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Assignment 5 Solutions Physics 106a: Assignment 5 (Solutions prepared by Alvin Li) 8 November, 2019; due 5pm Friday, 15 November in the \Ph106 In Box" in East Bridge mailbox. One Dimensional Systems: Central Forces and the Kepler Problem Reading: Hand and Finch Chapter 4. Problems 1. Binary-sun solar system: Consider a binary pair of identical suns of mass M orbiting in the x − y plane in an orbit centered at the origin. The gravitational constant is G. Now add a planet of small mass m with an initial condition on the z axis above the center of mass of the two suns and with a velocity in the z direction; by the symmetry of the system the small mass will remain on the z axis and suns will have equal z coordinates and their center of mass will also remain on the z axis. We can choose as coordinates describing the dynamics: the z coordinate of the planet z; the z coordinate of the two suns Z; and the polar coordinates (R; θ) giving the x; y coordinates of the suns (±R cos θ; ±R sin θ) | see figure. m z M Z R M R e (a) What is the Lagrangian of the system in terms of these coordinates and their time derivatives? [Solution] For each of the large suns i, the kinetic energy is 1 1 1 T = MR_ 2 + MZ_ 2 + MR2θ_2; (1) Mi 2 2 2 and for the small planet, the kinetic energy is 1 T = mz_2: (2) m 2 The potential energy comes from three contributions: The gravitational potential between M1 and M2 (separated by a distance of 2R), that between M1 and m (separated by a distance of p 2 2 p 2 2 R + (z − Z) ), and lastly that between M2 and m (separated by a distance of R + (z − Z) ). Note that the contributions from the latter two are identical. Mathematically, we have GM 2 GMm V = − − 2 × (3) 2R pR2 + (z − Z)2 | {z } GPE between M1 and M2 | {z } GPE between Mi and m Hence, the Lagrangian is given by L = 2 × TMi + Tm − V (4) 1 GM 2 2GMm = MR_ 2 + MZ_ 2 + MR2θ_2 + mz_2 + + (5) 2 2R pR2 + (z − Z)2 1 • (0.5 pt) : Writing down the kinetic energy of each of the large suns. • (0.5 pt) : Writing down the kinetic energy of the small planet. • (0.5 pt) : Writing down the potential between Mi and m. • (0.5 pt) : Writing down the potential between M1 and M2. • (1 pt) : Correct final expression for the Lagrangian. Note that if the student directly writes down the Lagrangian without the above derivations, full credit should still be given. Total sub-points : 3 (b) Using constants of the motion implied by translational and rotational symmetries find the coupled equations of motion forq; ¨ R¨ with q = z − Z. Your result will depend on the angular momentum l of the two suns as well as the parameters of the problem. [Solution] We notice that the Lagrangian: 1 GM 2 2GMm L = MR_ 2 + MZ_ 2 + MR2θ_2 + mz_2 + + (6) 2 2R pR2 + (z − Z)2 1 GM 2 2GMm = MR_ 2 + MZ_ 2 + MR2θ_2 + mz_2 + + (7) 2 2R pR2 + q2 does not depend on explicitly on the coordinate θ. This means that θ is an ignorable / cyclic coordinate, and hence the conjugate (generalized) momentum corresponding to θ, which, in this case, is the total angular momentum l of the system, is a conserved quantity. We have @L l = = 2MR2θ_ = constant (8) @θ_ The center of mass of the two suns lies on the z axis (and is of the same height as the two suns), which coincide with the planet. The overall center of mass of the system is given by m 2M Z = Q = z + Z: (9) com m + 2M m + 2M Together with the new coordinate q = z − Z; (10) we can solve for z and Z in terms of q and Q as ( 2M z = Q + m+2M q m : (11) Z = Q − m+2M q and hence we can write the Lagrangian as 1 GM 2 2GMm L = MR_ 2 + MZ_ 2 + MR2θ_2 + mz_2 + + (12) 2 2R pR2 + q2 m 2 1 2M 2 GM 2 2GMm = MR_ 2 + M Q_ − q_ + MR2θ_2 + m Q_ + q_ + + (13) m + 2M 2 m + 2M 2R pR2 + q2 m m 2 m 2M 2 GM 2 2GMm = MR_ 2 + M + Q_ 2 + M + q_2 + MR2θ_2 + + 2 m + 2M 2 m + 2M 2R pR2 + q2 (14) 1 Mm(m + 2M) GM 2 2GMm = MR_ 2 + (m + 2M) Q_ 2 + q_2 + MR2θ_2 + + (15) 2 p 2 | {z } (m + 2M) 2R R2 + q2 total mass 1 1 2Mm GM 2 2GMm = MR_ 2 + (m + 2M)Q_ 2 + q_2 + MR2θ_2 + + (16) 2 2 m + 2M 2R pR2 + q2 2 −1 1 1 1 1 GM 2 2GMm L = MR_ 2 + (m + 2M)Q_ 2 + + q_2 + MR2θ_2 + + (17) 2 2 2M m 