LECTURES INTRODUCTION TO CLASSICAL MECHANICS. Prof. N. Harnew University of Oxford HT 2017

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1 LECTURES INTRODUCTION TO CLASSICAL MECHANICS Prof. N. Harnew University of Oxford HT

2 OUTLINE : INTRODUCTION TO MECHANICS LECTURES Intro : programme for Hilary term (20 lectures) 11.1 Variable mass : a body acquiring mass 11.2 Example - the raindrop 11.3 Ejecting mass : the rocket equation 11.4 The rocket : horizontal launch 12.1 The rocket : vertical launch 12.2 The 1-stage vs. 2-stage rocket 12.3 Non-inertial reference frames Commonplace examples Example- accelerating lift 13.1 Magnetic Force on a Charged Particle 13.2 B Field only, v B 13.3 Motion under electric and magnetic fields 13.4 Kinetic energy in E & B fields B only, v B B and E

3 13.5 Cyclotron motion (E & B fields) 14.1 Differentiation of vectors wrt time The position vector in polar coordinates The velocity vector in polar coordinates The acceleration vector in polar coordinates 14.2 Angular momentum and torque 14.3 Angular velocity ω for rotation in a circle 15.1 Angular acceleration α for rotation in a circle 15.2 Angular motion : work and power 15.3 Correspondence between linear and angular quantities Reformulation of Newton s laws for angular motion Example: the simple pendulum 15.4 Moments of forces 15.5 Central forces A central force is conservative 16.1 Central force: the equation of motion 16.2 Motion under a central force Motion in a plane Sweeping out equal area in equal time

4 16.3 Central force : the total energy The potential term (inverse square interaction) Example 17.1 Effective potential U eff (r) for inverse square law 17.2 Examples Example 1 : 2-D harmonic oscillator Example 2 : Rotating ball on table 18.1 The orbit equation The ellipse geometry 18.2 Kepler s Laws Kepler III Planetary data 18.3 Elliptical orbit via energy (E min < E < 0) 19.1 Example: mistake in the direction of a satellite Orbits with the same energy 19.2 Impulse leaving angular momentum unchanged Orbits with the same angular momentum Mutual orbits

5 Example: binary star

6 Programme for Hilary term (20 lectures) Lectures 1-5 Rocket motion. Motion in B and E fields Lectures 6-10 Central forces (orbits) Lectures Rotational dynamics (rigid body etc) Lectures Lagrangian dynamics Plus 4 problem sets for your enjoyment 6

7 11.1 Variable mass : a body acquiring mass A body of mass m has velocity v. In time δt it it acquires mass δm, which is moving along v direction with velocity u The change in mass m is m + δm, the change in velocity v is v + δv Case 1: No external force. Change of momentum p 7 p = (m + δm)(v + δv) (mv + uδm) = 0 }{{}}{{} After Before mv + mδv + vδm + δmδv }{{} mv uδm = mδv + (v u) δm = 0 Ignore Divide by δt (time over which mass acquisition occurs) : p δt = m δv δm δt + (v u) }{{} δt = 0 Relative velocity, w As δt 0, m dv + w dm = 0 (in this case dv is -ve as expected)

8 A body acquiring mass - with external force Case 2: Application of an external force F NII : change of momentum = p = Fδt = mδv + w δm as before, where w = (v u) Divide by δt and let δt 0 m dv + w dm = F Note : ONLY in the case when u = 0 does d (mv) = F 8

9 11.2 Example - the raindrop An idealised raindrop has initial mass m 0, is at height h above ground and has zero initial velocity. As it falls it acquires water (added from rest) such that its increase in mass at speed v is given by dm/ = bmv where b is a constant. The air resistance is of the form kmv 2 where k is a constant. Formulate the equation of motion : m dv + w dm = F m dv + w dm = mg kmv 2 w = v (since u = 0) ; dm = bmv dv + (b + k)v 2 = g Terminal velocity : dv = 0 v T = g b+k 9

10 The raindrop, continued Calculate raindrop mass vs. distance dm dm dx = bmv = dm dx = bmv v dm m = bdx = bm Integrate : [log e m] m m 0 = [bx] x 0 m = m0 exp(bx) (Mass grows exponentially with x) 10 What is its speed at ground level? dv + (b + k)v 2 = g dv vh 0 vdv g (b+k)v 2 = dv dx dx = h 0 dx h = [ = v dv dx ( loge (g (b+k)v 2 ) 2(b+k) Solving : vh = g b+k [1 exp ( 2h(b + k))] )] vh 0

11 11.3 Ejecting mass : the rocket equation A body of mass m has velocity v. In time δt it ejects mass δm, with relative velocity w to the body Change of momentum p = δm(v w) + (m δm)(v + δv) }{{} mv After }{{} Before 11 = vδm wδm + mv + mδv vδm δmδv }{{} mv Ignore With external force : F = p δt As δt 0, δv δt dv & δm δt = m δv δt w δm δt dm Hence, again, F = m dv + w dm (as δm δt is +ve but dm is -ve) the rocket equation

