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Central Force Motion Central Force Problem Two Body Problem Central Force Problem Find the motion of two bodies interacting via a central force. Central Force Motion Examples: Gravitational force (Kepler problem): mm Fr()rG=− 12ˆ 8.01 1,2 r 2 W14D1 Linear restoring force: Fr1,2 (rkr) =− ˆ Two Body Problem: Center of Mass Coordinates Reduction of Two Body Problem Center of mass dd22rr Newton’s Second Law ˆˆ12 FF1,2rr==mm 122 2,1 2 mm11rr+ 2 2 dt dt Rrcm =−1 mm12+ 22 FF1,2ddrr 2,1 Divide by mass rrˆˆ==12 Relative Position Vector mdtmdt22 12 ′′ rrr=−=−12 rr 12 FF dd22()rr− r ()1,2−= 2,1 rˆ 12 = Subtract: m m dt22 dt Reduced Mass 12 rrr=− μ = m m /(m + m ) 12 1 2 1 2 Use Newton’s Third Law (in components) F =−F 1,2 2,1 Position of each object 2 2 m rr− 11 d r d r mm11rr+ 2 2 21() 2 μ μ ˆ ˆ rrR′ =− =− r = = r Summary ()F+ 1,2 r =⇒2 F1,2 r = μ 2 11cm 1 rr2′ =− mm++ mm m mm12 dt dt 12 12 1 m2 1 Interpretation of Solution: Reduction of Two Body Problem Motion about Center of Mass Reduce two body problem to one body of reduced mass μ moving Knowledge of rrr= 12′− ′ about a central point O under the determines the motion of influence of gravity with position each object about center of vector corresponding to the relative position vector from mass with position. object 2 to object 1 μ μ rr′ == rr′ =− Solving the problem means 1 m 2 m finding the distance from the d 2r 1 2 origin r(t) and angle θ(t) as F = μ functions of time 1,2 dt 2 Equivalently, finding r(θ) as a function of angle θ Concept Question: Angular Momentum Angular Momentum about O The angular momentum about the point O of the “reduced body” Torque about O: τΟ = rFO ×=×=1,2()rrFr rˆˆ 1,2 () r0 1. is constant. dr dθ 2. changes throughout the motion vr=+ˆ r θˆ Velocity dt dt because the speed changes. ⎛⎞dr dθ ˆ 3. changes throughout the motion LrvrO =×μμ =rrˆˆ ×⎜⎟ r + θ Angular Momentum ⎝⎠dt dt because the distance from O changes. 2 dθ ˆ 4. changes throughout the motion LkO = μr dt because the angle θ changes. dθ LL≡=μ r2 5. Not enough information to decide. z dt 2 2 6. We had the angular momentum quiz L2 1 ⎛ dθ ⎞ 1 ⎛ dθ ⎞ Key Relation: = μr 2 = μ r last Friday so I don’t need to think 2μ 2μ ⎝⎜ dt ⎠⎟ 2 ⎝⎜ dt ⎠⎟ about it anymore. 2 Recall: Potential Energy Concept Question: Energy Find an expression for the potential energy of the system The mechanical energy consisting of the two objects interacting through the central 1. is constant. forces given by 2. changes throughout the motion Gm m because the speed changes. a) Gravitational force Fr=− 12ˆ 1,2 r 2 3. changes throughout the motion because the distance from O b) Linear restoring force Fr1,2 =−krˆ changes. 4. is not constant because the orbit is rr== rr ffGm m Gm m ⎛⎞11 12ˆ 12 not zero hence the central force Gravitation: ΔUddrGmmgrav =−−22rr ⋅=−− =−12⎜⎟ − ∫∫rrrr⎜⎟ does work. rr==00 rr ⎝⎠fi rr= 5. Not enough information to decide. f 1 Δ=−−⋅=−Ukrdkrrrrˆ ()22 Linear restoring: spring∫ 2 f 0 rr= 0 Mechanical Energy and Force and Potential Energy Effective Potential Energy There are no non-conservative forces acting so the mechanical L2 energy is constant. Effective Potential Energy Ueff = 2 +U(r) 2 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 2 2μr Kinetic Energy 1 dr 1 dθ 1 dr L K = μ ⎜ ⎟ + μ ⎜ r ⎟ = μ ⎜ ⎟ + 2 ⎝ dt ⎠ 2 ⎝ dt ⎠ 2 ⎝ dt ⎠ 2μ dL22 L F 2 Repulsive Force rep =−23 = 11⎛⎞dr L2 dr2μμ r r Mechanical Energy EUrKU=++≡+μ ⎜⎟ 2 () effective effective 22⎝⎠dtμ r dU(r) 2 L Central Force Fr =− U = +U(r) Effective Potential Energy effective 2 dr 2μr 2 dU() r ⎛⎞LdUr2 () 1 ⎛⎞dr eff ˆˆ Effective Kinetic Energy Keffective = μ ⎜⎟ Effective Force Freff =− =⎜⎟3 − r 2 ⎝⎠dt dr⎝⎠μ r dr 3 Reduction to One Dimensional Central Force Motion: Motion constants of the motion Reduce the one body problem in two dimensions to a one body