## Physics Friday 16

Suppose we have two bodies, of mass M1 and M2, with M1<M2, with their centers separated by a distance R. These two bodies will orbit about their barycenter. From the basic definition of center of gravity, we see that body 1 will be at a distance  from the barycenter; body 2 will be at a distance . If both move in circular orbits about their barycenter, so that R is a constant, then they do so with (orbital) angular frequency .

Figure 1

Consider the plane of the orbit of these bodies via a rotating reference frame with origin at the barycenter that rotates with the bodies. We will use polar coordinates with body 2 on the  line (and thus body 1 is at ), with increasing θ in the direction of the rotation (Fig 1). A body of mass m at (r,θ), considered in this frame, will experience the fictitious centrifugal and coriolis forces:



In addition to these, we have the gravitational forces exerted by bodies 1 and 2. The displacement vector from body 1 to our mass m is:

Similarly for body 2:


Figure 2

Thus:


and similarly:


Now let us consider an object with mass , so that we can neglect any forces it exerts on bodies 1 and 2. Suppose the object occupies a position fixed relative to the two bodies; it orbits with the bodies at the same rate, and thus is at rest in the (non-inertial) rotating reference frame. Then it experiences no coriolis force; the net force on the object in the rotating frame is:




The r-component is:

and the θ-component is:


Now, for the body to remain in such an orbit, it must be in equilibrium in the rotating frame, meaning that the total force must be zero. Looking at the components above,  has solutions  and


which, using our formula for r1 and r2, we see , giving



So the θ-component gives that an equilibrium point must be either on the line through bodies 1 and 2 (and their barycenter), or equidistant from bodies 1 and 2. In the first such case,  or . We will now solve  for these cases:

(1) 


Here we must pay attention to the sign of :
(i) 

So:


Or, subsituting for r1 and r2:

Let . Then the above becomes:

This is a quintic equation in ρ, and must be solved numerically for  (). For positive (physical) values of M_1 and M_2, this has only one such solution.
(ii) 


Or, subsituting for r1, r2, and ρ:

This is also a quintic equation in ρ, and must be solved numerically for ; it also has one solution for the physical problem.

(2) 


Here we must again pay attention to signs; in this case, the sign of :
(i) 



This quintic equation has one solution for the physical problem, which can be shown to be a value of r not only greater than r1, but greater than r2.
(ii) 


This has no solution for r>0; thus there is no solution in this case.

Thus, we have found three points on the line through masses 1 and 2 where we have equilibrium. Now, we consider the  case, and solve for r. First, we find that  gives us the equation for the perpendicular bisector of the segment connecting bodies 1 and 2 (basic geometry):




Note that this gives
.

We will reexamine the forces involved; as we have already shown them to be purely radial when , we will only use this component. First,

and solving r in terms of d=d1=d2 and substituting,





We noted before that , so , and thus


For the forces to balance we need:




so that mass 1, mass 2, and our small mass thus form the vertices of an equilateral triangle. There are two such points for our small mass (one leading, one trailing).

Thus there are five equilibrium points where a small body can orbit about the barycenter of two larger (unequal) masses with the same orbital period as those two larger masses. These are known as the “Lagrange points” or “Lagrangian points,” and are numbered as in the figure below.

Figure 3

Lagrange points L1, L2, and L3 are unstable, but points L4 and L5 are stable. For example, the Trojan Asteroids are found clustered near the L4 and L5 points for Jupiter’s orbit about the sun.