Clearing the neighbourhood

"Clearing the neighbourhood around its orbit" is a criterion for a celestial body to be considered a planet in the Solar System. This was one of the three criteria adopted by the International Astronomical Union (IAU) in its 2006 definition of planet.^[1] In 2015, a proposal was made to use the criterion in extending the definition to exoplanets.^[2]

In the end stages of planet formation, a planet (as so defined) will have "cleared the neighbourhood" of its own orbital zone, meaning it has become gravitationally dominant, and there are no other bodies of comparable size other than its natural satellites or those otherwise under its gravitational influence. A large body that meets the other criteria for a planet but has not cleared its neighbourhood is classified as a dwarf planet. This includes Pluto, which is constrained in its orbit by the gravity of Neptune and shares its orbital neighbourhood with Kuiper belt objects. The IAU's definition does not attach specific numbers or equations to this term, but all the planets have cleared their neighbourhoods to a much greater extent (by orders of magnitude) than any dwarf planet, or any candidate for dwarf planet.

The phrase may be derived from a paper presented to the general assembly of the IAU in 2000 by Alan Stern and Harold F. Levison. The authors used several similar phrases as they developed a theoretical basis for determining if an object orbiting a star is likely to "clear its neighboring region" of planetesimals, based on the object's mass and its orbital period.^[3] Steven Soter prefers to use the term "dynamical dominance"^[4] and Jean-Luc Margot notes that such language "seems less prone to misinterpretation".^[2]

Clearly distinguishing "planets" from "dwarf planets" and other minor planets had become necessary because the IAU had adopted different rules for naming newly discovered major and minor planets, without establishing a basis for telling them apart. The naming process for Eris stalled after the announcement of its discovery in 2005, pending clarification of this first step.

Criteria

The phrase refers to an orbiting body (a planet or protoplanet) "sweeping out" its orbital region over time, by gravitationally interacting with smaller bodies nearby. Over many orbital cycles, a large body will tend to cause small bodies either to accrete with it, or to be disturbed to another orbit, or to be captured either as a satellite or into a resonant orbit. As a consequence it does not then share its orbital region with other bodies of significant size, except for its own satellites, or other bodies governed by its own gravitational influence. This latter restriction excludes objects whose orbits may cross but that will never collide with each other due to orbital resonance, such as Jupiter and its trojans, Earth and 3753 Cruithne, or Neptune and the plutinos.^[3] As to the extent of orbit clearing required, Jean-Luc Margot emphasises "a planet can never completely clear its orbital zone, because gravitational and radiative forces continually perturb the orbits of asteroids and comets into planet-crossing orbits" and states that the IAU did not intend the impossible standard of impeccable orbit clearing.^[2]

Stern–Levison's Λ

In their paper, Stern and Levison sought an algorithm to determine which "planetary bodies control the region surrounding them".^[3] They defined Λ (lambda), a measure of a body's ability to scatter smaller masses out of its orbital region over a period of time equal to the age of the Universe (Hubble time). Λ is a dimensionless number defined as

\Lambda ={\frac {m^{2}}{a^{\frac {3}{2}}}}\,k

where m is the mass of the body, a is the body's semi-major axis, and k is a function of the orbital elements of the small body being scattered and the degree to which it must be scattered. In the domain of the solar planetary disc, there is little variation in the average values of k for small bodies at a particular distance from the Sun.^[4]

If Λ > 1, then the body will likely clear out the small bodies in its orbital zone. Stern and Levison used this discriminant to separate the gravitionally rounded, Sun-orbiting bodies into überplanets, which are "dynamically important enough to have cleared its neighboring planetesimals", and unterplanets. The überplanets are the eight most massive solar orbiters (i.e. the IAU planets), and the unterplanets are the rest (i.e. the IAU dwarf planets).

Soter's µ

Steven Soter proposed an observationally based measure µ (mu), which he called the "planetary discriminant", to separate bodies orbiting stars into planets and non-planets.^[4] Per Soter, two bodies are defined to share an orbital zone if their orbits cross a common radial distance from the primary, and their non-resonant periods differ by less than an order of magnitude. The order-of-magnitude similarity in period requirement excludes comets from the calculation, but the combined mass of the comets turns out to be negligible compared to the other small Solar System bodies, so their inclusion would have little impact on the results. µ is then calculated by dividing the mass of the candidate body by the total mass of the other objects that share its orbital zone. It is a measure of the actual degree of cleanliness of the orbital zone. Soter proposed that if µ > 100, then the candidate body be regarded as a planet.

