Newton’s Great... Oversight
Galileo’s Falling Bodies and Lagrange’s Trojan Asteroids
With Their Tadpole and Horseshoe Orbits

3          TROJAN POINTS AND
        TROJAN BODIES

SECTIONS

3.1 Homographic Property - Maintaining Shape

3.2 Geometry

3.3 Maintaining An Equilateral Triangle Without Expansion and Contraction

3.4 Homographic, Equilateral Expansion and Contraction

3.5 Trojan Points

3.1         Homographic Property - Maintaining Shape

So, we have just seen that it is simply derivable that if all 3 bodies form an equilateral triangle, then the Earth and the lighter body will fall together at the same rate as the Earth and the heavier body, or rather that they will start to when all 3 are released simultaneously (we don’t even need calculus). We will see shortly that they will continue to do so.

Since we are trying to make a point here of how easily this oversight could have been avoided, even in the time of Newton, we will note that the relations among acceleration, velocity and distance were sufficiently well known long before Newton’s — and Leibniz’s — development of calculus. They were more than simple enough as far as Newton’s scientific contemporaries are concerned, even if calculus was eventually needed to provide a satisfying mathematical foundation for them.

Now we can note again that the equations we have just looked at do not actually distinguish the 3 bodies in the equilateral triangle, except by mass, and their falling rates are the same, independent of mass. It takes only a quick mental switch to see that, starting from an equilateral triangle, the “lighter body” and the “heavier body” will fall together at the same rate as the “Earth” and either one of the bodies. So with almost no extra calculational effort we now have the situation that, at least if they start at rest, all 3 bodies will fall together toward the common center of mass maintaining the shape of the equilateral triangle that they start in, where each of them is in either Lagrangian point L4 or L5 (Lagrange’s Trojan points) with regard to the other 2 bodies. Therefore they not only start to fall together at equal rates — i.e. each pair falling together — they continue to do so.

But the question remains whether they will remain in this equilateral triangle configuration under other circumstances.

Digression: perhaps you are wondering how it is known that they will fall toward their common center of mass. This follows simply from Newton’s well-known laws of motion which we have heretofore avoided mentioning. There is no external force acting on the 3 masses, so their center of mass, initially at rest, will stay at rest. We also know that they must all fall toward a common point as the equilateral triangle shrinks to a point. So, by an argument such as Newton or his contemporaries might have offered, this point must be their common center of mass.

The maintenance of the equilateral triangle as the 3 bodies move through space is the homographic property that Lagrange found them to have even when revolving around their common center of mass. The question comes up as to how simply and easily it can be shown that all 3 bodies will homographically maintain an equilateral triangle, even when revolving, even when revolving with expansion and contraction. Reminder: Lagrange already showed that this does happen, using his perturbation theory, but we are hoping to show here that Newton or even one of his lesser contemporaries could have discovered the Trojan points using only the mathematics of their time, preferably just algebra and trigonometry.

Some of you may have already objected that vector arithmetic had not yet been invented in their day, and this is correct in a strict sense. But they were familiar with surveying, architecture, engineering, the trigonometry to do related calculations, and with “something” corresponding to the associated “force vector components”, the mathematics of which even then included something more or less equivalent to rudimentary vector arithmetic.

SECTIONS

3.1 Homographic Property - Maintaining Shape

3.2 Geometry

3.3 Maintaining An Equilateral Triangle Without Expansion and Contraction

3.4 Homographic, Equilateral Expansion and Contraction

3.5 Trojan Points

3.2         Geometry

Here is a good place to remind ourselves of the:

3.2.1        Necessary and Sufficient Geometric Conditions

for the 3 point masses to remain in an equilateral triangle while revolving
(with no perturbing forces):

•  the angles formed by the lines joining the common center of mass to each body must remain the same

•  the ratios of the distances from the common center of mass to each body must remain the same

(Reminder: we have, with reasonable safety for our purposes, assumed that the masses will remain constant over time although this is not generally true for astronomical bodies, especially over long periods of time.)

