CSvax:Pucc-H:Physics:nisy@pur-ee.UUCP (09/29/83)
The gravitational stable points for the Earth and Moon are called LaGrange points.
RHB@MIT-MC@sri-unix.UUCP (10/03/83)
From: Robert H. Berman <RHB @ MIT-MC>
A "Modified Three Body problem" consistes of two finite mass bodies
and a third infinitesimal test body. The test body is used to find
gravitational equilibirum points. There are actually 5 equilibirum
points of which 2 are stable. They may be located by the following
intuitive arguments.
First, the two finite mass bodies (M_1 and M_2) revolve around
their center of mass with a definite angular velocity Om. Consider what
happens in a frame of reference rotating with that velocity. Pick the
origin of the frame to be the center of mass. Then the two bodies can
be placed at coordinates (-x_1,0) and (x_2,0) in the rotating frame.
Next, the acceleration of the test particle is the combined
gravitational force from the two bodies ** and ** the centripetal
acceleration in the rotating frame.
Equilirbirum points are found when there is no acceleration on
the test body. Stability is determined by what happens when small
displacements from equilbirum occur, namely whether they stay small or
grow larger. This can be determined by checking the "tidal" forces on
the test particle in equilbirum.
The first Lagranian equilibrium point is the orgin (0,0), denoted
as L1. The acceleration on the test particle is
0* Om**2 - GM_1/x_1**2 + GM_2/x_2**2 = 0 (because it is the CM)
The next two Lagrangian points are somewhere to the right of
(x_2,0) and to the left of (x_1,0). The diminshed gravitational pull
from the further body is compensated for by the increased centripetal
acceleration. Together, these points L1, L2 and L3 are known as the
"straight-line" solutions. They are all unstable because changes in
the centripetal force don't match changes in the gravitational forces.
The stable equilbirum points L4 and L5 are located symmetrically
at (0,y_L) and (0,-y_L). They are both stable, as small displacements
from the points continue to orbit around the points in epicyclic
orbits.rjnoe@ihlts.UUCP (10/31/83)
Yes, the center of mass of a two-body system is Lagrange libration point L1 if and only if the masses are equal. But the barycenter in any event is L2. Let m1 >= m2, u = m2/(m1+m2), and let the constant separation of the masses be 1. Choose cartesian coordinates such that the masses are always in the x-y plane and let this be a rotating reference frame so that m1 is always on the +X axis and m2 is always on the -X axis. Remembering that 0 < u <= .5, m1 is at (u,0) and m2 is at (u-1,0). The Lagrange libration points are: L1 (x1, 0) L2 (0, 0) L3 (x3, 0) L4 (u-.5, +sqrt(3)/2) L5 (u-.5, -sqrt(3)/2) where x1 and x3 (x1 < u-1 < 0 < u < x3) are roots of a fifth order polynomial with coefficients linear in u. x3 is easy to numerically approximate but x1 is difficult for small u. It doesn't really matter, since L1, L2, and L3 are always unstable libration points. L4 and L5 are stable iff 27u(1-u) < 1, i.e. iff u < 0.03852 (approximately). For the earth-moon system we have u = 0.01215 (approximately), so L4 and L5 are the stable Lagrange libration points, located equidistant from both the earth and moon (centers), which is also the earth - moon separation (these are equilateral triangles m1, m2, L4, and L5 make). L1 is 64.5e3 km from the moon's center, L3 is 381.7e3 km from the earth's center, and the barycenter L2 is only 4.7e3 km from the earth's center (toward the moon, this time) or about 1700 km beneath the earth's surface. -- Roger Noe ...ihnp4!ihlts!rjnoe
bill@utastro.UUCP (William H. Jefferys) (11/01/83)
yes, the center of mass of a two-body system is Lagrange libration point
L1 if and only if the masses are equal. But the barycenter in any
event is L2.
