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Open letter to science editors
A MODEL OF THE POLAR CONFIGURATION
I've been following David Talbott's material for a number of years, and the
first response, generally, to what he had proposed in the way of the planets
all lined up like a "shish-ka-bob" was, "That's nonsense, David. It's
against the laws of physics; you can't do that. Kepler's Laws and all that
sort of thing...You're dreaming. Forget about it." Well, when I got wind of
this I thought, "Well, my goodness, I don't believe that's the case at all.
believe that Kepler's Laws do not necessarily apply to this kind of a
configuration. And I also do not necessarily believe that the system would
be unstable or that the system would not maintain that configuration for a
long period of time." And so I've contacted him, and here I am.
(First slide) Now if we accept David's premise, these are the requirements
the model would have to fulfill: Number one, Saturn would definitely have to
be something more than the speck in the sky we see today. For that reason it
would have to be relatively close to the earth; much closer to the earth
at present. It has to remain in a fixed position, opposite the North Pole.
And in that case this would require a certain alignment of the earth's polar
axis, which would be a system which would mitigate against causing the axis
rotate because of the centrifugal forces and the requirements of a spinning
object to maintain its polar alignment. It should have a lighted crescent,
and he's talked about that fairly long. Venus and Mars have to be aligned
concentrically, with Venus in a position closer to Saturn, and Mars closer
earth, and Jupiter not visible. Jupiter does not enter into his mythological
model at all. And then he has the mountain-like apparition, and I have
nothing much to say about that. My competence has to do with the orbital
(Next, please.) So this is what I propose would be the configuration that
would fulfill these requirements, as mentioned by David. (Well, let me see
if I can work this. There we are.) We have the Sun and we have this array
of planets, and this is the array of planets orbiting. And what we have is
Jupiter, Saturn, Venus, Mars and Earth all rotating about a common center of
gravity with Jupiter being on the opposite side and the center of gravity,
course, being between Jupiter and Saturn. So then Jupiter would be behind
Saturn. And then the planets would be lined up. And Jupiter would not be
visible because of the angle of visibility. Now this array would orbit the
Sun as an array, but the group of planets would orbit in a clockwise
direction, while the array orbits in a counter-clockwise direction. And this
puts them in a retrograde motion. (I'll have a little something to say about
that afterwards.) The retrograde motion is quite significant. And the
planets are aligned...If this is synchronous-that is, if this spin or the
orbiting of these bodies is actually in the same period of rotation as the
orbit of the whole thing around the Sun-then the stack alignment will remain
in a permanent position. And this requires that the Earth's spin axis does
not have to precess. So the Earth's axis would be pointed along this line,
and it would always be pointed in space towards some point in space, and be
immovable relative to that point in space. I believe that's all on that one.
(Next one please.) I'm not going to bore you with the equations, but what I
show here is: Here is the center of gravity around which these planets are
orbiting. This would be Jupiter, Saturn, Venus, Mars and Earth, and these
will be orbiting in this direction, whereas this whole thing would be
in this direction around the Sun. And in order to determine the distances
between these things you have to solve a substantial number of algebraic
linear equations. To be exact there're ten equations for the five planets.
And so it's a task in itself just doing this. I've worn myself out, and my
computer, quite extensively, trying to solve these at times. Then the
distances solved from these equations, where "A" is the distance from the
center of gravity, "B" the distance to the center of gravity, "C" between
Venus and Saturn, "D" between Mars and Venus, and "E" between Earth and
And then these are the values you get out of these equations.
Now I don't vouch for the total accuracy of these things right now because
method of solving these equations was quite approximate. I would beat it to
death on my computer with great anxiety and terror! A man by the name of
Spedicato, who was supposed to be here today, a man from Italy, had taken
trouble of setting up a very elaborate scheme in which he could compute
numbers to a much greater precision than I could. And that precision was
really necessary in order to demonstrate that this system is indeed stable.
