Astronomy 340 Fall 2005 Class 3 13 September

Astronomy 340 Fall 2005 Class #3 13 September 2005

Announcements/Review o o o Kepler’s Laws n Elliptical orbits n Equal areas in equal times n P 2 goes as a 3 Newton’s Law to Kepler’s Laws n Kepler’s laws as consequences of inverse-square force law and the equation of relative motion Orbital elements n a = semi-major axis of orbit n e = eccentricity (deviation from circularity) n θ = “true longitude” = reference direction (θ=0) n ω-bar = longitude of pericenter n f = “mean anomaly” = angle between object and pericenter (θ = f + ω-bar)

Let’s go 3 -D o Coordinates n r = f(x, y, z) o o x = along major axis y = perpendicular to x, in plane z = perpendicular to plane Orbital elements n n n 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 o What’s restricted? n m 1 ~ m 2 >> m 3 n Neither energy nor momentum is conserved n Examples o o o Satellite orbiting Earth Asteroid near Jupiter Moons Ring particles Basic idea n Rotating reference frame in which the positions of the two large masses remain fixed and the frame has some fixed angular velocity, n. n Accelerations can be described as gradients of some function, U

Lagrangian Equilibrium Points o Where can we put another point such that it remains stationary in the rotating frame? n n o 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 F 1 and F 2 A little geometry and algebra… n n Equil. Points at the apex of an equilateral triangle with baseline connecting m 1 and m 2 these are L 4 and L 5 L 1, L 2, and L 3 lie in a line connecting m 1 and m 2