--
Roger Noe ...ihnp4!ihlts!rjnoe
************************************************************************
Roger, I am sorry, but you are wrong. I do know what I am talking about,
since I did my dissertation on the restricted problem of three bodies and
among my publications are approximately 20 scholarly papers on it. I teach
the subject nearly every year at the upper-division or graduate level. But
you (and the rest on the net) have a right to demand more than my credentials
as evidence that what I say is correct, so let me state my case.
Possibly you are just confused by the notation. Depending upon
whom you read, the terms L1, L2 and L3 can refer to any of the colinear
libration points. Only one of them is between the two masses (and
therefore can be at the barycenter). Let us call it L2, and let L1 be the
one on the far side of the Moon, and L3 the one on the far side of the Earth
(away from the Moon). Therefore it doesn't make sense to say as you do
that L1 is at the center of mass iff the masses are equal, and then to say
that L2 is always at the barycenter (center of mass). I say that only
L2 *can* be at the barycenter, and then only in the case that the masses
are equal.
You are right that the Earth-Moon barycenter is about 1700 km
beneath the Earth's surface. This in itself should be sufficient to
convince you that it can't be a libration point. Remember, the L2 point
is a place where it possible to place a test mass and have it
remain in place (relative to the two massive bodies). If one were
closer to the Moon than the L2 point (and still on the line joining
the Earth and Moon), one would be accelerated towards the Moon. But
that means that every time the Moon was overhead, we all ought to fly
off the Earth and be accelerated to the Moon, which doesn't happen.
Roger might object, that we are rotating with the Earth. I say that
if he were correct, then to launch a plane into space, it would
be sufficient merely to fly it at exactly the speed required to
compensate for the Earth's rotation (about 1000 mph at the equator),
in such a way as to keep the Moon directly overhead.
Since he says that we are closer to the Moon than the L2 point,
by his reckoning, the plane should be increasingly accelerated
towards the Moon. That would be a whole lot cheaper than building a
shuttle, wouldn't it? Unfortunately, it won't work, because the L2
point is nearer the Moon than the Earth.
Now for the mathematics. I am copying from Peter van de Kamp's *Elements
of Astrodynamics* (Freeman, 1964) with notational changes to make him
conform to my choice of L1, L2 and L3. At the libration point L2 which is
between the masses, the *gravitational* acceleration on the body is
given by
f = A*mu/(B + x2)**2 - B*mu/(A - x2)**2,
where
mu = G*(M + m),
A = M/(M + m), the distance of the Moon to the barycenter,
B = m/(M + m), the distance of the Earth center to the barycenter,
G = constant of gravity,
M = mass of Earth,
m = mass of Moon.
Units of distance have been chosen so that the distance from Earth to
Moon, A + B = 1.
The *centripetal* acceleration at L2 is given by
fc = v**2/x2
where
v = x2*V
and V is the relative [angular] velocity of m and M for which the general
expression for circular velocity takes the form
V**2 = mu. [Kepler's third law for unit distance]
Hence
fc = x2*mu.
The equilibrium contition is the equality of the gravitational
and centripetal accelerations. Equating the two expressions yields
A/(B+x2)**2 - B/(A-x2)**2 = x2.
The total mass, and hence, mu, does not, of course, figure in this
relation.
The locations of the other two libration points, L1 and L3 follow
from the equations
A/(x1+B)**2 + B/(x1-A)**2 = x1,
A/(x3-B)**2 + B/(x3+A)**2 = x3.
Here, x1 and x3 are the distances from the center of mass of M and m
to the libration points L1 and L3 respectively.
[End of extract from van de Kamp.]
Note the similarity of these equations. All are fifth order
polynomials.
Roger claims that the L2 point is at the barycenter, so that X2 = 0.
If he were right, then for any A and B such that A>0, B>0, A+B=1,
the relationship
A/B**2 - B/A**2 = 0
would be satisfied. I claim on the contrary that the only solution to
this equation under the stated circumstances is A = B = 0.5. For B<<1,
A is almost 1 and we have (approximately)
1/x2**2 - B/(1-x2)**2 = x2.