12 11.4 The rocket : horizontal launch Rocket equation: m dv + w dm = F = 0 (no gravity) Assume mass is ejected with constant relative velocity to the rocket w m dv = w dm dv = w dm m Initial/final velocity = vi, v f Initial/final mass = m i, m f vf v i dv = w m f m i dm m vf v i = w log e (m i /m f ) 12 This expression gives the dependence of rocket velocity as a function of its mass

13 The rocket : vertical launch Rocket equation: m dv + w dm = F Rocket rises against gravity F = mg Mass is ejected at constant velocity w relative to the rocket Rocket ejects mass uniformly: m = m 0 αt dm = α Now consider upward motion: vf mdv = ( mg + wα) v i dv = t f [ ] (m vf v i = g(t f t i ) w log 0 αt f ) e (m 0 αt i ) = [ g(t f t i ) w log e (m f /m i )] t i ( g + ) wα m 0 αt

14 Rocket vertical launch, continued The rocket starts from rest at t = 0; half the mass is fuel. What is the velocity and height reached by the rocket at burn-out at time t = T? ] ( )] v = (m [ gt w log 0 αt) e (m 0 ) = [ gt w log e 1 α m 0 t = dx What is the condition for the rocket to rise? dv > 0 dm At t = 0, m = m 0, = α : αw m 0 g > 0 w > m 0g α m = m0 αt ; at burnout t = T, m = m 0 2 α = m 0 2T 14 Maximum velocity is at the burn-out of the fuel: At t = T : v max = gt + w log e 2 Height : x 0 dx = T 0 x = gt w mo α [ gt w log e ( 1 α m 0 t )] Standard integral : loge z dz = z log e z z [( ) ( 1 α m 0 t (log e 1 α m 0 t After simplification : x = gt 2 )) 2 + wt (1 log e 2) ( )] T 1 α m 0 t 0

15 12.2 The 1-stage vs. 2-stage rocket A two stage rocket is launched vertically from earth, total mass M 0 = kg and carries an additional payload of m = 100 kg. The fuel is 75% of the mass in both stages; burn rate α = 500 kg s 1, thrust velocity w = 2.5 km s 1. The mass of the 2nd stage is 900 kg. i) Calculate the final speed for the equivalent single stage rocket ii) Find the final speed of the 2-stage rocket (i) Single stage rocket : Time to burn-out : m t = α T = 0.75(M 1 + M 2 )/500 [kg s 1 ] From before : vf = gt + w log e (m i /m f ) m i = M 1 + M 2 + m ; m f = 0.25(M 1 + M 2 ) + m Single stage : vf = 3.25 km s 1 Earth s escape speed : 1 2 mv 2 > GMem R E v esc = 2GME v esc = 11.2 km s 1 (i.e. v f < v esc ) 15 R E

16 Stage 1 (ii) The 2-stage rocket m t = α T = 0.75M 1/500 [kg s 1 ] v1 = gt + w log e (m i /m f ) Stage 2 m i = M 1 + M 2 + m ; m f = 0.25M 1 + M 2 + m After first stage :vf = 2.68 km s 1 m t = α T = 0.75M 2/500 [kg s 1 ] v2 = v 1 gt + w log e (m i /m f ) m i = M 2 + m ; m f = 0.25M 2 + m After second stage : 16 v 2 = = 5.48 km s 1 Need a 3rd stage, more thrust or less payload to escape

17 12.3 Non-inertial reference frames A frame in which Newton s first law is not satisfied - the frame is accelerating (i.e. subject to an external force) Recall Galilean transformation but now u varies with time : Galilean transformation Accelerating frame 17 Position r = r ut r = r u(t) Velocity v = v u v = v u(t) dv Acceleration = dv dv = dv du Force on mass F (r ) = F (r) = m dv F (r ) = F(r) m du So even if F(r) = 0, from NII, there is an apparent (or ficticious ) force acting in S of F (r ) = m du

18 Commonplace examples Accelerating train In the inertial frame Fx = T sin θ = ma Fy = T cos θ mg = 0 In the non-inertial frame F x = T sin θ F fictitious F y = T cos θ mg = 0 In NIF need to introduce F fictitious = ma to explain the displacement of the bob. 18 Mass rotating in a circle In the inertial frame Centripetal acceleration provided by tension in the string T = mrω 2 In the non-inertial frame The block is at rest and its acceleration is zero In NIF need F fictitious = mrω 2 (centrifugal force) to balance the tension in the string.