And I contend, as a result of the calculations that I made, that this
system in itself (not in the realm of the Sun), is a stable configuration.
It will do this, and it will stay in that configuration for a long time,
except for one thing. And that is this: (If I can push the right button
here...Yes...) Now we have the little tiny planet Mars, here. What we have
is generally two bowling balls, two golf balls and a marble. The two golf
balls and the marble won't have a great deal of influence on these two
bowling balls. They can bang it and beat it and do anything they want, but
it's not going to change their configuration. But Mars-being the smaller
planet in between these two larger ones, namely Earth and Venus-is like
having a steel bearing on a plate, positioned between two magnets. And you
have the plate, and you would tip the plate a little bit. And it would try
roll towards one, but then the magnet would try to pull it. And then you'd
have to tip the plate back the other way. So what I contend will happen, and
have yet to calculate, the equations of which I'm in the process of setting
to do now, leads me to be almost certain that this will set up a resonant
condition. That is, we have a synchronous orbit. But this thing can get
moving back and forth in sync with the synchronous orbit, and it's like a
shock absorber on your automobile going down a rough road. If your shocks
aren't very good your wheels will bounce and it gets what is called a
(Could I have the next one now?) So in order to demonstrate that this thing
would work, I have made the calculations of the in-line position, and it is
stable and it does work very well. Now the question is, "When you put that
in orbit around the Sun, will the Sun tear it apart?"
Now the first response of the astrophysical community to David's position
"Well, even if it were stable, when you put it in the Sun's gravity, the Sun
will just tear it apart. It can't possibly exist." And so I said, "Well,
I'll try to see what I can do about that, and maybe I can make it stay."
Well, hopefully Newton will do that for us. The first calculation was just
to take Jupiter and Saturn, which can be treated individually as a separate
sort of set. And you can sort of throw Earth and Venus and Mars out because
they don't have too much effect on this. And these are generally the
conclusions that all of my great fans are pointing out to me, which number,
think, a half! Now what I have done is set up a mathematical model,
calculating step by step in time the orbit of these two planets around their
mutual center of gravity as they orbit the Sun. And in doing so I'll later
give a little example of what the process of the mathematics is.
And I discover when you do this that the Sun does indeed have a significant
influence on the behavior of these two as they orbit. But the Sun's
influence is not of a total effect; it is one of a relative effect. It is a
condition called a "gradient." That is, as the planetsrotate each other (and
this is the Sun). They rotate about their center of gravity. At one point
they're at the inferior conjunction; at this point they're at the superior
conjunction. Now the Sun's gravity is different at this point than it is at
this point. And by virtue of that, then, the system is not in a constant
gravity field. Now, I've been a dynamicist and a dynamic modeler for a
years, and one of the things about the dynamic system is that when you put a
dynamic system, any kind of a rotating system, or any kind of a vibratory
system (except a pendulum; a pendulum is the exception) the motions of that
system-its frequency, its period and its response to external accelerations-
are independent of the gravity field in which it is functioning. It's just a
law of physics.
And so if the Sun's gravity were constant, that is, if this, for example,
so far away and the things were so close together as you would think of the
Moon and the Earth...The Moon and the Earth are very close together. And
they're quite far away from the Sun. And so there is not as much difference
between the gravity force at a distance away from the Sun as it is towards
the Sun. And so this says that that would not have as great an effect.
