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Introducing a third object into the Earth-Moon system

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Frank Scrooby

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Jul 7, 2004, 3:07:03 AM7/7/04
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Hi all

How massive an object could you introduce into the Earth-Moon without
hideously screwing everything up?

For purposes of a story I'd like to write I want a very dense object to
suddenly arrive in Earth orbit (about half-way between Earth and the Moon -
and no it is not a singularity - its nowhere near that dense!).

I am happy (more than happy) for it to trash any or all of Earth's
artificial satellites, de-orbit, impact, gravity sling, whatever.

I don't mind it having some environmental impact on Earth (like say an 1 mm
adjustment in tides) or the occasional (once monthly) stronger than usual
storm but on the whole I don't want it shredding the planet's atmosphere,
ripping up pieces of the crust or generally causing any extinction level
events.

So how big can this thing be? How many billions of tons?

And is half-way a convenient orbit?

Would it take this object one week to orbit the earth at this distance?

How long can this system remain stable with both the Earth and the Moon's
gravity wells affecting this object?

If there is an online resource where I can find these answers out please
point me in that direction.

Thanks and regards
Frank Scrooby


anon...@coolgroups.com

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Jul 7, 2004, 10:42:03 AM7/7/04
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From: "Frank Scrooby"

"
And is half-way a convenient orbit?

Would it take this object one week to orbit the earth at
this distance?
"

This contradicts the Third Law of Kepler!
A halfway orbit lasts 10 days. To get an orbit of 1 week,
you will need an orbit at about 40 % of the distance.

"
I don't mind it having some environmental impact on Earth
(like say an 1 mm adjustment in tides) or the occasional
(once monthly) stronger than usual storm but on the whole I
don't want it shredding the planet's atmosphere, ripping up
pieces of the crust or generally causing any extinction
level events.
"

The tides range up to 18 m.
I cannot decide whether the height of tide is proportional
to the mass of the body raising it or to the square root of
the mass. In any case I would expect that a body with the
density of Moon and, say, 140 km across would be in the
right range to cause tidal adjustments of up to 1 mm, if on
the orbit of Moon.

The tidal force increases with the inverse cube of distance,
and inverse square of orbital period. Thus a body 60 km
across with lunar density would be OK on the orbit with a
period of a week. Just how big did you want the density to be?

"
How long can this system remain stable with both the Earth
and the Moon's
gravity wells affecting this object?
"

And the gravity of the Sun, too.

Well, Jupiter, Saturn and Uranus have multiple big
sattellites. And these have presumably been stable for 4,5
milliards of years.

Hop David

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Jul 7, 2004, 1:39:20 PM7/7/04
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An L4 or L5 orbit is one of the more stable.
In such a configuration the object, moon and earth centers
would form the corners of an equilateral triangle. The object would be
about 400,000 kilometers from the earth and 400,000 kilometers from the moon
L4
/\
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
Earth---------------Moon
\ /
\ /
\ /
\ /
\ /
\ /
\ /
\ /
\ /
\/
L3


If I recall correctly an L4 or L5 object shouldn't be more than 1/26 the
mass of the moon.

Google "Lagrange Points".

--
Hop David
http://clowder.net/hop/index.html

Hop David

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Jul 7, 2004, 2:05:58 PM7/7/04
to

Frank Scrooby wrote:
> Hi all
>
> How massive an object could you introduce into the Earth-Moon without
> hideously screwing everything up?
>
> For purposes of a story I'd like to write I want a very dense object to
> suddenly arrive in Earth orbit (about half-way between Earth and the Moon -
> and no it is not a singularity - its nowhere near that dense!).
>
> I am happy (more than happy) for it to trash any or all of Earth's
> artificial satellites, de-orbit, impact, gravity sling, whatever.
>
> I don't mind it having some environmental impact on Earth (like say an 1 mm
> adjustment in tides) or the occasional (once monthly) stronger than usual
> storm but on the whole I don't want it shredding the planet's atmosphere,
> ripping up pieces of the crust or generally causing any extinction level
> events.
>
> So how big can this thing be? How many billions of tons?
>
> And is half-way a convenient orbit?
>
> Would it take this object one week to orbit the earth at this distance?

Sorry, had read your post too hurriedly before my first reply.

For convenience lets use units Lunar Distance (LD) for length and Months
(a month is a good approximation of the Moon's orbital period).

If the distance is n LD then its period n^(3/2) months.

For example if it is half a lunar distance away, then it will orbit the
earth in (1/2)^(3/2) months. Which is .3535 months or about ten days.


There are some problems. If this object arrives from outside of the
solar system, then it will be traveling a parabolic (or maybe even
hyperbolic) trajectory. On its arrival it will be traveling at least 12
km/sec wrt to the earth. In which case it would not be trapped by
earth's gravity (escape velocity is 11.2 km/sec at earth's surface).
You need some way to slow the object down upon arrival if it's going to
hang around our neighborhood.

>
> How long can this system remain stable with both the Earth and the Moon's
> gravity wells affecting this object?

Some orbits are much less stable than others. If you use an L4 or L5
orbit as I described in my first reply, it will stay around for quite
awhile.

If the object was slowed down by grazing earth's atmosphere, then its
perigee would graze earth's atmosphere every circuit. In this case it
would probably fall to the earth before too long.

>
> If there is an online resource where I can find these answers out please
> point me in that direction.
>
> Thanks and regards
> Frank Scrooby
>
>

George W. Harris

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Jul 7, 2004, 2:23:58 PM7/7/04
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Hop David <hopspageHA...@tabletoptelephone.com> wrote:

:There are some problems. If this object arrives from outside of the

:solar system, then it will be traveling a parabolic (or maybe even
:hyperbolic) trajectory. On its arrival it will be traveling at least 12
:km/sec wrt to the earth. In which case it would not be trapped by
:earth's gravity (escape velocity is 11.2 km/sec at earth's surface).
:You need some way to slow the object down upon arrival if it's going to
:hang around our neighborhood.

Would a close pass to the moon, killing much
of the object's velocity, work? Perhaps the object
could end up in a highly eccentric orbit with a period a
fraction of the moon's (say 1/2 or 2/3). Could that
possibly be stable?

--
They say there's air in your lungs that's been there for years.

George W. Harris For actual email address, replace each 'u' with an 'i'.

Mike Williams

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Jul 7, 2004, 2:25:57 PM7/7/04
to
Wasn't it Frank Scrooby who wrote:
>Hi all
>
>How massive an object could you introduce into the Earth-Moon without
>hideously screwing everything up?
>
>For purposes of a story I'd like to write I want a very dense object to
>suddenly arrive in Earth orbit (about half-way between Earth and the Moon -
>and no it is not a singularity - its nowhere near that dense!).
>
>I am happy (more than happy) for it to trash any or all of Earth's
>artificial satellites, de-orbit, impact, gravity sling, whatever.
>
>I don't mind it having some environmental impact on Earth (like say an 1 mm
>adjustment in tides) or the occasional (once monthly) stronger than usual
>storm but on the whole I don't want it shredding the planet's atmosphere,
>ripping up pieces of the crust or generally causing any extinction level
>events.
>
>So how big can this thing be? How many billions of tons?

It's possible to calculate the size of an object that causes a 1mm
change in tidal range if you know a sensible value for the current tidal
range. (Tidal ranges seem to vary so much due to local geography that
it's hard to get a firm figure for a global value.)

Let's suppose that the current tidal range is 5 metres. The tidal force
is (to a first approximation) proportional to the mass and inversely
proportional to the cube of the distance. So for an object at half the
distance to have no more than 1/5000 of the tidal effect it would have
to have a mass of no more than 1/40000 of the Moon. That's about 2
million billion tons (1.8 * 10^18 kilograms).

I guess it's possible that some of your other requirements may impose a
small mass limit, it's just that that one can be objectively calculated.

--
Mike Williams
Gentleman of Leisure

Erik Max Francis

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Jul 7, 2004, 3:16:50 PM7/7/04
to
Hop David wrote:

> If I recall correctly an L4 or L5 object shouldn't be more than 1/26
> the
> mass of the moon.

No, the secondary body (the Moon in your example) needs to be less than
(about) 1/26 the mass of the primary body (the Earth in your example).
The mass of the object actually attempting to reside in the Trojan point
must be of negligible mass compared to the other two.

--
__ Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
/ \ San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
\__/ My life was better before I knew you.
-- Edith Wharton (to Morton Fullerton)

Hop David

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Jul 7, 2004, 8:07:03 PM7/7/04
to

Erik Max Francis wrote:
> Hop David wrote:
>
>
>>If I recall correctly an L4 or L5 object shouldn't be more than 1/26
>>the
>>mass of the moon.
>
>
> No, the secondary body (the Moon in your example) needs to be less than
> (about) 1/26 the mass of the primary body (the Earth in your example).
> The mass of the object actually attempting to reside in the Trojan point
> must be of negligible mass compared to the other two.
>

Thanks for the correctio.

Chuck Stewart

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Jul 7, 2004, 10:02:06 PM7/7/04
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On Wed, 07 Jul 2004 12:38:46 -0700, Erik Max Francis wrote:

> You're going to have a hard time coming up with some mechanism that
> provides that deltavee naturally.

There's one proven method for a body of astronomical mass, but it would
require another largish body and one of _those_ coincidences... have your
inbound body slam into a smaller body. In the example above your target
body, a rocky/iron object, is impacted by a large comet as it's
traversing the Earth/Moon system.

Iknow... cheesy... but it would work :)

--
Chuck Stewart
"Anime-style catgirls: Threat? Menace? Or just studying algebra?"

Phillip Thorne

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Jul 7, 2004, 11:26:33 PM7/7/04
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Frank Scrooby asked:

>> How massive an object could you introduce into the Earth-Moon without
>> hideously screwing everything up?

ObVidSF: In the 1984 "Transformers" story "The Ultimate Doom," Evil
Decepticon Leader Megatron brings his home planet of Cybertron into
cislunar space, specifically to harvest energy from its tide-driven
meteorologic influence (wave generators, etc.) on Earth -- not that
the term "tidal effect" was used, this being a toon for pre-teens.

(Cybertron's exact physical properties are unclear in the show. It
seems to be smaller than Luna, but it retains an oxygen atmosphere,
and human characters brought to its surface seem to experience
Earth-normal gravity. Given that it consists of numerous metallic
levels, possibly built atop an asteroidal core -- like an inverse
Zebrowski Macrolife habitat -- it may contain gravity generators.)