2R pR2 + q2 | {z } 1 = µ 1 1 GM 2 2GMm = MR_ 2 + (m + 2M)Q_ 2 + µq_2 + MR2θ_2 + + (18) 2 2 2R pR2 + q2 where µ is the reduced mass of the system defined by 1 1 1 = + (19) µ 2M m Now, since the Lagrangian does not depend explicitly on the coordinate Q, Q is therefore a cyclic / ignorable coordinate, and hence the conjugate momentum, which is the linear momentum PQ along the z direction, is a conserved quantity. Mathematically, we have @L PQ = = (m + 2M)Q_ = constant; (20) @Q_ Using Lagrange equation d @L @L − = 0; (21) dt @q_i @qi we have the equation of motion for the R component as d GM 2 2GMmR _ _2 (2MR) − 2MRθ − 2 − 3 = 0 (22) dt 2R (R2 + q2) 2 2 ¨ _2 GM 2GMmR 2MR − 2MRθ + 2 + 3 = 0 (23) 2R (R2 + q2) 2 2 2 ¨ l GM 2GMmR 2MR − 3 + 2 + 3 = 0 (24) 2MR 2R (R2 + q2) 2 and the equation of motion for the q component as d 2GMmq (µq_) − − 3 = 0 (25) dt (R2 + q2) 2 2GMmq µq¨ + 3 = 0 (26) (R2 + q2) 2 Supplementary box 5.1: Conserved quantities for which coordinate? Some readers may try to rewrite the Lagrangian as P 2 1 l2 GM 2 2GMm L = MR_ 2 + Q + µq_2 + + + (27) 2(m + 2M) 2 4MR2 2R pR2 + q2 If they do so, when they apply the Euler-Lagrange equation they will find that the sign of l2 the term 4MR2 will be incorrect. What is wrong here? Readers should note that when we are going through the logic that \since the Lagrangian is not depending explicitly on a specific coordinate qi, and hence the conjugate momentum @L pi = corresponding to that coordinate is a conserved quantity", what we really mean @q_i is that the conjugate momentum is a conserved quantity in time (i.e. it is not changing with time). However, it does not immediately mean that the conjugate momentum is not changing with respect to the other coordinates! For example, consider the total angular momentum l = 2MR2θ:_ (28) 3 It is obvious that the angular momentum is changing with respect to the coordinate R. In particular, @l 2l = 4MRθ_ = 6= 0 (29) @R R And hence this explains why one will not get the correct answer if they substitute l into the Lagrangian before applying the Lagrange equation. Again, readers are reminded to be careful about the \conserved quantity" we are talking about here are only conserved in time. • (1 pt) : Stating that L does not depend on θ explicitly, and hence θ is ignorable / cyclic, and its conjugate momentum (the total angular momentum) is a conserved quantity. • (1 pt) : Solving l in terms of M, R and θ. • (2 pt) : Introducing the coordinate Q, and solve z and Z in terms of q and Q. • (1 pt) : Replace z, Z,_z and Z_ in the Lagrangian by q, Q,_q and Q_ . Note that if the student does not express the expression beforeq _2 as the reduced mass µ, credit should still be given. • (1 pt) : Stating that L does not depend on Q exlicitly, and hence Q is ignorable / cyclic, and its conjugate momentum (the linear momentum along the z direction) is a conserved quantity. • (1 pt) : Finding and expressing the equation of motion for r in terms of l and the other parameters. Note that if the term involving l does not have a correct sign, credit SHOULD NOT be awarded. • (1 pt) : Finding and expressing the equation of motion for q in terms of the other parameters. Total sub-points : 8 (c) In the limit of small planetary mass m M we can ignore the effect of the planet on the motion of the suns. Find the explicit solution for the motion of the suns R(t) for orbits with small eccentricity . Write your solution as circular motion plus a term proportional to . [Solution] m If M ! 0, the equation of motion for R can be written as 2 2 ¨ l GM 2GMmR 2MR − 3 + 2 + 3 = 0 (30) 2MR 2R (R2 + q2) 2 2 2 ¨ l G 2GR m R − 3 3 + 2 + 3 = 0 (31) M 2M R 2R (R2 + q2) 2 M | {z } ≈0 2 l2 G R¨ − + = 0: (32) M 2M 3R3 2R2 For circular orbit, we have R = R0 =constant and R_ = R¨ = 0, hence the equation of motion for R can be written as l2 G 3 3 + 2 = 0 (33) 2M R0 2R0 l2 R = (34) 0 GM 3 For a slightly non-circular orbit, we can write R(t) = R0(1 + f(t)) (35) where , the eccentricity, is a small quantity, and f(t) embeds the information of the time evolution of the orbit.
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