Margot's Π

Astronomer Jean-Luc Margot has proposed a discriminant, Π, that can categorise a body based only on its own mass, its semi-major axis, and its star's mass.^[2] Like Stern–Levison's Λ, Π is a measure of the ability of the body to clear its orbit, but unlike Λ, it is solely based on theory and does not use empirical data from the Solar System. Π is based on properties that are feasibly determinable even for exoplanetary bodies, unlike Soter's µ, which requires an accurate census of the orbital zone.

\Pi ={\frac {m}{M^{\frac {5}{2}}a^{\frac {9}{8}}}}\,k

where m is the mass of the candidate body in Earth masses, a is its semi-major axis in AU, M is the mass of the parent star in solar masses, and k is a constant chosen so that Π > 1 for a body that can clear its orbital zone. k depends on the extent of clearing desired and the time required to do so. Margot selected an extent of $2{\sqrt {3}}$ times the Hill radius and a time limit of the parent star's lifetime on the main sequence (which is a function of the mass of the star). Then, in the mentioned units and a main-sequence lifetime of 10 billion years, k = 807.^{[lower-alpha 1]} The body is a planet if Π > 1. The minimum mass necessary to clear the given orbit is given when Π = 1.

Π is based on a calculation of the number of orbits required for the candidate body to impart enough energy to a small body in a nearby orbit such that the smaller body is cleared out of the desired orbital extent. This is unlike Λ, which uses an average of the clearing times required for a sample of asteroids in the asteroid belt, and is thus biased to that region of the Solar System. Π's use of the main-sequence lifetime means that the body will eventually clear an orbit around the star; Λ's use of a Hubble time means that the star might disrupt its planetary system (e.g. by going nova) before the object is actually able to clear its orbit.

The formula for Π assumes a circular orbit. Its adaptation to elliptical orbits is left for future work, but Margot expects it to be the same as that of a circular orbit to within an order of magnitude.

Numerical values

Below is a list of planets and dwarf planets ranked by Margot's planetary discriminant Π, in decreasing order.^[2] For all eight planets defined by the IAU, Π is orders of magnitude greater than 1, whereas for all dwarf planets, Π is orders of magnitude less than 1. Also listed are Stern–Levison's Λ and Soter's µ; again, the planets are orders of magnitude greater than 1 for Λ and 100 for µ, and the dwarf planets are orders of magnitude less than 1 for Λ and 100 for µ. Also shown are the distances where Π = 1 and Λ = 1 (where the body would change from being a planet to being a dwarf planet).

