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From: msb@sq.uucp (Mark Brader)
Date: Tue, 24-Jan-89 22:59:38 EST
Message-ID: <1989Jan24.225938.19510@sq.uucp>
Newsgroups: sci.space
Subject: Re: Finding Lagrange's Libration Points
Summary: L4, L5 partly derived; comments on other issues
References: <11854@dartvax.Dartmouth.EDU> <11910@dartvax.Dartmouth.EDU> <1989Jan18.044744.18328@sq.uucp> <1989Jan20.180839.7800@utzoo.uucp> <1989Jan22.235911.23395@utzoo.uucp>
Reply-To: msb@sq.com (Mark Brader)
Organization: SoftQuad Inc., Toronto


Last week I wrote:
> I did figure out how to calculate L1, L2, and L3, but I don't even see
> why L4 and L5 exist.  I hope someone can explain it to me so that I do.

I received 6 mail messages, from Bob Ayers, Alan Paeth, Marc Ringuette,
and Steve Willner.  Both Marc and Steve gave me the critical hint that I
had missed, and I was able to complete the derivation.  Then before I
finished writing this followup, Henry Spencer posted a clearer version of
the same hint, and now I can't prove I had it before he helped. :-(

It turns out that a completely different approach is needed for the L4
and L5 points from what I used for the others: as Henry said, it is not
possible to say, just because the mass of the secondary is much smaller than
that of the primary, that you'll treat the whole system as revolving around
the center of the primary.

Accordingly, let us put the center of the one body at point P, and of
another at point S.  Somewhere on the line segment between these points
is the center of gravity of the two bodies; call that C.  Then the ratio
SC/PC is equal to the ratio of the masses of the body at P and that at S.

Now put the third body at point T, off to one side of the line PS.
Choose a point B beyond T on the line ST, such that ST/BT is also
equal to that ratio of masses, i.e., ST/BT = SC/PC.  Then CTS and PBS
are similar triangles, and so TC and BP are parallel.

The gravitational force on the body at T will be the resultant of two
forces, one toward P (that is in the direction TP), and one toward S
(that is in the direction TS, or what is the same thing, direction BT).
The ratio of these two forces will be the mass-ratio of P and S divided
by the square of the distance-ratio from T to P and to S.  That is, it
will be (ST/BT)(ST/TP)2, where 2 denotes squared as in my earlier article.

Now, so far nothing has been assumed about the relative distances of the
bodies.  Now add to the above considerations one more assumption:  that
T is equidistant from P and S.  Replacing ST by TP in the above formula
gives the ratio of forces as simply TP/BT.  But the forces are respectively
in the directions TP and BT.  Then the resultant force must be in the
direction BP, which, as shown above, is also the direction TC.

That is, if one body is equidistant from two others, then the resultant
gravitational force on it is directly toward the center of mass of the
two others.

But the center of mass of a three-body system is on a line between any
one of the bodies and the center of mass of the other two.  Therefore,
in a three-body system, if two of the bodies are equidistant from a third,
then the net force on the third is also directly toward the center of mass
of the system.

And so if all three bodies are equidistant -- forming an equilateral
triangle -- then the net force on each one is directly toward the center
of mass of the system, and so, no "sideways force" exists to move the
bodies out of the equilateral configuration.  And notice that this is true
no matter how equal or unequal the masses are.

Now, this does NOT prove that an equilibrium exists.  It only shows that
IF there is an equilibrium of the kind described, then it is exactly the
equilateral triangle formation.  For an equilibrium to exist, it must also
be true that when the three bodies are set revolving around their common
center of gravity, each has the same orbital period.

I'll leave the proof of this as an exercise for some reader more energetic
than I am.  I'm content to see it proved that the equilateral triangle is
in fact an exact one.

A second exercise is to correct the equations I gave in my earlier
article to allow for the fact that the orbits are really around the
center of mass; at least allow for the secondary's contribution to that.
In particular, consider that I computed the Earth-Sun L3 point as being
46.5 miles off the Earth's orbit (treated as circular).  The center of mass
of the Earth-Sun system is of course 280 miles from the center of the Sun.
How does this affect the L3 point?  I'm getting lazy.  You do it.

Now for a couple of other details.  I also wrote:
> Since r is 93,000,000 miles, the L1 and L2 points are each 936,000 miles
> from the Earth.  ...  Since the ratio of the Sun's and Earth's
> diameters is about 108, the Earth exactly eclipses the Sun's disk at
> a point only 861,000 miles beyond the Earth  ...  It's a curious
> coincidence how close L2 is to the exact-eclipse distance.

One of my correspondents suggested that that was no coincidence, but
happens because the Sun and Earth have similar densities.  Actually they
don't, but he was close.  If the Earth was exactly 3 times as dense as
the Sun, then the L2 point would be at the exact-eclipse distance... or
almost exactly so, anyway.  This number 3 is the factor of 3 in my
equation [10] last week, but that equation was only approximate.

And finally, Dave Alexander wondered how I could justify saying that
> The L in each of these positions stands for libration, as a body
> near those positions may librate or oscillate around them, and not
> for Lagrange.

... when 3 of the 5 positions are unstable.  The answer is that this is
what I have read in books that have described the points.  I don't remember
which ones offhand, though one was probably Clarke's "The Promise of Space".
Perhaps Lagrange named the points himself and was being modest but wanted to
keep his initial in there!

In any case, Henry tells us that the unstable libration points have some
(barely) stable orbits around them, and a body in such an orbit might be
considered to be librating, so the name is not really all that bad.

One of the messages from Alan Paeth pointed out that the gravity wells
around the L4 and L5 points have a curious comma-like shape, and they
can even extend around the "back" of the primary and join up!  So a body
orbiting near one of these points need not stay near it all the time to
remain captured.


My thanks to all of my correspondents.  It was fun.

Mark Brader, Toronto		"The singular of 'data' is not 'anecdote.'"
utzoo!sq!msb, msb@sq.com				-- Jeff Goldberg