SECTIONS

3.1 Homographic Property - Maintaining Shape

3.2 Geometry

3.3 Maintaining An Equilateral Triangle Without Expansion and Contraction

3.4 Homographic, Equilateral Expansion and Contraction

3.5 Trojan Points

3.3         Maintaining An Equilateral Triangle
Without Expansion and Contraction

We need to look at the equations relating to angular velocity and acceleration. We have the equation for the acceleration of an object moving at velocity v in an orbit of radius r:

Equations   4: relating angular velocity and linear acceleration

Assumptions 4: circular motion

Variables 4:

a is acceleration (radial, toward center of curvature)

v is velocity (tangential, perpendicular to radius)

r is the radius of curvature

q is the angular distance

 is the angular velocity (Newtonian notation)

m_i are masses

Eq. 4a:            

Note: in this last equation, the direction of the acceleration a for a mass revolving around a point is toward the center of the revolution/circle. (It is commonly known that this will also be toward the common center of mass, although we have not yet shown this here.) And we have the equation relating the velocity v in an orbit of radius r with the angular velocity q :

Eq. 4b:            

(Although we could have used a named constant for the angular velocity, we have slipped into both Newton’s —  — and Leibniz’s — dq /dt — calculus notation.) And so we have the equation relating the radial acceleration to the radius r with the angular velocity :

Eq. 4c:            

And we have that the above centripetal acceleration a of a mass in a circular orbit is linearly proportional to its distance r from the center around which it revolves, and that otherwise it is a function only of . Not shown here is the well-known result that the direction of this acceleration is radial, toward the center of revolution, which in our case will be the common center of mass. This is one of the fundamental equations used to calculate orbital velocities for satellites.

It is shown below — and also could have been easily shown in Newton’s time — that in our special circumstances of equidistant bodies, the acceleration due to gravity of each mass is toward their common center of mass. If we can show that this acceleration is proportional to the distance from that common center, then, combining this with the result just obtained (Eq. 4c), we know that the 3 masses can revolve around that common center of mass in such a way as to completely balance the accelerations due to gravity for a calculable constant angular velocity  (ignoring expansion-contraction for the moment).

In fact we know more than that: we know that since both the acceleration due to gravity and the acceleration due to revolution will only be radial, i.e. only along the lines toward the center of mass, that it is then impossible, barring other “perturbing” forces acting, for the angles among these lines to change. This last is one of the 2 necessary and sufficient conditions for the masses to remain in an equilateral triangle. And since both forces/accelerations (gravitational and rotationally induced centripetal-centrifugal) are both radial from the common center of mass and proportional to the radial distance r, then as those radial distances change, they will remain in the same proportions. This is the other of the 2 conditions.

If you objected that we have not quite pinned it down, you are partly correct: it is possible to give an initial velocity to each mass such that they will not remain in an equilateral triangle. But the point we are making here is that there does exist a set of initial conditions that will allow them to remain in an equilateral triangle, even when that triangle is rotating and expanding/contracting. This will be when the initial angular velocities are equal, which means that the initial tangential velocities are proportional to the radial distances (from the common center of mass), and the initial radial velocities are also in the same proportions as the radial accelerations and the radial distances. The radial/tangential velocities and their changes remain proportional to the radial/tangential accelerations which remain proportional to the radial distances. Thus the equilateral triangle formed by the masses will stay an equilateral triangle, of the same size as it rotates if the initial radial velocities are zero (trivially proportional), and expanding and contracting if there are non-zero initial radial velocities proportional to the radial distances. (You may think this needs calculus to pin down, but it follows very simply from the relations among acceleration, velocity and distance, well know long before Newton’s calculus.)

We know, of course, that the angular velocity , not to mention the distance r, will not remain constant if the triangle expands or contracts. This all means that if we can show — and we wish to do it simply — that the radial acceleration due to gravity of each mass is proportional to the (radial) distance from the common center of mass, we will have shown the dynamic homographic quality of the equilateral configuration of any 3 masses, and therefore the existence of at least “unstable equilibrium” at the Trojan points (i.e. with masses not perturbed from the equilateral triangle) and even the possible existence — that should have been noticed by Newton and his contemporaries — of Trojan planets/asteroids.