19 Example : accelerating lift First consider the lift in free-fall The ball is weightless (stationary or moves at constant velocity) according to an observer in the lift lift becomes an inertial frame (like in deep space) : NI. To this observer ( the ) fictitious acceleration balances the du gravitational acceleration 19

20 Observer in accelerating lift Lift plus passenger (total mass M) is now accelerated upwards with force F. Passenger drops ball mass m from height h (M >> m ). Total force on lift Ftot = F Mg = Ma Acceleration of lift a = F M g According to passenger in lift frame, downwards acceleration of ball is = ( F M g) + g = F M (downwards) Hence weight of ball appears to passenger to be (F/M) m Check this: If F = 0, free-fall, ball is weightless If F = Mg, lift is stationary, ball has weight m g 20 Time for ball to reach floor, use h = 1 2 at2 t = 2hM F

21 Observer watching from a frame outside the lift Observer watches ball fall with acceleration g and the lift rise with acceleration F M g Equate times when ball reaches floor: 2h t = 2(h h = ) g F/M g }{{}}{{} ball falling lift rising Solve for h h = Mgh F Substitute back t = as before. 2Mh F 21

22 13.1 Magnetic Force on a Charged Particle F = q v B F is the magnetic force q is the charge v is the velocity of the moving charge B is the magnetic field Because the magnetic force is perpendicular to the displacement (dw = F.dx), the force does no work on the particle 22 Kinetic energy does not change Speed does not change Only direction changes Particle moves in a circle if v B

23 F = q v B Newton s second law in components mẍ = Fx mÿ = Fy m z = Fz Simple case: v perpendicular to B 23 Magnetic force= centripetal force F = qvb = mv 2 R (magnitudes) where R is the radius of curvature R = mv qb = p qb p is the particle momentum

24 m r = qv B 13.2 B Field only, v B mẍ = Fx, mÿ = F y, m z = F z B field only B = Bz k v = (ẋ, ẏ, 0) i j k v B = ẋ ẏ ż 0 0 B z = ẏb zi ẋb z j mẍ = qbz ẏ; mÿ = qb z ẋ ẍ = ωẏ; ÿ = ωẋ where ω = qbz m ẏ = ωx + c; (ẏ = 0 at t = 0 c = 0) 24 ẍ +ω 2 x = 0 x = A 1 cos ωt +A 2 sin ωt At t = 0, x = 0 A1 = 0 x = A 2 sin ωt ẋ = A2 ω cos ωt ; at t = 0, ẋ = u 0 A 2 = u 0 ω At t = 0: r0 = (0, 0, 0) v0 = (u 0, 0, 0)

25 x = u 0 ω sin ωt From before : ẏ = ωx = u 0 sin ωt y = u 0 ω cos ωt + c At t = 0, y = 0 c = u 0 ω y = u 0 ω (cos ωt 1) x = R sin ωt (R = u 0 ω = u 0m y + R = R cos ωt B Field only, continued qb z ) Circle x 2 + (y + R) 2 = R 2 If at t = 0, v0 = (u 0, 0, w 0 ) The particle will spiral in circles about the z-direction: 25 z = w 0 t; x 2 + (y + R) 2 = R 2

26 13.3 Motion under electric and magnetic fields m r = qv B + qe mẍ = Fx, mÿ = F y, m z = F z B field B = Bz k E field E = Ex i i j k v B = ẋ ẏ ż 0 0 B z = ẏb zi ẋb z j mẍ = qbz ẏ + qe x ; mÿ = qb z ẋ ẍ = ωẏ + qex qbz m where ω = m ẏ = ωx (as before ẏ = 0 at t = 0) 26 ẍ + ω 2 x = qex m : solution x = x 1 + x 2 At t = 0: r0 = (0, 0, 0) v0 = (0, 0, 0) Complementary function : x1 = A 1 cos ωt + A 2 sin ωt Particular integral : x2 = qex mω 2

27 Electric and magnetic fields, continued x = A1 cos ωt + A 2 sin ωt + qex mω 2 Define a = qex mω 2 t = 0, x = 0 A1 + a = 0 t = 0, ẋ = 0 A2 = 0 x = a (1 cos ωt) From before : ẏ = ωx = a ω (1 cos ωt) y = a (sin ωt ωt) + c At t = 0, y = 0 c = 0 y = a (sin ωt ωt) a cos ωt = x a a sin ωt = y + aωt A circle rolling down the y axis : (x a) 2 + (y + aωt) 2 = a 2 27

28 B only, v B 13.4 Kinetic energy in E & B fields x = u 0 ω sin ωt ẋ = u 0 cos ωt y = u 0 ω (cos ωt 1) ẏ = u 0 sin ωt T = 1 2 m[u 0 cos ωt] m[ u 0 sin ωt] 2 = 1 2 mu B and E No energy change x = a (1 cos ωt) ẋ = a ω sin ωt y = a (sin ωt ωt) ẏ = a ω (cos ωt 1) T = 1 2 m(a ω)2 ( cos ωt) 28 = 1 2 m(2a ω)2 sin 2 ωt 2 W.D. by E field