Rank	Name	Margot's planetary discriminant Π	Soter's planetary discriminant µ	Stern–Levison parameter Λ ^{[lower-alpha 2]}	Mass (kg)	Type of object	Π = 1 distance (AU)	Λ = 1 distance (AU)
1	Jupiter	7004400000000000000♠4.0×10⁴	7005625000000000000♠6.25×10⁵	7009130000000000000♠1.30×10⁹	7027189860000000000♠1.8986×10²⁷	5th planet	7004640000000000000♠64,000	7006622000000000000♠6220000
2	Saturn	7003610000000000000♠6.1×10³	7005190000000000000♠1.9×10⁵	7007468000000000000♠4.68×10⁷	7026568460000000000♠5.6846×10²⁶	6th planet	7004220000000000000♠22,000	7006125000000000000♠1,250,000
3	Venus	7002950000000000000♠9.5×10²	7006130000000000000♠1.3×10⁶	7005166000000000000♠1.66×10⁵	7024486850000000000♠4.8685×10²⁴	2nd planet	7002320000000000000♠320	7003218000000000000♠2,180
4	Earth	7002810000000000000♠8.1×10²	7006170000000000000♠1.7×10⁶	7005153000000000000♠1.53×10⁵	7024597360000000000♠5.9736×10²⁴	3rd planet	7002380000000000000♠380	7003287000000000000♠2,870
5	Uranus	7002420000000000000♠4.2×10²	7004290000000000000♠2.9×10⁴	7005384000000000000♠3.84×10⁵	7025868320000000000♠8.6832×10²⁵	7th planet	7003410000000000000♠4,100	7005102000000000000♠102,000
6	Neptune	7002300000000000000♠3.0×10²	7004240000000000000♠2.4×10⁴	7005273000000000000♠2.73×10⁵	7026102430000000000♠1.0243×10²⁶	8th planet	7003480000000000000♠4,800	7005127000000000000♠127,000
7	Mercury	7002130000000000000♠1.3×10²	7004910000000000000♠9.1×10⁴	7003195000000000000♠1.95×10³	7023330220000000000♠3.3022×10²³	1st planet	7001290000000000000♠29	7001600000000000000♠60
8	Mars	7001540000000000000♠5.4×10¹	7003510000000000000♠5.1×10³	7002942000000000000♠9.42×10²	7023641850000000000♠6.4185×10²³	4th planet	7001530000000000000♠53	7002146000000000000♠146
9	Ceres	6998400000000000000♠4.0×10⁻²	6999330000000000000♠0.33	6996832000000000000♠8.32×10⁻⁴	7020943000000000000♠9.43×10²⁰	dwarf planet	6999160000000000000♠0.16	6998240000000000000♠0.024
10	Pluto	6998279999999999999♠2.8×10⁻²	6998800000000000000♠0.08	6997295000000000000♠2.95×10⁻³	7022129000000000000♠1.29×10²²	dwarf planet	7000170000000000000♠1.70	6999812000000000000♠0.812
11	Eris	6998200000000000000♠2.0×10⁻²	6999100000000000000♠0.10	6997215000000000000♠2.15×10⁻³	7022167000000000000♠1.67×10²²	dwarf planet	7000210000000000000♠2.10	7000112990000099999♠1.130
12	Haumea	6997780000000000000♠7.8×10⁻³	6998200000000000000♠0.02^[5]	6996241000000000000♠2.41×10⁻⁴	7021400000000000000♠4.0×10²¹	dwarf planet	6999579990000000000♠0.58	6999168000000000000♠0.168
13	Makemake	6997730000000000000♠7.3×10⁻³	6998200000000000000♠0.02^[5]	6996222000000000000♠2.22×10⁻⁴	7021400000000000000♠~4.0×10²¹	dwarf planet	6999579990000000000♠0.58	6999168000000000000♠0.168
Note: 1 light-year ≈ 7004632410000000000♠63,241 AU

Disagreement

Orbits of celestial bodies in the Kuiper belt with approximate distances and inclination. Objects marked with red are in orbital resonances with Neptune, with Pluto (the largest red circle) located in the "spike" of plutinos at the 2:3 resonance

Stern, currently leading NASA's New Horizons mission, disagrees with the reclassification of Pluto on the basis of its inability to clear a neighbourhood. One of his arguments is that the IAU's wording is vague, and that—like Pluto—Earth, Mars, Jupiter and Neptune have not cleared their orbital neighbourhoods either. Earth co-orbits with 10,000 near-Earth asteroids (NEAs), and Jupiter has 100,000 trojans in its orbital path. "If Neptune had cleared its zone, Pluto wouldn't be there", he has said.^[6]

However, Stern himself co-developed one of the measurable discriminants: Stern and Levison's Λ. In that context he stated, "we define an überplanet as a planetary body in orbit about a star that is dynamically important enough to have cleared its neighboring planetesimals ..." and a few paragraphs later, "From a dynamical standpoint, our solar system clearly contains 8 überplanets"—including Earth, Mars, Jupiter, and Neptune.^[3] Although he proposed this to define dynamical subcategories of planets, he still rejects it for defining what a planet essentially is, advocating the use of intrinsic attributes^[7] over dynamical relationships.