Well, the Moon does have an orbit that does that sort of thing-and it has
shown, it has been calculated, that it does do this. And this was a
distortion of Jupiter. It tended to want to bulge out on each side; it tends
to want to compress on the other. I started at the twelve o'clock
calculation...The fact that it doesn't come back to the twelve o'clock
calculation is related to two things. One, the input conditions that I use
were arbitrarily selected so it will take it a while to settle down. And
secondly, the thing is grossly sensitive to the magnitude of the integration
step that you use to make the calculation. And the integration step is
simply: How short a period of time do you calculate these things as you do
it step by step in time, because you can't make a smooth mathematical
representation, because the equations cannot be integrated directly-or, they
can be, but it takes Herculean intelligence to do it, and I'm not in that
(Let's have the next one.) The second calculation, again a trial
but here the idea was to see whether, if I put a tiny planet out here, which
would be a little farther distant away than Saturn and Jupiter were from the
center of rotation, and if I put the Earth out here at a position, it would
undergo a larger change in the gravity force, since the gravity force has a
lapse rate and drops off. The farther away you get from the Sun, the less
the gravity is. And so I tried this. And what I did was do something (of
course I'm beloved of the astrophysical community for doing anything like
this) which demonstrated that as long as these things are lined up I can
the mass of the two together. And as long as they stay in the oriented
position...and I determine the mass of this orbiting...the magnitude of this
lump mass gives the same gravitational effect as the two planets would be
together. Then I feel I've modeled it, in a sense. And I can determine if
putting another planet in is going to cause it to do something bad. Well the
answer is, "No, it doesn't do anything bad." It's a little farther away and
has a little more ellipticity. That is, it tends to want to bulge out a
little more on one side. But again this is still subject to the kind of
input conditions you introduce into the magnitude of the time step employed
in the integration procedure.
(Could I have the next one?) No. This one would represent the entire thing
without Mars. And I think you can consider that Mars would not have any
influence on the general configuration, or the general stability of the
orbits. Mars would, in my contention, get a resonant condition going and
would cause all three-that is Venus and Earth and Mars-to start oscillating
in a very undesirable way. This would cause either the Earth to approach
Mars or Mars to approach the Earth and then to go apart, for Mars to
Venus, and Venus to approach Mars. In other words, it would be sort of a
pulsating thing, synchronous with their orbiting around their center of
gravity and their orbit around the Sun. Let me have the next slide.
(Missing text due to change of audio tape)
...pattern and you have the Sun pulling on Saturn. And the resultant of
two acceleration factors is that little line right there. And that is the
force that is involved as a function of the angle. And so, you set up a
little mathematical subroutine in your computer program, and you fix it so
that whatever the position is of this angle, and the angle "I" is the angle
that I use as the variable of its rotates. For every angle "I" and every
distance between the two there is a difference of this gravitational vector.
And so you put this in a little calculational procedure. And when you get it
into this position, you call it forth and say, "Tell me what that vector
And you do it and it gives that, and it gives you the angle. Then you go to
the next step...
Now I'm having a hard time seeing this, but this is the velocity vector
triangles formed. For example, Saturn has a component of velocity clockwise
around this center of gravity. So you represent it as an arrow. That's the
arrow there. The center of gravity has a counterclockwise velocity vector
perpendicular to this axis going in this direction. And that's this one
right here. So those two vectors, then, are the two vectors which determine
which, in combination with the gravitational vector, make the calculation of
what the motion of these two things are, relative to each other and relative
to the angle "I."
Then to do this, I take the resultant vector and use that resultant vector,
and find, then, the displacement at a given step in time. Find this
displacement and then this displacement, which is the one of Saturn versus
inertial space. This is the displacement vectors of the center of gravity in
inertial space, and this is the vectors of Jupiter in inertial space. You
can take the vector sum of these two, and that will give you what the change
in this radius will be. And that's what gives you that swelling back and
forth. As this thing circles and the gravity changes, then these radii will
move back and forth.
(Let's see about this last one here. Oh, all right.) The next step-once
you've determined the length of this, then you can re-determine this
velocity. To do that I used two methods. One of the fears of a dynamicist in
making a time-dependent calculation is to get what you call a "boot-strap
operation." That is, you start to calculate something, which depends upon
its position. And then you change that position by integrating certain
components of velocity and acceleration, and you get a changed position
gives you a changed force acting on it. And this changed force changes
again. And what you end up doing is calculating a displacement, which
determines the input, which determines the placement, which determines the
input, which determines the displacement, which determines the input. And
you end up with a mathematical instability, not a physical instability. It's
liable to go anywhere. As a matter of fact, my suggestion has been, sometime
if you try to use some of the so-called end body models on this thing,
end up with Saturn in your garage!
