Planets evenly spaced on log scale

https://www.johndcook.com/blog/2018/04/05/solar-system-on-log-scale/

by tacon • 8 years ago

205 66 8 years ago

debt• 8 years ago

Was kind of hoping to see pictures of the planets themselves, evenly spaced on a log scale. Those line graphs were a little less exciting.

rcthompson5• 8 years ago

Unfortunately, I believe we only have decent photos for one of the mentioned solar systems.

EGreg• 8 years ago

Would have made for a great illustration, methinks!

bdamm2• 8 years ago

Graphs of exoplanet orbits exist and are easy to locate:

For example: http://www.openexoplanetcatalogue.com/planet/KOI-1843.03/

It isn't on a log scale but this example is easy to see as a counterexample to the OP as there are two planets out in large orbits that orbit close together with a large gap between the first planet.

topspin• 8 years ago

Resolving extrasolar planets so difficult that one can concisely list[1] all of the planets that have ever been resolved and adding an entry is a significant achievement. Directly imaging extrasolar planets is only possible given certain conditions (type of star, orbit, planet size/type, etc.) The vast bulk of extrasolar planets cannot be resolved directly with contemporary instruments.

[1] https://en.wikipedia.org/wiki/List_of_directly_imaged_exopla...

outworlder2• 8 years ago

We cannot even take a picture of Pluto that's bigger than a few pixels without sending a probe there. Not sure images from extrasolar systems, even if available, would be very exciting.

bdamm2• 8 years ago

It's an interesting conjecture, but what about the anomalies? There are over a hundred planet systems discovered by Kepler with various levels of confirmation. None of them are mentioned here. The essay only lists 9 systems. How do the rest compare? Are there anomalies in unconfirmed systems? Why were these systems chosen; was it a random selection?

rcthompson5• 8 years ago

The post explains exactly why these systems were chosen: because they have the most planets. This is every known system with 6 or more planets (according to the OP).

olympus3• 8 years ago

You need a system with more than two planets. A system with only two planets is going to be linearly spaced on any type of scale.

Once you've filtered out systems with two or fewer planets, you need orbital measurements with decent precision to tell whether a plot of planet spacing is actually linear on a log scale or not. Measuring a planet's orbit requires a decent amount of observation (since we can't measure star/planet mass directly and don't usually measure period directly), so unconfirmed systems likely don't have good measurements.

It's likely that most of the unmentioned systems get filtered out by one of the above two criteria.

grkvlt2• 8 years ago

Also, even if we detect a system with N > 3 exo-planets, we often cannot tell if there are more than N, but below the threshold of detectability, making the plots incomplete and incorrect.

kpwagner• 8 years ago

Fascinating that this relationship scales from solar systems in which the planets are close to the star to solar systems that are more spread out. I wonder if the mass of the star has an effect on the average distance that the mass of the rest of the solar system is from the star. I also wonder if the mass of the star has a correlation with the tightness of the logarithmic distance relationship exhibited in this article.

delecti• 8 years ago

Based on the graph for our system, it seems more likely that the mass of the planets affects the correlation than the mass of the star. Mars is a bit small (relative to earth) and is a slight dip from a straight line, whereas Jupiter which is quite large is a jump up above the straight line. If hypothetically the planets beyond Jupiter were all small and rocky then it seems they'd dip back down past the straight line again.

hinkley2• 8 years ago

I suspect some expert will come in and tell us that there is some fundamental of orbital mechanics which correlated to the observations. That is, stable planetary systems have orbits that correlate with the area of the orbit (the square of the distance from the star)

I recall one of the earliest breakthroughs in orbital mechanics was someone figuring out how the area of a slice of a elliptical orbit was the same anywhere in the orbit, when the angle of the arc is calculated as a unit of time and not degrees.

Not to say you’re wrong. I think you’re right, but the quality you’re attributing to the system is a behavior of the system that has an underlying link to physics that may already be well explored. Just nobody has bothered to put a pretty plot in front of us armchair types and students before.

vkou• 8 years ago

I recall one of the earliest breakthroughs in orbital mechanics was someone figuring out how the area of a slice of a elliptical orbit was the same anywhere in the orbit, when the angle of the arc is calculated as a unit of time and not degrees.

That would be Kepler's second law. It states that a line between the sun and the planet sweeps equal areas in equal periods of time.

rcthompson5• 8 years ago

There's an extra large jump from Mars to Jupiter, but that's because there's an asteroid belt between them, which might be an aborted planet. If you put the average distance of the asteroid belt as planet number 5, I bet it would look even more linear.

olympus3• 8 years ago

??? Adding a point in between two other points literally can't make something more linear, it can only make it less linear.

rcthompson5• 8 years ago

I mean that the asteroid belt would be counted as a 5th planet, and Jupiter would be 6, and so on.

afthonos• 8 years ago

The x-axis is the planet (+aborted planet) count. You’d be shifting all later planets over by one.

blacksmith_tb• 8 years ago

Some models[1] suggest that Neptune and Uranus formed closer to the sun, and then migrated to their current orbits later (explanations vary between Jupiter and Saturn pulling them out, or some extrasolar mass).