29 13.5 Cyclotron motion (E & B fields) Let the electric field vary with time as: cos ωt E = E 0 sin ωt, B = B z k 0 Can show by direct substitution x(t) = R [ω t sin ωt + sin ωt 1] y(t) = R [ω t cos ωt sin ωt] is a solution of the EOM where R = qe 0 mω 2 and ω = qbz m ω is the Cyclotron frequency (x + R) 2 + y 2 = R 2 [ (ωt) ] T = 1 2 m[ẋ 2 + ẏ 2 ] = 1 2 mr2 ω 4 t 2 particle accelerator 29

30 14.1 Differentiation of vectors wrt time Vectors follow the rules of differentiation: d a = dax i + day j + daz k = ȧ x i + ȧ y j + ȧ z k d da (a + b) = + db = ȧ + ḃ d d c (c a) = a + c da = ċ a + c ȧ d da db (a.b) =.b + a. = ȧ.b + a.ḃ d da (a b) = b + a db = ȧ b + a ḃ (order is impt.) Orthogonality of differentiated unit vectors d dˆr (ˆr.ˆr) = 2ˆr. = 0 (since ˆr.ˆr = 1) Therefore dˆr ˆr dˆr ˆθ Derivative of any unit vector gives a vector perpendicular to it. 30

31 The position vector in polar coordinates r = r0 (i cos θ + j sin θ) ˆr = (i cos θ + j sin θ) is a unit vector in the direction of r dˆr = [ i sin θ + j cos θ ] θ ˆθ = ( i sin θ + j cos θ) is a unit vector perpendicular to r ˆr = θ ˆθ also ˆθ = θ ˆr 31

32 The velocity vector in polar coordinates r = r ˆr v = ṙ = ṙ ˆr + r ˆr From before ˆr = θ ˆθ General case: v = ṙ = ṙ ˆr + r θ ˆθ For circular motion: Since ṙ = 0 v = r θ ˆθ = r ω ˆθ 32

33 The acceleration vector in polar coordinates From before v = ṙ = ṙ ˆr + r θ ˆθ a = v = r d (ṙ ˆr) = rˆr + ṙ θˆθ (since ˆr = θ ˆθ) d (r θ ˆθ) = r θ ˆθ + r θ ˆθ + ṙ θ ˆθ = r θ 2 ˆr + r θ ˆθ + ṙ θ ˆθ (since ˆθ = θ ˆr) General case : a = r = ( r r θ 2 ) ˆr + (2ṙ θ + r θ) ˆθ For circular motion: Since scalars r = ṙ = θ = 0 33 (no change in magnitudes of radius or azimuthal acceleration) a = r θ 2 ˆr = ω 2 r ˆr = v 2 r ˆr

34 Angular momentum and torque The definition of angular momentum (or the moment of momentum) J for a single particle : J = r p r is the displacement vector from the origin and p the momentum The direction of the angular momentum gives the direction perpendicular to the plane of motion Differentiate: dj = r dp + dr p dp Definitions of force and velocity: F = and v = dr dj = r F + v p this term= mv v = 0 Define torque τ = r F = dj For multiple forces : dj (cf. Linear motion F = dp ) = n i=1 r i F i = τ tot

35 Torque depends on the origin Torque wrt origin O τ o = r F Torque wrt point A τ A = r A F = r F R F = τ 0 R F Hence in general τ o τ A Same applies to angular momentum : J o J A 35

36 14.3 Angular velocity ω for rotation in a circle Definition of angular velocity : ṙ = ω r Note that ṙ is always r, so ω is defined for circular motion Define ˆn such that ˆθ = ˆn ˆr Recall v = ṙ = ṙr + r θˆθ For circular motion ṙ = 0 ; θ = ω ṙ = ω r = (ω ˆn) (r ˆr) = rω ˆθ Relationship between J and ω 36 J = r p = m r ṙ = m r (ω r) Recall vector identity a (b c) = (a.c) b (a.b) c J = m r 2 ω m (r ω) r r ω = 0 since the circular rotation is in a plane Hence J = Iω where I = mr 2 ; (generally I = i [m ir 2 i ] )

37 15.1 Angular acceleration α for rotation in a circle Angular velocity for rotation in a circle : ṙ = ω r ω = ω ˆn = θ ˆn Angular acceleration: α = ω Special case if α is constant dω = α ω = ω 0 + α t dθ = ω θ = θ 0 + ω 0 t α t2 Which should be recognisable equations! Relationship between τ and α for rotation in a circle 37 τ = d J = Iα

