Central Force Motion 8.01 W14D1Central Force Problem ( )1 21,22ˆm mr Gr= −F r Gravitational force (Kepler problem): Linear restoring force: Find the motion of two bodies interacting via a central force. Examples: ( )1,2ˆr kr= −F rReduction of Two Body Problem to One Body Problem Reduce two body problem to one body of reduced mass µ moving about a central point O under the influence of the central force with position vector corresponding to the relative position vector from object 2 to object 1 Solving the problem means finding the distance from the origin r(t) and angle θ(t) as functions of time Equivalently, finding r(θ) as a function of angle θ 21,22ddtµ=rFCentral Force Motion: Constants of the Motion Total mechanical energy E is conserved because the force is radial and depends only on r and not on θ Angular momentum L is constant because the torque about central point is zero The force and the velocity vectors determine the plane of motionTable Problem: Satellite in Elliptic Orbit A satellite of mass ms is in an elliptical orbit around a planet of mass m which is located at one focus of the ellipse. The satellite has a velocity va at the distance when it is furthest ra from the planet. The distance of closest approach is rp. What is the magnitude of the velocity vp of the satellite when it is closest to the planet? 5Equal Area Law and Conservation of Angular Momentum Change in area per time Angular momentum Equal area law ΔA = (1 / 2)rvθΔtL r vθµ=ΔAΔt=L2µΔAΔt=12vθrFor Most Force Laws the Most General Solution Does Not Involve Closed OrbitsWorked Example: Isotropic Harmonic Oscillator A special solution of the equation of motion for a linear restoring force is given by with where for the case shown in the figure with The solution for is an ellipse centered at the origin ˆ ˆ( ) ( ) ( )t x t y t= +r i j22ˆdkrdtµ= −rr x(t) = x0sin(ωt) y(t) = y0cos(ωt) y0< x0 rmax= x0 rmin= y0( )trDemo : Open Pendulum Orbits A pendulum moves in a precessing orbitConcept Problem: The orbit of the pendulum in the experiment precesses because 1) I pushed it in that direction. 2) The pendulum mass is spinning. 3) The Earth rotates. 4) The potential energy is not precisely cylindrically symmetric. 5) The potential energy is not precisely parabolic.Kepler’s Laws 1. Each planet moves in an elliptical orbit, with the sun at the focus of the ellipse. 2. A line from the sun to a given planet sweeps out equal areas in equal times. 3. The periods of the planets are proportional to the 3/2 power of the semi-major axis lengths of their orbits.Orbit Equation for Kepler Problem Solution (See Course Notes) Eccentricity 01 cosrrε θ=−201 2LrGm mµ= ε= 1 +2EL2µGm1m2( )2⎛⎝⎜⎜⎞⎠⎟⎟1/ 2= 1 −EEmin⎛⎝⎜⎞⎠⎟1/ 2 Emin= −12µ(Gm1m2)2L2Kepler’s Laws: Period Integral of Equal Area Law Period Period (T) squared is proportional to cube of the semi major axis (a) but depends on both masses 02TorbitdA dtLµ=∫ ∫2T ALµ=( )2 321 24 aTG m mπ=+Stars Nearby Galactic Center The UCLA Galactic Center Group, headed by Dr. Andrea Ghez, developed an animation of the orbitsof eight stars about the galactic center
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