Central Force Motion

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1 Cental Foce Motion

2 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

3 Two Body Poblem: Cente of Mass Coodinates Cente of mass m + m R cm = 1! m 1 + m 1 1 Relative Position Vecto = 1 " =!" 1! Reduced Mass µ = m 1 m / (m 1 + m ) Position of each object m + m m (! ) µ " =! R =! = = µ 1 1 cm 1! = " m 1 + m m 1 + m m1 m

4 Reduction of Two Body Poblem Newton s Second Law d d 1 F 1, ˆ = m 1 F ˆ = m,1 dt dt Divide by mass F 1, d 1 F,1 d ˆ = ˆ = m 1 dt m dt F 1, F,1 d ( 1! ) d ( )ˆ = = Subtact: m 1! m dt dt = 1! Use Newton s Thid Law (in components) F 1, =! F,1 1 1 d d Summay ( + ) F 1, ˆ =! F 1, ˆ = µ m1 m dt dt

5 Reduction of Two Body Poblem 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 d Equivalently, finding (θ) as a function of F 1, = µ angle θ dt

6 Intepetation of Solution: Motion about Cente of Mass Knowledge of =!"! 1 detemines the motion of each object about cente of mass with position. µ µ 1! == m1! = " m

7 Checkpoint Poblem: 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.

8 Angula Momentum about O Toque about O: = " F ˆ ˆ! O 1, ( ) = " F 1, ( ) = 0 # Velocity Angula Momentum Useful Relation: d v = ˆ + dt d!! ˆ dt ˆ " d d! ˆ ˆ # L O = $ µ v = $ µ % +!& ' dt dt ( d! L ˆ O = µ k dt d! L ) L z = µ dt L µ = 1 " µ # $ d! % µ dt & ' = 1 µ " d! % # $ dt & '

9 Recall: Potential Enegy Find an expession fo the potential enegy of the system consisting of the two objects inteacting though the cental foces given by Gm m a) Gavitational foce 1 F 1, =! ˆ b) Linea estoing foce F 1, =!kˆ Gavitation: = = f f Gm m Gm m! 1 1 " 1 1 d = $Gm m $ & ' =0 = 0 ( f i ) = f 1 k #U gav = $ * $ ˆ % d = $ * $ 1 Linea estoing:!u sping = " $ "k ˆ # d = ( f " 0 ) = 0

10 Checkpoint Poblem: 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.

11 Mechanical Enegy and Effective Potential Enegy Thee ae no non-consevative foces acting so the mechanical enegy is constant. 1! d $ 1! d' $ 1! d $ Kinetic Enegy L K = # + # = # + µ " dt & % µ " dt % & µ " dt & % µ 1! d " 1 L Mechanical Enegy E = µ $ % + + U ( ) # K +U & dt ' µ effective effective L U effective Effective Potential Enegy = µ + U () 1! d " Effective Kinetic Enegy K effective = µ # $ % dt &

12 Foce and Potential Enegy L U eff Effective Potential Enegy = + U () µ d L L Repulsive Foce F ep =! = d µ µ 3 Cental Foce F =! du () d du eff ( )! L ˆ # du ( ) " Effective Foce F eff = # = $ 3 % ˆ d & µ d '

13 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 du eff ( )! L du ( ) " d F eff = # ˆ = $ # ˆ = µ 3 % d & µ d ' dt

14 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 du eff ( )! L Gm m " 1 F eff = # ˆ = $ ˆ 3 % d & µ # ' d L Gm1m! 3 µ = dt µ

15 Cental Foce 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 The foce and the velocity vectos detemine the plane of motion

16 Cental Foce Linea Restoing Foce F =!kˆ 1, 1! d " 1 L 1 Enegy E = µ # $ + k = K effective +U % dt & + µ Angula Momentum d! L = µ dt Kinetic Enegy K = 1! d " effective µ # $ % dt & Effective Potential Enegy L 1 U = k effective + µ Repulsive Foce d! L " L F epulsive = # $ = % 3 d & µ ' µ Linea Restoing Foce =! du sping F sping =!k d effective

17 Enegy Diagam: Gaph of Effective Potential Enegy vs. Relative Sepaation Fo E > 0, the elative sepaation oscillates vaies between min!! max The effective potential has a minimum at 0 L 1 U = k effective + µ

