77Phys 325 Lecture 9 Tuesday 17 February, 2015Thus we have three coupled ODE's, one for eachcomponent of F = ma. &&r − r&θ2− r&φ2sinθ= −GM / r2r&&θ+ 2&r&θ− r&φ2sinθcosθ= 0r&&φsinθ+ 2r&θ&φcosθ+ 2&r&φsinθ= 0These equations are respectively the r , θ and φcomponents of F=ma, and describe either i) theposition r of a test mass in orbit around a fixed massM, or ii) the vector r = r1-r2 connecting two masses with total mass M.The second of these equations ( the θ component) is greatly simplified if we orient our coordinatesystem such that, at time zero, dθ/dt = 0 and θ = π/2. Then d2θ/dt2 = 0 and we may conclude thatθ = π/2 is a solution for all subsequent time. If the orbit starts in the equatorial plane of thecoordinate system, (and we may choose that plane so that it does) then it stays there. This is thesame conclusion we came to by recognizing that the direction of the angular momentum vector is aconstant.θ = π/2 then simplifies the third of these eqns (the φ component) , which becomes r&&φ+ 2&r&φ= 0It may be rewritten as 1rddt(r2&φ) = 0which implies that r2&φ= lis a constant ( l is interpreted as angular momentum per mass)The expression for l may be used to substitute for dφ/dt in the first of the above ODE's. We alsorecognize that θ = π/2 = constant. Then the first equation becomes &&r − l2/ r3= −GM / r2An uncoupled nonlinear equation for r(t).It is of the form discussed in case (d) in Lecture #2, of force in 1-d as a function of x. Theprocedure suggested there was to multiply by dr/dt to get78 &r&&r = (l2/ r3− GM / r2)&rWe then recognize that &&r&r =ddt12&r2And we would also recognize that (−GM / r2+ l2/ r3)&r = −ddtUeffwhere the effective potential (per mass) is defined by Ueff= −GM / r + l2/ 2r2Therefore ddt12&r2= −ddtUeffWe conclude that there is a conserved quantity e e =constant=12&r2+ Ueff=12&r2+ l2/ 2r2+ Ueff=12&r2+12r2&φ2− GM / rwhich may be interpreted as energy per massThe above expression may be solved for dr/dt toget &r = 2(e − Ueff(r))This indicates the radial part of the velocity is aunique (within a sign ±) function of r. Inparticular it indicates the turning points at whichdr/dt = 0. rmax and rmin The effective U is plotted here, for typical valuesof GM and l2.We can immediately see that, as long as e > 0, allinitial states escape to ∞. Some do so with one"bounce" off a minimum, some go directly.If on the other hand, e < 0, r(t) must be some periodic function of time. That function of time isimplicitly given by79t =dr2(e − Ueff(r))∫We conclude that, for e < 0, r(t) is periodic with some period P = 2dr2(e − Ueff(r))r minr max∫. Oncer(t) is found (and using the value of l) we can get φ from knowing dφ/dt = l/ r2. There is noreason to know, yet, that φ is also periodic with the same period…i.e the orbits are not yet shownto be closed. All we can conclude so far is that dφ/dt is periodic, with the same period.Eqn for r(φ) (instead of r(t)The above analysis has told us that, for parameters l≠0, and e< 0, the radial motion r(t) is periodic.And for e ≥ 0 the motion is unbounded. The precise functional dependence of r and φ on time isnot clear, but could be obtained with some work by evaluating the above integral for t(r). What isnot revealed is the shape of the orbit. For that we need r(φ), which is obtained with some tricksand a little calculus….A clever change of variables will give us a linear equation for the φ dependence of r. We defineu = 1/r; r= 1/uWe will seek, not u(t), but rather u(φ). To get a differential eqn for u(φ) we will need to use &&r − l2/ r3= −GM / r2 and change independent variables φ for t, and dependent variable u for r.Notice that: dr/dt = -(du/dt)/u2 = -(du/dφ)(dφ/dt)/u2 = - (du/dφ) r2 (dφ/dt) = -l (du/dφ)and d2r/dt2 = - l d/dt (du/dφ) = - l (dφ/dt) (d2u/dφ2) = - ( l2 /r2) (d2u/dφ2)= - ( l2 u2) (d2u/dφ2)so that d2r/dt2 − l2/r3 + G M / r2 = 0 becomes . . .- ( l2 u2 ) (d2u/dφ2) − l2u3 + G M u2 = 0On dividing by u2 and l2, it becomesd2u/dφ2 + u = GM/ l2 which is an ordinary differential equation for u(φ). Time t has been eliminated. This is a linearODE (of a sort that we will see lots of later) It being linear allows us to solve it in general. Thegeneral solution isu(φ) = D cos(φ−φo) + GM/ l280for constants of integration D and φο. ( Check it!) Thusr(φ) = 1/ [D cos(φ−φo) + GM/ l2 ]This is the shape of the trajectory.We now can see that r is a periodic function of φ, with period 2π. The orbits are therefore closedpaths ( at least if D is small enough that the denominator remains positive.) This was not obviousfrom our discussion above following the introduction of the effective potential U.With a further specification of the orientation of the coordinate system, we can choose the origin ofthe φ measurement such that r takes its minimum value at φ = 0, and D > 0 We then writer(φ) = 1/ [D cos(φ) + GM / l 2 ]But how is the constant of integration D related to the e and l that we have been using tocharacterize the orbits?We can identify the constant of integration D in terms of the energy e by observing thatrmin = 1/( D + GM/ l 2 )at which point dr/dt is of course zero ( because r is minimum )We use that e = (1/2) r2(dφ/dt)2 + (1/2) (dr/dt)2 - GM/r (we then drop the middle term andset r = rmin) and recall that dφ/dt is l/ r2. Then after bit of algebra(*) we conclude.D = [ (GM/ l2 )2 + 2e / l2 ]1/2Our expression for r(φ) is then rewritable in terms of the physical parameters l and e:r(φ) = ( l2/GM ) [ 1 + { 1 + (2e l 2/ G2M2) }1/2 cos(φ) ]-1= α / [ 1 + ε cos φ ] where ε is the eccentricity and α is the latus rectum(*) e = (1/2) r2(dφ/dt)2 + (1/2) (dr/dt)2 -GM/r= l2/(2r2) – GM/r= (l2/2)( D + GM/ l 2 )2 – GM( D + GM/ l 2 )which may be solved for D81This serves as a definition of the orbital parameters α and ε in terms of the physical parameters eand l. α = l2/GM ; ε = [ 1 + (2e l 2/ G2M2) ] 1/2(N.B. don't confuse energy e and eccentricity ε)The size and shape of every orbit is specified in terms of the geometric parameters α and ε, ( or ifyou prefer, the physical parameters l and e )It is easy to see that rmin occurs at φ = 0 where rmin = α/(1+ε)If ε < 1, rmax occurs at φ = π where rmax = α/(1−ε)( if ε ≥ 1, then rmax is ∞ )Also notice 1/rmax + 1/ rmin =