around the center of mass of the stars A &amp; B.</p>
<p>Assuming that stars A &amp; B have equal mass, they will orbit the center or mass or barycenter, at equal distances.  Assuming almost perfectly circular orbits, each star will always be almost exactly 25 units from the barycenter.</p>
<p>I stated that no planet of A or B could have a stable orbit more than 10 units from its star.  So no planet of A or B could ever get more 35 units from the barycenter of the two stars.</p>
<p>So a planet orbiting A or B could get between 215 and 285 units from planet AB I, with an average distance of about 251 units.</p>
<p>That means that a planet orbiting A or B could sometimes be closer to planet AB I than planet AB II is.</p>
<p>A planet orbiting A or B could get between 465 and 535 units from the planet AB II, with an average distance of about 500.646 units.  Which is shorter than the average distance between AB I and AB II.</p>
<p>A planet orbiting A or B could get between 965 and 1,035 units from the planet AB III, with an average distance of about 1,000.3124 units.  Which is shorter than the average distance between AB II and AB III and shorter than the averaage distance between AB I and AB III.</p>
<p>So apparently in a binary star system with planets in S-type orbits around one or each of the stars, as well as planets in P-type orbits around the center of mass of both the stars, the planets in S-type orbits will be closer - on the average, as defined in the question - to the planets in P-type orbits than any of the planets in P-type orbits will be  - on the average, as defined in the question - to each other.</p>
<p>Of course I have only demonstrated that for the specific orbits I have chosen for the system in my example.</p>

CC BY-SA 4.0
2021-11-01T21:11:41.997Z
2021-11-02T18:32:02.183Z
2021-11-02T18:32:02.183Z
M. A. Golding
<p>A modification to this question that might clarify it:</p>
<p>Is the average distance between a planet in a standard orbit (one planet per circular-ish orbit) and its primary, always less than the average distance between the planet and any other planet orbiting the primary?</p>
<p>The nearer a planet is to the primary, the nearer its average distance to any other planet is to the average distance between the primary to any other planet.</p>

CC BY-SA 4.0
2021-11-01T16:16:12.577Z
This doesn't address the aspect of how far an arbitrary pair of planets are from one another. This just states that if a star and planet are very close, the distance from anywhere to either the star or the planet is basically the same. The distance from Neptune to the Sun is nearly the same as from Neptune to Mercury, but that says nothing about how far it is from Neptune to anywhere else.
Nuclear Hoagie
CC BY-SA 4.0
2021-11-01T18:58:08.640Z
@NuclearHoagie I'll try to come up with how expand on this.   Essentially, the average distance from one planet and another will be the average between when both planets are one the same side of the primary and when they are on the opposite sides.  This average value will always be larger than the distance to the primary, and *I believe* that the farther from the primary the other planet is, the larger this value will be (regardless of where the initial planet is, relatively).  I'll see if I can get some values in a spreadsheet and confirm this.
Michael Richardson
CC BY-SA 4.0
2021-11-01T15:43:32.303Z
2021-11-01T15:43:32.303Z
Michael Richardson
<p>In our Solar system Mercury is not only the closest planet to Earth <em>on average</em> but also, for the same reason, the closest planet on average to all other planets (Jupiter, Neptune, etc...).</p>
<p>Does that always apply for any planetary system? For instance, in a binary system, is the planet closest to a given circumbinary planet always the innermost planet of one of the two stars? (or can it be the outermost planet, or even another circumbinary planet?)</p>
<p>From <a href="https://physicstoday.scitation.org/do/10.1063/PT.6.3.20190312a/full/" rel="noreferrer">Venus is not Earth’s closest neighbor</a>:</p>
<blockquote>
<p>Calculations and simulations confirm that on average, Mercury is the nearest planet to Earth—and to every other planet in the solar system.</p>
</blockquote>
<p>But this only applies to our solar system's planets, I'm interested in a broader generalization.</p>