On Wed, 07 Jul 2004, Erik Max Francis <m...@alcyone.com> responded in
detail:
>[Tidal amplitudes in water]
>The real problem you have here is how the object gets captured.
>[...]


>You're going to have a hard time coming up with some mechanism that
>provides that deltavee naturally.

The problem of capture velocities is neatly solved (er, entirely
bypassed) by shipping the planet via Space Bridge, an artificial
temporary wormhole that apparently compensates for relative
velocities.

The plausibility of the Heroic Autobots' eventual solution, of
detonating Megatron's freighter of Energon Cubes to propel Cybertron
out of Earth orbit, does not bear close scrutiny. :)

/- Phillip Thorne ----------- The Non-Sequitur Express --------------------\
| org underbase ta thorne www.underbase.org It's the boundary |
| net comcast ta pethorne site, newsletter, blog conditions that |
\------------------------------------------------------- get you ----------/

Stan

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Jul 8, 2004, 2:01:21 AM7/8/04
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Erik Max Francis <m...@alcyone.com> wrote:

}Turning a passby into a
}capture requires an application of deltavee on the intruder -- even if
}the object were headed so that its perigee is right at the 200 000 km
}distance mark, when it gets there something is going to have to
}circularize its orbit, because otherwise it will just head out again.

}You're going to have a hard time coming up with some mechanism that
}provides that deltavee naturally.

Hmmm...what about a tangential grazing of the moon, taking out only the
very top of a ridge or peak? It would leave it in an elliptical orbit,
but it does seem plausible (however improbable).

Stan.

Erik Max Francis

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Jul 8, 2004, 2:03:04 AM7/8/04
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Stan wrote:

> Hmmm...what about a tangential grazing of the moon, taking out only
> the
> very top of a ridge or peak? It would leave it in an elliptical
> orbit,
> but it does seem plausible (however improbable).

To shed the required speed, you're probably talking about something much
more like a major crash rather than a graze.

--
__ Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
/ \ San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis

\__/ There are countless planets, like many island Earths ...
-- Konstantin Tsiolkovsky

Mad Bad Rabbit

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Jul 8, 2004, 2:31:56 AM7/8/04
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Erik Max Francis <m...@alcyone.com> wrote:

> You're going to have a hard time coming up with some mechanism that
> provides that deltavee naturally.

Of course if it's an artificial object, it can just fire up the
obscenely-huge antimatter drive and insert itself into cislunar
orbit (no doubt frying all our satellites and providing a lovely
worldwide aurora show at the same time.)


--
>;K

Frank Scrooby

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Jul 8, 2004, 4:43:28 AM7/8/04
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Hi all


"Mad Bad Rabbit" <madbad...@yahoo.com> wrote in message
news:yo2dnVYzFeX...@texas.net...


I am afraid that I have not provide enough information, and that the premise
I am trying to create is probably so horribly flawed as to be unworkable.

The object in question is highly artifical, some leftover ammunition from a
cosmic conflict that was over before our star was born. The beings who
created it are long gone, extinct possibly at their own hand. When you use
Objects like these as ammunition your wars tend to be ... well ... highly
fatal.

The object and its twin (which is somewhere very far away) are the
containers for a worm-hole. It is not captured into the Earth moon system so
much as it 'appears'. One second people are looking up at a nice ordinary
night sky and suddenly there this black splotch occluding stars. That is as
much hand-waving as I want to do. I want everything else to be HARD SCIENCE.

I want this insanely dense, unwelcomed visitor to behave exactly like a
planetoid of its mass would. And I want it screw up Earth's locale exactly
like a previously unnoticed satellite of its mass would.

I am quite happy to have it wrecking satellite orbits, smashing global
telecoms, causing the evacuation (and possible subsequent destruction) of
ISS, etc. Depending on where it might be it might even vacuum up the Van
Allen radiation belts ;-)

I am not happy about it causing massive changes in Earth's tide or such
like. A one meter increase in the daily high tides at a place not so far
from where I am right now will result in the routine and probably permanent
flooding on the CBD of a city (the main street is a meter below sea level,
the beach front is the only thing keeping the sea out). There are some
people who would probably welcome this sort of thing. Its great for urban
renewal, and Hollywood blockbusters. And what happens in Durban South Africa
will happen in dozens cities all around the world.

I don't want to write a disaster story, so I would like to keep the Object's
effects on Earth's eco-and weather-system to a minimum. This might mean
making the Object smaller than I have estimated below.

It does not need to be in a regular orbit. Something high eliptical (sp?)
would be fine

My initial estimate on its mass and apparent density (based on what I want
it to be made out of it) are:

+- 14.804 billion billion tons with a apparent density of 28 billion tons
per cubic meter.

It appears to be a sphere 1000 meters in diameter. It is 100% black, except
when something touches it and then there is a bright flash of multicolored
light for a millisecond or two during which the whole object is hidden from
view.

By comparison

(according to http://www.nineplanets.org/earth.html)

Earht's mass is +- 5972 billion billion tons making this sucker just 0.24 %
of Earth's mass (if I got all my numbers right). Earth density is a
relatively modest 5.52 grams per cubic centimeter, which translates into
5.52 tons per cubic meter (if I get all my numbers right again).

The Moon's mass is +- 73.5 billion billion tons (same source), making the
Object's mass nearly 20% that of the Moon's. The Moon's Density is only
3.34 grams per cubic centimeter which translates to 3.34 tons per cubic
meters (standard disclaimer).

>
>
> --
> >;K


Can this be made to work?

Erik Max Francis

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Jul 8, 2004, 4:50:42 AM7/8/04
to
Frank Scrooby wrote:

> The object and its twin (which is somewhere very far away) are the
> containers for a worm-hole. It is not captured into the Earth moon
> system so
> much as it 'appears'. One second people are looking up at a nice
> ordinary
> night sky and suddenly there this black splotch occluding stars. That
> is as
> much hand-waving as I want to do. I want everything else to be HARD
> SCIENCE.

If it's highly dense, then it's not going to be that large, which means
that in cislunar space it's not going to occlude many stars.

The rest I already responded to with numbers.

--
__ Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
/ \ San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis

\__/ It seems like Karma's making his rounds
-- Kina

Mike Williams

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Jul 8, 2004, 7:42:23 AM7/8/04
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Wasn't it Frank Scrooby who wrote:
>The Moon's mass is +- 73.5 billion billion tons (same source), making the
>Object's mass nearly 20% that of the Moon's.

If you only want the tides to rise by 1 millimetre, and you want it to
be 20% of the Moon's mass, then it can't come any nearer than 10 times
the distance of the Moon.

[Tide is proportional to m/d^3. 1mm is 0.0002 times the lunar tide. So
0.0002 = 0.2/10^3.]

With current technology, we probably wouldn't be able to detect it.

Bryan Derksen

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Jul 8, 2004, 11:59:58 AM7/8/04
to
On Thu, 8 Jul 2004 12:42:23 +0100, Mike Williams
<nos...@econym.demon.co.uk> wrote:
>If you only want the tides to rise by 1 millimetre, and you want it to
>be 20% of the Moon's mass, then it can't come any nearer than 10 times
>the distance of the Moon.
>
>[Tide is proportional to m/d^3. 1mm is 0.0002 times the lunar tide. So
> 0.0002 = 0.2/10^3.]
>
>With current technology, we probably wouldn't be able to detect it.

We'd eventually notice its gravitational effects. Even if its tidal
effects on Earth slipped our attention, I expect it'd cause minor
changes in the Moon's orbit that we'd be able to notice pretty
clearly. Probably take at least a few months to become apparent,
though.

Hop David

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Jul 8, 2004, 12:10:26 PM7/8/04
to

George W. Harris wrote:
> Hop David <hopspageHA...@tabletoptelephone.com> wrote:
>
> :There are some problems. If this object arrives from outside of the
> :solar system, then it will be traveling a parabolic (or maybe even
> :hyperbolic) trajectory. On its arrival it will be traveling at least 12
> :km/sec wrt to the earth. In which case it would not be trapped by
> :earth's gravity (escape velocity is 11.2 km/sec at earth's surface).
> :You need some way to slow the object down upon arrival if it's going to
> :hang around our neighborhood.
>
> Would a close pass to the moon, killing much
> of the object's velocity, work? Perhaps the object
> could end up in a highly eccentric orbit with a period a
> fraction of the moon's (say 1/2 or 2/3). Could that
> possibly be stable?
>

From the moon's point of view, Vinf on the inward arm of the hyperbola
is the same as Vinf on the outward arm. The delta vee is a direction change.

The minimum 12 km/sec velocity I mentioned was if the object was moving
parallel to the earth at perihelion. The earth is moving 30 km/sec wrt
sun and the object is moving the same direction at sqrt(2)*30 km/sec wrt
the sun (about 42 km/sec). Changing the object's direction would make
matters worse in this scenario.

I don't see a lunar gravity assist taking enough velocity to enable capture.

Hop David

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Jul 8, 2004, 12:31:45 PM7/8/04
to

Frank Scrooby wrote:

> I am afraid that I have not provide enough information, and that the premise
> I am trying to create is probably so horribly flawed as to be unworkable.
>
> The object in question is highly artifical, some leftover ammunition from a
> cosmic conflict that was over before our star was born. The beings who
> created it are long gone, extinct possibly at their own hand. When you use
> Objects like these as ammunition your wars tend to be ... well ... highly
> fatal.
>
> The object and its twin (which is somewhere very far away) are the
> containers for a worm-hole. It is not captured into the Earth moon system so
> much as it 'appears'. One second people are looking up at a nice ordinary
> night sky and suddenly there this black splotch occluding stars. That is as
> much hand-waving as I want to do. I want everything else to be HARD SCIENCE.

Have you read "Impact Parameter" by Geoffrey Landis? It's a short story
in a collection of short stories of the same name.

Has some similarities to your story.

Landis' wormhole bends space in a fashion that seems similar to a black
hole. So it has an intense gravity field.

Here is a "Flatland" model of a wormhole
http://clowder.net/hop/etc./wormhol2.html

Note that you can enter it from all directions in the plane, Something A
Square or Yendred would probably find disorienting. In the same
fashion you could enter a wormhole in our space from any direction. Once
inside the hole, light geodesics would be helixes. You might be able to
see multiple images of your self shrinking into infinity, something like
a hall of mirrors (I believe you'd get a similar effect at 1.5
Schwarzchild radii from a blackhole).