Notes

↑ This expression for k can be derived by following Margot's paper as follows: The time required for a body of mass m in orbit around a body of mass M with an orbital period P is: $t_{clear}=P{\frac {\delta x^{2}}{D_{x}^{2}}}$ With $\delta x\simeq {\frac {C}{a}}\left({\frac {m}{3M}}\right)^{\frac {1}{3}},D_{x}\simeq {\frac {10}{a}}{\frac {m}{M}},P=2\pi {\sqrt {\frac {a^{3}}{GM}}},$ and C the number of Hill radii to be cleared. This gives $t_{clear}=2\pi {\sqrt {\frac {a^{3}}{GM}}}{\frac {C^{2}}{a^{2}}}\left({\frac {m}{3M}}\right)^{\frac {2}{3}}{\frac {a^{2}M^{2}}{100m^{2}}}={\frac {2\pi }{100{\sqrt {G}}}}{\frac {C^{2}}{3^{\frac {2}{3}}}}a^{\frac {3}{2}}M^{\frac {5}{6}}m^{-{\frac {4}{3}}}$ requiring that the clearing time t_clear to be less than a characteristic timescale t_* gives: $t_{*}\geq t_{clear}=2\pi {\sqrt {\frac {a^{3}}{GM}}}{\frac {C^{2}}{a^{2}}}\left({\frac {m}{3M}}\right)^{\frac {2}{3}}{\frac {a^{2}M^{2}}{100m^{2}}}={\frac {2\pi }{100{\sqrt {G}}}}{\frac {C^{2}}{3^{\frac {2}{3}}}}a^{\frac {3}{2}}M^{\frac {5}{6}}m^{-{\frac {4}{3}}}$ this means that a body with a mass m can clear its orbit within the designated timescale if it satisfies $m\geq {\left[{\frac {2\pi }{100{\sqrt {G}}}}{\frac {C^{2}}{3^{\frac {2}{3}}t_{*}}}a^{\frac {3}{2}}M^{\frac {5}{6}}\right]}^{\frac {3}{4}}={{\left({\frac {2\pi }{100{\sqrt {G}}}}\right)}^{\frac {3}{4}}{\frac {C^{\frac {3}{2}}}{{\sqrt {3}}{t_{*}}^{\frac {3}{4}}}}a^{\frac {9}{8}}M^{\frac {5}{8}}}$ This can be rewritten as follows ${\frac {m}{m_{Earth}}}\geq {{\left({\frac {2\pi }{100{\sqrt {G}}}}\right)}^{\frac {3}{4}}{\frac {C^{\frac {3}{2}}}{{\sqrt {3}}{t_{*}}^{\frac {3}{4}}}}{\left({\frac {a}{a_{Earth}}}\right)}^{\frac {9}{8}}{\left({\frac {M}{M_{Sun}}}\right)}^{\frac {5}{8}}{\frac {a_{Earth}^{\frac {9}{8}}M_{Sun}^{\frac {5}{8}}}{m_{Earth}}}}$ so that the variables can be changed to use solar masses, Earth masses, and distances in AU by ${\frac {M}{M_{Sun}}}\to {\bar {M}},{\frac {m}{m_{Earth}}}\to {\bar {m}},$ and ${\frac {a}{a_{Earth}}}\to {\bar {a}}$ Then, equating t_* to be the main-sequence lifetime of the star t_MS, the above expression can be rewritten using $t_{*}\simeq t_{MS}\propto {\left({\frac {M}{M_{Sun}}}\right)}^{-{\frac {5}{2}}}t_{Sun},$ with t_Sun the main-sequence lifetime of the Sun, and making a similar change in variables to time in years ${\frac {t_{Sun}}{P_{Earth}}}\to {\bar {t}}_{Sun}.$ This then gives ${\bar {m}}\geq {\left({\frac {2\pi }{100{\sqrt {G}}}}\right)}^{\frac {3}{4}}{\frac {C^{\frac {3}{2}}}{{\sqrt {3}}{{\bar {t}}_{Sun}}^{\frac {3}{4}}}}{\bar {a}}^{\frac {9}{8}}{\bar {M}}^{\frac {5}{2}}{\frac {a_{Earth}^{\frac {9}{8}}M_{Sun}^{\frac {5}{8}}}{m_{Earth}P_{Earth}^{\frac {3}{4}}}}$ Then, the orbital-clearing parameter is the mass of the body divided by the minimum mass required to clear its orbit (which is the right-hand side of the above expression) and leaving out the bars for simplicity gives the expression for Π as given in this article: $\Pi ={\frac {m}{m_{clear}}}={\frac {m}{a^{\frac {9}{8}}M^{\frac {5}{2}}}}{\left({\frac {100{\sqrt {G}}}{2\pi }}\right)}^{\frac {3}{4}}{\frac {{\sqrt {3}}{t_{Sun}}^{\frac {3}{4}}}{C^{\frac {3}{2}}}}{\frac {m_{Earth}P_{Earth}^{\frac {3}{4}}}{a_{Earth}^{\frac {9}{8}}M_{Sun}^{\frac {5}{8}}}}.$ which means that $k={\left({\frac {100{\sqrt {G}}}{2\pi }}\right)}^{\frac {3}{4}}{\frac {{\sqrt {3}}{t_{Sun}}^{\frac {3}{4}}}{C^{\frac {3}{2}}}}m_{Earth}P_{Earth}^{\frac {3}{4}}a_{Earth}^{-{\frac {9}{8}}}M_{Sun}^{-{\frac {5}{8}}}$ Earth's orbital period can then be used to remove a_Earth and P_Earth from the expression: $P_{Earth}=2\pi {\sqrt {\frac {{a_{Earth}}^{3}}{M_{Sun}G}}},$ which gives $k={\left({\frac {100{\cancel {\sqrt {G}}}}{\cancel {2\pi }}}\right)}^{\frac {3}{4}}{\frac {{\sqrt {3}}{t_{Sun}}^{\frac {3}{4}}}{C^{\frac {3}{2}}}}m_{Earth}{\left({\cancel {2\pi }}{\sqrt {\frac {\cancel {{a_{Earth}}^{3}}}{M_{Sun}{\cancel {G}}}}}\right)}^{\frac {3}{4}}{\cancel {a_{Earth}^{-{\frac {9}{8}}}}}M_{Sun}^{-{\frac {5}{8}}},$ so that this becomes $k={\sqrt {3}}C^{-{\frac {3}{2}}}(100t_{Sun})^{\frac {3}{4}}{\frac {m_{Earth}}{M_{Sun}}}$ Plugging in the numbers gives k = 807.
↑ These values are based on a value of k estimated for Ceres and the asteroids belt: k equals 1.53 × 10⁵ AU^1.5/M_⊕², where AU is the astronomical unit and M_⊕ is the mass of Earth. Accordingly, Λ is dimensionless.