We will not examine “stable equilibrium” at the Trojan points, where the positions, velocities or accelerations of the masses are perturbed. And of course, here we have continued to assume that the velocity of gravity is infinite. The more explicit one is about dependency on assumptions, the less deadly and more friendly those assumptions can be in the long run.

Referring to Figure 3, we will look at the equations that describe the acceleration due to gravity of one of the masses, m₃. We can simplify the equations greatly if we allow G = 1 as the gravitational constant and r = 1 as the distance among the each pair of the 3 masses. The accelerations a_3i of m₃ due to m_i are (scalar):

Equations   5: accelerations of the 3^rd of 3 masses, special case of equilateral triangle

Assumptions 5: same as 2 and 3, and

G = 1 and r = 1 (to simplify the calculations)

Variables 5:

G is the Gravitational constant (here = 1)

r is the radial (linear) distance (between the masses; here = 1)

m_i are masses

a_3i are accelerations of mass 3 toward mass i

a_3ix and a_3iy are the x and y axis direction components of a_3i

 is the angle with m₃ as the vertex from the y-axis to the common center of mass

Eq. 5a:                  and    

The components in the x and y directions are:

Eq. 5b:             |a_31x| = sin(30º)×|a₃₁| = sin(30º)×m₁     and

|a_32x| = sin(-30º)×|a₃₂| = -sin(30º)×m₂

and

Eq. 5c:             |a_31y| = cos(30º)×m₁     and

|a_32y| = cos(-30º)×m₂ = cos(30º)×m₂

If we sum the components in the x and y directions we get:

Eq. 5d:             |a_3x| = sin(30º)×m₁ + sin(-30º)×m₂ = sin(30º)×(m₁-m₂)     and

|a_3y| = cos(30º)×m₁+ cos(-30º)×m₂ = cos(30º)×(m₁+m₂)

The absolute value of the acceleration of m₃ is:

       so we get

Eq. 5e:            

And the direction of the acceleration of m₃ is:

     so

Eq. 5f:             

The general equations for a center of mass are:

Equations   6: general equations for a center of mass cm = (cm_x,cm_y), and special case of equilateral triangle

Assumptions 6: same as 2 and 3

Variables 6:

m_i are masses

x_i and y_i are x and y coordinates

cm_x and cm_y are x and y components of center of mass position

 is the angle from the y-axis of the line from m₃ to the common center of mass

Eqs. 6a:                 and    

NOTE VERY WELL that the center of mass of 2 point masses is collinear with them, i.e. on the straight line passing through both. Note also that if we calculate the center of mass of 2 of 3 masses, then “add” the 3^rd mass, in our special case of an equilateral triangle, the center of mass of all 3 will be on the straight line passing through both the 3^rd mass and the center of mass of the first 2. This means that — in our special case, and also when all 3 masses are collinear (!), but not in general — the force due to gravity on a 3^rd mass is toward both the center of mass of the first 2 masses and the center of mass of all 3 masses, neglecting the speed of gravity. There very well may be a way to use this result to give a simple yet complete demonstration of “stable equilibrium” (with those “tadpole” and “horsehoe” orbits; see Figure 5), avoiding abstruse or arcane mathematical theory... beautiful though it is. This is worthy of study. This is the kind of insight — here into the dynamics of Trojan points — that can come from questioning accepted scientific beliefs.

The center of mass of all 3 masses is:

Eqs. 6b:                 and

and just as a check, since we don’t really need the value, we will find the angle that the line from m₃ to the common center of mass makes with the y axis to compare it with Eq. 5f:

     and

  so

         so

Eq. 6c:            

which is the same as Eq. 5f. (Yes, we were playing a little fast and loose with the signs of and .)

At least in our special case of an equilateral triangle, this is a reasonably complete demonstration that the gravitational acceleration of each body is in the same direction as both the center of mass of the other 2 and the common center of mass of all 3 (again, assuming infinite speed of gravity).

We would now like to calculate the distance of mass 3 from the common center of mass.