George W. Harris

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Jul 8, 2004, 2:43:16 PM7/8/04
to
Hop David <hopspageHA...@tabletoptelephone.com> wrote:

:
:

Hmmm. Wouldn't conservation of energy
pretty much guarantee that the object would be
traveling at 42 km/sec when it crossed Earth's orbit,
regardless of perihelion (given that it crosses Earth's
orbit at all)? So if perihelion were just inside Earth's
orbit and it passed just in front of the Earth on the
outward leg, that could provide a significant deltaV,
could it not?

Here's another idea; solar braking. If this
object is tremendously dense, it could skim the
surface of the sun without being destroyed, which
could kill significant velocity. I doubt tidal effects
from a close solar pass would provide significant
deltaV, though.

--
"Intelligence is too complex to capture in a single number." -Alfred Binet

Erik Max Francis

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Jul 8, 2004, 4:04:17 PM7/8/04
to
"George W. Harris" wrote:

> Hmmm. Wouldn't conservation of energy
> pretty much guarantee that the object would be
> traveling at 42 km/sec when it crossed Earth's orbit,
> regardless of perihelion (given that it crosses Earth's
> orbit at all)?

Yes, if it originated from elsewhere in the Solar System, or even
outside (in which case its speed would be greater). The original poster
clarified his scenario, though; he's actually talking about something
that materializes in cislunar space, rather than actually having to get
there.

--
__ Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
/ \ San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis

\__/ We have always been space travellers.
-- Carl Sagan

George W. Harris

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Jul 8, 2004, 4:18:44 PM7/8/04
to
Erik Max Francis <m...@alcyone.com> wrote:

:"George W. Harris" wrote:
:
:> Hmmm. Wouldn't conservation of energy
:> pretty much guarantee that the object would be
:> traveling at 42 km/sec when it crossed Earth's orbit,
:> regardless of perihelion (given that it crosses Earth's
:> orbit at all)?
:
:Yes, if it originated from elsewhere in the Solar System, or even
:outside (in which case its speed would be greater). The original poster
:clarified his scenario, though; he's actually talking about something
:that materializes in cislunar space, rather than actually having to get
:there.

Fie on the original poster. The problem of
Earth capturing an object that starts out in a
parabolic/hyperbolic orbit around the sun is
interesting in its own right.

--
When Ramanujan was my age, he had been dead for ten years. -after Tom Lehrer

Hop David

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Jul 8, 2004, 6:44:19 PM7/8/04
to

If it's traveling a parabolic orbit, yes. However there's a wide variety
of delta vee's wrt to the earth. You can use traffic as an analogy.

Northbound car 30 mph and a northbound car 42 mph rear, difference in
velocity 12 mph. (this would be like a prograde asteroid with 0 degrees
inclination with a 1 A.U. perhihelion coming near the earth, it's
trajectory would be parallel.)

If you're northbound at 30 mph and a westbound car broadsides you at 42
mph, it's velocity from your point of view is sqrt(30^2 +42^2) mph or
about 52 mph. (An object could pass earth's orbit at nearly 90 degrees
if it's perihelion is very close to the sun. It could also pass at right
angles if it's inclination is high).

If you have a head on collision with a southbound car, it's a 30+42 or a
72 mph impact. (0 degrees inclination, 1 A.U. perihelion, retrograde orbit)

So the ranges of delta vee of an object crossing earth's orbit on a
parabolic trajectory range from 12 km/sec to 72 km/sec.

Altering the direction of a parabolic orbit of an object 0 degrees
inclination, 1 A.U. perihelion and a prograde orbit would only increase
delta vee wrt the earth.

So if perihelion were just inside Earth's
> orbit and it passed just in front of the Earth on the
> outward leg, that could provide a significant deltaV,
> could it not?

As I mentioned earlier the delta vee of a gravity assist changes the
direction of the velocity vector but not it's magnitude (speed) wrt to
the assisting body. So the delta vee is a direction change, not a change
in speed. The moon's frame of reference is nearly the same as the
earth's, it is only traveling aobut 1 km/sec wrt Earth.

>
> Here's another idea; solar braking. If this
> object is tremendously dense, it could skim the
> surface of the sun without being destroyed, which
> could kill significant velocity. I doubt tidal effects
> from a close solar pass would provide significant
> deltaV, though.
>

The point it slows down at would be it's perihelion. In the sun grazing
scenario, the kindest orbit, delta vee wise, would be one with a 700,000
km perihelion (sun's radius) and a 150,000,000 (1 A.U.) aphelion. At
aphelion the object is traveling about 1 km/sec wrt the sun. Wrt the
earth it's 29, maybe 28 km/sec.

Hop David

unread,
Jul 8, 2004, 7:11:07 PM7/8/04
to

George W. Harris wrote:
> Erik Max Francis <m...@alcyone.com> wrote:
>
> :"George W. Harris" wrote:
> :
> :> Hmmm. Wouldn't conservation of energy
> :> pretty much guarantee that the object would be
> :> traveling at 42 km/sec when it crossed Earth's orbit,
> :> regardless of perihelion (given that it crosses Earth's
> :> orbit at all)?
> :
> :Yes, if it originated from elsewhere in the Solar System, or even
> :outside (in which case its speed would be greater). The original poster
> :clarified his scenario, though; he's actually talking about something
> :that materializes in cislunar space, rather than actually having to get
> :there.
>
> Fie on the original poster. The problem of
> Earth capturing an object that starts out in a
> parabolic/hyperbolic orbit around the sun is
> interesting in its own right.
>

It's something I enjoy playing with.

You might like this resource:
http://neo.jpl.nasa.gov/cgi-bin/neo_ca?sort=vrel

Even with not very eccentric, low inclination orbits, you can see a lot
of near approachers fly by pretty fast.

But even 1991 VG would not be captured by earth's gravity. Even if it
were falling from infinity with zero starting velocity, it's orbit about
the earth would be parabolic. But even this slow asteroid would zoom by
the earth with 1.18 km/sec hyperbolic excess velocity. Maybe if it
grazed the earth's atmosphere and had Lunar gravity assists, it could be
captured.

Deimos and Phobos look like captured asteroids. How Mars got these moons
with nice circular orbits is a mystery to me.

Hop David
http://clowder.net/hop/index.html

George W. Harris

unread,
Jul 8, 2004, 7:26:18 PM7/8/04
to
Hop David <hopspageHA...@tabletoptelephone.com> wrote:

: So if perihelion were just inside Earth's


:> orbit and it passed just in front of the Earth on the
:> outward leg, that could provide a significant deltaV,
:> could it not?
:
:As I mentioned earlier the delta vee of a gravity assist changes the
:direction of the velocity vector but not it's magnitude (speed) wrt to
:the assisting body. So the delta vee is a direction change, not a change
:in speed. The moon's frame of reference is nearly the same as the
:earth's, it is only traveling aobut 1 km/sec wrt Earth.

I'm slow but eventually catch on. So about
the only way the object could end up in Earth orbit
would be lithobraking, which introduces a host of
other problems (unless it's lunar lithobraking).

--
"The truths of mathematics describe a bright and clear universe,
exquisite and beautiful in its structure, in comparison with
which the physical world is turbid and confused."

-Eulogy for G.H.Hardy

Jordan Abel

unread,
Jul 8, 2004, 7:48:20 PM7/8/04
to
Hop David wrote:

> But even 1991 VG would not be captured by earth's gravity. Even if it
> were falling from infinity with zero starting velocity, it's orbit about
> the earth would be parabolic.

In the sense that a line is a degenerate parabola... if we assume that
the earth is the only gravitational influence (and a perfectly spherical
cow), an object falling from infinity with zero starting velocity would
tend to fall downwards in a straight line towards earth.

Hop David

unread,
Jul 8, 2004, 8:06:18 PM7/8/04
to

Vinf is 0 regardless whether a parabola is degenerate or upstanding.

Chuck Stewart

unread,
Jul 8, 2004, 9:04:55 PM7/8/04
to
On Thu, 08 Jul 2004 23:26:18 +0000, George W. Harris wrote:

> I'm slow but eventually catch on. So about
> the only way the object could end up in Earth orbit
> would be lithobraking, which introduces a host of
> other problems (unless it's lunar lithobraking).

I still say hit it with a comet while crossing cislunar space.
Juggle the parameters right and the comet goes poof while the
(badly impacted) object is braked enough to be captured into the
desired Earth orbit.

If the object is handwaved into existence near the Earth then
the order of unlikelytude of two objects colliding in cislunar
space is lowered somewhat.

Nothing to be done for the cheeseitude, tho.

Jordan Abel

unread,
Jul 8, 2004, 9:50:49 PM7/8/04
to
Hop David wrote:

> Vinf is 0 regardless whether a parabola is degenerate or upstanding.

yes, but if it's _starting_ there it's being accelerated towards earth,
not towards some other point

i could be wrong, though. infinities are tricky to work with.

Geoffrey A. Landis

unread,
Jul 8, 2004, 3:50:27 PM7/8/04
to
In <40ED76F1...@tabletoptelephone.com> Hop David wrote:

> Have you read "Impact Parameter" by Geoffrey Landis? It's a short
> story in a collection of short stories of the same name.

>...

Cool. It's nice to see that not only do people occasionally read my
stories, they even pay attention.

--
Geoffrey A. Landis
http://www.sff.net/people/geoffrey.landis

Frank Scrooby

unread,
Jul 9, 2004, 5:23:49 AM7/9/04
to
Hi all

"Hop David" <hopspageHA...@tabletoptelephone.com> wrote in message
news:40ED76F1...@tabletoptelephone.com...
>
>
<much snipped>

> Have you read "Impact Parameter" by Geoffrey Landis? It's a short story
> in a collection of short stories of the same name.
>

I dont' believe I have read Geoffrey's story. Is it published online or in
paper.


> Has some similarities to your story.
>

Ahh, crap I'm a plagarist!

> Landis' wormhole bends space in a fashion that seems similar to a black
> hole. So it has an intense gravity field.
>

The 'Object' I have imagined has a relatively intense gravity field for its
volume, but mostly only because its hideously dense. Also it isn't really
one object but two (you only ever see one 'opening' but they behave as if
their individual masses are combined). But with only 20% of the mass of the
Moon its not really going to be a massive threat to interplanetary space,
not in the order of a real singularity.

> Here is a "Flatland" model of a wormhole
> http://clowder.net/hop/etc./wormhol2.html
>
> Note that you can enter it from all directions in the plane, Something A

This is a feature of my Object. I can not believe this. I thought I was
being completely original.

I am dreadful sorry, Geoffrey. I did not intend to plagarize. I promise.


