Central Force Problem. Central Force Motion. Two Body Problem: Center of Mass Coordinates. Reduction of Two Body Problem 8.01 W14D1. + m 2. m 2.

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1 Cental oce Poblem ind the motion of two bodies inteacting via a cental foce. Cental oce Motion 8.01 W14D1 Examples: Gavitational foce (Keple poblem): 1 1, ( ) G mm Linea estoing foce: ( ) k 1, Two Body Poblem: Cente of Mass Coodinates Cente of mass m11 + m Rcm 1 m + m 1 Relative Position Vecto 1 1 Reduced Mass μ m 1 m /(m 1 + m ) Position of each object m11 + m m ( 1 ) μ 1 1 Rcm 1 m + m m + m m μ m Reduction of Two Body Poblem Newton s Second Law Divide by mass Subtact: Use Newton s Thid Law (in components) 1,,1 Summay 1 1 d ( + ) 1, m m 1 d d m1 1,1 m 1, d d m m 1, 1,1 1 1,,1 d ( 1 ) d ( ) m1 m 1 d 1, μ 1

2 Reduction of Two Body Poblem Intepetation of Solution: Motion about Cente of Mass Reduce two body poblem to one body of educed mass μ moving about a cental point O unde the influence of gavity with position vecto coesponding to the elative position vecto fom object to object 1 Solving the poblem means finding the distance fom the oigin (t) and angle θ(t) as functions of time Equivalently, finding (θ) as a function of angle θ d μ 1, Knowledge of 1 detemines the motion of each object about cente of mass with position. μ 1 m 1 μ m Concept Question: Angula Momentum The angula momentum about the point O of the educed body 1. is constant.. changes thoughout the motion because the speed changes. 3. changes thoughout the motion because the distance fom O changes. 4. changes thoughout the motion because the angle θ changes. 5. Not enough infomation to decide. 6. We had the angula momentum quiz last iday so I don t need to think about it anymoe. Angula Momentum about O Toque about O: τ O 1, () 1, () 0 Ο d dθ Velocity v + θ d dθ L Angula Momentum O μv μ + θ dθ L O μ k dθ L Lz μ L Key Relation: μ 1 μ μ dθ 1 μ dθ

3 Recall: Potential Enegy ind an expession fo the potential enegy of the system consisting of the two objects inteacting though the cental foces given by a) Gavitational foce b) Linea estoing foce Gavitation: Linea estoing: Gm m 1 1, 1, k f f Gm1m Gm1m 1 1 Δ U gav d d Gmm f i f 1 Δ U sping k d k( f 0 ) 0 Concept Question: Enegy The mechanical enegy 1. is constant.. changes thoughout the motion because the speed changes. 3. changes thoughout the motion because the distance fom O changes. 4. is not constant because the obit is not zeo hence the cental foce does wok. 5. Not enough infomation to decide. Mechanical Enegy and Effective Potential Enegy Thee ae no non-consevative foces acting so the mechanical enegy is constant. Kinetic Enegy K 1 μ d + 1 μ dθ 1 μ d + L μ 1 d 1 L Mechanical Enegy E μ + + U() K + U μ Effective Potential Enegy Effective Kinetic Enegy L U μ + U() K 1 d μ oce and Potential Enegy Effective Potential Enegy Repulsive oce Cental oce Effective oce ep U eff L μ +U() d L L d μ μ 3 du () d dueff () L du() eff 3 d μ d 3

Reduction to One Dimensional Motion Reduce the one body poblem in two dimensions to a one body poblem moving only in the adial diection but unde the action of two foces: a epulsive foce and the