References

↑ "IAU 2006 General Assembly: Result of the IAU Resolution votes". IAU. 24 August 2006. Retrieved 2009-10-23.
1 2 3 4 5 Margot, Jean-Luc (2015-10-15). "A Quantitative Criterion for Defining Planets". The Astronomical Journal. 150 (6): 185–191. arXiv:1507.06300. Bibcode:2015AJ....150..185M. doi:10.1088/0004-6256/150/6/185.
1 2 3 4 Stern, S. Alan; Levison, Harold F. (2002). "Regarding the criteria for planethood and proposed planetary classification schemes" (PDF). Highlights of Astronomy. 12: 205–213, as presented at the XXIVth General Assembly of the IAU–2000 [Manchester, UK, 7–18 August 2000]. Bibcode:2002HiA....12..205S.
1 2 3 Soter, Steven (2006-08-16). "What Is a Planet?". The Astronomical Journal. 132 (6): 2513–2519. arXiv:astro-ph/0608359. Bibcode:2006AJ....132.2513S. doi:10.1086/508861.
1 2 Calculated using the estimate for the mass of the Kuiper belt found in Iorio, 2007 of 0.033 Earth masses
↑ Rincon, Paul (25 August 2006). "Pluto vote 'hijacked' in revolt". BBC News. Retrieved 2006-09-03.
↑ "Pluto's Planet Title Defender: Q & A With Planetary Scientist Alan Stern". Space.com. 24 August 2011. Retrieved 2016-03-08.