> Square or Yendred would probably find disorienting. In the same
> fashion you could enter a wormhole in our space from any direction. Once
> inside the hole, light geodesics would be helixes. You might be able to
> see multiple images of your self shrinking into infinity, something like
> a hall of mirrors (I believe you'd get a similar effect at 1.5
> Schwarzchild radii from a blackhole).

I had decided that humans who utilize the wormhole (and that is the main
thrust of the story I want to write - there is something much more
interesting and tempting than a left-over, decaying, 10-billion old,
forgotten munition that now acts as an instanteous interstellar transport
system at the other end of the wormhole) will not experience at all anything
when going through the wormhole. I.e. They see the approaching 'event
horizon' of the object. It passes directly through their ship, and their
space suits. If they don't blink, faint, or haven't pinched their eyes shut
in prayer they will even see the moment of contact between their eyes and
the Object, and then they see the destination. They will have absolutely no
awareness of any time having passed in between those moment. This is not
like sleep, or being unconscious or even blinking. It is like not having
missed any time at all. People who try counting during the passage will find
themselves finishing saying the last number they were counting at the moment
of contact. Stuff like that. Computers and atomic clocks, especially those
that synchronize themselves with equipment back in Earth's solar system will
record the loss of a couple of 100ths of a millisecond, if they are that
precise.

Geoffrey A. Landis

unread,
Jul 9, 2004, 12:07:15 PM7/9/04
to
Hop David" <hopspageHA...@tabletoptelephone.com> wrot:> <much
snipped>
>
>> Have you read "Impact Parameter" by Geoffrey Landis? It's a short
>> story in a collection of short stories of the same name.


In <ppGdnRggZM6...@is.co.za> Frank Scrooby replied:


> I dont' believe I have read Geoffrey's story. Is it published online
> or in paper.

It's in my collection (which is hardcover only, alas. I haven't gotten
a paperback deal on it yet; it's hard to sell collections in mass market.)

The short-story is independently available from fictionwise:
http://www.fictionwise.com/servlet/mw?authorid=28&template=author.htm&
action=view

>> Has some similarities to your story.

>> Landis' wormhole bends space in a fashion that seems similar to a
>> black hole. So it has an intense gravity field.


Frank Scrooby continued:


> Ahh, crap I'm a plagarist!

Well, it's not plagiarizing to independently come up with a similar idea.

I had the thought of gravitational lensing from a workhole from the
original Morris and Thorne paper, where the wormhole external metric is
identical to a Schwartzschild metric-- i.e., from the outside it looks
like a black hole. Later, after I learned a lot more about wormholes, I
discovered that the external metric can be any of many different
possibilities, even negative mass solutions.

In any case, my plot was entirely about the reaction of astronomers on
Earth, and if you're talking about people travelling through the
wormhole to explore, you'll be looking at a different plot entirely.

Hop David

unread,
Jul 9, 2004, 2:53:07 PM7/9/04
to

I see what you mean..

In an ideal two body scenario where the bodies are completely motionless
wrt to one another they'd approach each other along a straight line.

But more practically speaking, straight geodesics don't exist. Our space
isn't flat with masses like the planets, stars, galaxies etc. bending
space. "Falling from infinity" is certainly an imprecise and inaccurate
term to describe NEOs as most asteroids don't go further than 6 A.U.
from the earth.

Even so, those using orbital mechanics will use Vinf for hyperbolic
excess velocity. Vinf is a very good approximation for velocity wrt
earth once outside Earth's Sphere Of Influence (SOI). Prussing & Conway
define the radius of a planet's Sphere Of Influence as (Mp/Ms)^(2/5) * Rsp

Mp = Mass of planet
Ms = Mass of Sun
Rsp = distance between planet and sun.

Using this definition, Earth's SOI has a radius about 925,000 kilometers

In a flat space an object at this distance traveling .077 km/sec wrt
earth would have a perigee in low earth orbit and it's apogee would be
right back to 925,000 km, the very edge of Earth's SOI.

Jordan Abel

unread,
Jul 9, 2004, 3:39:51 PM7/9/04
to
Geoffrey A. Landis wrote:

> original Morris and Thorne paper, where the wormhole external metric is
> identical to a Schwartzschild metric-- i.e., from the outside it looks
> like a black hole.

If it looks like a black hole, how does anyone on the outside know it's
not a black hole? other than to accidentally (who's going to go into a
black hole on purpose?) fall in and then come back to tell about it,
that is.

James Nicoll

unread,
Jul 9, 2004, 3:42:31 PM7/9/04
to
In article <20040709120...@newsread.grc.nasa.gov>,

Geoffrey A. Landis <Geoffrey_...@sff.net> wrote:
>
>It's in my collection (which is hardcover only, alas. I haven't gotten
>a paperback deal on it yet; it's hard to sell collections in mass market.)

Happily, it is worth getting in hardcover.
--
"The keywords for tonight are Caution and Flammable."
Elvis, _Bubba Ho Tep_

LukeCampbell

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Jul 9, 2004, 4:13:27 PM7/9/04
to

If the wormhole has the Schwartzschild metric down past the event
horizon, then you don't come back after falling in - ever. If you are
lucky, the wormhole on the other side does not have an event horizon and
you can get out. Otherwise, you are stuck.

If the Schwartzschild metric stops before the event horizon, it does not
look like a black hole.

Note that I have not read the story by Landis, so I cannot comment on
specific things that happen in the story or the author's representation
of this system.

Luke

--
To email me, take out the trash.

pervect

unread,
Jul 9, 2004, 5:06:01 PM7/9/04
to
On Thu, 8 Jul 2004 10:43:28 +0200, "Frank Scrooby" <X...@Xer.com> wrote:


>I am not happy about it causing massive changes in Earth's tide or such
>like. A one meter increase in the daily high tides at a place not so far
>from where I am right now will result in the routine and probably permanent
>flooding on the CBD of a city (the main street is a meter below sea level,
>the beach front is the only thing keeping the sea out). There are some
>people who would probably welcome this sort of thing. Its great for urban
>renewal, and Hollywood blockbusters. And what happens in Durban South Africa
>will happen in dozens cities all around the world.

Tidal responses are very complex, but some insight can be gained by
looking at the simpler 'driving terms'.

Tidal acceleration is directly proportional to mass, and inversely
proportional to distance^3

So if your object has .2 of the mass of the moon, but is twice as
close, it will produce a tidal acceleration of about 1.6 that of the
moon.

That would probably qualify as a major disruption.

If you keep it at the same distance as the moon, with a mass of 20% of
the moon, it would have a tidal acceleration of about 20% of that of
the moon. Presumably the response would be around 20% as well, but
that's very approximate, as I mentioend the earth's tidal response is
rather tricky.

Assuming intelligent guidance, a convenient place to stick it would be
at one of the stable Lagrange points as someone else has mentioned.
I'm not sure about long term (astronomical) stability, but over the
short term it should be OK here.

If you want to have the body approach closer than the moon, you'll
have to decrease its mass, or live with the higher tidal effects.

I believe there are several orbital simulators available on the web,
you might start your search at

http://www.burtleburtle.net/bob/physics/solar.html

this would allow you to numerically simulate how the body would act
and what effect it would have on the orbits of the earth and moon.


jimirwin

unread,
Jul 9, 2004, 8:51:21 PM7/9/04
to
pervect wrote in rec.arts.sf.science:

>
> I believe there are several orbital simulators available on the web,
> you might start your search at
>
> http://www.burtleburtle.net/bob/physics/solar.html
>
> this would allow you to numerically simulate how the body would act
> and what effect it would have on the orbits of the earth and moon.
>
>

I ran the Solex program, which is supposedly one of the most accurate
integrators. I put an object instantaneously at about 40% the distance
to the moon, with an approximately circular orbit, and varying masses.

I was surprised that the earth-moon-wormhole system remained relatively
stable (with respect to the earth) for several years even with a
wormhole mass equal to 100% of the moon's mass. The moon's orbital
period became shorter by about 40 hours, and the year became longer by
about 4 hours, but neither the moon nor the wormhole escaped the system,
although the moon's orbit was becoming quite noticably more eccentric at
the end of 4 years.

At 10% of the moon's mass, the moon's orbital period decreased by about
6 hours. The year was about a half hour longer.

At 1% of the moon's mass, the moon's orbital period remained
approximately the same, but I'm sure there would still be long term
effects. I only ran this simulation for a couple of months.

--
Jim Irwin
http://www.holoscenes.com

Jordan Abel

unread,
Jul 10, 2004, 4:46:09 PM7/10/04
to
jimirwin wrote:
> I was surprised that the earth-moon-wormhole system remained relatively
> stable (with respect to the earth) for several years even with a
> wormhole mass equal to 100% of the moon's mass. The moon's orbital
> period became shorter by about 40 hours, and the year became longer by
> about 4 hours, but neither the moon nor the wormhole escaped the system,
> although the moon's orbit was becoming quite noticably more eccentric at
> the end of 4 years.

Did you try putting it in with the same initial velocity as the velocity
of the CM of the earth-moon system? [that might be more neutral to the
length of the year, and possibly also to the length of the month, than
one that's not moving]

jimirwin

unread,
Jul 10, 2004, 5:52:19 PM7/10/04
to
Jordan Abel wrote in rec.arts.sf.science:

> Did you try putting it in with the same initial velocity as the velocity
> of the CM of the earth-moon system? [that might be more neutral to the
> length of the year, and possibly also to the length of the month, than
> one that's not moving]
>
>

No, because if it had the same velocity as the CM of the earth-moon system,
it would quickly plummet to the earth rather than orbit. It takes a little
more than a day to fall to earth under those initial conditions. I chose
an initial velocity that resulted in an approximately circular orbit around
earth, in the plane of the ecliptic, in the same angular direction as the
moon's orbit. Because I placed it initially to the outside of earth's
orbit, it gave the system a little more energy and made the year longer.
If I had placed it to the inside of earth's orbit, it would have made the
year shorter. No matter where it starts, if it's massive, it's going to
perturb the system in some manner, either by changing the plane or the
eccentricity of the earth's orbit, or some combination of those.

Geoffrey A. Landis

unread,
Jul 10, 2004, 4:47:47 PM7/10/04
to
Geoffrey A. Landis wrote:

>>> ...original Morris and Thorne paper, where the wormhole external
>>> metric
>>> is identical to a Schwartzschild metric-- i.e., from the outside it
>>> looks like a black hole.

