Sect. 3.8: Motion in Time, Kepler Problem We’ve seen: Orbital eqtn for r -2 force law is fairly straightforward. Not so, if want r(t) & θ(t). Earlier: Formal solution to Central Force problem. Requires evaluation of 2 integrals, which will give r(t) & θ(t) : (Given V(r) can do them, in principle.) t(r) = ∫dr({2/m}[E - V(r)] - [ 2  (m 2 r 2 )]) -½ (1) –Limits r 0  r, r 0 determined by initial condition –Invert this to get r(t) & use that in θ(t) (below) θ(t) = ( /m)∫(dt/[r 2 (t)]) + θ 0 (2) –Limits 0  t, θ 0 determined by initial condition Need 4 integration constants:E,, r 0, θ 0 Most cases: (1), (2) can’t be done except numerically

Look at (1) for 1/r 2 force law: V(r) = -k/r. t(r) = (m/2) ½ ∫dr[E + (k/r) - { 2  (m 2 r 2 )}] -½ (1´) –Limits r 0  r, r 0 determined by initial condition For θ(t), instead of applying (2) directly, go back to conservation of angular momentum: = mr 2 θ = constant  dθ = ( /mr 2 )dt or dt = (mr 2 / )dθ (3) Put orbit eqtn results r(θ) into (3) & integrate: –We had [α/r(θ)] = 1 + e cos(θ - θ´) –With e = [ 1 + {2E 2  (mk 2 )}] ½ –And 2α = [2 2  (mk)]

 (3) becomes: t(θ) = ( 3 /mk 2 ) ∫dθ[1 + e cos(θ - θ´)] -2 (4) –Limits θ 0  θ We had: t(r) = (m/2) ½ ∫dr[E + (k/r) - { 2  (m 2 r 2 )}] -½ (1´) (1´) & (4) are not difficult integrals. Can express them in terms of elementary functions (tabulated!). However, they are complicated. Also, inverting to give r(t) and θ(t) is non-trivial. This is especially true if one wants high precision, as is needed for comparison to astronomical observations!

Time: Parabolic Orbit Even though, of course, we’re primarily interested in elliptic orbits, its instructive to first evaluate (4) for a parabolic orbit:  e = 1, E = 0 –From table of planetary properties from earlier: Halley’s Comet, e =  1  Orbit is  parabolic. Results we are about to get are  valid for it. In this case, (4) becomes: t(θ) = ( 3 /mk 2 )∫dθ[1 + cos(θ - θ´)] -2 (4´) –Limits θ 0  θ –Measure θ from distance of closest approach (perihelion). That is θ = 0 at r = r min  θ 0 = θ´ = 0

So, we want to evaluate: t(θ) = ( 3 /mk 2 )∫dθ[1 + cosθ] -2 (5) (Limits 0  θ) –Use trig identity: 1 + cosθ = 2cos 2 (½θ) So: t(θ) = [( 3 )/(4mk 2 )]∫d θ sec 4 (½θ) (5´) (Limits 0  θ) –Change variables to x = tan(½θ) Gives: t(θ) = [( 3 )/(2mk 2 )]∫dx (1+x 2 ) (5´´) (Limits 0  tan (½θ))  t(θ) = [( 3 )/(2mk 2 )][tan(½θ) + (⅓)tan 3 (½θ)] (6)

t(θ) = [( 3 )/(2mk 2 )][tan(½θ) + (⅓)tan 3 (½θ)] (6) (-π < θ < π) To understand what the particle is doing at different times, look at orbit eqtn at the same time as (6): [α/r(θ)] = 1 + cosθ (7)  At t  -  (θ = - π), particle approaches from r   At t = 0 ( θ = 0), particle is at perihelion r = r min At t  +  (θ = + π), particle again approaches r   (6) gives t = t(θ). To invert and get θ = θ(t): –Treat (6) as cubic eqtn in tan(½θ). Solve & compute arctan or tan -1 of result.  θ = θ(t) To get r(t), substitute resulting θ = θ(t) into (7): [α /r(t)]= [α /r{θ(t)}] = 1 + cos[θ(t)] –Same if do integral & invert to get r(t) (E=0: parabola) t(r) = (m/2) ½ ∫dr[(k/r) - { 2  (m 2 r 2 )}] -½

Time: Elliptic Orbit Elliptic orbit: [α/r(θ)] = 1 + e cosθ (θ´ = 0) (1) –With e = [1 + {2E 2  (mk 2 )}] ½ –And 2α = [2 2  (mk)] Rewrite (2): r = [a(1- e 2 )]/[1 + e cosθ] (2) a  Semimajor axis  (α)/[1 - e 2 ] = (k)/(2|E|) –Its convenient to define an auxiliary angle: ψ  Eccentric Anomaly (elliptic orbits only!)  By definition: r  a(1 - e cosψ) (2´) –(2) & (2´)  cosψ = (e + cosθ)/(1 + e cosθ) cosθ = (cos ψ -e)/(1- e cosψ) –ψ goes between 0 & 2π as θ goes between -π & π –Perihelion, r min occurs at ψ = θ = 0. –Aphelion, r max occurs at ψ = θ = π.