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[5] This expression for k can be derived by following Margot's paper as follows: The time required for a body of mass m in orbit around a body of mass M with an orbital period P is: $t_{clear}=P{\frac {\delta x^{2}}{D_{x}^{2}}}$ With $\delta x\simeq {\frac {C}{a}}\left({\frac {m}{3M}}\right)^{\frac {1}{3}},D_{x}\simeq {\frac {10}{a}}{\frac {m}{M}},P=2\pi {\sqrt {\frac {a^{3}}{GM}}},$ and C the number of Hill radii to be cleared. This gives $t_{clear}=2\pi {\sqrt {\frac {a^{3}}{GM}}}{\frac {C^{2}}{a^{2}}}\left({\frac {m}{3M}}\right)^{\frac {2}{3}}{\frac {a^{2}M^{2}}{100m^{2}}}={\frac {2\pi }{100{\sqrt {G}}}}{\frac {C^{2}}{3^{\frac {2}{3}}}}a^{\frac {3}{2}}M^{\frac {5}{6}}m^{-{\frac {4}{3}}}$ requiring that the clearing time t_clear to be less than a characteristic timescale t_* gives: $t_{*}\geq t_{clear}=2\pi {\sqrt {\frac {a^{3}}{GM}}}{\frac {C^{2}}{a^{2}}}\left({\frac {m}{3M}}\right)^{\frac {2}{3}}{\frac {a^{2}M^{2}}{100m^{2}}}={\frac {2\pi }{100{\sqrt {G}}}}{\frac {C^{2}}{3^{\frac {2}{3}}}}a^{\frac {3}{2}}M^{\frac {5}{6}}m^{-{\frac {4}{3}}}$ this means that a body with a mass m can clear its orbit within the designated timescale if it satisfies $m\geq {\left[{\frac {2\pi }{100{\sqrt {G}}}}{\frac {C^{2}}{3^{\frac {2}{3}}t_{*}}}a^{\frac {3}{2}}M^{\frac {5}{6}}\right]}^{\frac {3}{4}}={{\left({\frac {2\pi }{100{\sqrt {G}}}}\right)}^{\frac {3}{4}}{\frac {C^{\frac {3}{2}}}{{\sqrt {3}}{t_{*}}^{\frac {3}{4}}}}a^{\frac {9}{8}}M^{\frac {5}{8}}}$ This can be rewritten as follows ${\frac {m}{m_{Earth}}}\geq {{\left({\frac {2\pi }{100{\sqrt {G}}}}\right)}^{\frac {3}{4}}{\frac {C^{\frac {3}{2}}}{{\sqrt {3}}{t_{*}}^{\frac {3}{4}}}}{\left({\frac {a}{a_{Earth}}}\right)}^{\frac {9}{8}}{\left({\frac {M}{M_{Sun}}}\right)}^{\frac {5}{8}}{\frac {a_{Earth}^{\frac {9}{8}}M_{Sun}^{\frac {5}{8}}}{m_{Earth}}}}$ so that the variables can be changed to use solar masses, Earth masses, and distances in AU by ${\frac {M}{M_{Sun}}}\to {\bar {M}},{\frac {m}{m_{Earth}}}\to {\bar {m}},$ and ${\frac {a}{a_{Earth}}}\to {\bar {a}}$ Then, equating t_* to be the main-sequence lifetime of the star t_MS, the above expression can be rewritten using $t_{*}\simeq t_{MS}\propto {\left({\frac {M}{M_{Sun}}}\right)}^{-{\frac {5}{2}}}t_{Sun},$ with t_Sun the main-sequence lifetime of the Sun, and making a similar change in variables to time in years ${\frac {t_{Sun}}{P_{Earth}}}\to {\bar {t}}_{Sun}.$ This then gives ${\bar {m}}\geq {\left({\frac {2\pi }{100{\sqrt {G}}}}\right)}^{\frac {3}{4}}{\frac {C^{\frac {3}{2}}}{{\sqrt {3}}{{\bar {t}}_{Sun}}^{\frac {3}{4}}}}{\bar {a}}^{\frac {9}{8}}{\bar {M}}^{\frac {5}{2}}{\frac {a_{Earth}^{\frac {9}{8}}M_{Sun}^{\frac {5}{8}}}{m_{Earth}P_{Earth}^{\frac {3}{4}}}}$ Then, the orbital-clearing parameter is the mass of the body divided by the minimum mass required to clear its orbit (which is the right-hand side of the above expression) and leaving out the bars for simplicity gives the expression for Π as given in this article: $\Pi ={\frac {m}{m_{clear}}}={\frac {m}{a^{\frac {9}{8}}M^{\frac {5}{2}}}}{\left({\frac {100{\sqrt {G}}}{2\pi }}\right)}^{\frac {3}{4}}{\frac {{\sqrt {3}}{t_{Sun}}^{\frac {3}{4}}}{C^{\frac {3}{2}}}}{\frac {m_{Earth}P_{Earth}^{\frac {3}{4}}}{a_{Earth}^{\frac {9}{8}}M_{Sun}^{\frac {5}{8}}}}.$ which means that $k={\left({\frac {100{\sqrt {G}}}{2\pi }}\right)}^{\frac {3}{4}}{\frac {{\sqrt {3}}{t_{Sun}}^{\frac {3}{4}}}{C^{\frac {3}{2}}}}m_{Earth}P_{Earth}^{\frac {3}{4}}a_{Earth}^{-{\frac {9}{8}}}M_{Sun}^{-{\frac {5}{8}}}$ Earth's orbital period can then be used to remove a_Earth and P_Earth from the expression: $P_{Earth}=2\pi {\sqrt {\frac {{a_{Earth}}^{3}}{M_{Sun}G}}},$ which gives $k={\left({\frac {100{\cancel {\sqrt {G}}}}{\cancel {2\pi }}}\right)}^{\frac {3}{4}}{\frac {{\sqrt {3}}{t_{Sun}}^{\frac {3}{4}}}{C^{\frac {3}{2}}}}m_{Earth}{\left({\cancel {2\pi }}{\sqrt {\frac {\cancel {{a_{Earth}}^{3}}}{M_{Sun}{\cancel {G}}}}}\right)}^{\frac {3}{4}}{\cancel {a_{Earth}^{-{\frac {9}{8}}}}}M_{Sun}^{-{\frac {5}{8}}},$ so that this becomes $k={\sqrt {3}}C^{-{\frac {3}{2}}}(100t_{Sun})^{\frac {3}{4}}{\frac {m_{Earth}}{M_{Sun}}}$ Plugging in the numbers gives k = 807.