CC BY-SA 4.0
2021-11-01T00:44:45.033Z
"...Mercury is not only the closest planet to Earth..."  Hmm... Mercury is not the closest planet to Earth. Venus is the planet that gets closest to Earth.
uhoh
CC BY-SA 4.0
2021-11-01T01:23:02.897Z
To follow up on uhoh's comment, Mercury is not the closest planet to any planet expect Venus.  There is an oddity where Mercury is -On Average- the closest planet to Earth because Venus' orbit carries it to the far side of the Sun further than does Mercury's.  Is that what you're asking about?
JohnHunt
CC BY-SA 4.0
2021-11-01T02:54:46.457Z
@JohnHunt Yes, I'm talking about average. Mercury is indeed the closest planet (on average) to Earth, and to all other 7 planets as well.
Gerardo Furtado
CC BY-SA 4.0
2021-11-01T02:56:18.417Z
@uhoh https://physicstoday.scitation.org/do/10.1063/PT.6.3.20190312a/full/
Gerardo Furtado
CC BY-SA 4.0
2021-11-01T03:15:53.557Z
@GerardoFurtado Thanks and *very cool!* I've made some edits to your question to reflect what's in your comments, and looking forward to seeing some answers as well! Feel free to edit further.
uhoh
CC BY-SA 4.0
2021-11-01T03:24:05.443Z
@uhoh Thanks, your edit is fine. Actually, since this subject is not exactly clear (as I found out because your and John's comments!) I could make this question initially only about simple planetary systems (that is, *"is the innermost planet always the closest planet to the others giving all orbital distances and speeds?"*), then asking a new question for binary systems, which was my original curiosity.
Gerardo Furtado
CC BY-SA 4.0
2021-11-01T03:33:58.667Z
The way I thought you ment was if the planet of the sun-side of a planet always is closer than the closest planet on the peripheral side? In other words why is  Mercury closer to Venus than Earth. Venus closer to Earth than Mars. Mars closer to Jupiter than Saturn. Jupiter closer to Saturn than Uranus. Saturn closer to Uranus than Neptune.  And if there is a cosmic rule prescribing this phenomen universally?
Constantthin
CC BY-SA 4.0
2021-11-01T13:04:12.203Z
not worthy of an answer I think, but a theoretical example for which it's not the case is when 2 planets have exactly the same orbital period. Their distance will be fixed and could possibly have a shorter distance to each other than to the innermost planet. But this is an exception and I think in all other cases it's true
Ivo
CC BY-SA 4.0
2021-11-01T13:25:08.177Z
**Warning**  make sure you identify what the parameter is that you are averaging over!  I suspect most of us expect the average with respect to time.   But if one chooses to take the average over, say,  angle of the orbits ( rho -theta coordinates),  things will come out rather differently.
Carl Witthoft
CC BY-SA 4.0
2021-11-01T18:14:46.227Z
@Constantthin it's talking about the average over the path. The places when planet B is in-line sun-side and opposite side cancel each other out to average the same as the planet A's orbital radius irrespective of the radius of B. In the places when planet A, the sun and planet B form a triangle, the AB distance is the longest side of a triangle where the other two sides are the orbital radii of A and B, so the bigger the orbital radius of B the bigger that triangle is, and so the bigger the average distance between A and B.
Pete Kirkham
CC BY-SA 4.0
2021-11-01T22:38:58.673Z
@IvoBeckers that's in my opinion worthy of an answer, because that's precisely one of scenarios that, as a layman, I imagined for breaking this "rule". So, is it really possible two planets having the same orbital period?
Gerardo Furtado
CC BY-SA 4.0
2021-11-02T00:15:13.200Z
@Pete Kirkham. Granted. But wouldn’t you agree that my version has an interesting slant to?
Constantthin
CC BY-SA 4.0
2021-11-02T10:35:44.857Z
@Constantthin it's certainly more interesting than a question whose answer is the same for every planet
Pete Kirkham
CC BY-SA 4.0
2021-11-02T10:42:23.077Z
See also "Which Planet is the Closest", by CGP Grey: https://www.youtube.com/watch?v=SumDHcnCRuU
HenricF
CC BY-SA 4.0
2021-11-04T05:39:15.857Z
@CarlWitthoft For low eccentricity, aren't they close to the same thing? Unless you're measuring the angle from the planet, rather than from the star.
Acccumulation
CC BY-SA 4.0
2021-11-04T05:41:54.273Z
"Mercury is the nearest planet to Earth—and to every other planet in the solar system." Nonsense. Earth is the closest planet to Earth.
Acccumulation
CC BY-SA 4.0
2021-11-04T12:27:16.613Z
@Acccumulation LOL
Carl Witthoft
<exoplanet><orbital-mechanics><planetary-systems><solar-system-evolution>
Is the closest planet to another planet always the innermost planet?
Gerardo Furtado
# Is the closest planet to another planet always the innermost planet?

(Asked by: username_0 on 2021-11-01)

In our Solar system Mercury is not only the closest planet to Earth *on average* but also, for the same reason, the closest planet on average to all other planets (Jupiter, Neptune, etc...).

Does that always apply for any planetary system? For instance, in a binary system, is the planet closest to a given circumbinary planet always the innermost planet of one of the two stars? (or can it be the outermost planet, or even another circumbinary planet?)

From [Venus is not Earth’s closest neighbor](https://physicstoday.scitation.org/do/10.1063/PT.6.3.20190312a/full/):

> Calculations and simulations confirm that on average, Mercury is the nearest planet to Earth—and to every other planet in the solar system.

But this only applies to our solar system's planets, I'm interested in a broader generalization.

**Tags:** exoplanet, orbital-mechanics, planetary-systems, solar-system-evolution

> **Comments:**
> **username_1**: "...Mercury is not only the closest planet to Earth..."  Hmm... Mercury is not the closest planet to Earth. Venus is the planet that gets closest to Earth.  
> **username_2**: To follow up on username_1's comment, Mercury is not the closest planet to any planet expect Venus.  There is an oddity where Mercury is -On Average- the closest planet to Earth because Venus' orbit carries it to the far side of the Sun further than does Mercury's.  Is that what you're asking about?  
> **username_0**: @username_2 Yes, I'm talking about average. Mercury is indeed the closest planet (on average) to Earth, and to all other 7 planets as well.  
> **username_0**: @username_1 https://physicstoday.scitation.org/do/10.1063/PT.6.3.20190312a/full/  
> **username_1**: @GerardoFurtado Thanks and *very cool!* I've made some edits to your question to reflect what's in your comments, and looking forward to seeing some answers as well! Feel free to edit further.  
> **username_0**: @username_1 Thanks, your edit is fine. Actually, since this subject is not exactly clear (as I found out because your and John's comments!) I could make this question initially only about simple planetary systems (that is, *"is the innermost planet always the closest planet to the others giving all orbital distances and speeds?"*), then asking a new question for binary systems, which was my original curiosity.  
> **username_3**: The way I thought you ment was if the planet of the sun-side of a planet always is closer than the closest planet on the peripheral side? In other words why is  Mercury closer to Venus than Earth. Venus closer to Earth than Mars. Mars closer to Jupiter than Saturn. Jupiter closer to Saturn than Uranus. Saturn closer to Uranus than Neptune.  And if there is a cosmic rule prescribing this phenomen universally?  
> **username_4**: not worthy of an answer I think, but a theoretical example for which it's not the case is when 2 planets have exactly the same orbital period. Their distance will be fixed and could possibly have a shorter distance to each other than to the innermost planet. But this is an exception and I think in all other cases it's true  
> **username_5**: **Warning**  make sure you identify what the parameter is that you are averaging over!  I suspect most of us expect the average with respect to time.   But if one chooses to take the average over, say,  angle of the orbits ( rho -theta coordinates),  things will come out rather differently.  
> **username_6**: @username_3 it's talking about the average over the path. The places when planet B is in-line sun-side and opposite side cancel each other out to average the same as the planet A's orbital radius irrespective of the radius of B. In the places when planet A, the sun and planet B form a triangle, the AB distance is the longest side of a triangle where the other two sides are the orbital radii of A and B, so the bigger the orbital radius of B the bigger that triangle is, and so the bigger the average distance between A and B.  
> **username_0**: @username_4Beckers that's in my opinion worthy of an answer, because that's precisely one of scenarios that, as a layman, I imagined for breaking this "rule". So, is it really possible two planets having the same orbital period?  
> **username_3**: @username_6. Granted. But wouldn’t you agree that my version has an interesting slant to?  
> **username_6**: @username_3 it's certainly more interesting than a question whose answer is the same for every planet  
> **username_7**: See also "Which Planet is the Closest", by CGP Grey: https://www.youtube.com/watch?v=SumDHcnCRuU  
> **username_8**: @CarlWitthoft For low eccentricity, aren't they close to the same thing? Unless you're measuring the angle from the planet, rather than from the star.  
> **username_8**: "Mercury is the nearest planet to Earth—and to every other planet in the solar system." Nonsense. Earth is the closest planet to Earth.  
> **username_5**: @username_8 LOL  

