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4 Classical Dynamics 4 Classical Dynamics 4.1 Newtonian gravity 4.1.1 Basic law of attraction Two p oint masses, with mass M and M , lying at r and r , attract one 1 2 1 2 another. M feels a force from M 1 2 GM M 1 2 F = force on M from M = r 12 1 2 12 3 r 12 where r = r r =vector from 2 to 1 and r = jr j 12 1 2 12 12 M1 F r1 12 F21 M2 Coordinate r2 Origin Similarly M feels a force from M 2 1 GM M GM M 2 1 1 2 F = force on M from M = r =+ r = F 21 2 1 21 12 12 3 3 r r 21 12 where r = r r is the vector from 1 to 2. Wehave used r = r , and 21 2 1 21 12 jr j = jr j.Thus, the gravitational forces are equal and opposite. This is in 21 12 rd accordance with Newton's 3 law and ensures that the comp osite system of (M + M ) do es not suddenly start moving as a whole and violating Newton's 1 2 st 1 law. Notice also that the gravitational force at M from M is directed exactly 1 2 at M . It is a \central" force: F is parallel to r . This remark causes us 2 12 12 to digress and mention::: 60 rd 4.1.2 The little known co dicil to Newton's 3 law Many b o oks, e.g. Marion, state Newton's laws in something like this form: 1. A b o dy remains at rest or in uniform motion unless acted up on bya force. 2. A b o dy acted up on by a force moves in such a manner that the time rate of change of momentum equals the force. 3. If two b o dies exert forces on each other, these forces are equal in mag- nitude and opp osite in direction. This statement of 3 is WRONG! The correct statementis 0 3.Iftwo b o dies exert forces on each other, these forces are equal in mag- nitude and opp osite in direction AND ALONG THE SAME LINE. Newton knew this, but mo dern text b o ok writers have garbled it. B B OK NOT OK! A A (a) (b) If (b) were p ossible, then we could take the two particles A and B , nail one to the hub of a wheel, the other to the center, and watch as the wheel accelerated up to in nite angular velocity. 61 Thus, while one often hears ab out the \central force" nature of gravita- tion, this is actually a prop ertyofany action-at-a-distance, classical force law. You mightwonder ab out the magnetic force b etween twomoving par- ticles in electromagnetism. This do es not seem to b e a central force. The explanation is that the electromagnetic eld itself carries momentum and angular momentum. Classical electromagnetism cannot b e written as a pure action-at-a-distance theory. That is, there is more to Maxwell's equations than just the Coulomblaw. 4.1.3 Gravitational Potential The force on M , due to M 's presence can b e written in terms of the gradient 1 2 of a gravitational potential: F = force at M due to M 12 1 2 = force at p osition r due to mas at r 1 2 ! ! !! @ GM @ GM @ GM 2 2 2 = M ; ; 1 @x jr r j @y jr r j @z jr r j 1 1 2 1 1 2 1 1 2 = M r V (r ) 1 1 1 @ @ @ where r is the gradient op erator at M ; . V (r) is the gravi- ; ; 1 1 @x @y @z 1 1 1 tational p otential at r , 1 GM 2 V = : jr r j 1 2 The p otential at p oint r from a numb er of masses; at r ; r ; r ;:::;r is, in 1 2 3 n general N X GM i V = jr r j i i=1 where the sum excludes any mass exactly at r (do not include the in nite self-energy!). 62 4.2 The 2-b o dy problem Although every mass in the universe exerts a force on every other mass, there is a useful idealization where the mutual interaction of just two masses dominates their motion. The equations of motion are then GM M 1 2 M r = (r r ) 1 1 1 2 3 jr r j 1 2 GM M 1 2 M r = (r r ) : 2 2 2 1 3 jr r j 1 2 These apply, for example, to binary stars at large separations, but small enough that the tidal e ects of other stars in the galaxy can b e neglected. Or, the orbit of the Earth around the Sun, to the extent that the e ects of the Mo on and other planets can b e ignored. 