[6] These values are based on a value of k estimated for Ceres and the asteroids belt: k equals 1.53 × 10⁵ AU^1.5/M_⊕², where AU is the astronomical unit and M_⊕ is the mass of Earth. Accordingly, Λ is dimensionless.

[IAU_definition-1] "IAU 2006 General Assembly: Result of the IAU Resolution votes". IAU. 24 August 2006. Retrieved 2009-10-23.

[Margot_2015-2] 1 2 3 4 5 Margot, Jean-Luc (2015-10-15). "A Quantitative Criterion for Defining Planets". The Astronomical Journal. 150 (6): 185–191. arXiv:1507.06300. Bibcode:2015AJ....150..185M. doi:10.1088/0004-6256/150/6/185.

[Stern_2002-3] 1 2 3 4 Stern, S. Alan; Levison, Harold F. (2002). "Regarding the criteria for planethood and proposed planetary classification schemes" (PDF). Highlights of Astronomy. 12: 205–213, as presented at the XXIVth General Assembly of the IAU–2000 [Manchester, UK, 7–18 August 2000]. Bibcode:2002HiA....12..205S.

[Soter_2006-4] 1 2 3 Soter, Steven (2006-08-16). "What Is a Planet?". The Astronomical Journal. 132 (6): 2513–2519. arXiv:astro-ph/0608359. Bibcode:2006AJ....132.2513S. doi:10.1086/508861.

[est-7] 1 2 Calculated using the estimate for the mass of the Kuiper belt found in Iorio, 2007 of 0.033 Earth masses

[8] Rincon, Paul (25 August 2006). "Pluto vote 'hijacked' in revolt". BBC News. Retrieved 2006-09-03.

[Stern_Interview-9] "Pluto's Planet Title Defender: Q & A With Planetary Scientist Alan Stern". Space.com. 24 August 2011. Retrieved 2016-03-08.