Back to time for the elliptic orbit: t(r) = (m/2) ½ ∫dr[E + (k/r) - { 2  (m 2 r 2 )}] -½ (3) –Limits r 0  r, r 0  r min = perihelion distance –Rewrite (3) using eccentricity e  [1 + {2E 2  (mk 2 )}] ½ a  Semimajor axis  (α)/[1 - e 2 ] = (k)/(2|E|), α  [ 2  (mk)]  (3) becomes: t(r) = -(m/2k) ½ ∫rdr[r - (r 2 /2a) – (½)a(1-e 2 ) ] -½ (3´) By definition: r  a(1 - e cosψ) (2´) Change integration variables from r to ψ  (3´) is: t(ψ) = (ma 3 /k) ½ ∫dψ (1 - e cosψ) (4) Limits 0  ψ Given t(ψ), combine with (2´) to get t(r). Invert to get r(t).

t(ψ) = (ma 3 /k) ½ ∫dψ (1 - e cosψ) (4) Limits 0  ψ r  a(1 - e cosψ) or cosψ = (1-r/a)/e (2´)  ψ = cos -1 [(1-r/a)/e] (excludes e = 0!) Convenient to define ω  (k/ma 3 ) ½ –Will show ω  frequency of revolution in orbit.  ωt(ψ) = ∫dψ (1 - e cosψ) (Limits 0  ψ) Integrates easily to ωt(ψ) = ψ - e sinψ (4´) Note that: sin ψ = [1 - cos 2 ψ] ½ Combining, (4´) becomes: ωt(r) = cos -1 [(1-r/a)/e] + [1 - {(1-r/a)/e} 2 ] ½ (4´´) –Inverting this to get r(t) can only be done numerically!

Time: Orbit Period Back briefly ωt(ψ) = ∫dψ (1 - e cosψ) ω  (k/ma 3 ) ½ –If integrate over full range, ψ = 0 to 2π  t  τ = period of orbit. This gives, τ = 2π(m/k) ½ a 3/2  2π/ω τ 2 = [(4π 2 m)/(k)] a 3 Same as earlier, of course! Kepler’s 3 rd Law! Clearly, ω  frequency of revolution in orbit: ω  2π/τ

Elliptic Orbit Back to general problem: ωt(ψ) = ψ - e sinψ (4´) ω  (k/ma 3 ) ½ (4´)  Kepler’s Equation Recall also: r  a(1 - e cosψ)  [a(1- e 2 )]/[1 + e cosθ] Terminology (left over from medieval astronomy): ψ  “eccentric anomaly”. Medieval astronomers expected angular motion of planets to be constant (indep of time). That is, they expected circular orbits (r =a & e = 0 above).  Deviations from a circle were termed “anomalous”! For similar reasons θ  “true anomaly”. Still use these terms today. From earlier table, eccentricities e for MOST planets are very small! Except for Mercury (e = ) & Pluto (e = ) all planet’s have e < 0.1. Several have e < Earth (e = 0.017), Venus (e = 0.007), Neptune (e = 0.010)  Orbits are  circles & “anomalies” are small!

Summary: Motion in time for elliptic orbits: ωt(ψ) = ψ - e sinψ, ω  (k/ma 3 ) ½ Kepler’s Equation Also: r  a(1 - e cosψ)  [a(1- e 2 )]/[1 + e cosθ] Combining (e  0): ωt(r) = cos -1 [(1-r/a)/e] + [1 - {(1-r/a)/e} 2 ] ½ –Inverting this to get r(t) can only be done numerically! Comparing 2 eqtns for r gives: cosψ = (e + cosθ)/(1 + e cosθ) or cosθ = (cosψ -e)/(1-e cosψ) Using some trig identities, this converts to: tan(½θ) = [(1-e)/(1+e)] ½ tan(½ψ) Use this to get θ once ψ is known.

tan(½θ) = [(1-e)/(1+e)] ½ tan(½ψ) Use this to get θ once ψ is known. Solving this & Kepler’s Equation ωt(ψ) = ψ - e sinψ, ω  (k/ma 3 ) ½ to get ψ(t), θ(t) & r(t): A classic problem, first posed by Kepler. Many famous mathematicians worked on it, including Newton. To study motion of bodies in solar system & to understand observations on such bodies one needs this solution to very high accuracy! Goldstein: “ The need to solve Kepler’s equation to accuracies of a second of arc over the whole range of eccentricities fathered many developments in numerical mathematics in the eighteenth and nineteenth centuries.” More than 100 methods of solution have been developed! Some are in problems for the chapter! (# 2,3,25,27)

Sect. 3.9: Laplace-Runge-Lenz Vector Conserved Quantities (1 st integrals of motion) in the Central Force Problem (& so in Kepler r -2 force problem): Total mechanical energy: E = (½)m(r 2 + r 2 θ 2 ) + V(r) = const = (½)mr 2 + [ 2  (2mr 2 )] + V(r) Total angular momentum: L = r  p = const (magnitude & direction!)  3 components or 2 components + magnitude: Constant magnitude   p θ  mr 2 θ = const. Can show: There is also another conserved vector quantity  Laplace-Runge-Lenz Vector, A