4.2.1 Conservation laws for 2-b o dy orbits For a 2-b o dy system, wewant to solve for 6 functions of time r (t) and r (t), 1 2 and 12 constants will app ear in the solution (equivalentto2vectors of initial p ositions and 2 vectors of initial velo cities). It is most ecient to approach this problem by nding constants of the motion (to reduce the number of equations to solve) and bycho osing a convenient set of co ordinates in which to work. Add the two equations of motion given ab ove. Then, since the forces are equal and opp osite, the right hand side sums to zero, and we are left with M r + M r =0: 1 1 2 2 Let the center of mass of the system b e at R ; then by the de nition of center of mass as the mass-weighted average p osition, M r + M r 1 1 2 2 R = (M + M ) 1 2 63 we get (M + M )R =0 or R =0 : 1 2 Thus, the lo cation of the center of mass of the system is unaccelerated, since there is no external force on the system. It follows that the center of mass moves at a constantvelo city: dR _ R = v = constant dt R = R + v t: 0 0 This relation expresses the \Conservation of Linear Momentum." R and 0 V are the rst 6 \constants of motion." 0 The next logical thing to do is to change to center-of-mass co ordinates 0 0 r ; r , that is, to co ordinates relative to the center of mass. 1 2 M1 CoM ′ r1 r2′ M2 r 1 R r 2 O The relevant relationships are: 0 r = R + r 1 1 0 r = R + r 2 2 R = R + v t: 0 0 Substituting these into the equations of motion, we get GM M 1 2 0 0 0 M r = (r r ) 1 1 1 2 0 0 3 jr r j 1 2 64 GM M 1 2 0 0 0 M r = (r r ) : 2 2 2 1 0 0 3 jr r j 1 2 Note that the equations preserve exactly the form they had b efore: We could have just written down the answer without calculation. We can arbitrarily cho ose co ordinates relative to the CoM, b ecause the CoM frame is a go o d inertial frame.Anychange of origin (R ) and/or change of velo city(V )is 0 0 allowed by (so-called) \Gallilean invariance." So let us cho ose to do this: co ordinates are, henceforth, measured relative to the CoM; and we drop the primes. In CoM system, GM M 1 2 (r r ) M r = 1 2 1 1 3 jr r j 1 2 GM M 1 2 M r = (r r ) 2 2 2 1 3 jr r j 1 2 with M r + M r =0 by de nition. 1 1 2 2 We can now construct further integrals of the motion. Consider the total _ _ angular momentum ab out the CoM: L = M r r + M r r . Then 1 1 1 2 2 2 dL _ _ _ _ = M (r r + r r )+M (r r +r r ) 1 1 1 1 1 2 2 2 2 2 dt = M r r +M r r : 1 1 1 2 2 2 Substitute the equations of motion for M r and M r , then 1 1 2 2 GM M dL 1 2 fr (r r )+r (r r )g = 1 1 2 2 2 1 3 dt jr r j 1 2 GM M 1 2 = (r r + r r ) 1 2 2 1 3 jr r j 1 2 = 0 i.e. the total angular momentum of the system is conserved: L = constant. 65 That is, the angular momentum vector is xed in direction and magnitude (b ecause no external torques are acting). L M1 r r 2 1 r1 CoM r 2 M 2 The angular momentum provides 3 more integrals of the motion, so we are nowupto9. We can also lo ok at the total energy of the system: E = (kinetic energy) + (p otential energy) ! 1 GM M 1 1 2 _ _ _ _ M r r + M r r + : = 1 1 1 2 2 2 2 2 jr r j 1 2 To see that this is conserved, write GM M d dE 1 2 _ _ = M r r + M r r + jr r j : 1 1 1 2 2 2 1 2 2 dt jr r j dt 1 2 Now d d 1=2 jr r j = (r r + r r 2r r ) 1 2 1 1 2 2 1 2 dt dt _ _ _ _ _ _ r r + r r r r r r (r r )(r r ) 1 1 2 2 1 2 1 2 1 2 1 2 = = : jr r j jr r j 1 2 1 2 Therefore ! ! dE GM M GM M 1 2 1 2 _ _ (r r ) + r (r r ) = r M M r + r + 1 2 2 2 1 1 2 1 2 1 3 3 dt jr r j jr r j 1 2 1 2 = 0 using the equations of motion.
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