4 Reduction to One Dimensional Motion Reduce the one body poblem in two dimensions to a one body poblem moving only in the adial diection but unde the action of two foces: a epulsive foce and the cental foce Cental oce Motion: constants of the motion Total mechanical enegy E is conseved because the foce is adial and depends only on and not on θ Angula momentum L is constant because the toque about oigin is zeo dueff () L du() d eff μ 3 d μ d The foce and the velocity vectos detemine the plane of motion Cental oce Enegy Angula Momentum Kinetic Enegy Linea Restoing oce 1, k 1 d 1 L 1 E μ + + k K + U μ dθ L μ 1 d K μ Enegy Diagam: Gaph of Effective Potential Enegy vs. Relative Sepaation o E > 0, the elative sepaation oscillates vaies between min max Effective Potential Enegy Repulsive oce Linea Restoing oce L U μ + 1 k d L L epulsive d μ μ sping du sping d 3 k The potential has a minimum at 0 4

Goup Poblem: Lowest Enegy Solution The potential enegy is L U μ + 1 k ind the adius and the enegy fo the lowest enegy obit. What type of motion is this obit?

of cicula obit 0 L μ + k 3 0 0 0 L μk ( ) L k 0 U 1/4 μ 1/ Obit Equation: Isotopic Hamonic Oscillato A special solution of the equation of motion fo a linea estoing foce d μ k is given by () t x() t

5 Goup Poblem: Lowest Enegy Solution The potential enegy is L U μ + 1 k ind the adius and the enegy fo the lowest enegy obit. What type of motion is this obit? Goup Poblem: Lowest Enegy Obit Solution The lowest enegy state coesponds to a cicula obit whee the adius can be found by finding the minimum of the potential enegy 0 du d Radius of cicula obit Enegy of cicula obit 0 L μ + k L μk ( ) L k 0 U 1/4 μ 1/ Obit Equation: Isotopic Hamonic Oscillato A special solution of the equation of motion fo a linea estoing foce d μ k is given by () t x() t i+ y() t j with x(t) x 0 sin(ωt) y(t) y 0 cos(ωt) whee fo the case shown in the figue with y 0 < x 0 x min y 0 max 0 The solution fo () t is an ellipse centeed at the oigin Keple Poblem: Gavitation Reduced Mass μ m 1 m m 1 + m Enegy μ Angula Momentum dθ L μ Kinetic Enegy K 1 d μ Effective Potential Enegy U L Gm1m μ Repulsive oce centifugal d L L d μ μ 3 Gavitational foce gavitational du gavitational Gm1m d 1 d 1 L Gm1m E μ + K + U 5

6 Obit Equation fo Keple Poblem Solution (See Couse Notes) 0 1 ε cos θ Case 1: Hybebolic Obit E > 0, ε >1 Case : Paabolic Obit E 0, ε 1 Enegy Diagam Eccenticity EL ε 1+ μ Gm 1 m 1 μ(gm 1 m ) L ( ) 1/ 1 E L 0 μgm m 1 1/ Case 3: Elliptic Obit < E < 0, 0 < ε < 1 Case 4: Cicula Obit E, ε 1 U L Gm m 1 μ Goup Poblem: Lowest Enegy Obit The potential enegy is U L μ Gm m 1 Make a gaph of the potential enegy as a function of the elative sepaation. ind the adius and the enegy fo the lowest enegy obit. What type of motion is this obit? Goup Poblem: Lowest Enegy Obit Solution The lowest enegy state coesponds to a cicula obit whee the adius can be found by finding the minimum of the potential enegy 0 du d Radius of cicula obit 0 Enegy of cicula obit L μ + Gm m U L 0 μgm 1 m ( ) μ ( Gm m 1 ) 0 L 6

7 Keple s Laws 1. The obits of planets ae ellipses; and the cente of sun is at one focus. The position vecto sweeps out equal aeas in equal time Equal Aea Law and Consevation of Angula Momentum Change in aea pe time Angula momentum ΔA (1 / )v θ Δt ΔA Δt 1 v θ L μv θ 3. The peiod T is popotional to the length of the majo axis A to the 3/ powe 3/ T A Equal aea law ΔA Δt L μ 7

Central Force Motion

Central Force Motion Cental Foce Motion Cental Foce Poblem Find the motion of two bodies inteacting via a cental foce. Examples: Gavitational foce (Keple poblem): m1m F 1, ( ) =! G ˆ Linea estoing foce: F 1, ( ) =! k ˆ Two