I used two methods. One method was to find the acceleration component
tangential to this point and the acceleration component tangent to this
point, and then you vectorially subtract those two. Then that becomes the
resultant acceleration vector of this point. Then you can compute that. But
if you do, you end up possibly with a boot-strap operation. So to avoid
that, I found that if I do that, let's say, for about a 90 degree movement
this angle, I find that if I go back and assume for a minute that there was
equilibrium condition formed that is, assumed that the velocity is related
this radial distance-and apply, then, the same law of equilibrium that was
applied in the line of planets, I can see what velocity it gives. And it
turns out that that velocity is less than a half a percent different from
other one. And so it becomes a convenient way to keep the whole thing tied
to that radius and prevents the boot-strap operation.
Now in some of the larger programs, the larger end body computer codes, they
go to something like double precision, and in some cases triple precision,
which means in order to make their calculations, in some cases, they'll have
the parameters out to as much as ten, fifteen significant figures, which is
well beyond the scope of my dealing with this thing. I want to stay within a
realm of precision which will serve my purpose of not determining what is
precise location of this at any precise moment of time. But does it hang
together? Well, so what I would say in conclusion is this. On the basis of
my calculations and with only modest approximations and only modest
assumptions-no assumptions, only modest approximations-I maintain that the
system is stable and will remain stable for a prolonged period of time,
such time as the planet Mars disrupts the whole thing and causes it not to
fall apart because it's unstable in the beginning; it's unstable because
is not going to let it be too stable. If there's anything that can possibly
start Mars in a resonant mode it'll tear up the whole situation, and it will
be disastrous. Something's going to hit something or come very close to
One other conclusion about this is that I noticed that the spin of the
Venus today is in retrograde. It is spinning backwards from everything else
in the solar system. And it's been sort of a mystery. Why does Venus spin
backwards? It spins backwards at a very, very slow period, at a period of
something like 250 - 260 days, which is nearly a year. It not only spins
backwards, but every time it comes into conjunction with the Earth it gives
the Earth its same face. It aims its...It moons us! (Laughter...) So I took
the trouble of saying, "Well my goodness. In all my calculations I have
assumed that the distance between the Sun and this orbiting system (which
would be the distance from the Sun to the center of gravity) was roughly
the distance is from the Sun to the Earth today. Now that would put, then,
the Earth in this system...This is exaggerated scale, of course, this
distance here is nowhere near that large. We're talking something which is
eight to ten percent of this distance. So it's exaggerated here only to show
my procedure. If Venus was orbiting in a contrary way to the center of
gravity, it would, in a period of time which would be fairly substantial,
align itself. It would eventually cause its rotation, its rotational period,
to equal that of the orbital period exactly the way our Moon does. Our Moon
has a spin which is totally synchronous with the Earth, so we only see one
side of the Moon. We never can see the other side. Now what would happen
then if it were orbiting here and we're in a period long enough so that it
would have that period of 250 to 260 days? Just pretend for a minute that it
didn't change. When the thing blew up or tore apart or did whatever it was
going to do...Obviously it can't be totally stable or it we'd still be in
configuration today. That's why I tell everybody when they say, "Why, your
system isn't stable!" And I say, "Well you know it can't be too much or we'd
still be there. We'd still be looking at Saturn right now."
Anyway, so what we do here is say that...Let's pretend that it has the same
spin period as the period of this thing around the Sun. That will tell us
from Kepler's Law how far it is from the Sun. It turns out to be something
close to the Earth's orbit, but between the Earth's orbit and Venus' orbit.
