1. Astronomy 340 Fall 2005
Class #3
13 September 2005

2. Announcements/Review
Kepler’s Laws
Elliptical orbits
Equal areas in equal times
P2 goes as a3
Newton’s Law to Kepler’s Laws
Kepler’s laws as consequences of inverse-square force law and the equation of relative motion
Orbital elements
a = semi-major axis of orbit
e = eccentricity (deviation from circularity)
θ = “true longitude” = reference direction (θ=0)
ω-bar = longitude of pericenter
f = “mean anomaly” = angle between object and pericenter (θ = f + ω-bar)

3. Let’s go 3-D Coordinates Orbital elements r = f(x,y,z)
x = along major axis
y = perpendicular to x, in plane
z = perpendicular to plane
Orbital elements
Reference plane = ecliptic
i = inclination of orbital plane to reference frame
“line of nodes” = intersection of orbital and reference planes
Ω = longitude of ascending node (angle from reference to ascending node)
ω = “argument of pericenter” (angle between Ω and pericenter)

4. “Restricted” 3-body Problem
What’s restricted?
m1 ~ m2 >> m3
Neither energy nor momentum is conserved
Examples
Satellite orbiting Earth
Asteroid near Jupiter
Moons
Ring particles
Basic idea
Rotating reference frame in which the positions of the two large masses remain fixed and the frame has some fixed angular velocity, n.
Accelerations can be described as gradients of some function, U

5. Lagrangian Equilibrium Points
Where can we put another point such that it remains stationary in the rotating frame?
Must be at a fixed distance from origin, b
Centrifugal acceleration must be balanced by a force in b-direction that is some combination of F1 and F2
A little geometry and algebra…
Equil. Points at the apex of an equilateral triangle with baseline connecting m1 and m2  these are L4 and L5
L1, L2, and L3 lie in a line connecting m1 and m2