Newton’s 2 nd Law for a central force: (dp/dt) = p = f(r)(r/r) (1) Cross product of p with angular momentum L: p  L = p  (r  p) = p  [r  (mr)] (1)  p  L = [mf(r)/r][r  (r  r)] Use a  (b  c) = b(a  c) - c(a  b)  p  L = [mf(r)/r][r(r  r) - r 2 r] Note that r  r = rr. L = const  p  L = d(p  L)/dt Combining these gives: d(p  L)/dt = - [mf(r)r 2 ][(r/r) - (rr/r 2 )] Or: d(p  L)/dt  - [mf(r)r 2 ][d(r/r)/dt] (2) Valid for a general central force!

General central force: d(p  L)/dt = - [mf(r)r 2 ][d(r/r)/dt] (2) Look at (2) in case of r -2 force: f(r) = -(k/r 2 )  - mf(r)r 2 = mk For r -2 forces, (2) becomes: d(p  L)/dt = [d(mkr/r)/dt] Or: d[(p  L) - (mkr/r)]/dt] = 0 (2´) Define Laplace-Runge-Lenz Vector A A  (p  L) - (mkr/r) (3) (2´)  (dA/dt) = 0 or A = constant (conserved!)

Summary: For r -2 Central Forces (Kepler problem) the Laplace-Runge-Lenz Vector A  (p  L) - (mkr/r) (3) is conserved (a constant, a 1 st integral of the motion). Question: That A is conserved is all well and good, but PHYSICALLY what is A? What follows is more of a GEOMETRIC interpretation than a PHYSICAL interpretation. –By relating A to elliptic orbit geometry, perhaps the physics in it can be inferred.

A  (p  L) - (mkr/r) Definition  A  L = 0 (L is  to p  L; r is  L = r  p)  A = fixed (direction & magnitude) in the orbit plane.

A  (p  L) - (mkr/r) A  L = 0 A = fixed in orbit plane θ  angle between r & fixed A direction  A  r = Ar cos θ = r  (p  L) - mkr Identity: r  (p  L) = L  (r  p) = L  L  2  Ar cosθ = 2 - mkr Or: (1/r) = (mk/ 2 ) [1 + (A/mk)cosθ] (1) Identify θ as orbital angle: A is in perihelion direction (θ = 0, A || r min ) see diagram. (1)  Another way to derive that, for the Kepler Problem, orbit eqtn is a conic section!

The Laplace-Runge-Lenz Vector A = (p  L) - (mkr/r)  (1/r) = (mk/ 2 ) [1 + (A/mk) cosθ] (1) (1)  The orbit eqtn is a conic section. Also, A is in the perihelion direction. Earlier, we wrote: (α/r) = 1 + e cosθ (2) α  [ 2  (mk)]; e  [ 1 + {2E 2  (mk 2 )}] ½ Comparison of (1) & (2) gives relation between A and the eccentricity e (& thus between A, energy E, & angular momentum ): A  mke = mk[ 1 + {2E 2  (mk 2 )}] ½

Physical interpretation of Laplace-Runge-Lenz Vector, A = (p  L) - (mkr/r) Direction of A is the same as the perihelion direction: (A || r min ). Magnitude of A: A  mke = mk[ 1 + {2E 2  (mk 2 )}] ½ (3) For the Kepler problem, we’ve found 7 conserved quantities: 3 components of vector angular momentum, L 3 components of Laplace-Runge-Lenz Vector, A 1 scalar energy E

A  mke = mk[ 1 + {2E 2  (mk 2 )}] ½ (3) 7 conserved quantities: –3 components of L, 3 components of A, energy E Recall original problem: 2 masses, 3 dimensions  6 degrees of freedom  6 independent constants of motion.  The 7 quantities aren’t independent. Reln between them is (3):  Reducing the number of independent ones to 6. All 7 also are functions of r & p which describe the orbit in space. None relate to the initial conditions of the orbit (r(t=0)). Mathematically, one const of motion must contain such initial condition info. There must be a const (say time when r = r min ) indep of the 7 listed above  The 6 consts resulting from using (3) on the 7 cannot all be indep either!  There must be one more reln between them. This is supplied by orthogonality of A & L: A  L = 0

 For Kepler (r -2 force) problem, we have 5 indep consts of motion containing orbital info (+ one containing initial condition info). Usually choose these as: 3 components of angular momentum L, energy E, and magnitude of Laplace- Runge-Lenz Vector, A Question: Is there a similar conserved quantity to A for the general central force problem (or for specific central forces which are not r -2 forces)? –Answer: Yes, sometimes, but these usually have no simple interpretation physically. –Can show: such a quantity exists only for force laws which lead to closed orbits.  Also: the existence of such a quantity is another means to show that the orbit is closed. Bertrand’s theorem: Happens for power laws forces only for f  r -2 & f  r (Hooke’s “Law”).