Jordan Abel replied:

>> If it looks like a black hole, how does anyone on the outside know
>> it's not a black hole? other than to accidentally (who's going to go
>> into a black hole on purpose?) fall in and then come back to tell
>> about it, that is.

LukeCampbell elucidated:

> If the wormhole has the Schwartzschild metric down past the event
> horizon, then you don't come back after falling in - ever. If you are
> lucky, the wormhole on the other side does not have an event horizon
> and you can get out. Otherwise, you are stuck.
>
> If the Schwartzschild metric stops before the event horizon, it does
> not look like a black hole.

The Morris-Thorne wormhole has the Schwarzschild metric in to a distance
r larger than the event horizon, has a shell of exotic matter comprising
the wormhole throat, and then has a Schwartzschild metric metric on the
other side.

Joseph Hertzlinger

unread,
Jul 11, 2004, 1:35:20 PM7/11/04
to
On Wed, 07 Jul 2004 12:16:50 -0700, Erik Max Francis <m...@alcyone.com>
wrote:

> Hop David wrote:
>
>> If I recall correctly an L4 or L5 object shouldn't be more than 1/26
>> the
>> mass of the moon.
>
> No, the secondary body (the Moon in your example) needs to be less than
> (about) 1/26 the mass of the primary body (the Earth in your example).
> The mass of the object actually attempting to reside in the Trojan point
> must be of negligible mass compared to the other two.

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

If I did the arithmetic correctly, in a system where two of the bodies
have masses of 1 and 1/81, the third must have a mass of either
greater than 25.281 or less than 0.27.

--
http://hertzlinger.blogspot.com

Erik Max Francis

unread,
Jul 11, 2004, 4:47:32 PM7/11/04
to
Joseph Hertzlinger wrote:

> 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
>
> If I did the arithmetic correctly, in a system where two of the bodies
> have masses of 1 and 1/81, the third must have a mass of either
> greater than 25.281 or less than 0.27.

The stability criterion is usually given as (_Orbital Motion_, A.E. Roy,
p. 139):

mu = (1/2) - (23/108)^(1/2),

where mu is the mass of the secondary body expressed in units of the
primary body; that's 0.03852, or 1/25.96.

I'm not sure where you get your stability criterion (involving all three
bodies) from; for one thing, it gives seemingly absurd results (at least
by your solution, I didn't solve it myself), such as a Trojan point
being stable where the test body is more than 25 times massive than the
primary -- that's considerably more massive than Neptune!

--
__ Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
/ \ San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis

\__/ Forever we / Infinitely
-- Sandra St. Victor

Aaron Denney

unread,
Jul 11, 2004, 5:03:09 PM7/11/04
to
On 2004-07-11, Erik Max Francis <m...@alcyone.com> wrote:
> Joseph Hertzlinger wrote:
>
>> 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
>>
>> If I did the arithmetic correctly, in a system where two of the bodies
>> have masses of 1 and 1/81, the third must have a mass of either
>> greater than 25.281 or less than 0.27.
>
> The stability criterion is usually given as (_Orbital Motion_, A.E. Roy,
> p. 139):
>
> mu = (1/2) - (23/108)^(1/2),
>
> where mu is the mass of the secondary body expressed in units of the
> primary body; that's 0.03852, or 1/25.96.
>
> I'm not sure where you get your stability criterion (involving all three
> bodies) from; for one thing, it gives seemingly absurd results (at least
> by your solution, I didn't solve it myself), such as a Trojan point
> being stable where the test body is more than 25 times massive than the
> primary -- that's considerably more massive than Neptune!

Well, in that case, the trojan point is the primary, the primary is the
secondary, and the secondary is the trojan point.

--
Aaron Denney
-><-

Erik Max Francis

unread,
Jul 11, 2004, 5:05:46 PM7/11/04
to
Aaron Denney wrote:

> Well, in that case, the trojan point is the primary, the primary is
> the
> secondary, and the secondary is the trojan point.

His lesser stability limit still puts the test body at much more massive
than the secondary body (1/81 vs. 0.27), which isn't stable at all, so
there's clearly something wrong with the equation he's using. (I've
never seen an equation like that, so I suspect this is just a
misapplication of something.)

Joseph Hertzlinger

unread,
Jul 11, 2004, 5:53:34 PM7/11/04
to
On Sun, 11 Jul 2004 13:47:32 -0700, Erik Max Francis <m...@alcyone.com> wrote:

> Joseph Hertzlinger wrote:
>
>> 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
>>
>> If I did the arithmetic correctly, in a system where two of the bodies
>> have masses of 1 and 1/81, the third must have a mass of either
>> greater than 25.281 or less than 0.27.

Actually, the 0.27 should have been 0.027.

> The stability criterion is usually given as (_Orbital Motion_, A.E. Roy,
> p. 139):
>
> mu = (1/2) - (23/108)^(1/2),
>
> where mu is the mass of the secondary body expressed in units of the
> primary body; that's 0.03852, or 1/25.96.
>
> I'm not sure where you get your stability criterion (involving all three
> bodies) from; for one thing, it gives seemingly absurd results (at least
> by your solution, I didn't solve it myself), such as a Trojan point
> being stable where the test body is more than 25 times massive than the
> primary -- that's considerably more massive than Neptune!

I got it from an article in Volume 5 of of "What's Happening in the
Mathematical Sciences" by Barry Cipra. In the limiting case where m_3
goes to zero, it becomes the usual formula.

--
http://hertzlinger.blogspot.com

pervect

unread,
Jul 11, 2004, 9:21:36 PM7/11/04
to

For what it's worth, I did some numerical simulations for several
cases

case #1, m1=.98, m2=.01, m3=.01, should be stable according to the
quoted formula, the orbits were right on top of each other ploted for
100 complete cycles.

case #2, m1=.96, m2=.02, m3=.02, should be unstable, the orbital plots
were not on top of each other, I got a fuzzy "ring" instead for the
same simulation period (100 orbits). (However, I should note that
nothing was "flung off").

case #3, m1=.97,m2=.02, m3=.01, should be stable, the orbits were
again right on top of each other.

I used maple vers 7 to do the simulations and plots (it's a very old
version of maple BTW).

I attempted to do a more rigorous treatment with eigenvalues, but
failed due to numerical problems :-(.

Tenatively, my simulations results seem to be in agreement with
Joseph's formula.

The formula's I used were:


# let m1 be the most massive, m2 the next, m3 the least
# bodies are arranged in equilateral triangle

# let d = length of leg of triangle
>
# let m1 be at (-a,-b), a and b small when m1 large
# let m2 be at (d-a,-b)
# let m3 be at (d/2-a,sqrt(3)*d/2-b)

from center of mass considerations (to put the c of m at the origin)

a := .5*d*(2*m2+m3)/(m1+m2+m3);
b := .5*m3*3^(1/2)*d/(m1+m2+m3);

# potential energy

V :=
-G*m1*m2/sqrt((x2-x1)^2+(y2-y1)^2)-G*m1*m3/sqrt((x3-x1)^2+(y3-y1)^2)-G*m2*m3/sqrt((x3-x2)^2+(y3-y2)^2);

I chose G=1 for ease of simulation, which makes w =
sqrt(G*(m1+m2+m3)/d^3) = 1 when d=1 and m1+m2+m3=1.

The system of diffeq's and initial conditions:

> diff(vx1(t),t) = -subs(x1=x1(t),x2=x2(t),x3=x3(t),y1=y1(t),y2=y2(t),y3=y3(t),diff(V,x1)/m1),
> diff(vx2(t),t) = -subs(x1=x1(t),x2=x2(t),x3=x3(t),y1=y1(t),y2=y2(t),y3=y3(t),diff(V,x2)/m2),
> diff(vx3(t),t) = -subs(x1=x1(t),x2=x2(t),x3=x3(t),y1=y1(t),y2=y2(t),y3=y3(t),diff(V,x3)/m3),
> diff(vy1(t),t) = -subs(x1=x1(t),x2=x2(t),x3=x3(t),y1=y1(t),y2=y2(t),y3=y3(t),diff(V,y1)/m1),
> diff(vy2(t),t) = -subs(x1=x1(t),x2=x2(t),x3=x3(t),y1=y1(t),y2=y2(t),y3=y3(t),diff(V,y2)/m2),
> diff(vy3(t),t) = -subs(x1=x1(t),x2=x2(t),x3=x3(t),y1=y1(t),y2=y2(t),y3=y3(t),diff(V,y3)/m3),
> vx1(t) = diff(x1(t),t),
> vx2(t) = diff(x2(t),t),
> vx3(t) = diff(x3(t),t),
> vy1(t) = diff(y1(t),t),
> vy2(t) = diff(y2(t),t),
> vy3(t) = diff(y3(t),t),
> x1(0) = -a,
> y1(0) = -b,
> x2(0) = d-a,
> y2(0) = -b,
> x3(0) = d/2-a,
> y3(0) = evalf(sqrt(3)*d/2-b),
> vx1(0)=-b*w,
> vy1(0)=a*w,
> vx2(0)=-b*w,
> vy2(0)=-(d-a)*w,
> vx3(0)=evalf((sqrt(3)*d/2-b)*w),
> vy3(0)=-(d/2-a)*w

Solving and plotting
dsol := dsolve(deqs,numeric,range=0..200*Pi,maxfun=0);

odeplot(dsol,[x2(t),y2(t)]);
odeplot(dsol,[x3(t),y3(t)]);

Frank Scrooby

unread,
Jul 12, 2004, 5:01:12 AM7/12/04
to
Hi all


"James Nicoll" <jdni...@panix.com> wrote in message
news:ccmsf7$d7k$1...@panix2.panix.com...


> In article <20040709120...@newsread.grc.nasa.gov>,
> Geoffrey A. Landis <Geoffrey_...@sff.net> wrote:
> >
> >It's in my collection (which is hardcover only, alas. I haven't gotten
> >a paperback deal on it yet; it's hard to sell collections in mass
market.)
>
> Happily, it is worth getting in hardcover.

I shall have to order it online. Unfortunately book stores here are still in
the dark ages. If it isn't tied in with a current movie release or it isn't
something that been in continuous print for the last 20 years they simply
won't stock it. And don't even think about trying to use them to order
something.