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

19 Obit Equation: Isotopic Hamonic Oscillato A special solution of the equation of motion fo a linea estoing foce d µ =!k ˆ dt is given by ( t ) = x( t ) î + y( t ) ĵ with x(t) = x 0 sin(!t) y(t) = y 0 cos(!t) whee fo the case shown in the figue with y 0 < x 0 min = y 0 max = x 0 The solution fo ( t ) is an ellipse centeed at the oigin

20 Summay: Keple Poblem m 1 m Reduced mass µ = m1 + m Angula Momentum L = µ d! dt! $ L 1 1! d $! d' $ 1! d $ 1 K = µ # # " dt & % " # dt % & & µ " # " % dt & % µ Kinetic Enegy µv = + & = + Potential enegy U () =! Gm 1 m 1! d $ 1 L Gm µ 1 m # " dt & + % µ Enegy E = K + U () = ' () 1 Effective Kinetic Enegy K eff = µ (d / dt ) Effective Potential Enegy L Gm 1 m U =! eff µ Effective Repulsive Foce du ep L F =! = ep d µ 3 du gav Gm 1 m d Gavitational Foce F =! =! gav = K eff + U eff

21 Enegy Diagam: Gaph of Effective Potential Enegy vs. Case 1: Hybebolic Obit E > 0 Case : Paabolic Obit E = 0 Case 3: Elliptic Obit E min < E < 0 Relative Distance Case 4: Cicula Obit E = E min = U effective L! Gm 1 m µ

22 Checkpoint Poblem: Lowest Enegy Obit The effective potential enegy is L Gm 1 m U = effective! µ Make a gaph of the effective potential enegy as a function of the elative sepaation. Find the adius and the enegy fo the lowest enegy obit. What type of motion is this obit?

23 Checkpoint Poblem: Cicula Obital Mechanics A double sta system consisting of one sta of mass m 1 and a second sta of mass m ae obiting each othe such that the elative sepaation emains constant = R. a. Find the atio of kinetic to potential enegy b. Suppose the obits emain cicula but the elative sepaation R inceases. Do the following quantities incease, emain the same, o decease: angula momentum, velocity, kinetic enegy, potential enegy, enegy, and eccenticty?

24 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 3. The peiod T is popotional to the length of the majo axis A to the 3/ powe T A 3/

25 Equal Aea Law and Consevation of Angula Momentum Change in aea!a = (1 / )v "!t pe time!a 1 = v "!t Angula momentum L = µv! Equal aea law!a L =!t µ

26 Obit Equation Solution Obit Equation d L Gm1m µ =! 3 dt µ d d du 1 du Change of Vaiables: u! 1 " = = " d! du d! u d! d! L L Angula momentum condition: = = u dt µ µ d d d! d 1 du L = " = # Chain ule: u dt d! dt dt u d! µ d d u d" L d u L =! =! u d" Second deivative: dt d" dt µ µ d L Gm m L =! 1 = u 3! Gm 1 m u One dimensional foce equation µ dt µ 3 µ d u L L Gm 1 m d u µgm 1 m Result:! u =! # =!u + 3 d" µ µ µ d" L

27 Obit Equation Inhomogenous hamonic oscillato equation d u µgm 1 m + u = d! L with angle independent solution µgm 1 m u0 = L Solution: $ Acos("! " 0 )' u! u 0 = Acos("! " 0 ) # u = u 0 & 1+ % u 0 ( ) Change vaiables back with: 1 1 L =! 0 = u u 0 µgm 1 m 1 1 u! u 0 = Acos("! " 0 ) # = (1+ 0 Acos("! " 0 )) 0 Constants fixed by conditions: choose (!," 0 )! 0 = " A =! / 0 Conclusion: = 0 0 = 1! 0 Acos(") 1! # cos(")

28 Eccenticity Obit Equation 0 = 1! " cos(#) Neaest Appoach (θ=π): min = 0 (1+!) Futhest Appoach (θ=0): = max 0 (1! ") a = 1 ( + ) = 0 min max Semi-majo axis: (1! " ) Enegy at neaest appoach: E = U eff ( min Eccenticity: ) = L ( 1+! ) Gm 1 m (1+!) µ(gm 1 m ) Gm 1 m " # E = (! " 1) = " L µ0 0 1/ " % 1/ L E ) E % µ(gm m 1! = $1+ ' (! = " 1) ' $ µ Gm 1 ) µ ' #$ E # ( m 0 & & E 0 = L a