---

## 10 answers

### Answer by username_9 on 2021-11-01. Score: 14

For a planet with another planet at their [Lagrange L4 or L5 point](https://en.wikipedia.org/wiki/Lagrange_point#L4_and_L5_points), those would be each others' closest neighbours; as you can see from the simulations, Mercury has an average distance a few % more than the average distance between the Sun and the planet (e.g. 1.04 AU for Earth), and the Lagrange points form equilateral triangles with the largest planet and the Sun, so for Earth that would be at a (constant) distance of 1 AU.

Of course, whether the smallest of the two would count as a planet is [debatable](https://en.wikipedia.org/wiki/Clearing_the_neighbourhood). They could be quite large, e.g. Jupiter (317.8 Earth masses) could have a Trojan weighing 317.8 / 24.96 = 12.7 Earth masses.

> **Comments:**
> **username_5**: Yep, depends on whether you allow objects at a L4 or L5 to be called planets rather than moons (or even something less "major" )  
> **username_10**: A planet in the L4 or L5 point of a bigger planet has not cleared its neighbourhood and can be a dwarf planet at most.  
> **username_11**: Do you have a source/explanation for the possible Jupiter Trojan of 12.7 Earth masses? Very interesting (especially for sci-fi ideas), though I don't understand the math you've done  
> **username_9**: @DrakeP 24.96 is the minimum mass ratio: https://en.wikipedia.org/wiki/Lagrange_point#cite_ref-exact_stability_threshold_7-0  
> **username_9**: @username_10 exactly, that's why I linked to the Wikipedia article about [Clearing the neighbourhood](https://en.wikipedia.org/wiki/Clearing_the_neighbourhood). Pluto-fanatic might disagree with that. I don't have a particular opinion on this matter.  
> **username_12**: @username_9 I think you're reading it incorrectly.  The M1 is the main body (Sol) and M2 is the secondary body (Jupiter).  The ratio of 24.96 is required for there to be L4 and L5 points at all, it doesn't describe the mass of objects that could be in a stable orbit there. "*indeed, the third body need to have negligible mass*"  
> **username_9**: @JasonGoemaat ... you're right, thanks for pointing out! Any idea where we can find a maximum mass / minimum ratio for M2/M4?  

---

### Answer by username_13 on 2021-11-02. Score: 11

As has been mentioned earlier, the fully general version of this question requires rather tricky analysis to deal with orbits that aren't circular and coplanar. It's even trickier if you want to include the gravitational interactions between the planets, although in most cases those small perturbations won't be large enough to affect which planet has the smallest mean distance to a given planet.

Fortunately, for our Solar System, we can use [JPL Horizons](https://ssd.jpl.nasa.gov/horizons/) to find distances between the planets. I'll show some results derived from Horizons for the mean distance between Earth & Mercury and Earth & Venus later in this answer. But first, I'd like to give a brief analysis of the simple coplanar circular orbits version of this problem.

In our simple flat circular solar system, the planets all orbit the sun S in the same direction. The orbit of our "home" planet H has radius 1 and period 1. The planets obey Kepler's laws, so they move around S with constant orbital speed. A planet P with an orbital radius of $r$ has a period of $T=r^{3/2}$.

[![circular orbits](https://i.sstatic.net/mEHdJ.png)](https://i.sstatic.net/mEHdJ.png)

Relative to H, P has a [synodic period](https://en.wikipedia.org/wiki/Orbital_period#Synodic_period) of
$$T\_s = \frac T{|T-1|}$$

Let the angle at HSP be $\theta$. Then $$\theta=\frac{2\pi t}{T\_s}$$
where $t$ is time. So to find the mean distance between H & P we need to average over the synodic period.

Let HP be $s$. Then by the cosine rule,
$$s^2 = 1 + r^2 - 2r\cos\theta$$
Therefore, the mean distance $m$ between H & P is
$$m = \frac1{2\pi}\int\_0^{2\pi} \sqrt{1 + r^2 - 2r\cos\theta}\,d\theta$$

That's not an easy integral, but it can be converted to a [complete elliptical integral of the second kind](https://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_second_kind).

Using Sage, we can plot $s$ and compute $m$ for a few values of $r$, over two synodic cycles. The blue dotted horizontal line shows the mean distance.

[![animated plot of s](https://i.sstatic.net/Xuu5v.gif)](https://i.sstatic.net/Xuu5v.gif)

When $r=0$, $m$ has its minimum value of 1, and $m$ grows as $r$ does. For small $r$, the graph of $s$ is *almost* a sine wave ranging from $1-r$ to $1+r$. But as $r$ grows, we see that the lower part of the graph is decidedly more "pointy" than the upper part, which shifts the mean value upwards. The pointiness reaches a peak at $r=1$, after which it gradually diminishes. (Note that $r=1$ means that H & P are in the same orbit, thus $T\_s$ is infinite, so the $r=1$ frame isn't really possible, it's just the limiting case).

You can use [this script](https://sagecell.sagemath.org/?z=eJxVkEFrwzAMhe_9FSKX2qnXJYHuEAgMdh6ENbfSFjdxOoNjZ7Kz0Y3-98ntCqsvtp6ePj0coIJPiWwe5nz2rG1QKNsw61QPe4bVi7NB28lNfm10p5CxTMCKC_BBjV5_qypb5gLILicTqFhxXs6ATk9g_4GB5bAA3BXwAEWKaes8K9JRp4Hzi28gn50GhbqVZh8DHFEa1guIq3LON9n2YhyRmix5k52efJkIQP5ff1XSkjhcxYaouMuXq2vlqWzgEeTBs4aS5HejNS13HQ03d_L6ZF2n26j7a6MmzGhcuKUr6COCDkZV_S0X_OBZQAxDz6FcPvVnAhxRd0Zb5atk0NZhcuMtKog620TcQDhWxHsrLrIPJ2In5Z_dv7svVvNfJLt4Rg==&lang=sage) to generate a static plot for smallish values of $r$.