38 Work linear motion : dw = F ds 15.2 Angular motion : work and power W = F ds Work angular motion : τ = r F ds = dθ r (dθ out of page) dw = F ds = F (dθ r) Power : = (r F) dθ (scalar triple product) W = τdθ = τ ω Linear motion : P = F v Rotational motion : P = τ ω 38

39 15.3 Correspondence between linear and angular quantities Linear quantities are re-formulated in a rotating frame: Linear/ translational quantities Angular/ rotational quantities Displacement, position: r [m] Angular displacement, angle: θ [rad] Velocity: v [m s 1 ] Angular velocity: ω [rad s 1 ] Acceleration: a [m s 2 ] Angular acceleration: α [rad s 2 ] Mass m [kg] Moment of inertia: I [ kg m 2 rad 1 ] Momentum: p [kg m s] 1 Angular momentum: J [kg m 2 s 1 ] Force F [N = kg m s 2 ] Torque: τ [kg m 2 s 2 rad 1 ] Weight F g [N] Moment [N m] Work dw = F dx [N m] Work W = τ dθ [N m] 39

40 Reformulation of Newton s laws for angular motion 1. In the absence of a net applied torque, the angular velocity remains unchanged. 2. Torque = [moment of inertia] [angular acceleration] τ = Iα This expression applies to rotation about a single principal axis, usually the axis of symmetry. (cf. F = ma). More on moment of inertia comes later. 3. For every applied torque, there is an equal and opposite reaction torque. (A result of Newton s 3rd law of linear motion.) 40

41 Example: the simple pendulum Derive the EOM of a simple pendulum using angular variables: τ = r F = mgr sin θ ẑ J = r mv = mrv ẑ v = r θ v = r θ dj = mr v ẑ = (mr 2 θ) ẑ (since ẑ is a constant vector) dj = τ mr 2 θ = mgr sin θ θ + g r sin θ = 0 41

42 15.4 Moments of forces Simple example : ladder against a wall If no slipping, torques (moments) must balance About any point: n i=1 r i F i = τ tot = 0 Moments about O mg L 2 cos θ = N 2L sin θ Also balance of forces in equilibrium mg = N 1 and F s = µn 1 = N 2 General case: body subject to gravity. Total moment : M = V r g ρ dv mass term + n i=1 r i F i external forces S r (pn ds) surface pressure 42

43 15.5 Central forces Central force: F acts towards origin (line joining O and P) always. F = f (r) ˆr only Examples: Gravitational force F = GmM ˆr r 2 Electrostatic force F = q 1q 2 ˆr 4πɛ 0 r 2 43

44 A central force is conservative A force F is conservative if it meets 3 equivalent conditions: 1. The curl of F is zero : F = 0 2. Work over closed path W F dr = 0, independent of path C 3. F can be written in terms of scalar potential F = U Equivalence of 1 & 2 from Stokes theorem S ( F) da = C F dr = 0 Equivalence of 1 & 3 from vector calculus identity : ( U) = 0 For a central potential, take the grad of U(r) : In cartesians U(r) = U( x 2 +y 2 +z 2 ) ˆx +... (ŷ and ẑ terms) Chain rule U x = U r x r x : U(r) = x x 2 +y 2 +z 2 U(r) r ˆx +... xˆx+yŷ+zẑ Since U(r) = ˆr U(r) = x 2 +y 2 +z2 r ˆr f (r)ˆr = F(r) The grad of the scalar potential has only one non-vanishing component which is along ˆr ( central force). Hence condition (3) is satisfied central force is conservative force. 44

45 16.1 Central force : the equation of motion Recall the acceleration in polar coordinates a = r = ( r r θ 2 ) ˆr + (2ṙ θ + r θ) ˆθ If F = f (r) ˆr only, then Fθ = 0 F θ = m(2ṙ θ + r θ) = 0 F r = m( r r θ 2 ) = f (r) Consider d (r 2 θ) = 2rṙ θ + r 2 θ 45 Hence 1 r d (r 2 θ) = 0 (r 2 θ) = constant of motion The angular momentum in the plane : J = m r v = m r (ṙ ˆr + r θ ˆθ) = (mr 2 θ) ˆn where r ˆr = 0 and ˆn = ˆr ˆθ Torque about origin : τ = dj = r F = 0 (F acts along r) Angular momentum vector is a constant of the motion

46 16.2 Motion under a central force Motion in a plane J = m r v Angular momentum is always perpendicular to r and v J is a constant vector ; J r = 0 ; J v = 0 46 Motion under a central force lies in a plane

47 Sweeping out equal area in equal time Central force example : planetary motion : Fr = GMm r 2 Angular momentum is conserved J = mr 2 θ = constant 47 da 1 2 r 2 dθ da da = 1 2 r 2 θ = J 2m = constant (Kepler 2nd Law) Orbit sweeps out equal area in equal time