So it is something less than it is today, which would make the Earth a
bit warmer, but the seasons would be kind of the same. In other words there
would be a seasonal variation very much as there is today, and it would fix
the position of this. So then I made another calculation. I said, "Well,
what is the total energy of these planets in the solar system today?" So you
can take Saturn out in its orbit, way out, Jupiter, Mars, Venus and Earth.
Take their kinetic and potential energy condition as of today, and see what
position it would have to be, relative to the Sun, in order to produce that
same energy. And the energy balance occurs at about the distance that we
talking about-where Venus would be, in order to put itself into that
retrograde spin condition. Which gives another sort of credibility to the
fact that it could indeed exist in this manner. I think that's about all.
Moderator: Well done, Sir, time-wise. Are there questions or comments?
Well, all right, go ahead. Better go to the microphone.
Grubaugh: I don't need the microphone. I have a booming...
Questioner 1: I'm somewhat mathematically challenged, as are most of us
So if you don't mind, I'll refer to this one as the ferris wheel. You have
Jupiter and Saturn orbiting about a common gravitational or ...(???)...
center which, in effect, is not unlike an Earth-Moon system today. But there
are also some Lagrangian points beyond the Moon's orbit, which would be like
beyond Saturn's orbit. Now wouldn't the Venus-Mars-Earth system be in that
Lagrangian area-whether it's L-3 or whatever, I'm not quite sure-beyond
Saturn, if they are orbiting together following Saturn, as it were, in its
Grubaugh: As they orbit around Saturn, the Earth moves farther away and is
the position beyond Saturn at one point, and then as it gets around to the
other side it's between the Sun and Saturn.
Questioner 1: No, no, no...It's orbiting beyond Saturn, but you never see
Grubaugh: Oh, Jupiter.
Questioner 1: ...because in the Earth-Moon system...If you're in, I think
it's position L-3, I'm not quite sure...You're beyond the Moon. You never
see the Earth, but you're in that position. And you go around with the Moon.
Now I see this in that configuration, but these Lagrangian points are
Grubaugh: The what now?
Questioner 1: Lagrangian.
Moderator: The Lagrangian points are unstable.
Grubaugh: Oh, this has nothing to do with the Lagrangian points.
Questioner 1: You are describing Lagrangian points.
Grubaugh: No, Lagrange point is only one.
Questioner 1: You are. I see the Earth, Mars and Venus in a Lagrangian
system, and this is the description I get.
Moderator: Would you explain what is a Lagrangian point?
Grubaugh: Lagrange was a mathematician-astronomer who derived various
positions in which a planet could orbit the Sun, relative to another planet.
In other words, there was like a three planet system. And in his, Lagrange
points are not orbiting. You would have Jupiter and Saturn orbiting the Sun
in some synchronous manner, not one orbiting the other while it orbits. And
so there is no Lagrange point here. The only Lagrange point that would be
similar would be when this thing arrives in the superior conjunction
where Saturn would be beyond Jupiter. That would be a Lagrange position.
But for that to be a Lagrange position it would have to stay with that same
angle, relative to the Sun. And that's not what it does. In this model it
doesn't do that. It is orbiting as it orbits the Sun. And so a Lagrange
point is only a stable orbit of two planets around the Sun, not orbiting
other. And Lagrange does not apply to the Moon and the Earth system. The
Moon-Earth-Sun is not applicable to the Lagrangian points. There is no
Lagrange point that corresponds to the Moon orbiting the Earth orbiting the
Sun. I think that answers it.
Questioner 1: Well it sounds reasonable. But I think there may a...
Grubaugh: Could I get some help, Bob, from you? Would you back me up on
Questioner 2: Let's see, I have six points but I can't cover them all now.
On the Lagrange points-I think the last questioner's point was well taken.
