PHY5200 F06

Chapter 8: Two-Body Central Force Problems

Reading

Taylor 8.1-8.3 (today) and 8.3-8.5 (Wednesday).

Recall:

Fr = mr[ddot] - mr(φ[dot])²
Fφ = mrφ[ddot] + 2mr[dot]φ[dot]

We will now look at the general problem of two bodies interacting by a central force. This is the problem that Newton developped the calculus and his mechanics to solve, specifically for the orbits of planets and comets about the Sun. Many similar problems arise in Physics: the orbits of binary stars, including neutron stars and black holes; the interaction of two atoms in a diatomic molecule such as H2, N2, O2, or CO; and the behavior of an electron bound to a nucleus. The last two are properly treated only with quantum mechanics, but by studying the problem using classical mechanics we will (1) learn some important concepts about the problem that are applicable in quantum mechanics, and (2) understand how classical mechanics fails in such situations. The problem of orbiting neutron stars or black holes also is beyond the validity of classical mechanics, and must be treated with general relativity.

Taylor uses the language of Lagrange mechanics to arrive at the equations of motion in this chapter. I will derive the equations of motion using Newtonian mechanics.

Outline of the Two-Body Central Froce Problem

Recall that in chapter 4 we derived that the motion of a system of particles can be separated into the problem of the motion of the center of mass of the system subjected to the sum of all external forces and the internal motion of the system relative to the center of mass. Consider the two objects to be point-like, that is, small enough that we never have to consider their size. Let their positions be r1 and r2, and their masses m1 and m2. A two body central force is automatically conservative. Consider the two objects to be point-like, that is, small enough that we never have to consider their size. Let their positions be given by the vectors r1 and r2, then the potential is U(r1, r2) = U(|r1-r2|) = U(r), where r is just the separation of the two objects. The corresponding forces are F12 = -∇1U and F21 = -∇2U = ∇1U = -F12. Assuming there are no other forces, internal or external, the equations of motion are

m1r1[ddot] = F12
m2r2[ddot] = F21

CM and Relative Coordinates

The center of mass is given by R = (m1r1 + m2r2) / (m1 + m2). Calling M = m1 + m2, we can write this as MR = m1r1 + m2r2. The equation of motion for the center of mass is

MR[ddot] = m1r1[ddot] + m2r2[ddot] = F12 + F21 = F12 - F12 = 0

That is, the total external forces are zero, therefore the center of mass moves at constant velocity.

The internal motion is the topic of interest. This can be investigated using relative coordinates r'1 = r1 - R and r'2 = r2 - R. In terms of the relative coordinates, the equations of motion become

m1r'1[ddot] = F12
m2r'2[ddot] = F21

But notice that the relative coordinates are really the same vector with different scaling

r'1 = r1 - R = (1/M)[(m1 + m2)r1 - m1r1 - m2r2] = (m2/M)(r2 - r1) = (m2/M)r12
r'2 = r2 - R = (1/M)[(m1 + m2)r2 - m1r1 - m2r2] = (m1/M)(r1 - r2) = -(m1/M)r12

where I've used r12 = r2 - r1. Now it is easy to see that the two equations of motion for the relative coordinates are actually the same equation written in terms of the separation vector r:

m1r'1[ddot] = (m1m2/M)r12[ddot] = F12
m2r'2[ddot] = -(m1m2/M)r12[ddot] = F21 = -F12

And even more fortuitously, the force is also a function of just the separation distance between the two objects. So, the two original equations of motion in terms of r1 and r2 transform into two nicer equations, one in terms of the CM coordinate R and the other in terms of the separation vector r12 which I'll call simply r from now on.

MR[ddot] = 0
(m1m2/M)r[ddot] = F12

Reduced Mass

The quantity m1m2/M has the units of mass and is called the reduced mass of the system, μ = m1m2 / (m1 + m2). The reduced mass has some interesting properties. It is always less than either of the original masses, μ<m1 and μ<m2. If one mass is much greater than the other, say m1<<m2, as happens with the earth orbiting the sun, or an electron orbiting a nucleus, then the reduced mass is approximately equal to the smaller mass, μ≅m1. If the two original masses are equal, then the reduced mass is just half the mass of either, m1 = m2 = m, then μ = m/2. And, if one mass is fixed, or if the problem involves just one mass and a fixed origin of force, then this solution is directly applicable, simply by setting the mass of the moving object equal to μ.


© Robert Harr 2006