Here's a plot of $m$ vs $r$, computed using elliptic integrals.

[![general mean distance plot](https://i.sstatic.net/5gvSW.png)](https://i.sstatic.net/5gvSW.png)

And here's the [plotting script](https://sagecell.sagemath.org/?z=eJxNzbEOgjAYBOCdp7h0akkRQV1M-ggurkZJhR_zSymkhUR9etE4eNPlyyXXUItWBrVPsKSDwTYNuQzIUKhL-dVA0xw87DX-HCnIOR4nriuqZdZ9pESOkZNkdMMkWw251tgojWfP3izdPihWzl7JRXMSR9vwHIWGOJD1aDhO1tckznq5_JtWkV9kilWx07gFbhx7ikb09j4Eod5KsDee&lang=sage)

---

As promised, here's a Horizons plot of the Earth-Mercury distance, over 12 synodic cycles. The starting & ending points are close to local maxima. The time interval is divided into 1390 equal steps (which are slightly less than 1 day).

|  | Julian day number | Date & time |
| --- | --- | --- |
| From: | 2458159.229166667 | A.D. 2018-Feb-09 17:30:00.0000 |
| To: | 2459549.125000000 | A.D. 2021-Nov-30 15:00:00.0000 |

Mean distance = 154762296 km

[![Earth-Mercury distance](https://i.sstatic.net/9GMjg.png)](https://i.sstatic.net/9GMjg.png)

And here's the Earth-Venus distance, over 12 synodic cycles. The time interval is divided into 879 equal steps (slightly less than four days).

|  | Julian day number | Date & time |
| --- | --- | --- |
| From: | 2456953.250000000 | A.D. 2014-Oct-22 18:00:00.0000 |
| To: | 2460466.604166667 | A.D. 2024-Jun-05 02:30:00.0000 |

Mean distance = 170272478 km

[![Earth-Venus distance](https://i.sstatic.net/1jlNw.png)](https://i.sstatic.net/1jlNw.png)

Mercury's orbit is rather eccentric, so there's quite a variation in the heights of the maxima & minima, and in the length of the synodic cycle. To get a more accurate value for the mean distance we should really average over a lot more cycles.

The eccentricity of Venus's orbit is quite low, so its graph is much closer to the simple circular orbit plots above.

Although the mean distance of Mercury is certainly smaller than the mean distance of Venus, by 15510182 km, clearly Venus has the lower minimum distance to Earth.

---

Those last two plots were created using this [Horizons Sage / Python script](https://sagecell.sagemath.org/?z=eJx1V3tz2kgS_59PMVFyi5SALGC9t-aKusWOXOuseRSw2cphSjWgAeToddIIQ1z-7tfdIwnJ3qMKpOnX9OPXPYOmaWwmZOKJg2AJD3eCuVxytk2igP0eJd6PKEzZNkoYZ5InOyHZOnJPDQYfHrrwZdE6FcnBC3dsI0IpEqLHfiSZJ03WING_Ek9KEbL1iU1HrDtD6a7V7ZjWldm5amjgRaPhBXGUSJaIFnz_m4lUpg3yY5uFGxlFfspyET_JnA3f7AVoZYnPBkzbSxmn_YuLNHXNx9g3Q55ycxcdLnjsXezzQJyt5wsTKFoDfpyDSFIvClEdPEEX1jwVziZwkaRpjdHwD9uxp7_bIyA0v9nzZmMxvL63ncW3qY2kr_bNYjID8uTPhfPn-G4xR-ofo_bnZmNm3zrT--GYBO2b-7vp4u5Gkeff5gtl80vXsqxmA-04N5PZDGnjydgmknM_vLbv54qmKJ_t-8XQWRSkm_lX53YyGw0XpYOT6y_O5-FiWNUip5HwSzPP9W9lCvWAH1Pvhxh4odR7htFwxZZthdzsHUSCrqreyovbYikQJD6iGH9FbPSpxCprW0zbuw8f5pPbxnORzZfGzWQ0Go4_owvPyuALOG-PFzZF_KyMA22-GM4WzuJuRO4-02ZEnkyr1Cgmoj115nf_KYgCibC3DXurKNEvgBLwC0CZcZRKHTDTIpwPnpuA7YDLZp81pTjK5kuLIUhS4HhhnCFDb0IMzRZGaLwYZPSMnESYqeDJZq8n2nB6x77as_ndZNx_SD_qD_OPhkZgNtG2Ye6SKIv1jrLhbUsz7wasgkeVT_zECRZlq_3FkxBaps9wh0LJS9lz_g5OrzPJ5B5o6SbxYolc7Nrnit0XzchzErxyXP93ajx8eICq2br5kV7tiV11vfA4QMPjKBSvndTGE0adYs_u5prxilu3o8oisyQkU0TzvVCkqlAkaaax70mi6kor3-ch1MzHyAt14i0v-51fVobRqNgMyjwvO_0Vgt1DdPGNJGwHHJQVCAda5-pKK7A90C4t67ceUsAcQW-gda3Or-1bsW5fsc4_-z1LU9BHRrfTHkeHds9incu-pTgCOJ3elUUmAGGCSD2rxaQXAKpuuZ_CfHMjmQ4WSQavmyw5iOI98iNwIxGucgHbstdtsSfPlfvBz7DBYVeY4JmMnCzGPRQp70MoUgizV8ctGCCA7BtvyjUXvthIxmGYCp5KBnVg0ZbVtLQ35VLwoRNi8HZImCkcJLFuFAkt17kZSulZCPNYXYlyVcKNdvq_iIuYiPciEAlIqFMrykL33Ru3q07j4y22cuYyVSy92Woay367s6ImgnMnJJEVCb8nYhI9FeR-gfLoyfhXPVXv2RRZKnT2E8XMvmS-x1H1lAKJUFJF-C2ceX1A00e0vbRWy353VWuBRdRnrBQAL0ki31E6MOKg5fFUXK6K6Ki38iV677WKAESYQQZBQkdrFaC8Z_ZRYtewDfdF6PKEDJV8oOIaezZ6grmKDZf7UFc_h9uCNt_tZRt7oaWuG_mjnVRNP7ot5gAHGiVBv7d-xKWeGudqwJarUrwMGY72GDzVdTRA2kbdpRH_LlQnSh7EackBqHnsH2XDssGAWedM4OcUgMl9gGnMA8-hgkOmt6rLtlgAm4MoaOViWrsCS3IaTbnsE4zUT-y07PbrNsAj8rPuRa74Ca4nMAdBETwqNjABETDyrLodCqmWF5mWSXEBJQT9AqzXmee77Fr88EQ-A1SOZHwAbz3oab1MtgpHwmg7wfdggUAIg1sHWcVS-mfcIeBgoSsVQxF5IrDzrBKZcLeA_B2wyGDpHL4rQUyyNux4ph0xh5JdsN6Zht64J2CASx9BBI3RswZNBGYeJ7pQ8sgfyK9-srAu8NXBImzrngzYp2ugLdyxWwUPhVovFTYFpAPTjiGDIXSFrIJBGHg6BkM0ZR0psDBWdZyoLBYVxBHzeptzht8UpEhmUd-RwEb0AP3hRrC1kE8CbuUSLtJwq_eKYvsRqhZ1xgmUgwouGDUOjp6cFaDlgUrfBUQCkm20VB1cW62-_4A9o1rf7G1f2PegvJ68Z3Mal-jXLuHxHg5AuadlUDOAI5w0pmALF5hqP2qRGCV07-WLVXG60m9L3TfkyRcDra8V0NglnkuMwXKZQWQEyCwf8yIFG1rAH6NEWxUa_ChSx-dr4YOOdp500Iva94CBYFXEoRO9Y3Yu8Za5oxX-GHkQgDu8--iUJ63m_yv38380kvsOh5kaBnDgDjRfbGWZRU-d5mdUkv0Yrk4y1SE8ODPO8ZU1BYfJrcvahuWB_AroZHJNbeTEXO51hda6blnVoetWTrtT4IUAT_j7AcWbmrjEewC-8WNxaTjREAUR6EN6tkkPu9AyrctyauBgg2FbluqVi5RVGuE0ArHZtpA9ilRLA-77WqvSVvk9bO3zzXcoA9yfpbepp1rdcLRqcI8Z3KN4GuO9Co4zLyLOkcI8gu-t1yEfy5CPKuS_SYMaksooTKWjysGRcnBRz4kCEdzqpaMUHPJCt8xfIV25jQt1mSz8TvfRkz4tywv3y0rmzJQfhN4s_j2bwG0a_wP4fqRW&lang=sage). You can use it to create general distance plots. Any body that Horizons knows about can be given as the target, although only major bodies can be given as the center. There's an earlier version of the script [here](https://space.stackexchange.com/a/55061/38535), along with brief instructions.

