Phys 325 Lecture 9 Tuesday 12 February, 2019 NB Midterm next Thursday 2/21. Sample exam will be posted soon Formula sheet on sample exam will be same as formula sheet on actual exam You will be permitted one side of one page of hand written notes that you prepared for yourself. As described alst time, for satellite motion around a fixed mass M, or for the relative position between two masses that sum to M, we have three coupled ODE's, one for each component of F = ma.!! r r θ! 2 r!φ 2 sinθ = GM / r 2 r!! θ + 2!r θ! r!φ 2 sinθ cosθ = 0 r!! φ sinθ + 2r θ!!φ cosθ + 2!r!φ sinθ = 0 These equations are respectively the r, q and f components of F=ma, and describe either i) the position r of a mass in orbit around a fixed mass M, or ii) the vector r = r 1-r 2 connecting two masses with total mass M. The second of these equations ( the q component) is greatly simplified if we orient our coordinate system such that, at time zero, dq/dt = 0 and q = p/2. Then d 2 q/dt 2 = 0 and we may conclude that q = p/2 is a solution for all subsequent time. If the orbit starts in the equatorial plane of the coordinate system, (and we may choose that plane so that it does) then it stays there. This is the same conclusion we came to by recognizing that the direction of the angular momentum vector is a constant. q = p/2 then simplifies the third of these eqns (the f component), which becomes r!! φ + 2!r!φ = 0 It may be rewritten as 1 d r dt (r 2!φ) = 0 which implies that r 2!φ is a constant r 2!φ = l (! is interpreted as angular momentum per mass) The expression for! may be used to substitute for df/dt in the first of the above ODE's. We also recognize that q = p/2 = constant. Then the first equation becomes!! r l 2 / r 3 = GM / r 2 An uncoupled nonlinear equation for r(t). 77
It is of the form discussed in case (d) in Lecture #2, of force in 1-d as a function of x. procedure suggested there was to multiply by dr/dt to get The!r!! r = (l 2 / r 3 GM / r 2 )!r We then recognize that!! r!r = d 1 dt!r 2 2 And we would also recognize that ( GM / r 2 + l 2 / r 3 )"r = d dt U eff where the effective potential (per mass) is defined by U eff = GM / r + l 2 / 2r 2 Therefore d 1 2 dt!r 2 = d dt U eff We conclude that there is a conserved quantity e e = constant = 1 2!r 2 +U eff = 1 2!r 2 + l 2 / 2r 2 GM / r = 1 2!r 2 + 1 2 r 2!φ 2 GM / r which may be interpreted as energy per mass (of the satellite) The above expression may be solved for dr/dt to get!r = 2(e U eff (r)) This indicates the radial part of the velocity is a unique (within a sign ±) function of r. In particular it indicates the turning points at which dr/dt = 0. rmax and rmin The effective U is plotted here, for typical values of GM and! 2. We can immediately see that, as long as e > 0, all initial states escape to. Some do so with one "bounce" off a minimum, some go directly. The above may be integrated to give time as a function of r: 78
t = dr 2(e U eff (r)) Which is probably not analytically doable. If it were done and inverted for r(t), then using the value of! we can get f from knowing df/dt =!/ r 2. dr If e < 0, we can conclude that r(t) must be periodic period P = 2. Once r(t) is 2(e U eff (r)) found (and using the value of!) we can get f from knowing df/dt =!/ r 2. There is no reason to know, yet, that f is also periodic with the same period i.e the orbits are not yet shown to be closed. All we can conclude so far is that, if e < 0, df/dt is periodic, with the same period. r max r min Eqn for r(f) (instead of r(t)) The above analysis has told us that, for parameters"! 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 f on time is not clear, but might be obtained with some work by evaluating the above integral for t(r). What is not revealed by that is the shape of the orbit. For that we need r(f), which is obtained with some tricks and a little calculus. A clever change of variables will give us a linear equation for the f dependence of r. We define u = 1/r; r= 1/u We will seek, not u(t), but rather u(f). To get a differential eqn for u(f) we will need to use!! r l 2 / r 3 = GM / r 2 and change independent variables f for t, and dependent variable u for r. Notice that: dr/dt = -(du/dt)/u 2 = -(du/df)(df/dt)/u 2 = - (du/df) r 2 (df/dt) = -! (du/df) and d 2 r/dt 2 = -! d/dt (du/df) = -! (df/dt) (d 2 u/df 2 ) = - (! 2 /r 2 ) (d 2 u/df 2 ) = - (! 2 u 2 ) (d 2 u/df 2 ) so that d 2 r/dt 2 -! 2 /r 3 + G M / r 2 = 0 becomes... - (! 2 u 2 ) (d 2 u/df 2 ) -! 2 u 3 + G M u 2 = 0 On dividing by u 2 and! 2, it becomes d 2 u/df 2 + u = GM/! 2 79