48 16.3 Central force : the total energy Total energy = kinetic + potential : E = T + U(r) = 1 2 mv 2 + U(r) = constant v = ṙˆr + r θˆθ v 2 = (ˆr + r θˆθ) (ˆr + r θˆθ) v 2 = ṙ 2 + r 2 θ 2 (since ˆr ˆθ = 0) E = 1 mṙ 2 + 1mr θ2 + U(r) No external torque: angular momentum is conserved J = mr 2 θ = constant E = 1 2 mṙ 2 + J2 + U(r) 2mr 2 Potential energy for a central force U(r) = r r ref F dr = r r ref f (r) dr 48

49 The potential term (inverse square interaction) F = A r 2 ˆr f (r) = A r 2 [Attractive force for A > 0 signs are important!] r U(r) = r ref F dr = r r ref f (r) dr U(r) = A r + A r ref Usual to define U(r) = 0 at r ref = U(r) = A r Newton law of gravitation : F = GMm r 2 ˆr U(r) = GMm r 49

50 Example A projectile is fired from the earth s surface with speed v at an angle α to the radius vector at the point of launch. Calculate the projectile s subsequent maximum distance from the earth s surface. Assume that the earth is stationary and its radius is a. 50

51 Example : solution U(r) = GMm r J = m r v = mav sin α Energy equation : E = 1 2 mṙ 2 + J2 2mr 2 + U(r) E = 1 2 mṙ 2 + ma2 v 2 sin 2 α 2r 2 GMm r At r = a : E = 1 2 mv 2 GMm a. At maximum height : ṙ = mv 2 GMm a ( v 2 2GM a = ma2 v 2 sin 2 α 2r 2 max GMm r max (1) ) r 2 max + 2GM r max a 2 v 2 sin 2 α = 0 Solve and take the positive root Note from Equ.(1) : When ṙ 0 as rmax, the rocket just escapes the earth s gravitational field 1 i.e. 2 mv 2 GMm 2GM a 0, v esc = a (independent of α)

52 17.1 Effective potential Energy equation : E = 1 2 mṙ 2 + J2 2mr 2 + U(r) Define effective potential : Ueff (r) = J2 2mr 2 + U(r) then E = 1 2 mṙ 2 + U eff (r) Note this has the same form as a 1-D energy expression : E = 1 2 mẋ 2 + U(x) the analysis becomes 1-D-like problem since J = const Allows to predict important features of motion without solving the radial equation 1 2 mṙ 2 = E U eff (r) U eff (r) < E LHS is always positive The only locations where the particle is allowed to go are those with U eff (r) < E 52

53 U eff (r) for inverse square law Ueff (r) = J2 GmM 2mr 2 r Ueff (r) < E tot for all r Three cases : Etot < 0 : Bound (closed) orbit with r 1 < r < r 2 Etot has minimum energy at r = r 0 : du eff dr = 0, circular motion with ṙ = 0 Etot > 0 : Unbound (open) orbit with r > r 3 53

54 17.2 Examples Example 1 : 2-D harmonic oscillator F = kr (ignore the natural length of the spring) Energy equation : E = 1 2 mṙ 2 + U eff (r) Ueff (r) = J2 2mr kr 2 For circular motion ṙ = 0 : E min when U eff r r0 = 0 J2 mr kr 0 = 0 where J = mv 0 r 0 mv Leads to 0 2 r 0 = k r 0 as expected Including the natural length: F = kr F = k(r a) U = 1 2 kr 2 U = 1 2k(r a)2 Leads to mv 2 0 r 0 = k (r 0 a) 54

55 Example continued Ueff (r) = J2 2mr kr 2 For general motion : F = kr mẍ = kx mÿ = ky Solution for B.C s at t = 0: x = r 2, y = 0, ẋ = 0 x = r 2 cos ωt y = r 1 sin ωt where ω 2 = k m Ellipse: ( x r 2 ) 2 + ( y r 1 ) 2 = 1 55

56 Example 2 : Rotating ball on table Two particles of mass m are connected by a light inextensible string of length l. The particle on the table starts at t = 0 at a distance l/2 from the hole at a speed v 0 perpendicular to the string. Find the speed at which the particle below the table falls. Energy equation : E = 1 2 mṙ 2 + J2 + U(r) 2mr 2 ẏ = ṙ, U(r) = mgy E = 1 2 mṙ mẏ J2 2mr 2 mg(l r) At t = 0 : J = mv 0 l 2, E = 1 2 mv 2 0 mg l 2 Solution : ṙ 2 = gl+v (lv 0) gr. r 2 Condition for the particle on the table to move in circular motion ṙ = 0, Equate forces mv 2 0 l/2 = mg gives v 2 0 gl =

57 Example 2 continued : effective potential Effective potential : Ueff = J2 2mr 2 mg(l r) Closed orbit with rmin < r < l/2 Ball never passes though hole in absence of friction, minimum radius r = r min 57