You haven't distinguished between stable and unstable equilibriums. You have
described equilibriums, but you have done nothing about stability which
to be addressed, and I hope that you have taken a close look at Victor
Grubaugh: Oh yes, but Victor Slabinski's critique was a specific condition
which Saturn and Jupiter were not orbiting each other. He has the position
where Saturn and Jupiter are orbiting the Sun in sync. And in order for them
to orbit the Sun in sync, they would have to have Jupiter and Saturn in the
same orbit, and they would have to be separated at a distance, equal to the
distance from both of them to the Sun. So it would have to be an equilateral
triangle which is not applicable to what we're talking about.
Questioner 2: Well, there is a question of definition there, because if you
take the perspective of a distant inertial observer, your configuration is
not really orbiting at all. It's remaining...
Grubaugh: Oh no it's orbiting, absolutely.
Questioner 2: You're describing it going around the Sun, but they're
maintaining a fixed direction in inertial space.
Grubaugh: It doesn't maintain a fixed direction relative to the Sun. It
maintains a fixed direction relative to the...
Questioner 2: ...Relative to distant stars which means that...
Grubaugh: In order to do that it has to orbit; you have to have orbits. They
have to orbit the center of gravity.
Questioner 2: Well, the laws of inertia seem to refer to inertial space with
the direction of the fixed stars, which means that they're not moving with
respect to the fixed stars. And I think there would be a problem there.
Ultimately let me get to the point about the way all dynamicists check their
models with a numerical integration, as you know. But as you also know, this
model does not hold up under a numerical integration, which is the only way
you can tell if you're faithful to the Newtonian laws.
Grubaugh: Let me answer that, I think, as carefully as possible. This is a
numerical integration. This is all Newton's laws. It is Newton's laws of
equilibrium, and it is a numerical integration. Now what you're referring
to, I think, is the fact that...My great number of fans that I've developed
the news-and I have a large number of fans...a large number of fans...They
say, "Grubaugh, give me some numbers." And I say, "Get your own numbers."
"No, we have to have input conditions."
And I say, "Oh you need input conditions? Well, what are we going to give
for input? Well, I'll give you my input. I have a relative...I guess you
would call it a Larian (sp?) type of system in which I'm using as my
variables...are the component of velocity relative to the center of motion
the group as they...plus the movement of that center of gravity as it moves
around the Sun." Now in making that calculation it ends up that you have...
The only two independent variables of that calculation are the velocity of
thing relative to the center of gravity and the velocity of the center of
gravity as an input condition. Now if you take those two and add them
vectorially, that becomes the dependent variable. And the dependent variable
is not an operable variable. So when I gave these numbers to several of my
friends, they said, "Well, you don't understand. You just used...That's the
I said, "Well, if you use that velocity you're going to put Saturn in your
garage." And they went ahead and they did it anyway and they said, "It's
unstable." So I've gone to another end-body model with another researcher.
And instead of putting my numbers in, which are lousy, they're not numbers
that you can use in that model...If you're going to integrate and determine
position of an object by the means of a boot-strap integration, you have to
have the exact input conditions or you'll get garbage. If you don't have the
input conditions, it will integrate itself and add itself up and indeed you
will end up flying apart. And they end up with absurdities. Like the thing
disintegrating in a quarter of a cycle. And that's absurd. It can't...even
if it's unstable it wouldn't do that. It would take it at least an orbit or
so and, the way it would become unstable and disintegrate would be to spiral
outward. And finally to spiral out to a point where the gravitational forces
that would hold it to its point of orbit would disappear. I've told these
people and they choose to ignore me.
Questioner 2: Well there's a little difficulty in communication because
you're using terms with a slightly different meaning than is conventional in
the field. For example, your model is what we call an analytic theory, not a
numerical theory. If it were a numerical theory you'd have to...
Grubaugh: I'm a dynamicist, not an astrophysicist.
Questioner 2: In dynamical terminology you have to have precise positions,
velocities and masses for each of the bodies in your system...
Grubaugh: Absolutely. And if they're not correct when you do the numerical
integration you get the wrong answer.