The script uses the Horizons range data (& its time derivative) to compute a series of cubic Bézier curves passing through the data points. It determines the mean distance by exact integration of the Bézier curves.

> **Comments:**
> **username_14**: This is outstanding analysis, and (at least to me), is the most intuitive explanation and demonstration of why Mercury is (on average) the closest planet to any other planet in our solar system.  However, I don't see how it actually answers the original question.  
> **username_13**: Thanks, @Connor. True, I didn't attempt to answer the OP's question about possible complicated stellar systems that don't follow the "innermost is closest on average" rule, I just mentioned that it's quite hard to do (analytically). But in compensation I did provide actual data for Mercury's eccentric orbit. Also, my Horizons script *could* be used on bodies with more eccentric & inclined orbits. (Horizons can provide data for 1,145,682 asteroids and 3,771 comets).  
> **username_13**: FWIW, I primarily wrote this answer to complement the existing answers (which do discuss various "oddball" systems), and to add to the explanation from the Physics Today article which the OP linked.  
> **username_14**: It is an excellent complement to the existing answers, +1!  

---

### Answer by username_15 on 2021-11-01. Score: 7

Not for [double planets](http://en.wikipedia.org/wiki/Double_planet), for which each body is always the nearest planet to the other.

Pluto and Charon disputably fit the bill. [This 2006 International Astronomical Union report](http://iau.org/static/public_press/iau_po_report.pdf) states

> Q: Is Pluto a planet?
>
> A: Yes. In fact, Pluto’s large companion named
> Charon is also large enough and massive enough to satisfy the
> definition of “planet”. Because Pluto and Charon are gravitationally
> bound together, they are actually now considered to be a “double
> planet.”

Even if one doesn't consider them planets, an Earth-Theia-like collision in another system could conceivably produce a pair with a barycenter outside each body.

[The wiki](http://en.wikipedia.org/wiki/Double_planet) also states

> ...the Moon currently migrates outward from Earth at a rate of
> approximately 3.8 cm (1.5 in) per year; in a few billion years, the
> Earth–Moon system's center of mass will lie outside Earth, which would
> make it a double-planet system.

> **Comments:**
> **username_16**: -1 for cherry-picking the resolution that was described, but not passed.  
> **username_4**: Actually any 2 planets with the same orbital period around the sun, which double planets also have I guess, potentially have a shorter distance to each other than to the innermost planet, I think  
> **username_17**: @username_4Beckers Sharing an orbit requires a very specific spatial configuration. Aside from double planets, you'd need to have a body at the L4 or L5 Lagrange points as described in username_9's answer, but both bodies might not be considered planets.  
> **username_15**: @username_16: That's why I wrote "disputably". It depends on the definition of "planet" and is subject to change at any time.  

---

### Answer by username_18 on 2021-11-01. Score: 7

The reason Mercury is the closest planet to all of the planets on average is in part because of the fact that, because of Kepler's laws, period scales with the semimajor axis of an orbit, and so big orbits are going to be slow and spend lots of time far away from any given object because the orbits are so slow. At opposition, for any two objects, they will be (given a near-circular orbit) the sum of the two semi-major axes away from each other. So naturally, if a planet has a small semimajor axis it's going to move fast around and be a lot closer a lot more of the time than another object with a larger semimajor axis and slower orbit.

This principle is easily expandable to other systems, as Kepler's laws apply to any planetary system. As for the binary system, dynamics close to a binary star system can get pretty complicated so I'm not sure what would happen exactly, but inasmuch as the binary system could be approximated as one mass because of distances, then this would still hold.

