PHY5200 F06

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 = 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 sides of this expression by r[dot], then each side is 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

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] = Fr + 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

or

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) = rmin/cos(φ - δ)

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

The Kepler Orbits

The interesting case is for an attractive 1/r² force, F = -k/r². For the gravitational force, k = Gm1m2, and the force is always attractive. For the coulomb force, k = k|q1q2| = |q1q2|/4φε0, and is attractive if the charges have opposite sign.

Changing to the variable u, the force is F(1/u) = -ku² and the equation of motion looks like

u''(φ) = -u(φ) + μk/l²

Notice that the term involving the force becomes constant. This is a special property of the 1/r² force and yields orbits that close on each rotation. To solve, change variables again to w(φ) = u(φ) - μk/l² so that w''(φ) = u''(φ) and the differential equation becomes w''(φ) = -w(φ). This is the same differential equation we saw above, and the solution is

w(φ) = Acos(φ - δ)

The phase shift δ can be set to zero by choosing an appropriate axis from which to measure φ. With this, we can change variables back and find

u(φ) = μk/l² + Acosφ = μk/l²(1 + εcosφ)

where we've replaced the constant A by the constant ε. This change seems arbitrary at this point, but be patient. Finally, we get back to r

r(φ) = (l²/μk) / (1 + εcosφ) = C / (1 + εcosφ)

This is our solution for the orbit under a 1/r² force. We have an expression for r as a function of the angle φ and the constants ε and C. C depends on the angular momentum, reduced mass, and force constant. ε is related to the constant of integration, A, so depends on initial conditions. We'll explore the nature of these orbits.

The Orbits

Note that if ε<0, then we have the same result as for |ε| and φ going to φ+π. Therefore, we only need to consider the cases for &epsilon≥0. Also, if ε<1, then the denominator is never equal to zero. Physically this means that r never goes to infinity; therefore, solutions with &epsilon<1 correspond to bounded orbits. If ε≥1, then the denominator can become zero for certain values of φ and r goes to infinity; these solutions correspond to unbounded orbits. The case of ε=1 is the transition from bounded to unbounded orbits. We'll see that the bounded orbits have E<0, the unbounded orbits have E≥0, and E=0 is the transition energy. Let's begin by understanding the characteristics of the bounded orbits.

The Bounded Orbits

When ε<1, the orbits are ellipses, that is, the expression for the orbit can be cast in the form

(x+d)²/a² + y²/b² = 1

This will be left as an exercise for you. There is a displacement, d, in the x direction because the ellipse is centered at (x,y) = (-d, 0) rather than at (0,0). But the origin is a focus of the ellipse.

The separation, r, oscillates in synch with cosφ, with a minimum value, rmin = c/(1 + ε) when φ = 0, and a maximum value rmax = c/(1 - ε) when φ = π. For orbits of planets, comets, or other bodies about the sun, the location of minimum separation is called the perihelion (peri-, Greek for around or near and -helion, Greek for sun) and the location of maximum separation is called the aphelion (ap- Greek for away). For orbits around the earth, these locations are called perigee and apigee (gee Greek for earth). If ε=0, then r = c is constant and the orbit is circular.

The Orbital period; Kepler's Third Law

Relation Between Energy and Eccentricity


© Robert Harr 2006