Questioner 2: ...And the configuration, which, if you perturbed the initial
condition slightly and it flies apart, it's called unstable. Even if it's in
equilibrium, it's called unstable. And that means that in physical reality
the slightest force that came along, it would come apart and fly apart
and faster as it went.
Grubaugh: Well let me say this: ...If you had Saturn and Jupiter orbiting
their center of gravity...If you bring them farther and farther apart, so
that the gradient of the Sun's gravity is sufficient to cause it to be
disrupted and cause that little...where I showed that orbiting distortion.
When that distortion becomes larger it would be essentially unstable, which
have done. I took the model and ran it with Saturn and Jupiter and put the
distance out to (instead of one-tenth), out to about 40%. In other words,
instead of having the distance between Jupiter and Saturn being one-tenth of
the distance to the Sun, I moved it out so that it was four-tenths of that
distance to the Sun. And it blows up. It just goes unstable as hell. It
starts out and it goes up and it won't center, and then it goes on, it
spirals away. It just won't stay. And that's an instability. But up until
then, it's stable as hell. As a matter of fact, I can't disturb it enough to
cause it to go unstable.
Questioner 2: Yes. Well one last quick comment, then we'll have some more
discussion later, I'm sure. There would be spectacular transits or eclipses
of the Sun under these conditions. And also you'd get 6 month days on
certain points on the Earth. Why aren't those in the record too? (Then I'll
Grubaugh: Oh well, yes. The answer to that is yes, if the things are in
perfect plane. And, of course, I would assume that they're in perfect plane
because I haven't modified my equations to allow for any slope of the
eccentricity. But as you know today, if the Moon and the Earth were in
absolute plane or condition, there would be an eclipse every month. And so
you have to say, "Well, yes, once a year the Sun is going to be behind
Jupiter and Saturn." Once a year. Now I will just say that the likelihood of
these things being perfectly coplanar is very slim. They probably will have
some slight edge, some slight turn. Not only that, but they won't be
absolutely perfectly aligned, probably. The equations of the stack itself
tends to pull them in a line. It's like having them with a rubber band
between them. You try to pull one out of line and it will twang right back
into position because the gravity forces tend to make it more stable and
bring it together. The fact is, I'm worried about the stability of that
stack itself, because I'm going to have trouble, I think, putting in a kind
a nudge to get Mars to do something bad. And if I can't do something bad I'm
not sure what I'm going to...We'll have to tell David that maybe Mars isn't
going to come back and forth. I kind of feel that it will-that it will set
some kind of a synchronous condition. But to go back and repeat: The
problem with using the end-body models is that they do not work in reverse.
You can't say, "I have an orbiting system of these things and you tell me
the input conditions are to put into that model in order to make it an
orbiting system. It doesn't exist. You have to put in numbers and then it
calculates it. And I'll say it again: If you don't put it in, if you're
doing double precision or triple precision, if you give it the wrong number
you're going to get the wrong answer. And that's what they're doing. And I
found one model, now, where what we have done is adjusted. Let's say we take
the conditions that I have given and said you can't use, and tried it, and
course in the model just flies away. It won't even go 20 degrees. It goes
20degrees and flies off. Well, now, wait a minute. We've got bad influence.
Let's change the inputs just a little bit. So we change the inputs a little
bit...ah!...It doesn't do so bad. Finally we get this adjusted right in the
place-"Home sweet home."
Talbott: This question of the Lagrange point troubles me in terms of a
conversation we had and I just want...It's a fundamental theoretical issue
here and the possibility of actually an error that is ultimately
to the system. When I asked you many months ago, "Now how far would we have
to put the Moon away from the Earth for the Moon to revolve around the
so to speak, in one year?" You said, "Well now, curiously, that's a Lagrange
Grubaugh: No. It is not.
Talbott: It is not?
Grubaugh: No. It is not.
Talbott: Okay, then that answers the question. Because if, indeed it was,
then it's revolving...