which is an ordinary differential equation for u(f). Time t has been eliminated. This is a linear ODE (of a sort that we will see lots of later) It being linear and constant coefficient allows us to solve it in general. The general solution is u(f) = D cos(f-fo) + GM/! 2 for constants of integration D and fo. ( Check it!) Thus r(f) = 1/ [D cos(f-fo) + GM/! 2 ] This is the shape of the trajectory in polar coordinates. We now can see that r is a periodic function of f, with period 2p. The orbits are therefore closed paths ( at least if D is small enough that the denominator remains positive.) This was not obvious from 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 fo of the f measurement such that r takes its minimum value at f = 0, and D > 0 We then write r(f) = 1/ [D cos(f) + GM /! 2 ] But how is the constant of integration D related to the e and! that we have been using to characterize the orbits? We can identify the constant of integration D in terms of the energy e by observing from the above r(f) at f = 0 that rmin = 1/( D + GM/! 2 ) at which point dr/dt is of course zero ( because r is minimum ) We use that e = (1/2) r 2 (df/dt) 2 + (1/2) (dr/dt) 2 - GM/r (we then drop the middle term and set r = rmin) and recall that df/dt is!/ r 2. Then after bit of algebra(*) we conclude. D = [ (GM/! 2 ) 2 + 2e /! 2 ] 1/2 (*) e = (1/2) r 2 (df/dt) 2 + (1/2) (dr/dt) 2 -GM/r =! 2 /(2r 2 ) GM/r = (! 2 /2)( D + GM/! 2 ) 2 GM( D + GM/! 2 ) which may be solved for D Our expression for r(f) is then rewritable in terms of the physical parameters! and e: 80
r(f) = (! 2 /GM ) [ 1 + { 1 + (2e! 2 / G 2 M 2 ) } 1/2 cos(f) ] -1 = a / [ 1 + e cos f ] where e is the eccentricity and a is the latus rectum This serves as a definition of the orbital parameters a and e in terms of the physical parameters e and!. a =! 2 /GM ; e = [ 1 + (2e! 2 / G 2 M 2 ) ] 1/2 (N.B. don't confuse energy e and eccentricity e!! ) The size and shape of every orbit is specified in terms of the geometric parameters a and e, ( or if you prefer, the physical parameters! and e ) It is easy to see that rmin occurs at f = 0 where rmin = a/(1+e) If e < 1, rmax occurs at f = p where rmax = a/(1-e) ( if e 1, then r max is See below.) Also notice 1/rmax + 1/ rmin = 2/a The latus rectum is the harmonic mean of rmin and rmax. At f = p/2, we see that r = a. Special cases Consider the case (c) e = -G 2 M 2 / 2! 2, then r(f) = a = (! 2 /GM ) = constant. This is a circular orbit. For specified!, e cannot be less than this. This case has eccentricity e = 0 ------ Consider the case (a) e = 0, then e = 1 r(f) = (! 2 /GM ) / [ 1 + cosf ] this is a parabola ( as can be seen if you change to Cartesian coordinates ) It has its closest approach to the origin at f = 0. As f p, r. e = 0 implies it has zero speed at r=. 81
----- Consider the case (b) e > 0. This case has e > 1. The curve is a hyperbola. The mass µ has a non-zero velocity, even at r =. For this case there is a critical angle f ( f is measured from the +x axis) at which r goes to fcrit = b + p/2. (where b is defined in the figure.) f crit is less than p and greater than p/2. It is the direction along which r. At large r, the particle is moving along an asymptotic straight line trajectory parallel to the dotted line in the figure. From the formula for r(f), we see that this happens at f = fcrit = arccos(-{1+(2e! 2 /G 2 M 2 ) } -1/2 ) = arccos(-1/e) The angle -f crit is the direction along which the trajectory comes from. It then departs to in direction +f crit. This describes how an object comes from, swings round the central point, and then goes out in a new direction, finally having changed its direction of motion by a net angle 2b. This is the kind of trajectory followed by objects falling towards the sun from outside the solar system. Like our recent visitor Oumuamua, or some comets. If you know e>0 and!, then you know e and you know this critical angle, and then you know how much a comet's trajectory is diverted as it passes near the sun. cf HW4A.2 Finally, Consider the case (d) 0 > e > - G 2 M 2 / 2! 2, this has 0 < e < 1 and gives an ellipse. r min is the perigee and is at r min = (! 2 /GM ) [ 1 +{ 1 + (2e! 2 /G 2 M 2 ) } 1/2 ] -1 = a/(1+e) This is achieved at f = 0 r max is the apogee and is at r max = (! 2 /GM ) [ 1 - { 1 + (2e! 2 /G 2 M 2 ) } 1/2 ] -1 = a/(1-e) This is achieved at f = p. It is conventional to define a = a/[1-e 2 ] = GM/2 e is the semi-major axis ( the larger of the two radii from the center of the ellipse) a = (rmax + rmin ) / 2 82