Let u = 1-x2, then this becomes (assuming u small, as is the case)
1 - B/u**2 = 1 - u
or
u**3 = B.
Therefore, for small B the distance from the center of the Moon
to L2 is of the order of the cube root of B. For the Earth-Moon
system, this means that the L2 point is about 90000 km from the Moon,
less than 25% of the distance between the Earth and Moon.
I hope that this detailed explanation will clear this up. PLEASE
FOLKS, I provided one reference in my previous posting and another
one here. Both of them are by acknowledged experts in the field (one
of the authors of the Astronomical Journal reference, Victor Szebehely,
is "Mr. Restricted Problem", author of a monumental treatise on the
subject and the winner of the American Astronomical Society's Brouwer
Prize for his work in this field.) Please read them carefully before
posting any more claims to the net that the barycenter is a libration
point (except for the equal mass case). After doing so, if you have a
*proof* that Celestial Mechanicians have been wrong about this for the
past 200 years, post it and we will be happy to read it.
Bill Jefferys 8-%
Astronomy Dept, University of Texas, Austin TX 78712 (Snail)
{ihnp4,kpno,ctvax}!ut-sally!utastro!bill (uucp)
utastro!bill@utexas-20 (ARPANET)
s rjnoe@ihlts.UUCP (Roger Noe) (11/03/83)
Bill jumped on a typographical error. My article had a mistake in the second line. What I first said was: >Yes, the center of mass of a two-body system is Lagrange libration point >L1 if and only if the masses are equal. But the barycenter in any >event is L2. Let m1 >= m2, u = m2/(m1+m2), and let the constant separation ... The obvious error is that I typed "L1" above where I meant "L2". That clears THAT up. I still maintain that the barycenter is a point known as L2, where there would be an unstable libration point if the unequal masses were points. I never said that L2 was a libration point (except when the masses are equal). I merely stated that the point existed and had mathematical significance in and of itself. Bill, do you really think I'm so ignorant of physics that I can say in the same article that a point beneath the surface of the earth is a libration point, stable or otherwise? Hardly! We're all very impressed with your credentials (and your arguments) but you should realize that some other people do know what they speak of once in a while. I particularly take exception to your unfounded speculations of the type "But Roger might think . . ." followed by some absurd contortion of well-understood laws of celestial mechanics which you know as well as I do. Please read more carefully next time before you post a followup to this newsgroup and don't put words into the mouths of others. And if you have something to say that's not of interest to all readers of this newgroup, use mail. -- Roger Noe ...ihnp4!ihlts!rjnoe
rjnoe@ihlts.UUCP (Roger Noe) (11/03/83)
Bill jumped on a typographical error. My article had a mistake in the second line. What I first said was: >Yes, the center of mass of a two-body system is Lagrange libration point >L1 if and only if the masses are equal. But the barycenter in any >event is L2. Let m1 >= m2, u = m2/(m1+m2), and let the constant separation ... The obvious error is that I typed "L1" before where I meant "L2". That clears THAT up. I still maintain that the barycenter is a point sometimes called L2. This peculiar nomenclature has NOTHING to do with libration. I never said that the BARYCENTER was a libration point (except when the masses are equal). I merely stated that the point existed and had mathematical significance in and of itself. Bill, do you really think I'm so ignorant of physics that I can say in the same article that a point beneath the surface of the earth is a libration point, stable or otherwise? Hardly! We're all very impressed with your credentials (and your arguments) but you should realize that some other people do know what they speak of once in a while. I particularly take exception to your unfounded speculations of the type "But Roger might think . . ." followed by some absurd contortion of well-understood laws of celestial mechanics which you know as well as I. Bill, please read more carefully next time before you post a followup to this newsgroup and don't put words into the mouths of others. And if you have something to say that's not of interest to all readers of this newgroup, use mail. -- Roger Noe ...ihnp4!ihlts!rjnoe