PHY5200 F06

Chapter 8: Two-Body Central Force Problems

Reading

Taylor 8.3-8.5 (today) and 8.5-8.7 (Friday).

Recall:

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

The Equations of Motion

Using the reduced mass, the equations of motion read

MR[ddot] = 0

where

M = m1 + m2  and R = (1/M)(m1r1 + m2r2)

and

μr[ddot] = F12

where

μ = m1m2 / (m1 + m2)  and r = r1 - r2

Angular Momentum

In chapter 3 we learned that when objects interact only by a central force, their angular momentum is conserved. To demonstrate this, let's look at various angular momenta in this problem, and their time derivatives. Consider the angular momenta of object 1 and object 2 relative to the center of mass:

L'1 = r'1×p'1
L'2 = r'2×p'2

and the angular momentum of the center of mass

Lcm = R×P

The time derivatives of these are:

dLcm/dt = Vcm×P + R×dP/dt = R×Fext = 0
dL'1/dt = v'1×p'1 + r'1×dp'1/dt = r'1×F12 = 0
dL'2/dt = v'2×p'2 + r'2×dp'2/dt = -r'2×F12 = 0

So all three of these angular momenta are conserved. The conservation of the center of mass angular momentum will be mainly ignored, except to point out that it doesn't mean that the CM rotates around the orgin of coordinates. This would contradict the result that the total momentum vector is conserved. Rather, it is just that R×P is constant, and this can occur for straight line motion (draw a figure).

More important to us is that the individual angular momenta about the CM are conserved. And both point in the same direction as we can see by writing the relative position and momentum vectors in terms of the separation vector r:

L'1 = r'1×p'1 = (m2/M)r×(m1m2/M)r[dot]
L'2 = r'2×p'2 = -(m1/M)r×-(m1m2/M)r[dot] = (m1/m2)L'1

Therefore, relative to the CM, the objects move in the plane perpendicular to angular momenta. The total angular momentum is the sum of the individual angular momenta

L' = L'1 + L'2 = (m1m2/M²)(m2r×r[dot] + m2r×r[dot]) = r×μr[dot]

This is the expression for the angular momentum of a particle of mass μ with coordinates r and velocity r[dot].

Let's also look at the energy ... This again confirms that using the reduced mass and relative separation allows us to treat the problem in a general manner, and the solution can be applied to a number of different circumstances.

The Equation of Motion in Cylindrical Polar Coordinates

We will now make use of this fact to write the equation of motion in a particular coordinate system, namely cylindirical polar coordinates. Since the motion occurs in a plane, cylindirical polar coordinates with the CM as origin, and the z axis aligned with the angular momentum vector, perpendicular to the plane of motion. With this choice, the motion occurs in the r-φ plane; there is no motion in the z direction. Of course, I'm speaking of the motion relative to the CM, the CM can move in the z direction so that objects move in the z direction. In cylindrical polar coordinates, the equation of motion separates into three equations, one in r, one in φ, and one in z:

μr[ddot] - μr(φ[dot])² = Fr = -∂U/∂r
μrφ[ddot] + 2μr[dot]φ[dot] = Fφ = 0
μz[ddot] = Fz = 0

The equation in z is written for completeness, but simply expresses the fact that there is no motion in z (zero acceleration means constant velocity, but since the initial velocity is zero, there is no motion in z).

The angular momentum is constant. It is useful to evaluate those in polar coordinates as well.

L' = r×μr[dot] = μ(rrhat)×(r[dot]rhat + rφ[dot]φhat) = μr²φ[dot]zhat = l

I've used the symbol l to represent the total momentum (relative to the CM) of the system. From this expression, we can solve for φ[dot] in terms of r, φ[dot] = l/μr², and use this to simplify the r and φ equations of motion.

μr[ddot] - l²/μr³ = Fr = -∂U/∂r
μrφ[ddot] + 2lr[dot]/r² = Fφ = 0

The r equation can be rewritten in the form

μr[ddot] = Fr + l²/μr³ = -∂U/∂r - ∂(l²/2μr²)/∂r = -∂Ueff/∂r

where I've introduced the effective potential Ueff = U(r) + l²/2μr² in order to make this look like a one-dimensional problem of motion in r subject to this effective potential. If we multiply both isdes of this expression by r[dot], then each side is can be seen to be a time derivative:

μr[dot]r[ddot] = d/dt(½μr[dot]²) = -∂Ueff/∂r r[dot] = -dUeff/dt

therefore

(d/dt)(½μr[dot]² + Ueff) = 0

The quantity in parentheses is the energy in the CM system, and this says that the energy is constant.

Energy Considerations: Effective Potential and Orbits, Bounded and Unbounded

Show Ueff versus r for several potentials, ½kr², -k/r

The Equation for the Orbit

E = T + U = ½μ(r[dot]² + r²φ[dot]²) + U(r)
E = T + U = ½μr[dot]² + l²/2μr² + U(r)

The expression now looks like a one-dimensional kinetic energy in the coordinate r, with the effective potential Ueff = U(r) + l²/2μr².


© Robert Harr 2006