> --
> "The keywords for tonight are Caution and Flammable."
> Elvis, _Bubba Ho Tep_


Regards
Frank


James Nicoll

unread,
Jul 12, 2004, 7:25:56 AM7/12/04
to
In article <HYKdnSqZAKH...@is.co.za>, Frank Scrooby <X...@Xer.com> wrote:
>Hi all
>
>
>"James Nicoll" <jdni...@panix.com> wrote in message
>news:ccmsf7$d7k$1...@panix2.panix.com...
>> In article <20040709120...@newsread.grc.nasa.gov>,
>> Geoffrey A. Landis <Geoffrey_...@sff.net> wrote:
>> >
>> >It's in my collection (which is hardcover only, alas. I haven't gotten
>> >a paperback deal on it yet; it's hard to sell collections in mass
>market.)
>>
>> Happily, it is worth getting in hardcover.
>
>I shall have to order it online. Unfortunately book stores here are still in
>the dark ages. If it isn't tied in with a current movie release or it isn't
>something that been in continuous print for the last 20 years they simply
>won't stock it. And don't even think about trying to use them to order
>something.
>

Ah, been there. One local store took over a year to get _THe Praxis_
in for me.

www.goldengryphon.com

Dr John Stockton

unread,
Jul 12, 2004, 11:24:37 AM7/12/04
to
JRS: In article <sTeIc.1812$sV2....@newsread2.news.atl.earthlink.net>,
seen in news:rec.arts.sf.science, Joseph Hertzlinger <jcyclespersecondlo
ngis...@nine.reticulatedcom.com> posted at Sun, 11 Jul 2004 17:35:20 :


By <URL:http://www.merlyn.demon.co.uk/js-demos.htm#FZ> , the
"expression" Y=1 ; Z=1/81 ; 27*(X*Y+Y*Z+Z*X) - (X+Y+Z)*(X+Y+Z)
has a zero at X = +2.5281289e+01 and one at X = +2.7352657e-02

Your arithmetic is OK, apart from the decimal point location; but
neither answer is the usual one, for which the test mass is zero.

For that,
Y=1 ; Z=0 ; 27*(X*Y+Y*Z+Z*X) - (X+Y+Z)*(X+Y+Z)
has a zero at X = +4.0064206e-02 which is the reciprocal of 24.95994
or thereabouts; and indeed has another zero at X = +2.4959936e+01 .

I have earlier noted 24.9599 as a reputably-quoted value (possibly seen
in this newsgroup), in
<URL:http://www.merlyn.demon.co.uk/gravity3.htm#L15>, now modified
(green box).


EMF cited : mu = (1/2) - (23/108)^(1/2), the reciprocal of which is
25.95993579437713

The difference between those figures is remarkably close to 1.0; ISTM
that the original sources should be examined to see precisely what is
being calculated in each case.

With Y=1, Z=0, the expression 27*(X*Y+Y*Z+Z*X) - (X+Y+Z)*(X+Y+Z)
becomes 27*X - (X+1)*(X+1) ,

0 = 27*X - (X+1)*(X+1)
=> X^2 +2*X - 27*X + 1 = 0 X^2 - 25*X + 1 = 0
X = (25 +- (625 - 4)^1/2) / 2

And (X+1)*mu is exactly equal to 1, on multiplying out.

--
© John Stockton, Surrey, UK. ?@merlyn.demon.co.uk Turnpike v4.00 MIME. ©
Web <URL:http://www.merlyn.demon.co.uk/> - FAQqish topics, acronyms & links;
some Astro stuff via astro.htm, gravity0.htm; quotes.htm; pascal.htm; &c, &c.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.

pervect

unread,
Jul 12, 2004, 5:35:34 PM7/12/04
to
On Mon, 12 Jul 2004 16:24:37 +0100, Dr John Stockton
<sp...@merlyn.demon.co.uk> wrote:

>Your arithmetic is OK, apart from the decimal point location; but
>neither answer is the usual one, for which the test mass is zero.
>
>For that,
> Y=1 ; Z=0 ; 27*(X*Y+Y*Z+Z*X) - (X+Y+Z)*(X+Y+Z)
>has a zero at X = +4.0064206e-02 which is the reciprocal of 24.95994
>or thereabouts; and indeed has another zero at X = +2.4959936e+01 .
>
>I have earlier noted 24.9599 as a reputably-quoted value (possibly seen
>in this newsgroup), in
><URL:http://www.merlyn.demon.co.uk/gravity3.htm#L15>, now modified
>(green box).
>
>
>EMF cited : mu = (1/2) - (23/108)^(1/2), the reciprocal of which is
>25.95993579437713
>
>The difference between those figures is remarkably close to 1.0; ISTM
>that the original sources should be examined to see precisely what is
>being calculated in each case.

Note that one derivation of the (linear) stability of the lagrange
points is online at

http://scienceworld.wolfram.com/physics/LagrangePoints.html

(It may be a bit hard to follow, I would suggest doing an independent
derivation of the equations of motion by writing down the Lagrangian
in rotating coordinates, and replacing "mean motion" by "angular
velocity".)

It's assumed in this derivation that the mass of the primary is 1-u
and the mass of the secondary is u, i.e. that the sum of the primary
and secondary masses is 1.

The resulting equation is 27 u (1-u) < 1

This is the _exact_ equation that Joseph Hertzlinger's formula gives
when m3 = 0 and m1+m2=1, so it does indeed result in the usual answer
when m3=0.

Unfortunately I don't have the source Joseph Hertzlinger cites as the
origin of the formula in question, but so far it seems to be producing
sensible results.

Dr John Stockton

unread,
Jul 13, 2004, 11:07:47 AM7/13/04
to
JRS: In article <sTeIc.1812$sV2....@newsread2.news.atl.earthlink.net>,
seen in news:rec.arts.sf.science, Joseph Hertzlinger <jcyclespersecondlo
ngis...@nine.reticulatedcom.com> posted at Sun, 11 Jul 2004 17:35:20 :
>
>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

Have you a formula for N equal bodies? The above, seemingly applicable,
rules out a Klemperer Triangle.

Geoffrey A. Landis

unread,
Jul 13, 2004, 2:07:10 PM7/13/04
to
On 7/13/04 11:07 AM, Dr John Stockton wrote:
> Have you a formula for N equal bodies? The above, seemingly
> applicable, rules out a Klemperer Triangle.

Are they stable?

Dr John Stockton

unread,
Jul 14, 2004, 10:36:57 AM7/14/04
to
JRS: In article <20040713140...@newsread.grc.nasa.gov>, seen in
news:rec.arts.sf.science, Geoffrey A. Landis
<Geoffrey_...@sff.net> posted at Tue, 13 Jul 2004 18:07:10 :

>On 7/13/04 11:07 AM, Dr John Stockton wrote:
>> Have you a formula for N equal bodies? The above, seemingly
>> applicable, rules out a Klemperer Triangle.
>
>Are they stable?

That is what the formula should tell us.

If JH's 3-body formula is correct and applicable, three equal bodies are
not stable.

If 'Ringworld' is to be believed, three or more are stable, with or
without a central mass; in particular, five without.

If 'The City and the Stars' and 'Against the Fall of Night' are to be
believed, six around a central mass are stable.

EMF wrote :

where mu is the mass of the secondary body expressed in units of
the primary body; that's 0.03852, or 1/25.96.

I suspect mu is the reduced mass, so the secondary is 1/24.96 of the
primary.

pervect

unread,
Jul 14, 2004, 10:01:46 PM7/14/04
to
On Wed, 14 Jul 2004 15:36:57 +0100, Dr John Stockton
<sp...@merlyn.demon.co.uk> wrote:


>That is what the formula should tell us.
>
>If JH's 3-body formula is correct and applicable, three equal bodies are
>not stable.
>
>If 'Ringworld' is to be believed, three or more are stable, with or
>without a central mass; in particular, five without.
>
>If 'The City and the Stars' and 'Against the Fall of Night' are to be
>believed, six around a central mass are stable.

I'm not sure how much numerical simulations can be trusted, but my
numerical simulations for three equal masses are very unstable. The
situation for m1=.96,m2=.02,m3=.02 is less unstable, the moons never
escape (at least in the time simulated), but the radius of their orbit
"squegs" (sorry, I can't think of a better word to describe it, it's
an electronics term.)

I'm not sure why my attempts to analyze the stability of the system
via it's eigenvalues are failing - true, there's a twelfth-degree
polynomial equation involved, but with computer algebra that shouldn't
be impossible. In any event,so far I'm not getting good results.

I'm definitely curious about the method used by JH's source, I'll have
to see if I can order it via a library loan (but it may be too new for
that).

I haven't simulated the other cases myself, but there's some
simulation results on the webpage

http://burtleburtle.net/bob/physics/kempler.html

that includes the 5 world and 6 world cases (it's still an open
question as to how much one trusts simulations, esp "other peoples").

As far as "Against the Fall of Night" goes, I don't recall the six
world case. I do recall that in the original version of the story,
the moon spiraled in towards the earth and had to be destroyed - this
unlikely fate was fixed in Benford's & Clarke's re-write of the story
(which has the same name as Clarke's original) - were the six worlds
in the original version of the story, or the Clarke-Benford re-write?

Urban Fredriksson

unread,
Jul 15, 2004, 1:24:55 AM7/15/04
to
In article <1lmbf01pntc1m5nvc...@4ax.com>,
pervect <per...@invalid.invalid> wrote:

>As far as "Against the Fall of Night" goes, I don't recall the six
>world case.

It wasn't worlds in that story, it was stars. Placed as a
sort of marker if I don't misremember.
--
Urban Fredriksson http://www.canit.se/%7Egriffon/
1) What is happening will continue to happen
2) Consider the obvious seriously
3) Consider the consequences - Asimov's "Three Laws of Futurics", F&SF, Oct 74

Dr John Stockton

unread,
Jul 15, 2004, 1:21:30 PM7/15/04
to
JRS: In article <1lmbf01pntc1m5nvc...@4ax.com>, seen in
news:rec.arts.sf.science, pervect <per...@invalid.invalid> posted at
Wed, 14 Jul 2004 19:01:46 :

>On Wed, 14 Jul 2004 15:36:57 +0100, Dr John Stockton
><sp...@merlyn.demon.co.uk> wrote:
>
>>That is what the formula should tell us.
>>
>>If JH's 3-body formula is correct and applicable, three equal bodies are
>>not stable.
>>
>>If 'Ringworld' is to be believed, three or more are stable, with or
>>without a central mass; in particular, five without.
>>
>>If 'The City and the Stars' and 'Against the Fall of Night' are to be
>>believed, six around a central mass are stable.