Let me know if this explanation is confusing; I can edit and reword this to make it clearer if necessary.

Edit: Upon further consideration and with the help of the comments, one needs to consider what kind of average we're looking at here. In our solar system, a lot of the orbits are near-circular, and in situations like these, circular approximations can yield the possibility of purely spatial averages because of uniform angular velocity. However, as seen by Kepler's laws, orbits are ellipsis, and technically except in the case of a perfectly circular orbit, have varying angular velocities (and by extension tangential velocities). For an orbit with an eccentricity that can't be approximated by a circular orbit, a temporal average must be considered as opposed to just a spatial one.

A good example of this is given by d\_e in the comment; he says

> Let's imagine 3 planets - one is very close to the Sun. the other 2 are very far (in aphelion) from the Sun and have about the same semimajor axis - and the same direction of the apsides line. Now, the trick is to make them have very very high eccentricity - so the time they spend near the sun is very short; therefore most of the time the planets will spend in their aphelion (which their aphelion, as we said is near to each other)

For circular orbits (or near thereunto) a simple thought experiment can make the reasoning for the mercury conclusion a bit clearer. Let's take mercury, Saturn and Uranus for example. The semi-major axes for each of these is (roughly) .4, 10 and 20 AU (again there's appreciable rounding going on here, but this is a heuristic example). Assuming circular orbits, Saturn will only ever be 9.6 or 10.4 AU away from mercury at all points in its orbit,which means the average would have to fall in there. For Saturn and Uranus, they will be about 10 AU at there closest, and 30 at their farthest. Given the range here, it's not difficult to imagine that the average will likely be a decent amount larger than 10.4, the absolute maximum possible distance average between Saturn and Mercury.

While this example is almost completely qualitative, I hope it's useful for believing the math and quantitative analysis presented by the link posted in the question.

For non-circular orbits, like d\_e illustrated quite well, this reasoning does not necessarily extend, and so the answer would be **no**, this does not extend generally. Each system would have to be analyzed individually. But for a group of near-circular orbits, this seems to be the case.

I have left my previous, faulty answer here as context for the comments below, which are quite illustrative of some important principles.

> **Comments:**
> **username_19**: I can't see how it is easily expandable to other systems. Let's imagine 3 planets - one is very close to the Sun. the other 2 are *very* far (in aphelion) from the Sun and have about the same semimajor axis - and the same direction of the apsides line. Now, the trick is to make them have very very high eccentricity - so the time they spend near the sun is very short; therefore most of the time the planets will spend in their aphelion (which their aphelion, as we said is near to each other). I believe in this case it will not generalize.  
> **username_20**: This orbital period argument doesn't make sense. A planet having a fast orbital period doesn't mean it's going to be "a lot closer a lot more of the time" to other planets. The orbital periods determine how *frequently* the orbits will line up, but not what overall proportion of time the bodies will spend close to or far from each other.  
> **username_20**: If you could magically speed up or slow down Mercury's orbit without changing the orbit's shape, it would *still* be the planet with the shortest time-averaged distance to Earth, or to other planets.  
> **username_20**: If you read the article linked in the question, you'll see that the math doesn't involve orbital periods at all.  
> **username_21**: There was a CGP Grey video about this: [Which Planet is the Closest?](https://www.youtube.com/watch?v=SumDHcnCRuU)  
> **username_18**: Perhaps I emphasized the wrong thing here; Its not the orbital period of mercury that matters as much as how long farther planet's orbital periods are; the fact that they are so long and also different from other planets means that they're going to be spending a lot of time away from each other, which will, in turn, affect the average. Sorry guys I don't think this is as clear as I wanted it to be, will edit in a bit when I have more time.  
> **username_18**: @username_20supportsMonica The assumptions in the articles linked are assuming circular orbits, which is a good approximation for the bodies in our solar system, but not for even moderately eccentric orbits, and like is illustrated by username_19, time dependence comes up because orbital velocities aren’t constant in elliptical orbits. Although I would revise my statement and say perhaps the period itself is less important, and perhaps what’s happening during the orbit time-wise. For a circular orbit the average can be purely spatial, but for these elliptical orbits, it would have to be a time average  
> **username_18**: Ok edited for clarity and to reflect good points mentioned in comments  
> **username_22**: This answer is embarrassingly wrong, from start to finish. How on earth does it have four upvotes?  
> **username_18**: @username_22 Care to actually add some useful insight? Maybe elaborate a bit more on points you think should be clarified or corrected  
> **username_23**: Kepler's laws don't have too much to do with this.  The result would be true even if all the planets moved at constant speed.  I'm finding it difficult to spot anything in this answer that's either useful or correct.  Downvoted.  Sorry.  
> **username_18**: @DawoodibnKareem First Kepler law states all orbits are elliptical, which leads to the second law saying that the orbital speed is not constant. You can look at username_19’s example for why this matters; in a highly eccentric orbit, the vast majority of the time it will be extremely far away from the object, making the average distance different than if it spent the same amount of time at all points of the orbit. Does this make sense? This of course is for temporal averages, since purely spatial averages are useless in this kind of scenario; it wouldn't mean anything.  
> **username_18**: This further leads to our conclusion, which is that the results in the paper linked by OP cannot be generalized, since they assume constant orbital speeds which allow spatial averages. username_5 also mentions this as a warning in the comments of the OP's post  
> **username_23**: No, a circular orbit complies with Kepler's laws, and has constant orbital speed.  The law that you've cited in your opening paragraph - that period scales with the semimajor axis - does not in any way affect the result about which planet is closest to any other.  
> **username_18**: @DawoodibnKareem If you read the end of the answer, you’ll see that "I have left my previous, faulty answer here as context for the comments below, which are quite illustrative of some important principles." The stuff post the ‘edit’ are the adjusted answer.  
> **username_23**: You have asserted without proof (or any kind of supporting argument) that the result for circular orbits does not extend to non-circular orbits; even though your first attempt at answering implied that it did.  In fact, your original answer was closer to the truth than the edit, even though its reasoning was faulty.  There are only a few edge cases where the result doesn't hold, such as double planets.  In a solar system where no pair of planets are gravitationally bound to each other, and no pair of planets are the same distance from the sun, the result DOES apply.  
> **username_18**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/131098/discussion-between-justin-tackett-and-dawood-ibn-kareem).  