58 18.1 The orbit equation Note that the derivation of this is off syllabus Acceleration in polar coordinates a = r = ( r r θ 2 ) ˆr + 1 r d (r 2 θ) ˆθ If F = f (r) ˆr only, then Fθ = 0. J = mr 2 θ = constant. F r = m( r r θ 2 ) = α r 2 (gravitational force, α = GmM) Hence r = J2 α = u3 J 2 u2 α m 2 r 3 mr 2 m 2 m (where u = 1 r ) (1) ṙ = dθ dr dθ = r = d dθ (ṙ) = J dr m r 2 dθ = J m ( ) J du m dθ d dθ d(1/r) dθ = J m du dθ = ( J2 u 2 ) d 2 u m 2 dθ 2 Substituting in Eq (1) : ( J 2 )u 2 d 2 u = u3 J 2 u2 α m 2 dθ 2 m 2 m d 2 u dθ 2 = u + m α J 2 d 2 u dθ 2 = u + 1 r 0 (r 0 = J2 m α ) 58

59 The orbit equation continued d 2 u dθ 2 = u + 1 r 0 where u = 1 r and r 0 = J2 m α Solution is 1 r = 1 r 0 + C cos(θ θ 0 ) where C, θ 0 = constants r r(θ) = 0 1+e cos(θ θ 0 ) e = eccentricity (e = C r 0 ) This is in the form of an ellipse. Also have a link between angular momentum and the ellipse geometry (J 2 = m αr 0 ). More precisely a conic section, which includes hyperbola, parabola and circle. Choose major axis as x axis θ 0 = 0 r(θ) = r 0 1+e cos θ Equivalent form of ellipse : ( x a )2 + ( y b )2 = 1 b = a (1 e 2 ) a = r 0 (1 e 2 ) 59

60 The ellipse geometry Example of a rotating planet : the sun is at the ellipse focus F r r(θ) = 0 ( x ) 2 ( 1+e cos θ a + y ) 2 b = 1 Closest approach θ = 0 perigee r min = r 0 1+e Furthest approach θ = π apogee r max = r 0 1 e (ṙ = 0 in both cases) rmax + r min = 2a = 2r 0 1 e 2 r 0 = a(1 e 2 ) xc + r min = a x c + r 0 1+e = a x c = a a(1 e) = a e At point A, r 2 = x 2 c + b 2 ; cos θ = xc r ; r = r 0 1 ex c/r (r 0 + ex c ) 2 = x 2 c + b 2 ( a(1 e 2 ) + e 2 a ) 2 = e 2 a 2 + b 2 60 b = a (1 e 2 ) also r min = a(1 e) & r max = a(1 + e)

61 18.2 Kepler s Laws KI: The orbit of every planet is an ellipse with the sun at one of the foci. [Already derived] KII: A line joining a planet and the sun sweeps out equal areas during equal intervals of time. [Already derived] KIII: The squares of the orbital periods of planets are directly proportional to the cubes of the semi-major axis of the orbits. 61

62 Kepler III The squares of the orbital periods of planets are directly proportional to the cubes of the semi-major axis of the orbits This is trivial for a circle : mr 0 ω 2 = mr 0 ( 2π T )2 = GmM r 2 0 r 3 0 = T 2 GM 4π 2 For an ellipse: da = 1 2 r 2 θ = J 2m = constant Integrate A = T 0 J 2m A2 = ( J 2m )2 T 2 From before : r0 = a(1 e 2 ), b = a (1 e 2 ) a = b2 r 0 Area of an ellipse : A = πab A 2 = π 2 a 3 r 0 Putting it all together T 2 a 3 62

63 Planetary data Kepler-III The squares of the orbital periods of planets are directly proportional to the cubes of the semi-major axis of the orbits 63 IMPRESSIVE!

64 18.3 Elliptical orbit via energy (E min < E < 0) E = 1 2 mṙ 2 + J2 α 2mr 2 r At turning points ṙ = 0 r = r min or r = r max E = J 2 2mr 2 α r r 2 + α E r J2 2mE = 0 r = α 2E ± [ ( α 2E )2 + J2 2mE ] 1 2 rmin,max = ( ) α 2EJ 2 2E [1 ± (1 + mα 2 ) 1 2 ] }{{} e r max = α (1 + e) } 2E {{} = a(1 + e), r min = α (1 e) } 2E {{} = a(1 - e) Consistent with the orbit equations. NICE! 64 Also: (1 + 2EJ2 ) 1 mα 2 2 ) = e E = mα2 (e 2 1) 2J 2 = α 2r 0 (e 2 1) where r 0 = J2 mα and e = eccentricity of ellipse.

