If the mass of 3rd body is negligible compared to the 2nd, it is
relatively easy to prove that L4 and L5 are stable if M1/M2 is larger
than about 25 and unstable otherwise. But what about non-negligible
mass of 3rd body?
L4 and L5 is not actually the only stable 1:1 mean motion resonance.
There is Cruithne and another asteroid on the orbit of Earth, with
horseshoe orbits. And there are Janus and Epimetheus, with horseshoe
shared orbits - and mass ratio of just 3,6.
Is there any mathematical demonstration about how large the M2/M3 ratio
could be for stable L4/L5? And how does the orbit behave around the
limit for stability?
> Just when is the L4/L5 resonance stable?
>
> If the mass of 3rd body is negligible compared to the 2nd, it is
> relatively easy to prove that L4 and L5 are stable if M1/M2 is larger
> than about 25 and unstable otherwise. But what about non-negligible
> mass of 3rd body?
In a past thread someone derived an equation relating the stability
criterion including all three masses (though I can't speak to its
accuracy). You can probably dig it up with a Google search.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM, Y!M erikmaxfrancis
Come not between the dragon and his wrath.
-- King Lear (Act I, Scene I)
> Unsuccessfully. Nothing credible showed up. The Lagrange solution
> applies only to restricted three-body problem. This does not prove
> absence of stable solutions for the unrestricted problem...
This is the article by Joseph Hertzlinger I was referring to:
http://groups.google.com/group/rec.arts.sf.science/msg/fed748de2c10622a
He later gives a citation for where he got the source material, though I
don't think anyone had a copy of the book he referenced so couldn't
check it separately.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM, Y!M erikmaxfrancis
A man can stand a lot as long as he can stand himself.
-- Axel Munthe
And it is:
In order to be stable, the masses of the three bodies must obey the
formula:
27*(m_1*m_2 + m_2*m_3 + m_3*m_1) < (m_1 + m_2 + m_3)^2
Rearranging gives:
27 * [m_1*(m_2+m_3)+m_2*m_3]<[m_1+(m_2+m_3)] ^2
So, no constraints on the mass ratio of secondary and tertiary...
> In order to be stable, the masses of the three bodies must obey the
> formula:
>
>
> 27*(m_1*m_2 + m_2*m_3 + m_3*m_1) < (m_1 + m_2 + m_3)^2
>
> Rearranging gives:
> 27 * [m_1*(m_2+m_3)+m_2*m_3]<[m_1+(m_2+m_3)] ^2
>
> So, no constraints on the mass ratio of secondary and tertiary...
What makes you say that? The term m_2/m_3 never appears in that
relation, but both m_2 and m_3 do, so the result is certainly dependent
on their relative sizes.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM, Y!M erikmaxfrancis
The little I know, I owe to my ignorance.
-- Sacha Guitry
Not your question, but have you seen:
http://www.cse.ucsc.edu/~charlie/3body/
Assuming m_2 + m_3 = 1, and writing m_2 = r, m_3 = (1-r), 0 <= r <= 1,
your version becomes:
27 * (m_1 + r - r^2) < (m_1 + 1)^2
where 0 <= r-r^2 <= 1/4. Specifically, r-r^2 <= 1, so a sufficient
(but not necessary) condition for stability is:
m_1 > 26 * (m_2 + m_3)
So, indeed, as long as the mass of the primary body is more than 26
times the combined mass of the secondary and tertiary bodies, their
mass ratio doesn't seem to matter.
In fact, the actual limit is somewhat lower, since we know that r-r^2
is quite a bit less than one. Numerically, the lowest value of m_1 /
(m_2 + m_3) to guarantee stability seems to be a bit less than 25.23.
Conversely, if the ratio m_1 / (m_2 + m_3) is less than about 24.96,
the system is unstable no matter what the mass ratio of the smaller
bodies is. Only in that rather narrow range does the m_2 / m_3 ratio
matter, with equal masses being less stable than unequal.
The exact limits are actually [25 + sqrt(25^2 + 23)]/2 =~ 25.2279 for
the m_2 = m_3 = 1/2 case, and [25 + sqrt(25^2 - 4)]/2 =~ 24.9599 for
the m_2 = 1, m_3 = 0 case, as can be seen by solving the quadratic
equation m_1^2 - 25*m_1 + 1 - 27*(r-r^2) = 0 for m_1 with r constant.
--
Ilmari Karonen
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