PHY5200 F07

Chapter 8: Two-Body Central Force Problems

Reading

Taylor 8.5-8.7 (today) and 8.7-8.8 (Monday).

Recall:

μr[ddot] - l²/μr³ = F_r = -∂U/∂r

μrφ[ddot] + 2lr[dot]/r² = F_φ = 0

The r equation can be rewritten in the form

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

where I've introduced the effective potential U_eff = 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 sides of this expression by r[dot], then each side is a time derivative:

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

therefore

(d/dt)(½μr[dot]² + U_eff) = 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

Now we will find the equation for the orbit. We continue to think in terms of a general central force, but the following technique is certainly motivated by the specific case of a 1/r² force. We begin with the equation of motion

μr[ddot] = F_r + l²/μr³

and massage it into a more convenient form. Begin by making a change of variables to u=1/r (or r=1/u), and, because it is convenient to solve for the spatial shape of the orbit (r as a function of φ) rather than the coordinates as a function of time, transform the time derivative to a φ derivative using the chain rule

d/dt = (dφ/dt)d/dφ = φ[dot]d/dφ = (l/μr²)d/dφ = (lu²/μ)d/dφ

The time derivative of r becomes

r[dot] = dr/dt = (lu²/μ)d(1/u)/dφ = -(l/μ) du/dφ

and the second time derivative becomes

r[ddot] = d(r[dot])/dt = (lu²/μ)d/dφ(-(l/μ) du/dφ) = -(l²u²/μ²) d²u/dφ²

Substituting these changes into the equation of motion, it becomes

-(l²u²/μ)d²u/dφ² = F(1/u) + lu²u³/μ

u''(φ) = -u(φ) - (μ/l²u²(φ))F(1/u)

where the double prime is shorthand for the second derivative with respect to φ.

Example: Free Particle

Find the orbit for a free particle (no force).

Letting F=0, the equation becomes

u''(φ) = -u(φ)

We've seen this equation before but in the form of a second time derivative rather than a φ derivative. The solution is the same, just use φ rather than t

u(φ) = Acos(φ - δ)

Of course, we want to know r, not u, so change variables again

r = 1/u = 1/(Acos(φ - δ) = r_min/cos(φ - δ)

This odd looking function is nothing more than the equation for a straight line in polar coordinates. Sketch to demonstrate.