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### Answer by username_14 on 2021-11-01. Score: 6

Not necessarily
The authors of the article linked above assume that the planets involved are

in roughly circular, concentric, and coplanar orbits

Suppose Mercury and Earth had circular co-planar orbits with Mercury at 0.4 AU and Earth at 1 AU from the Sun.  With a quick program and Monte Carlo analysis, we get the same result as the above referenced article author's that the average distance from the Earth to Mercury is 1.04 AU.
Alter the above scenario slightly so that Mercury now has a stellar orbital inclination offset from Earth by 90 degrees, i.e. their orbital planes are perpendicular.  Now a Monte Carlo analysis reveals an average distance of 1.0584 AU.
So, clearly (notwithstanding orbital instabilities) we could have an innermost planet with an inclination perpendicular to the ecliptic, which on average is further away from Earth than the next planet out with a co-planar inclination.
Add in the possibility for a highly eccentric orbiting innermost planet (where semi-major axis is our measure of stellar proximity) and we can get an even more glaring discrepancy in average distance since the vast amount of time for such a planet is spent far away from the star at apoapsis.
Note:
A specific exoplanet counterexample of this kind could be exceedingly difficult to find as:

planets tend to evolve in near co-planar orbits.
Such a system would almost certainly be unstable.
Perpendicular orbital planes would make detection of both planets via transit methods extremely rare.

Code co-planar:
n=1e7;
theta = rand(1,n)*2*pi;
points =[0.4*sin(theta)-1;0.4*cos(theta)];
aveDist = mean(sqrt(points(1,:).^2+points(2,:).^2));

Code perpendicular:
n=1e7;
theta = rand(1,n)*2*pi;
phi = rand(1,n)*2*pi;
points =[0.4*cos(phi).*sin(theta)-1;0.4*sin(phi).*cos(theta);0.4*cos(theta)];
aveDist = mean(sqrt(points(1,:).^2+points(2,:).^2+points(3,:).^2));

> **Comments:**
> **username_12**: "Such a system would almost certainly be unstable." - I think that's putting it mildly.  Say Mercury's orbit were tilted at 90 degrees to get the value 1.056.  For a planet *further out* to be closer on average it would still have to have an orbit within  0.47 au (Mercury is 0.39, Venus is 0.73).  I think that is close enough to Mercury to make instability a certainty, even if Mercury's orbit weren't tipped 90 degrees.  
> **username_14**: @JasonGoemaat Agreed.  Though I suppose all n-body systems are unstable, it is just a matter of the time-scale.  The point I make in this post is that even if we are unlikely to ever see a counter-example, we can't rule it out based purely on the orbital geometry.  

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### Answer by username_12 on 2021-11-08. Score: 4

The page you link to and the paper refer to PCM (the Point Circle Method), and the PCM treats the orbits of two objects as circular, concentric, and coplanar. If planets in a system had odd (and probably unstable) orbits, this might not be the case.

The PCM method treats the planet's position at any given time as a uniform probabilistic distribution. That means that if you look at a random point in time over the last billion years, the each planet has about the same probability of being at any point in it's orbit. What that gives you is the ability to treat one planet as a point (P), and calculate an average for all points in the other planet's circular orbit as if they are equally likely (C).

In general then for any two planets you look at, the average distance between them will get smaller as the inner planet's orbit gets smaller. If you made a list of the closest planets to any other, they would always be in order of closest distance to the sun (smallest orbit).

In the picture below there is the test planet (outer light blue), and two inner planets that are green and blue where we want to find which one is closest on average.

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

PCM uses a static position for the test planet (outer light blue) and calculates the average distance to each point on the circles representing the other planets' orbits.

You don't need to do any calculations to tell that the average distance will be smaller for the inner planet though. If we take any angle around the circle and plot a vertical line (on the right), we can see that the outer planet will be further to the side at that point in it's orbit. It will be closer 1/2 the time and further 1/2 the time vertically than the outer planet, but in all cases except for when they're aligned with our test planet, the outer planet will have a larger horizontal component to the distance.


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### Answer by username_19 on 2021-11-06. Score: 4

[This is complementary answer to username_18ackett answer. I felt somewhat compelled in light of the comments (even after edits), and to illustrate my example (referenced in the answer) with a diagram.]

I say there are 2 naïve situations where the innermost planet is not the closet. In the attached illustration there are only 2 planets and I consider the distance to the Sun (As we might assume the innermost planet is so close to the Sun - so it won't make any difference to take the distance to the Sun instead).

1. **In the first example the period (and therefore the semi-major axis) and the pehilion angle of the outer planets are the same. They differ only in eccentricity.** We can see clearly see that the distance between the planets is always shorter than the distance to the Sun. (I used to same orbit phase - but we can allow some small difference - also in the pehilion location)[![enter image description here](https://i.sstatic.net/B6hk1.gif)](https://i.sstatic.net/B6hk1.gif)

One can justly claim that this setting is not possible in real-life - and I'll agree with him. The value in my answer is primarily to demonstrate that there is nothing universal here that can be deduced from the Keplerian system; rather, that every system has its own characteristics.

2. The second example that was given in username_18ackett answer, includes **the two planets to have very high eccentricity and about the same semi-major axis and pehilion angle.** I admit I thought - in the numbers below given - the effect would be stronger; nevertheless even here it is quite easy to see that the average distance between the planets is shorter than the the distance to the Sun. Note that the diagram is not on scale: the X-axis is longer - had it on scale the orbits would seem much more like strong ellipses. At most of the time the planets are at X>3 while the sun is at 0. this enough to establish I believe my claim here. [![enter image description here](https://i.sstatic.net/UHpHM.gif)](https://i.sstatic.net/UHpHM.gif)

> **Comments:**
> **username_19**: FWIW, the actual numbers in case of someone doubts  (using different periods of time, and gap between measure to measure) is: Green planet to the Sun: ~3.51. Green planet to blue planet: ~2.08  

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### Answer by username_24 on 2021-11-04. Score: 3

Lets take a more general look at the situation in systems with one star:

What is the average distance to a planet? Break this down into two components, the distance along a line from the planet you are measuring from (which I will call O), through the barycenter of the planetary system which I will call X, and the distance the planet is offset from this line which I will call Y.