b = a/[1-e 2 ] 1/2 =! / [ 2 e ] 1/2 is the semi minor axis These are the horizontal and vertical radii from the center of the above ellipse. See figure. =================== Why do we know these are conic sections? (ellipses, hyperbolas etc?) A little algebra starting with the formula in polar coordinates can convert it to the more familiar Cartesian coordinates: α r = 1+ ε cosφ r(1+ ε cosφ) = α x 2 + y 2 + εx = α x 2 + y 2 = α εx x 2 + y 2 = α 2 2αεx + ε 2 x 2 x 2 (1 ε 2 ) + 2αεx + y 2 = α 2 which, being of the class of arbitrary quadratic forms + linear forms = constant, is a conic section. In particlar, consider e=0, so that the above becomes x 2 +y 2 = a 2, a circle. Or consider e=1, so that the above becomes y 2 +2ax=a 2. Obviously a parabola. ======== What about Kepler's laws that he inferred from observations of the planets' motions? We've confirmed Kepler's first law about ellipses. What about Kepler's other two laws? ( equal areas in equal times? Period ~ semimajor axis 3/2? ) Kepler's second law is Equal areas in equal times. We can show that it is merely conservation of angular momentum: Clearly da = (1/2) r 2 df. So da/dt = (1/2) r 2 df/dt =!/2 Area is swept out at a rate!/2, constant. We see that this Kepler law is merely a consequence of conservation of angular momentum. It would apply regardless of whether or not the force law is inverse square. 83
What about Kepler's third law? The period is the amount of time to sweep out the full area pab of the ellipse. P= πab l / 2 where a and b are the semi axes of the ellipse. Using and a = GM/2 e b =! / [ 2 e ] 1/2 ( so b 2 = a! 2 / GM ) permits us to write P 2 = 4π 2 a 2 b 2 = 4π 2 a 3 l 2 GM which is Kepler's 3d law. A planet's year P is proportional to 3/2 power of its semi-major axis. Lectures have now covered all material pertaining to HW4, except for HW4B.3 started on HW4. Next time tides. Please get Also - Office hours this week have been moved, see email. The sample exam will soon be available on the course website. The exam is next Thursday in class 2/21.. see schedule page for details on coverage Exam is closed book you may bring a one page sheet of notes with which to supplement the formulas that will be given (and are in the sample exam). Material covered is Lectures 1-10. 84
Phys 325 Miniquiz 9 12 February 2019 name A Keplerian orbit of an satellite is given, in polar coordinates centered on the earth, by α r(φ) = 1+ ε cos(φ) where e and a are the orbit's eccentricity and latus rectum. As a function of e and a, at what angle f and what distance r is its closest approach to the earth? Answer: Minimum r is at the for which the denominator is maximum, i.e. f = 0. This corresponds to r = a/(1+e). In terms of e and a, at what value of y does the orbit cross the y-axis? This is where f = p/2, so r = y = a For e = 1/2 and a = 2, what is r max? Sketch the trajectory. Answer. We know minimum r is at f = 0, r = a/(1+e) = 4/3 We know that at f = ±p/2, r = a = 2 We see that at f = p, r = a/(1-e) = 4 ( This is r max.) These points have been marked with asterisks in the plot and then connected smoothly. This is an ellipse. For e = 1.5 and a = 2, at what angle does r go to? Sketch the trajectory ( Be careful, the formula for r says it is at some critical angle and negative at others ) Answer. We know minimum r is at f = 0, where r = a/(1+e) = 4/5 We know that at f = ±p/2, r = a = 2 These points have been marked with asterisks in the plot and then connected smoothly. We see that at f crit = arccos(-1/e) = ±132 degree where r = This is the direction along which r goes to. The orbit is a hyperbola. If you use the formula to ask what r is when f=p, you get 4. This is unphysical. The satellite never gets to f=p. It asymptotes at r=, f = f crit. 85