# Mass Needed to Clear an Orbital Neighborhood

In 2006 the IAU deemed that Pluto was no longer a planet because it fails to "clear" the neighborhood around its Kuiper Belt orbit. Presumably, this is because Pluto (1.305E22 kg) has insufficient mass to do the job. How massive must a body in Pluto's orbit (semi-major axis 39.5 AU) be to "clear" its orbit? Would Mars (6.24E23 kg) or Earth (5.97E24 kg) be declared planets if they were in Pluto's orbit?

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It's probably unfair to ask this type of quantitive question. Let's rephrase: What are the factors and methodologies in determining the mass of a body in Pluto orbit that satisfies IAU's planet criterion? – Michael Luciuk Mar 30 '11 at 1:28
if you want to change your question, you can "edit" it directly rather than leaving a comment. (Good question, btw) – Mark Eichenlaub Mar 30 '11 at 2:50

Dear Michael, there is an article dedicated to this criterion:

http://en.wikipedia.org/wiki/Clearing_the_neighbourhood

The relevant quantity is $$\Lambda / \Lambda_E = \frac{M^2 / P}{M_E^2 / P_E}$$ where $M$ is the mass of the would-be planet and $P$ is the orbital period. The subscript $E$ means that the value is for the Earth. The definition above guarantees the ratio is $1$ for the Earth and $1.08$ for its evil sister, Venus. Uranus and Neptune are close with 2.51 and 1.79, respectively. It's 8,500 and 300 for Jupiter and Saturn, 0.01 and 0.006 for Mercury and Mars. Pluto, much like Ceres, Eris, Makemake, and Haumea have figures between $10^{-9}$ and $10^{-7}$ - too small.

The "threshold" was chosen to be $1/153,000$ (by Stern and Levison in 2000, I don't understand where the threshold came from - probably some model of how the planets are clearing the neighborhood as a function of $\Lambda$) so the non-planets are beneath it. All the explanations should be in this paper:

http://www.boulder.swri.edu/~hal/PDF/planet_def.pdf

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Thank you Lubos for the references. From the WIKI site, it appears that a body 11.7x Pluto's mass, about twice that of our Moon, would "clear" it's orbital neighborhood. – Michael Luciuk Mar 30 '11 at 13:23
@Michael: Well...note that the required mass depends on the length of the semi-major axis. – dmckee Mar 30 '11 at 16:46