Now, what is the average X? While the planet will spend more time near it's apoapsis than it's periapsis this is independent of the orbit of the planet you are measuring from and thus will average to zero. Thus the average X is always the distance from O to the barycenter. This is the same for **all** planets.

This leaves the Y. Once again, the shifting angles between O and the target average things out--it's going to vary from zero to the average orbital radius of the target. My calculus is far too rusty to sum this but it doesn't matter--it's clearly dependent on the orbital radius and nothing else.

This we have X fixed and Y varying at the orbital radius. Thus sqrt(x^2 + y^2) will likewise be ordered by orbital radius although the relationship is no longer linear. Thus the average distance to a planet will be ordered by the orbital radius of the planet. The inner planet is always the closest.

Now, there are some special cases that have been presented in other answers: double planets, L4 & L5 objects. However, one of the requirements to be a planet is to have cleared one's orbit--and all such cases mean you have two objects in the orbit. Thus they can't be planets to be a closer planet.


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### Answer by username_25 on 2021-11-01. Score: 3

In a binary star system, there are two types of orbits which planets can have.

A planet in an S-type or non circumbinary orbit will orbit one of the two stars. Thus the distance between the two stars must be at least several times as great as the semi-major axis of the planet's orbit, if the planet is to have a stable orbit.

A planet in an p-type or circumbinary orbit will orbit both of the two stars. Thus the semi-major axis of the planet's orbit must be at least a few times as great as the distance between the two stars for the planet to have a stable orbit.

It is possible for a binary star system to have planets in S-type orbits around one of the stars, or both of the stars, and to have planets in P-type orbits around both of the stars.

Each planet around a star will have a forbidden zone around its orbit where no other planet can orbit because of gravitational interactions. The closer two planets are to their star, the smaller their forbidden zones will be, and the closer the two orbits can be.

If a binary system has two stars, A and B, and planets in S-type orbits around each of them, it seems obvious to me that the innermost planet around A will be - on average, as the question asks,-closest to each of the other planets orbiting around A. And similarly the innermost planet orbiting around B will be, on average, as the question asks, closest to each of the other planets orbiting around B.

If the separation between A and B must be at least, for example, at 5 times the distance of the orbit of the outermost planet around either star, we can make a picture of the system.

Assume that the outermost planet around each of the stars orbits at 10 units, and the separation of the two stars in their orbits is 50 units. Obviously the separation between the outermost planets of each star will vary between 40 to 60 units, with an average separation of about 50.99 units.

But an inner planet orbiting star A at 5 units, for example, planet A I, will always be between 5 and 15 units from the outermost planet orbiting Star A at 10 units, planet A II, and the average Separation between the two planets will be about 11.180 units according to the Pythagorean theorm.

So any planet orbiting star A will always be closer to any planet orbiting star A than any planet orbiting star B can ever get, even at its closest.

Assume that there are also planets in P-type or circumbinary orbits around the center of mass of the two stars A & B. Imagine that planet AB I orbits at 250 units and planet AB II orbits at 500 Units and planet AB III orbits at 1,000 units.

The distance bween AB I and AB II will always be between 250 and 750 units with an average of 559.01 units.

The distance between AB II and AB III will always be between 500 and 1,500 units with an average of 1118.03 units.

The distance between AB I and AB III Will always be between 750 and 1,250 units, with an average of 1,030.77 units.

So that indicates that the innermost planet orbiting in a P-type orbit around the center of gravity of the two stars A and B, namely planet AB I, will be the closest - on the average, as the question asks - to the other planets orbiting. in P-type orbits around both of the stars.

So that leaves the question of whether any of the planets in S-type orbits around only one of the stars, A or B, will be closer - on the average, as the question asks - to the planets in P-type orbits around the center of mass of the stars A & B.

Assuming that stars A & B have equal mass, they will orbit the center or mass or barycenter, at equal distances. Assuming almost perfectly circular orbits, each star will always be almost exactly 25 units from the barycenter.

I stated that no planet of A or B could have a stable orbit more than 10 units from its star. So no planet of A or B could ever get more 35 units from the barycenter of the two stars.

So a planet orbiting A or B could get between 215 and 285 units from planet AB I, with an average distance of about 251 units.

That means that a planet orbiting A or B could sometimes be closer to planet AB I than planet AB II is.

A planet orbiting A or B could get between 465 and 535 units from the planet AB II, with an average distance of about 500.646 units. Which is shorter than the average distance between AB I and AB II.

A planet orbiting A or B could get between 965 and 1,035 units from the planet AB III, with an average distance of about 1,000.3124 units. Which is shorter than the average distance between AB II and AB III and shorter than the averaage distance between AB I and AB III.

So apparently in a binary star system with planets in S-type orbits around one or each of the stars, as well as planets in P-type orbits around the center of mass of both the stars, the planets in S-type orbits will be closer - on the average, as defined in the question - to the planets in P-type orbits than any of the planets in P-type orbits will be - on the average, as defined in the question - to each other.

Of course I have only demonstrated that for the specific orbits I have chosen for the system in my example.


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### Answer by username_26 on 2021-11-01. Score: 2

A modification to this question that might clarify it:

Is the average distance between a planet in a standard orbit (one planet per circular-ish orbit) and its primary, always less than the average distance between the planet and any other planet orbiting the primary?

The nearer a planet is to the primary, the nearer its average distance to any other planet is to the average distance between the primary to any other planet.

> **Comments:**
> **username_17**: This doesn't address the aspect of how far an arbitrary pair of planets are from one another. This just states that if a star and planet are very close, the distance from anywhere to either the star or the planet is basically the same. The distance from Neptune to the Sun is nearly the same as from Neptune to Mercury, but that says nothing about how far it is from Neptune to anywhere else.  
> **username_26**: @NuclearHoagie I'll try to come up with how expand on this.   Essentially, the average distance from one planet and another will be the average between when both planets are one the same side of the primary and when they are on the opposite sides.  This average value will always be larger than the distance to the primary, and *I believe* that the farther from the primary the other planet is, the larger this value will be (regardless of where the initial planet is, relatively).  I'll see if I can get some values in a spreadsheet and confirm this.  

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