>As far as "Against the Fall of Night" goes, I don't recall the six


>world case. I do recall that in the original version of the story,
>the moon spiraled in towards the earth and had to be destroyed - this
>unlikely fate was fixed in Benford's & Clarke's re-write of the story
>(which has the same name as Clarke's original) - were the six worlds
>in the original version of the story, or the Clarke-Benford re-write?

I wrote bodies, not worlds. I refer to the Clarke stories; not to the
first editions, but almost certainly to the final versions. The bodies
are stars, visited on the Universe-spanning trip in which IIRC Vanamonde
appeared.

I've read Clarke and Benford once; the mistake will not be repeated.



--
© John Stockton, Surrey, UK. ?@merlyn.demon.co.uk Turnpike v4.00 MIME. ©

Web <URL:http://www.merlyn.demon.co.uk/> - FAQish topics, acronyms, & links.
Proper <= 4-line sig. separator as above, a line exactly "-- " (SonOfRFC1036)
Do not Mail News to me. Before a reply, quote with ">" or "> " (SonOfRFC1036)

pervect

unread,
Jul 16, 2004, 5:26:00 PM7/16/04
to
On Wed, 14 Jul 2004 19:01:46 -0700, pervect <per...@invalid.invalid>
wrote:


>I'm not sure why my attempts to analyze the stability of the system
>via it's eigenvalues are failing

OK, I found my error. At least I'm getting sensible results, and they
appear to match JH's!

The trick that I was missing is that one has to transform the three
massive body problem into an equiavelent 2 body problem by using
what's known as Jacobi coordinates. Otherwise, the coordinates are
not truly independent. It's necessary for the coordinates to be
independent because I'm using the Lagrangian method.

The old variables x1,x2,x3,y1,y2,y3 in terms of the new variables

(rx,ry,px,py) are:

x3 := (m1+m2)/(m1+m2+m3)*px;
y3 := (m1+m2)/(m1+m2+m3)*py;
x2 := m1/(m1+m2)*rx - m3/(m1+m2+m3)*px;
y2 := m1/(m1+m2)*ry - m3/(m1+m2+m3)*py;
x1 := -m2/(m1+m2)*rx - m3/(m1+m2+m3)*px;
y1 := -m2/(m1+m2)*ry - m3/(m1+m2+m3)*py;

By formulating the problem in ths manner, we automatically constrain

m1x1+m2x2+m3x3=0,m1y1+m2y2+m3y3=0

and reduce the original 6 varaible problem to a 4 variable problem.

letting vx1=(d/dt)x1,vy1=(d/dt)y1,vrx=(d/dt)rx, etc etc, we can write
the total kinetic energy T as

T = 1/2(m1(vx1^2+vy1^2)+m2(vx2^2+vy2^2)+m3(vx3^2+vy3^2))


a straighforward substitution and computer-algebraic simplificaiton
expresses the kinetic energy T in the new variables vrx,vry,vpx,vpy as

T = (1/2)*(m1*m2)/(m1+m2)*(vrx^2+vry^2)+
(1/2)*m3*(m1+m2)/(m1+m2+m3)*(vpx^2+vpy^2)

The equations of motion are derived by using Lagrange's equation, with
L = T-V, using the new variables rx,ry,px,py. We also have to convert
V to the new variables, this turns out to be not too hard, I'll omit
the details since I suspect not many people will care anyway.

After a *whole lot* of computer algebra, doing another conversion to a
co-rotating system of coordinates along the way, we wind up with the
eigenvalue equation for L (setting G=1 and the distance between
bodies=1 along the way as well):
2 2 2 2 4
eq := 1/4 (27 m2 m1 + 4 m2 L + 27 m2 m3 + 81 m2 m3 m1 + 8 m2 L

2 2 2 2
+ 35 m2 L m3 + 35 m2 m1 L + 27 m2 m1 + 27 m2 m3

2 2 2 2 2 4
+ 27 m3 m1 + 27 m1 m3 + 4 L m3 + 35 L m3 m1 + 8 m1 L

4 2 2 6 2 2 2 2 /
+ 8 L m3 + 4 m1 L + 4 L ) m3 L m1 m2 /
/

2
(m1 + m2 + m3)

which has solutions

0,0,+/-sqrt(m1+m2+m3)*i,

+/- sqrt(2)/2 *
sqrt(+/-sqrt((m1+m2+m3)^2-27*(m1*m2+m1*m3+m2*m3))-(m1+m2+m3));

and it's this last solution that gives the eigenvalues a positive real
part if (m1+m2+m3)^2 - 27*(m1m2+m1m3+m2m3) < 0.

So the overall conclusion is that three bodies of equal mass are NOT
stable, and we still have no analytical answer for the 5 or six body
case (and I'm not going to attempt one, either). It's encouraging (or
perhaps just lucky) that totally naieve computer simulations pick up
the instability (for m1=.96,m2=.02,m3=.02) even though there's no
written guarantee anywhere about what a random numerical method does
to the phase space of the differential eq's.

pervect

unread,
Jul 16, 2004, 9:57:38 PM7/16/04
to
On Thu, 15 Jul 2004 18:21:30 +0100, Dr John Stockton
<sp...@merlyn.demon.co.uk> wrote:

>
>I wrote bodies, not worlds. I refer to the Clarke stories; not to the
>first editions, but almost certainly to the final versions. The bodies
>are stars, visited on the Universe-spanning trip in which IIRC Vanamonde
>appeared.
>
>I've read Clarke and Benford once; the mistake will not be repeated.

OK, I wasn't even sure which version you were talking about (much less
the details of the scenario), but now I know. I didn't have the same
negative reaction you did to the Benford version, BTW - I rather liked
the increased accuracy of the astronomy, of course accuracy doesn't
uniformly make for a good story.

6 bodies would require 5 second order equations of 2 variables each,
for a 20th degree polynomial for the eigenvalues after the reduction
of the problem to the equivalent 5 body problem. Also required would
be 100 entries in a 10x10 matrix to compute (the second partial
derivative of the potential with respect to each of the 10
coordinates). This sounds like A Bit Too Much Work, even if it's
possible (numerical issues may raise their head). Of course, there
may be a better method than the straightforwards one I'm using.

So far the track record of numerical simulations has actually been
very good as far as determining "stability" goes even using general
purpose alogirithms. I know I've seen the 5 body case simulated, I
don't think I've seen six (except with a central body which isn't the
case you want). Of course, there's always a nagging doubt about
numerical results, esp. if a general purpose integrator was chosen
(like the one I'm using), rather than one with better theoretical
properties (like a symplectic integrator). Another less theoretical
issue than the integrator choice is that people have differing ideas
of what the term "stability" means. I've seen results that have been
called stable based on eyeballing numerical results that I would have
called unstable eyeballing the same numerical results.

Joseph Hertzlinger

unread,
Jul 18, 2004, 1:18:43 AM7/18/04
to
On Wed, 14 Jul 2004 19:01:46 -0700, pervect <per...@invalid.invalid>
wrote:

> I'm definitely curious about the method used by JH's source, I'll


> have to see if I can order it via a library loan (but it may be too
> new for that).

It just covered some of the results of research. It didn't contain the
derivation.

--
http://hertzlinger.blogspot.com

Dr John Stockton

unread,
Jul 18, 2004, 9:11:41 AM7/18/04
to
JRS: In article <p50hf0t4l1pkn8400...@4ax.com>, seen in

news:rec.arts.sf.science, pervect <per...@invalid.invalid> posted at
Fri, 16 Jul 2004 18:57:38 :

>6 bodies would require 5 second order equations of 2 variables each,
>for a 20th degree polynomial for the eigenvalues after the reduction
>of the problem to the equivalent 5 body problem. Also required would
>be 100 entries in a 10x10 matrix to compute (the second partial
>derivative of the potential with respect to each of the 10
>coordinates). This sounds like A Bit Too Much Work, even if it's
>possible (numerical issues may raise their head). Of course, there
>may be a better method than the straightforwards one I'm using.


For a rosette, there are N bodies of equal mass m moving symmetrically
about a centre of mass M, for either M=0 or M>0.

If the bodies are perfectly placed with identical velocities (measured
with respect to their radii from the centre) the situation remains
symmetrical and must be easy enough to compute, especially for the case
where they have circular orbit velocity.

One can envisage a set of points in those perfect orbits, and express
the actual positions and velocities of N*m & M as offsets from that.

I'd try expressing those in terms of multiples of the N'th harmonic, so
to speak; using a series in sin & cos of 2 pi k / N or something like
that.

But I don't recall enough mechanics to attempt it.


>So far the track record of numerical simulations has actually been
>very good as far as determining "stability" goes even using general
>purpose alogirithms. I know I've seen the 5 body case simulated, I
>don't think I've seen six (except with a central body which isn't the
>case you want).

Both are of interest. The cases in those particular books are of no
special intrinsic importance.


> Of course, there's always a nagging doubt about
>numerical results, esp. if a general purpose integrator was chosen
>(like the one I'm using), rather than one with better theoretical
>properties (like a symplectic integrator). Another less theoretical
>issue than the integrator choice is that people have differing ideas
>of what the term "stability" means. I've seen results that have been
>called stable based on eyeballing numerical results that I would have
>called unstable eyeballing the same numerical results.

I'd call the situation stable if, for bodies started with arbitrary
small deviations from the ideal positions & velocities, the subsequent
deviations from the ideal remained within proportionate bounds.

I suspect that once the bodies are no longer bound to within something
less than their "fair share" of the pie, the ultimate motion will appear
chaotic (but perhaps not for very small N>2) and bodies may be ejected
to infinity (but not for small N>2).



--
© John Stockton, Surrey, UK. ?@merlyn.demon.co.uk Turnpike v4.00 MIME. ©

pervect

unread,
Jul 19, 2004, 1:47:35 AM7/19/04
to
On Sun, 18 Jul 2004 14:11:41 +0100, Dr John Stockton
<sp...@merlyn.demon.co.uk> wrote:


>If the bodies are perfectly placed with identical velocities (measured
>with respect to their radii from the centre) the situation remains
>symmetrical and must be easy enough to compute, especially for the case
>where they have circular orbit velocity.

