PHY5200 F07

Chapter 8: Two-Body Central Force Problems

Reading

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

Recall:

u''(φ) = -u(φ) - (μ/l²u²(φ))F(r=1/u)

where u = 1/r and the double prime is shorthand for the second derivative with respect to φ.

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) = A cos(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 + A cosφ = μk/l²(1 + ε cosφ)

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 + ε cosφ) = C / (1 + ε cosφ)

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.


© 2007 Robert Harr