1: https://en.wikipedia.org/wiki/Neptune#Formation_and_migratio...

dr_zoidberg• 8 years ago

The asteroid belt doesn't have nearly enough mass to be/have been a planet[0]. Largest thing there is Ceres, followed by Vesta, of which only the first qualifies as a dwarf planet.

There was an hypothesis about it being the remnants of a destroyed planet[1], but that was mostly an idea to support the Titius-Bode law[2], which was disproved a few hundred years ago.

[0] Currently about 3x Ceres mass, though it may have about Earth mass early in its history. See https://en.wikipedia.org/wiki/Asteroid_belt#Formation and https://en.wikipedia.org/wiki/Asteroid_belt#Evolution

[1] https://en.wikipedia.org/wiki/Phaeton_(hypothetical_planet)#...

[2] https://en.wikipedia.org/wiki/Titius%E2%80%93Bode_law

mdpopescu• 8 years ago

... the Titius-Bode law[2], which was disproved a few hundred years ago

Er... the Wikipedia page you're citing at [2] has this to say:

"Results from simulations of planetary formation support the idea that a randomly chosen stable planetary system will likely satisfy a Titius–Bode law."

and

"96% of these exoplanet systems adhere to a generalized Titius–Bode relation to a similar or greater extent than the Solar System does."

stromgo• 8 years ago

The qualifier "dwarf" doesn't mean that the celestial body is too small to be considered a planet, it means that it hasn't cleared the neighborhood around its orbit. (Blame the IAU if you find this counterintuitive). This means that your conclusion is wrong: the asteroid belt actually has enough mass to be a planet.

rcthompson5• 8 years ago

If the asteroid belt originally had the mass of Earth, then I'd say it's fair to count it as a planet in this context, assuming that the orbits of the planets were established fairly early and have stayed about the same since then.

outworlder2• 8 years ago

Is it possible that this was a planet, but most of the mass was ejected out of the system?

grkvlt2• 8 years ago

That's why the second graph on the page is 'Our Solar System with Ceres' showing exactly that.

perlgeek2• 8 years ago

One has to be careful with inspecting log graphs by mere looks. What looks like a small difference from expected to actual value in log scale can be very large difference in terms of actual values.

So it bothers me a bit that there is no actual fit and error margins.

liberte82• 8 years ago

What looks like a small deviation is only actually a large difference on a linear scale, in these graphs. On a logarithmic scale, it’s a small difference. That’s the point of scaling it logarithmically in the first place, to show how closely the planets line up on this scale.

perlgeek2• 8 years ago

The purpose of plotting it on a logarithmic scale is to demonstrate a power law, not necessarily to obscure a deviation from the linear-in-log power law.

floil• 8 years ago

Moreover, x and y are not independent on this plot -- x is just the rank of y. A graph constructed this way can never dip down. So a linear-looking upward trend is basically guaranteed.

petters• 8 years ago

Completely agree. I made a plot with a uniform distribution: https://www.strandmark.net/Planets.html

So I am not sure there is anything going on here.

wpietri• 8 years ago

Mar's Law: "Everything is linear if plotted on a log-log scale with a fat magic marker."

It's #6 on the excellent (and surprisingly software-applicable) "Akin's Laws of Spacecraft Design": http://spacecraft.ssl.umd.edu/akins_laws.html

joveian• 8 years ago

Those are good. My favorites:

16. The previous people who did a similar analysis did not have a direct pipeline to the wisdom of the ages. There is therefore no reason to believe their analysis over yours. There is especially no reason to present their analysis as yours.

21. (Larrabee's Law) Half of everything you hear in a classroom is crap. Education is figuring out which half is which.

29. (von Tiesenhausen's Law of Program Management) To get an accurate estimate of final program requirements, multiply the initial time estimates by pi, and slide the decimal point on the cost estimates one place to the right.

30. (von Tiesenhausen's Law of Engineering Design) If you want to have a maximum effect on the design of a new engineering system, learn to draw. Engineers always wind up designing the vehicle to look like the initial artist's concept.

amelius• 8 years ago

Something tells me that the mass of the planets should be in the conjecture somewhere, because small objects have little meaning and perhaps would not even classify as planets.

mikekij• 8 years ago

...sort of like what you’d expect inside a simulation.