29 Constants of the Motion: Enegy and Angula Momentum Angula momentum L = ( µgm m ) 1/ 0 1 whee 0 is the adius of the cicula obit Enegy: µ(gm 1 m ) E = E (1! " ) =! (1! " ) min L whee E min is the enegy of the cicula obit 1 Gm 1 m µ(gm 1 m ) E min = (U effective ) = U =! =! = 0 gav = 0 0 L and ε is the eccenticity 1/ " % EL " E %! = $1+ ' = 1( ' $ µ(gm m ) ' # $ E # min & 1 & 1/

30 Obit Equation Solution: = 0 1 #! cos " Radius of cicula obit 0 = L µgm m 1 Enegy of cicula obit E min =! 1 µ(gm 1 m ) L Eccenticity "! = $ 1+ # $ EL ( ) µ Gm 1 m % ' & ' 1/ " = $ 1( # E E min % ' & 1/

31 Obit Classification: Case 1: Hybebolic Obit ε > 1 E > 0 Case : Paabolic Obit ε = 1 E = 0 Case 3: Elliptic Obit 0 < ε < 1 E min < E < 0 Case 4: Cicula Obit ε = 0 E = E min U effective = L! Gm 1 m µ

32 Popeties of Ellipse Eccenticity! = ( 1+ EL / µ(gm 1 m ) )1/ Semi-Majo axis Gm 1 m a =! E Semi-Mino axis b = a 1! " 0 = 1! " cos# Aea A =! ab =! a 1" # Location of the cente of the ellipse x 0 =!a

33 Popeties of an Elliptic Obit Enegy E =! Gm 1 m a Angula Momentum L = µgm 1 m a(1! " ) Neaest Appoach (θ=π): min = a(1! ") L L = Speed v p = = 1! " cos(#) µ min µa(1! ") L 0 Futhest Appoach (θ=0): = 0 = a(1+!) µgm m max 1 L L Speed v a = = µmax µa(1 +! ) (! = 1+ EL / µ(gm 1 m ) )1/

34 Keple s Laws: Equal Aea Aea swept out in time Δt ( ) " A 1 # "! $ "! " = % & + " t ' " t ( " t da dt 1 Equal Aea Law: d! dt d! dt L µ = = da dt = L µ = 1 G(m 1 + m )a(1! " ) = constant

35 Keple s Laws: Peiod Aea A =! ab =! a 1" # Integal of Equal Aea Law T µ da = dt L!! obit 0 Peiod µ µ! a 1" # T = A = L L Peiod squaed popotional to cube of the majo axis but depends on both masses 4µ 4! µ a 4 (1" # ) 4! a 3 T =! a 4 (1" # ) = = L µgm 1 m a(1" # ) G (m 1 + m )

36 Stas Neaby Galactic Cente The UCLA Galactic Cente Goup, headed by D. Andea Ghez, developed an animation of the obitsof eight stas about the galactic cente

37 Astonomical Data Obsevation data is given in tems of the semi-majo axis a and eccenticity ε Example: obit of stas aound cente of galaxy 4! a 3 m = GT 1AU = 1.50! m Sta Peiod (ys) Eccenticity Semi-majo axis Peiapse (AU) Apoapse (AU)! (10 3 ac sec ) S0-15. (0.68/0.76) (0.0063) 10.7 (4.5) (3.9) 181 (73) S (6.8/13) (0.019) 191 (4) 87 (17) 970 (560) S (35/11000) (0.065) 340 (0) 301 (41) 5100 (3600) Numbes in paentheses ae the eos on the given quantities.

38 Checkpoint Poblem: Black Hole Mass 1. Find the mass of the black hole at the cente of Milky Way Galaxy using Keple s 3d law.. What is the atio of the mass of the black hole to one sola mass? 3. What is the atio of the mass of the sun to the mass of the eath? 4. How do these atios compae? Mass of eath: 6 x 10 4 kg 3 4! a Mass of sun: x kg m = GT I AU = 1.5 x m G= 6.7 x N m kg -

39 MIT OpenCouseWae SC Physics I: Classical Mechanics Fo infomation about citing these mateials o ou Tems of Use, visit:

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.

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. 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