Grubaugh: You asked me...I'll tell you the question you asked me. "How far
would you have to put them apart so they would orbit the Sun in exactly the
same period?" The two of them orbiting the Sun at the same period. The
answer is they would have to be one Earth diameter distance from the Sun
apart, in other words, a 60degree triangle.
Talbott: Okay, then. And that is a completely different position than the
position at which it would revolve...the Earth and the Moon would revolve
around that very center in one year. That's a completely different position.
And it might...
Grubaugh: ...It's totally different from a Lagrange point. It has nothing to
do with a Lagrange point.
Talbott: Okay. That does answer it.
From audience: Is that true, Tom, by the way? (Laughter)
Talbott: Excuse me, just a second. The other night I asked him that
question. When I asked Tom that question the other night, he said, "No.
That position in which it would revolve in one year is the Lagrange point."
Grubaugh: No it isn't. No it isn't. (Laughter from audience.) It is
Grubaugh: I might amplify that a little bit. Facts are the enemy of truth.
Questioner 4: Somebody said that this discussion is "far out!" I thought I
heard you say in your model that the Jupiter-Saturn system was orbiting the
Grubaugh: No. Saturn and Jupiter...Well, it depends on what you want to call
retrograde. That is, I say that Saturn-Jupiter, they align, so the stack is
orbiting its center of gravity in a clockwise orientation and the whole
is orbiting the Sun counterclockwise. And that makes one or the other
Questioner 4: I misunderstood. I thought you said that it was clockwise.
Grubaugh: Well sometimes my tongue gets a little bit loose and you have to,
sort of, correct me. What I'm saying and I'll repeat it: The stack is
orbiting its center of gravity in a clockwise rotation, a clockwise orbit.
The whole system is orbiting the Sun counterclockwise.
Questioner 4: All right. Then I did hear you right. Then when the system
flew apart, because it has come apart...
Grubaugh: Well, when I...If I put the wrong numbers in an end-body model.
Questioner 4: Have you given thought to why these planets are now
Grubaugh: I think it fell apart because of the...I think because it's set
...It's my opinion, I don't know, I wasn't there, but I would guess that
since there is a good chance of a very severe instability resulting from
(Missing text due to change of tape)
Questioner 5: Yes, actually I hope this doesn't require a technical answer.
I wouldn't understand it anyway. But the innermost planet, Mercury, would
that be of such small mass as to have no implications on the dynamics...
Grubaugh: It is not in the model, and I wouldn't know where to put it in. If
David will tell me what the myth's say about where Mercury should be, I'll
put it in there somewhere and see what it does. See if it sticks. We'll try
Moderator: Charles, do you want to make your announcement as well as ask
Questioner 6: Yes, if possible. I'll ask my question first, then make my
announcement. I'm interested in the effects on the gravity of the Earth of
the model that you created. That is, Ted Holden was talking about the
dinosaurs, and he spoke of them as being in an environment which implies
surface gravity. And...Have you...Would this model, in some way, I don't
Grubaugh: Yes, you can calculate the change. There would be a substantial
tide as a result of all these planets acting on the Earth at these close
distances. Yes. The Earth would have a very substantial tide. A tide, I'd
say, depends upon what you consider to be tidal. If it's Earth and there's
not too much water...Let's say, if you assume that there's a lot of water
there, then the water would be pretty much piled up on the North end...There
would be an early substantial permanent tide. Now if there is not much
then it wouldn't make much difference. But there would be a substantial
Earth tidal effect and the effect of gravity at that point...
Questioner 6: I just want to know in a general sense.
Grubaugh: My immediate reaction is to say that the gravity would be less,
I hesitate to say that. I'm not sure, Charles. I'm not sure that these
gravity accelerations of a particle on the surface of the Earth-those
acceleration components-would vectorially add into a lesser gravity. I kind
of think it will. But if I say it is...I have some fans who will tell me,
"Boy, you're wrong again, Grubaugh."