Eq. 6d:            

If we now take the ratio of the acceleration (Eq. 5e) to the distance (Eq. 6d), we get:

Equations   7: ratio of the absolute value (scalar) of the gravitational acceleration of m₃ toward the common center of mass to the distance of m₃ from the common center of mass

Assumptions 7: same as 2 and 3, and

3 masses are in an equilateral triangle

Variables 7:

r_i are the radial (linear) distances (between the masses)

m_i are the masses

a_3i are accelerations of mass 3 toward mass i

q is the angular distance

 is the angular velocity

Eq. 7a:            

and the symmetry among the 3 masses means that interchanging any 2 still gives us the same ratio:

Eq. 7b:            

If we combine Eq. 4c () and Eq. 7b we get:

Eq. 7c:

(NOTE: we would really need to deal with the problem of physical units to do all this right.)

So we have shown that, as long as we are dealing only with circular orbits around the common center of mass — and by implication zero radial velocities (think polar coordinates with the origin at the common center of mass) — there exists an angular velocity that will “balance” the gravitationally induced, radial accelerations and the rotationally induced, centripetal (or “center seeking”), radial accelerations. I.e. we have a homographic solution for circular orbits, i.e. with at least “unstable equilibrium” at the Trojan points. It remains to look at expansion and contraction.

SECTIONS

3.1 Homographic Property - Maintaining Shape

3.2 Geometry

3.3 Maintaining An Equilateral Triangle Without Expansion and Contraction

3.4 Homographic, Equilateral Expansion and Contraction

3.5 Trojan Points

3.4         Homographic, Equilateral Expansion and Contraction

All that we need for the more general case is to have initial radial velocities be proportional to the radial distances. (Note: they can all be negative and still be proportional to the radial distances.) If the initial radial velocities are proportional to the corresponding initial (radial) distances from the common center of mass/center of rotation of the equilateral triangle (thus the center of revolution of the bodies), and the initial angular velocities are equal, making the initial tangential velocities also proportional to the radial distances, then the changes in the radial distances will be proportional, and the already proportional radial distances will remain proportional.

Since the initial radial forces/accelerations are proportional to radial distance, (and since the initial angular accelerations are equal — and non-zero since the radial velocities are non-zero — and thus the tangential accelerations are proportional to the radial distances), then changes in the radial velocities will be proportional to the radial distances, radial distances will remain proportional, and, even with expansion and contraction, the equilateral triangle will homographically remain equilateral. (This really needs calculus to at least feel more rigorous about it, but the arguments would have been considered both simple and cogent enough by Newton and his contemporaries, and even in their style, the intents here.)

Thus there exist initial conditions which will allow homographic expansion and contraction of the equilateral triangle formed by the 3 point masses revolving around their common center of mass.

SECTIONS

3.1 Homographic Property - Maintaining Shape

3.2 Geometry

3.3 Maintaining An Equilateral Triangle Without Expansion and Contraction

3.4 Homographic, Equilateral Expansion and Contraction

3.5 Trojan Points

3.5         Trojan Points

Thus we have demonstrated the existence of Trojan points using only algebra, trigonometry and Newton’s law of gravity. Or have we?! Actually we have demonstrated a purely theoretical result, that 3 gravitational bodies (“point” masses) in an equilateral triangle are (“passively”) “stable” in this configuration even when the triangle is rotating, if they are not “perturbed” by other forces. We have not yet shown that they will be at a “stable equilibrium” in the usual sense, i.e. that they will tend to “return” to (or “stay close” to) the equilateral configuration if they are “perturbed” away from it, or if they were not initially in this configuration but “close” to it. (Actually, “return” is not strictly correct; the asteroid or what-have-you will tend to orbit the Trojan point. See next section.)

This is where Lagrange’s perturbation theory comes in: so far, only it shows that Trojan points have the ability to trap and keep bodies with small enough Trojan-point-relative velocities (i.e. if their relative velocities are below an “escape velocity”, much the same as with rockets and satellites). We really need to show stability, but will not attempt to do so here.

To be fair we will note that one can do a simplified version of perturbation theory, by guessing (we don’t know ahead of time e.g. what the size/shape/etc of a region of stability around the Trojan point might be, so guessing) at certain particular and small perturbations to be applied to one of the bodies, and calculating what happens to all 3 bodies. I.e. we can try small perturbations of one of the bodies in various directions from its Trojan point, and see if it will keep wandering away, or circle back, or whatever. This is what perturbation theory does, but with much bigger mathematical guns.