Unfortunately, we have to ask the question of what happens if the
starting conditions are not perfectly symmetrical. And if the system
is unstable, even a very small initial asymmetry can grow so large
that (for instance) one of the bodies is ejected. The symmetry
argument doesn't work here. For a visual aid, look at what happens to
the Kempler Rosette (last figure) on

http://burtleburtle.net/bob/physics/kempler.html

To make life even worse, the very first step, reducing the N body
problem to the equivalent N-1 body problem, pretty much totally
destroys the symmetry of the problem anyway, at least according to the
only algorithm I know about for doing it. (This is the use of the
so-called Jacobi normal coordinates, which I found rather fortuitously
in "An Exactly Conservative integrator for the N-body problem" by
Kotovych and Bowman, you can probably find this on the WWW like I
did). I presume Jacobi normal coordinates are "well known" in spite
of the fact that I never heard of them before in my life :-). They
are definitely, neat, though, a very nice "trick". I definitely did
*not* get good results when I omitted this step -- having more
variables in the equations than actual physical degrees of freedom
appears to be a Really Bad Idea if one is attempting to compute the
eigenvalues of the system.

Basically, once the first hurdle is over (one has the correct number
of variables to describe the state of the system that matches the
number of physical degrees of freedom that the system actually has),
the idea is to write down the differential equations which describe
the time evolution of the system in terms of these variables. One
could use any method one desires to derive them (though I would
suggest the Lagrangian method). Having written the equations down,
one then linearizes them about the equilibrium solution. The
linearized equations for the "error solution" then must have solutions
of the form

error-coord1 = C1 *exp(lambda*t)
error-coord2 = C2 *exp(lambda*t)
...
error-coordn = CN * exp(lamba*t)

and one finds that only a few values of lambda (which is in general a
complex number) allow non-trival solutions (i.e. solutions in which
not all the Ci are zero) which satisfy the differential equation.

If any lambda has a positive real part, the equations are definitely
unstable (according to the linear analysis, anyway) because the error
grows without bound. It turns out the case where lambda=0 turns up in
my solution of the 3-body problem and actually deserves a lot more
consideration than I gave it BTW.

I also have the nagging feeling that there must be some better way of
taking advantage of the almost-symmetrical nature of the problem than
the method I use, BTW, but it's not clear how this can be done, it
would require a totally different approach than the eigenvalue
approach I'm using, I think. Maybe something using Lie groups... or
maybe not, I don't know.

>I'd call the situation stable if, for bodies started with arbitrary
>small deviations from the ideal positions & velocities, the subsequent
>deviations from the ideal remained within proportionate bounds.

Me too.

Dr John Stockton

unread,
Jul 19, 2004, 11:16:42 AM7/19/04
to
JRS: In article <jvkmf0lr7ej0rojcl...@4ax.com>, dated
Sun, 18 Jul 2004 22:47:35, seen in news:rec.arts.sf.science, pervect
<per...@invalid.invalid> posted :

>On Sun, 18 Jul 2004 14:11:41 +0100, Dr John Stockton
><sp...@merlyn.demon.co.uk> wrote:
>
>
>>If the bodies are perfectly placed with identical velocities (measured
>>with respect to their radii from the centre) the situation remains
>>symmetrical and must be easy enough to compute, especially for the case
>>where they have circular orbit velocity.
>
>Unfortunately, we have to ask the question of what happens if the
>starting conditions are not perfectly symmetrical.


That is why you should have read the next paragraph before responding.

--
© John Stockton, Surrey, UK. ?@merlyn.demon.co.uk / ??.Stoc...@physics.org ©


Web <URL:http://www.merlyn.demon.co.uk/> - FAQish topics, acronyms, & links.

Correct <= 4-line sig. separator as above, a line precisely "-- " (SoRFC1036)
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pervect

unread,
Jul 24, 2004, 3:42:54 AM7/24/04
to
OK, here's the Really Easy way to figure out the stability of the
three body system (via computer).

Step 1: Write down the Hamiltonian of the system. It turns out NOT to
be essential to reduce the three body problem to a two body problem in
spite of my earlier remarks. It may be desirable to do this if one is
doing the problem by hand or if one wants the extra insight as to
which eigenvalues can be eliminated from the system.

Perosnal note - usually I find Hamilton's method to be a pain as
compared to Lagrange's method, this is a case where the pain appears
to be worthwhile.

For the three body system, the Hamiltonian takes the form

H := 1/2*(px[1]^2+py[1]^2)/m[1]+w*(px[1]*y[1]-py[1]*x[1])+
1/2*(px[2]^2+py[2]^2)/m[2]+w*(px[2]*y[2]-py[2]*x[2])+
1/2*(px[3]^2+py[3]^2)/m[3]+w*(px[3]*y[3]-py[3]*x[3])+V

where V =
-G*m[1]*m[2]/(x[1]^2-2*x[1]*x[2]+x[2]^2+y[1]^2-2*y[1]*y[2]+y[2]^2)^(1/2)
-G*m[1]*m[3]/(x[1]^2-2*x[1]*x[3]+x[3]^2+y[1]^2-2*y[1]*y[3]+y[3]^2)^(1/2)
-G*m[2]*m[3]/(x[2]^2-2*x[2]*x[3]+x[3]^2+y[2]^2-2*y[2]*y[3]+y[3]^2)^(1/2)

Step 2: Figure out the inital conditions that solve the system

For the three body problem, we know this is an equilaterial triangle
revolving around the center of mass, even for the case where the mass
of all three bodies isn't equal.

For N>3, things are not so easy, unfortunately, when the masses are
not equal.

Step 3: Check that the inital conditions actually do solve the system.
Since the solution is assumed to be a stationary this means that
dx[i]/dt = dy[i]/dt = dpx[i]/dt = dpy[i]/dt = 0. (It's possible to
relax this condition somewhat for "uniform motion", but we don't need
to do it here.)

*note all derivatives other than d/dt below are partial derivatives

Using Hamilton's equations, this means (i = 1..3)

0 = dx[i]/dt = dH/dpx[i];
0 = dy[i]/dt = dH/dpy[i];

0 = dpx[i]/dt = -dH/dx[i];
0 = dpy[i]/dt = -dH/dy[i];

where the partial derivatives of H are evaluated at the chosen initial
conditions. (Note the pesky minus signs in the second set of eq's).

Next step - linearize the equations. This involves taking sets of
second partial derivatives of H, i.e. we expand

dH/dpx[1] = sum of d^2H / dpx[1] d<some variable> * <some variable>

where <some variable> takes on the values

x[1] y[1] x[2] y[2] x[3] y[3] px[1] py[1] px[2] py[2] px[3] py[3]

The result is a 12x12 matrix, but a lot of the entries are zero. It'd
be very tedious to compute and arrange all 144 entries by hand, but
that's what computers are for.....

By arranging the rows of this matrix in the proper order, and by
putting the pesky minus sign in the lower half of the matrix where
needed, the matrix transforms the vector V

x[1] y[1] x[2] y[2] x[3] y[3] px[1] py[1] px[2] py[2] px[3] py[3]

into its own derivative

d/dt[V]


Because each solution is of the form A*exp(lambda*t) (we know this
because the equations are linear, and we know the equations are linear
because we linarized them ourselves), d/dt V = lambda*A*exp(lambda*t)
= lambda*V

Thus we've reduced solving the differential equation to solving the
standard eigenvalue problem

lambda*<VECTOR> = <MATRIX><VECTOR>


For the Hamiltonian above representing the three body problem with
masses m[1],m[2],m[3], and the appropriate initial condtions, the
characteristic polynomial of the 12x12 matrix of second partial
derivatives is

1/4*lambda^2*(lambda^2+m[3]+m[1]+m[2])^3*
(4*lambda^4+(4*m[1]+4*m[2]+4*m[3])*lambda^2+27*m[1]*m[3]+27*m[3]*m[2]+27*m[1]*m[2])

The roots of the characteristic polynomial give the eigenvalues which
are the allowed values of lambda.

If we do chose some of the alternatives I mentioned to reduce the size
of the matrix, we can get rid of the two zero eignvalues (the lambda^2
term), and two of the repeated triple eigenvalues of
(lambda^2+m[1]+m[2]+m[3]). The remaining eigenvalues can't be
eliminated.

The stability criterion for an undamped system like this is is that
lambda^2 must be real and negative (lambda must be purely imaginary).

If we consider the last term as a quadratic polynomial in lambda^2, we
get the appropriate condition

(m[1]+m[2]+m[3])^2 -27*(m[1]*m[3]+m[3]*m[2]+m[1]*m[2]) >= 0

to make lambda^2 real (and negative).

Failure to meet this condition will generate errors which increase
exponentially with time.

T

unread,
Aug 18, 2004, 12:13:59 PM8/18/04
to
Frank Scrooby wrote:
<snip>
> By comparison
>
> (according to http://www.nineplanets.org/earth.html)
>
> Earht's mass is +- 5972 billion billion tons making this sucker just 0.24 %
> of Earth's mass (if I got all my numbers right). Earth density is a
> relatively modest 5.52 grams per cubic centimeter, which translates into
> 5.52 tons per cubic meter (if I get all my numbers right again).
>

Wait, wait, wait-
5.52 tons per cubic meter =
(5.52*2000) =

11040 pounds for a cube one meter to a side.

???


TBerk

John Schilling

unread,
Aug 18, 2004, 3:00:32 PM8/18/04
to
T <tb...@sbcglobal.net> writes:

>> (according to http://www.nineplanets.org/earth.html)

>???


This is correct. A cube of "Average Planet Earth Material" one meter
to a side, will weigh 11,040 pounds.

Note, for example, that a cube of cast iron one meter to a side will
weigh 17,420 pounds. A cube of basalt (the most common igneous rock)
one meter to a side, will weigh about 6,000 pounds depending on exact
composition. The Earth is essentially made of a layer of basalt on
top of a layer of iron, so 11,040 pounds per cubic meter for the planet
as a whole is about right.

And for the pedants, the densities I gave for cast iron and basalt are
at room temperature and pressure. Most of the Earth's interior is quite
hot, which reduces the density slightly, but also compressed by the
weight of Sagans of tons of rock sitting on top of it, which increases
the density slightly.


--
*John Schilling * "Anything worth doing, *
*Member:AIAA,NRA,ACLU,SAS,LP * is worth doing for money" *
*Chief Scientist & General Partner * -13th Rule of Acquisition *
*White Elephant Research, LLC * "There is no substitute *
*schi...@spock.usc.edu * for success" *
*661-718-0955 or 661-275-6795 * -58th Rule of Acquisition *

Sea Wasp

unread,
Aug 18, 2004, 5:27:52 PM8/18/04
to

Yes.

Remember that the average density of earth includes a core made of
mostly solid/liquid iron.

So imagine a cube one meter on a side, made of, say, half soil and
half solid steel. Yes, it's that heavy.


--
Sea Wasp
/^\
;;;
Live Journal: http://www.livejournal.com/users/seawasp/

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