This paper discusses the forces on stars in a globular cluster. Consider Figure 1; the symbols are defined as follows:
Mg = mass of the stars of a globular ciuster Internal to g that of a particular star
m = mass of that particular star
mp = mass of the nearest neigboring stars
xg = distance of the star from the mass center of the globular g cluster
xp = distance of the star from the mass center of the nearest neighboring stars
xpg = distance of the mass center of the nearest neighboring stars from the mass center of the globular cluster
xpo = equilibrium distance of the star from the mass center of the nearest neigboring stars
x = distance of the star from the mass center of the nearest neighboring stars, relative to the equilibrlum distance
Recall that in the Reciprocal System two forces are acting on the star:
- Gravitation of the star by the cluster as a whole—this produces an inward motion.
- Progression of the star away from its nearest neighbors—this produces an outward motion.
My goal in this paper is to derive the expression for the net force acting on the star, to find the equilibrium position (xpo) of the star, and to determine whether or not this position is stable.
Nehru’s recent paper  provides the starting point. Some additional symbols are needed:
dog = gravitational limit of the globular cluster
dop = gravitational limit of nearest neigboring star:
yg = non-dimenslonal distance of the star from the mass center of the globular cluster
yp = non-dimensional distance of the star from the mass center of the nearest neighboring stars
vog = “zero-point speed” of the star relatlve to the globular cluster
vop = “zero-point speed” of the star relative to the nearest neighboring stars
vng = net inward gravitational speed of the star
vnp - net outward progression speed of the star
vn = net speed of the star
G = “universal” gravitational constant
Mo = mass of the sun
ag = acceleration from gravitation of the globular ctuster
ap = acceleration from progression away from the nearest neigbors
an = net acceleration of the star
In this notation,
|dog = 3.77 * (Mg / Mo)½ (ly)||
|yg = xg /dog = (x + xpo + xpg ) / dog||
|vog = (2 * G * Mg / dog )½||
|vng = vog * (1 / yg½ - yg½) (Inward)||
|dop = 3.77*(mp / Mo)½ (ly)||
|yp = xp / dop = (x + xpo) / dop||
|vop = (2 * G * mp /dop )½||
|vnp = (½) * vop * (yp - 1/yp ) (outward)||
|vn = vnp - vng||
Differentiating the velocity expressions with respect to time gives the accelerations:
|ag = G * Mg * (1/xg² - 1/dog ²) (inward)||
|ap = G * mp * (½) * (xp / dop³ - dop / xp³) (outward)||
|an = ap - ag||
|an = 0||
|h = mp / (2 * dop³)||
|i = Mg * (1/dog² - 1/xg²)||
|j = (1/2) * mp * dop||
Then, in terms of xpo, at equilibrium,
|h * xpo4 + i * xpo ³ - 1 = 0||
a quartic equation.
The appendix gives a simple computer program written in BASICA to solve equation 17 numerically. (An attempt to solve the equation analytically using the MU MATH AI program failed). A sample run with Mg = 200*Mo , mp = 2*Mo , xg = 40 ly, dog = 53.32 ly, and dop = 5.33 ly produced xpo = 9.29 ly.. Another sample run with Mg = 30000*Mo , mp = 200*Mo ,xg = 400 ly, dog = 652.98 ly, and dop = 33.32 ly produced xpo = 178.94 ly. Input parameters that are physically impossible produce negative distances.
Now let’s turn to the question of the stability of this positlon, xpo The net force acting on the star in terms of the distance from equilibrium, x, is
|F = m * G * ((1/2) * mp * ((xpo + x)/dop³ - dop / (xpo + x)³)|
|- Mg * (1 / (xpo + x + xpg )² - 1 / dog²))||
Differentiating F with respect to x gives
|dF / dx = m *G * ((1/2) * mp * (1/dop³ + 3 * dop / (xpo + x)4)|
|+ 2 * Mg / (xpo + x + xpg )³)||
If x is positive, dF / dx is positive and hence F increases with x.
If x is negative, dF / dx is still positive. Thus
|- dF / dx < 0||
This is the definition of instability. Hence, xpo is a point of unstable equilibrium. But there is one saving grace: the forces near this point are quite small, so sudden changes in position are precluded.
Globular clusters continually grow by accretion until eventually being absorbed into galaxies. The stars in the clusters must keep changing their temporary equilibrium positions.
- K. V. K. Nehru, “The Gravitational Limit and the Hubble’s Law,” presented at the 1986 Convention of ISUS.