Astronomy 340 Fall 2005 Class #3 13 September 2005

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)

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)

“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

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