65 r(θ) = Elliptical orbit via energy, continued r 0 1+e cos θ Total energy in ellipse parameters E = α ( 2r 0 e 2 1 ) e = 0, r = r0, E = α 2r 0 motion in a circle If 0 < e < 1, E < 0 motion is an ellipse If e = 1, E = 0 r(θ) = r 0 1+cos θ motion is a parabola If e > 1, E > 0 65 motion is a hyperbola

66 19.1 Example: mistake in the direction of a satellite Mistake is made in boosting a satellite, at radius R, into circular orbit : magnitude of velocity is right but direction is wrong. Intended to apply thrust to give velocity v0 along circular orbit. Instead thrust at angle θ wrt direction of motion. Energy of orbit is right, angular momentum is wrong. 66 What is the perigee and apogee of the resulting orbit? (Points B &C) Conservation of angular momentum, points A & B J = mv 0 R sin( π 2 θ) = mv Br B Energy at A = energy at perigee B 1 2 mv 0 2 α R = 1 2 mṙ 2 + J2 α 2mrB 2 r B where α = GMm

67 19.1 Example continued At point B, ṙ = 0, energy conservation becomes 1 2 mv 0 2 α R = m2 v0 2R2 cos 2 θ α 2mrB 2 r B Equate forces for circular motion to get v0 : mv0 2 R = α v 2 R 2 0 = α mr Sub for v 2 0 : energy conservation becomes α 2R α R = αr cos2 θ α 2rB 2 r B 1 2R = R cos2 θ 1 2rB 2 r B ( 2Rr 2 B ) r 2 B 2Rr B + R 2 cos 2 θ = 0 rb = R R 2 R 2 cos 2 θ, also r C = R + R 2 R 2 cos 2 θ r B = R(1 sin θ) Perigee r C = R(1 + sin θ) Apogee 67

68 Orbits with the same energy U eff 68 r

69 19.2 Impulse leaving angular momentum unchanged Example: A satellite in circular orbit has been given an impulse leaving J unchanged. The kinetic energy is changed by T = βt 0. Describe the subsequent motion. If J is not changed, impulse must be perpendicular to the direction of motion, with angular part of the velocity unchanged. E = 1 2 mṙ 2 + J2 α 2mr 2 r Circular orbit: ṙ = 0, J = mr 0 v 0 (1) E initial = 1 2 mv 2 0 α r 0 Equate forces : mv 2 0 r 0 = α r 2 0 v 2 0 = α mr 0 (2) 69

70 19.2 Example continued New orbit (elliptical): Enew = 1 2 βmv 0 2 α r 0 Equate energies: subsequent motion described by: 1 2 βmv 2 0 α r 0 = 1 2 mṙ 2 + J2 2mr 2 α r (3) Now solve for rmin, r max. Set ṙ = 0 From (1), (2), (3) (β 2) r 2 + 2r 0 r r 2 0 = 0 rmin,max = r 0± r 2 0 +(β 2) r 2 0 (β 2) Example: β = rmax = r 0 r min = r 0 Changes in orbit as a result of impulse 70

71 Orbits with the same angular momentum E ellipse = α 2 a E circle = α 2 r 0 71

72 19.3 Mutual orbits The two bodies make a mutual elliptical orbit on either side of the C of M (origin) in a straight line through the C of M Relative position vector : r = r2 r 1 Definition of C of M about O : m1 r 1 + m 2 r 2 = 0 72

73 Mutual orbits continued Internal forces: F12 = m 1 r 1, F 21 = m 2 r 2 Then r = r 2 r 1 = F 21 m 2 F 12 m 1 But F 12 = F 21 Hence r = F21 ( 1 m m 2 ) Define 1 µ = 1 m m 2 µ = m 1m 2 m 1 +m 2 µ is the reduced mass of the system Hence µ r = F21 and µ r = Gm 1m 2 r 2 r 1 2 ˆr = G µ (m 1+m 2 ) r 2 r 1 2 ˆr 73 Therefore Newton s Second Law for mutual motion can be re-written in terms of the position of the second body with respect to the first. The second body has the reduced mass which orbits round the first body with an effective mass equal to the sum of the two masses.

74 Example: binary star A binary star consists of two stars bound together by gravity moving in roughly opposite directions along a nearly circular orbit. The period of revolution of the starts about their centre of mass is 14.4 days and the speed of each component is 220 km s 1. Find the distance between the two stars and their masses. 74 For single star : v = ( r 2 ) ω = r 2π 2 T r = vt π = m Mutual motion : µ ( r r θ 2 )ˆr = Gm 1m 2 ˆr r 2 For circular motion : r = ṙ = 0, r θ 2 = constant = rω 2 Equating forces : rµω 2 = Gm 1m 2 = Gµ(m 1+m 2 ) r 2 r 2 (m1 + m 2 ) = r 3 ω 2 G m1 = m 2 = kg ; m 1 = m 2 (symmetry)