Retra• 8 years ago

A simulation is very general concept. You may as well have said "sort of like what you'd expect to hear with your ears."

Obviously you hear things with your ears. And equally obviously, you can expect things in a simulation.

rdiddly• 8 years ago

Anyone care to speculate based on physics why this might have worked out this way?

yonkshi• 8 years ago

Indulge my wild guess (I'm not a physicist):

If the proto solar system's matter distribution was a Gaussian-like distribution, it would make sense that matter density droped in a logarithmic scale, and thus planets formed droped logarithmically w.r.t radius.

More intuitively.. further away from the sun has less probability for rocks/particles to interact with each other.

Someone• 8 years ago

In our solar system, the small planets are closest to the sun.

Jupiter, with 72% of all planetary mass, is more than twice as heavy as all other planets, combined.

Saturn, Uranus and Neptune take most of the remaining 28% (https://en.wikipedia.org/wiki/List_of_Solar_System_objects_b...)

(Of course, there also is way more volume that far away)

clord• 8 years ago

Another wild guess: a sphere of gas collapsed to a rotating disk. If you take a sphere and squash it flat, the middle has most of the mass, and it declines to zero as you get farther. I would guess the pattern of falloff of a squashed sphere is roughly logarithmic. That pattern gets quantized into planets, and so we see the original spherical distribution in our planets.

jacquesm• 8 years ago

That makes a lot of sense.

alan• 8 years ago

We might be able to start using this as a way to specify planets, rather than the arbitrary number-by-discovery. Try StarName(with catalog info)#log(approximate radius of orbit)

Earth would we Sol#0 Kepler-90#-1.3, Kepler-90#-1.1, out to Kepler-90#0 for the outermost one on his graph.

That way, finding a new planet doesn't require either renumbering to specify where the planet is relative to the star. (BTW, I'm not real happy about using AU as the base measurement. It might be better to use Megameters since you're less likely to end up with negative log)

saagarjha• 8 years ago

What would you do for highly elliptical orbits?

jwiley• 8 years ago

Some more discussion on Reddit:

https://www.reddit.com/r/askscience/comments/58plow/why_are_...

wl• 8 years ago

Mar's law: Everything is linear when plotted on a log-log scale with a fat marker.

(I know this is a lin-log plot)

antognini2• 8 years ago

As a former dynamicist, the underlying reason for this correlation is that planetary systems with many planets are always on the edge of dynamical instability. In other words, these kinds of planetary systems with lots of planets are always packed with as many planets as they can support over any given lifetime. It then turns out that the stability criteria imply that the system cannot be any more packed than having approximately log-equal spacing. In fact, if you take a star and put planets around it with a log uniform distribution and evolve it for a while, you find that eventually the planets that remain roughly fit the Bode-Titus law [1]. (I don't think the reasons for this are entirely well understood.)

One consequence of this is that our own solar system has been around for about 5 billion years, and is therefore probably unstable on timescales of 5-50 billion years. Indeed, long-term simulations indicate that there is a few percent chance that Mercury will get kicked out of the solar system or crash into Venus or the Sun before the Sun dies in ~5 billion years [2] [3].

[1] https://arxiv.org/abs/astro-ph/9710116

[2] http://www.scholarpedia.org/article/Stability_of_the_solar_s....

[3] http://adsabs.harvard.edu/abs/2009Natur.459..817L

yummybear• 8 years ago

I wonder what effects Mercury crashing into the sun would have - large solar flares? Nothing at all?

dbasedweeb3• 8 years ago

Nothing much. The Sun is most of the mass in the Solar system, by over 97%. Most of the rest of that mass is Jupiter, followed at quite a distance by Saturn. Jupiter has more than 5700 times the mass of Mercury. In other words, next to the Sun, Mercury is a teeny little speck that would barely even leave a “splash” as it roasted to a cinder in its corona. It would be the most underwhelming event you can imagine, for Sol. For Mercury it would be an exciting journey of vaporizing outer layers, being torn to little shreds, and those shreds vaporizing completely (all before it contacted anything like a “surface”).

antognini2• 8 years ago

I doubt there would be much of an effect on anything. Something like that is unlikely to interact much with the Sun's magnetic fields (so it wouldn't cause any solar flares). It would probably produce a relatively large sunspot. Comet Shoemaker-Levy produced observable spots on Jupiter and the relative mass of Mercury to the Sun is much larger than Shoemaker-Levy to Jupiter (and the relative diameter is much larger as well). I bet that there would be a measurable increase in the metallicity immediately after impact. This would last probably about an hour before convection caused the debris from the planet to be spread evenly throughout the Sun's convection zone (at which point it would be unobservable).

rhelzerm• 8 years ago

Anything is evenly spaded on a log scale...

Planets evenly spaced on log scale

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