problem moving only in the radial direction but under the action of two forces: a repulsive force and the central Total mechanical energy E is conserved because the force is radial and depends only on r and not on θ force Angular momentum L is constant because the torque about origin is zero The force and the velocity vectors determine the plane of motion dU() r ⎛⎞LdUr22() dr eff ˆˆ Freff =− =⎜⎟32 − r =μ dr⎝⎠μ r dr dt Energy Diagram: Graph of Linear Restoring Force Effective Potential Energy vs. Central Force Fr1,2 =−krˆ Relative Separation 2 2 Energy 111⎛⎞dr L 2 For E > 0 , the relative EkrKU=++=+μ ⎜⎟ 2 effective effective 222⎝⎠dtμ r separation oscillates dθ Angular Momentum Lr= μ 2 varies between dt 2 Kinetic Energy 1 ⎛⎞dr rmin ≤ r ≤ rmax Keffective = μ ⎜⎟ 2 ⎝⎠dt Effective Potential Energy L2 1 U = + kr 2 The effective potential effective 2μr 2 2 has a minimum at r0 Repulsive Force dL⎛⎞22 L Frepulsive =−⎜⎟ = dr2μ r23μ r Linear Restoring Force ⎝⎠ dUspring Fspring =− =−kr dr 4 Group Problem: Lowest Group Problem: Lowest Energy Solution Energy Orbit Solution The effective The lowest energy state corresponds to a circular potential energy is orbit where the radius can be found by finding the 2 minimum of the effective potential energy L 1 2 U = + kr 2 effective 2 dU 2 effective L 2μr 0 = =− + kr ⇒ dr 3 0 μr0 r =r0 Find the radius and 1/4 the energy for the Radius of circular orbit ⎛ L2 ⎞ r0 = ⎜ ⎟ lowest energy orbit. ⎝ μk ⎠ What type of motion 1/ 2 is this orbit? ⎛ L2k ⎞ Energy of circular orbit E = U = min ()effective ⎜ ⎟ r =r0 μ ⎝ ⎠ Orbit Equation: Isotropic Kepler Problem: Gravitation Harmonic Oscillator m m Reduced Mass μ = 1 2 A special solution of the equation of m + m 1 2 motion for a linear restoring force d 2r Energy 11dr2 L2 Gm m μ =−kr rˆ ⎛⎞ 12 2 EKU=+−=+μ ⎜⎟ 2 effective effective dt 22⎝⎠dtμ r r ˆˆ dθ is given by rij()txtyt=+ () () Angular Momentum Lr= μ 2 dt x(t) = x sin(ωt) 1 dr 2 with 0 ⎛⎞ Kinetic Energy Keffective = μ ⎜⎟ 2 ⎝⎠dt y(t) = y0 cos(ωt) 2 L Gm12 m Effective Potential Energy Ueffective =−2 where for the case shown in the 2μrr dL⎛⎞22 L figure with F =− = r = x Repulsive Force centrifugal ⎜⎟23 y0 < x0 rmin = y0 max 0 dr⎝⎠2μ rμ r The solution for r()t is an ellipse dU gravitational Gm12 m centered at the origin Gravitational force Fgravitational =− =− dr r 2 5 Orbit Equation for Kepler Energy Diagram Problem Case 1: Hyberbolic Orbit Solution (See Course Notes) E > 0, ε >1 r r = 0 Case 2: Parabolic Orbit 1cos− ε θ E = 0 , ε =1 1/2 ⎛ ⎞ 1/2 2EL2 ⎛ E ⎞ Case 3: Elliptic Orbit ε = ⎜1+ ⎟ = 1− Eccentricity 2 ⎜ ⎟ Emin < E < 0 , 0 < ε < 1 ⎜ ⎟ ⎝ Emin ⎠ ⎝ μ()Gm1m2 ⎠ 2 Case 4: Circular Orbit 2 L 2 1 μ(Gm1m2 ) E = E , ε =1 E =− r0 = min L Gm12 m min 2 μGm m U =− 2 L 12 effective 2μrr2 Group Problem: Lowest Group Problem: Lowest Energy Orbit Energy Orbit Solution The effective potential The lowest energy state corresponds to a circular energy is orbit where the radius can be found by finding the 2 L Gm1m2 minimum of the effective potential energy Ueffective = 2 − 2μr r dU 2 effective L Gm1m2 0 = =− 3 + 2 ⇒ Make a graph of the dr μr0 r0 r =r0 effective potential energy L2 as a function of the Radius of circular orbit r = relative separation. Find 0 μGm1m2 the radius and the energy for the lowest energy 2 orbit. What type of motion μ(Gm1m2 ) is this orbit? Energy of circular orbit Emin = U effective =− ()r =r 2 0 2L 6 Kepler’s Laws Equal Area Law and Conservation 1. The orbits of planets are ellipses; and the center of of Angular Momentum sun is at one focus Change in area ΔA = (1 / 2)rvθ Δt per time ΔA 1 = v r Δt 2 θ Angular Lrv 2. The position vector sweeps out equal areas in momentum = μ θ equal time 3. The period T is proportional to the length of the Equal area law ΔA L = major axis A to the 3/2 power TA∼ 3/2 Δt 2μ 7.
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