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(m2r2)])-½ (1)– Limits r0 r, r0 determined by initial condition– Invert this to get r(t) & use that in θ(t) (below)

θ(t) = (/m)∫(dt/[r2(t)]) + θ0 (2)– Limits 0 t, θ0 determined by initial condition

• Need 4 integration constants: E, , r0, θ0

• Most cases: (1), (2) can’t be done except numerically

• Look at (1) for 1/r2 force law: V(r) = -k/r. t(r) = (m/2)½∫dr[E + (k/r) - {2(m2r2)}]-½ (1´)– Limits r0 r, r0 determined by initial condition

• For θ(t), instead of applying (2) directly, go back to conservation of angular momentum: = mr2θ = constant

dθ = (/mr2)dt or dt = (mr2/)dθ (3)• Put orbit eqtn results r(θ) into (3) & integrate:

– We had [α/r(θ)] = 1 + e cos(θ - θ´) – With e = [ 1 + {2E2(mk2)}]½ – And 2α = [22(mk)]

(3) becomes: t(θ) = (3/mk2) ∫dθ[1 + e cos(θ - θ´)]-2 (4)– Limits θ0 θ

• We had: t(r) = (m/2)½∫dr[E + (k/r) - {2(m2r2)}]-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 = 0.967 1 Orbit is parabolic. Results we are about to get are

valid for it. • In this case, (4) becomes:

t(θ) = (3/mk2)∫dθ[1 + cos(θ - θ´)]-2 (4´)– Limits θ0 θ– Measure θ from distance of closest approach (perihelion). That

is θ = 0 at r = rmin θ0 = θ´ = 0

• So, we want to evaluate: t(θ) = (3/mk2)∫dθ[1 + cosθ]-2 (5)

(Limits 0 θ)– Use trig identity: 1 + cosθ = 2cos2(½θ)

So: t(θ) = [(3)/(4mk2)]∫d θ sec4 (½θ) (5´)(Limits 0 θ)

– Change variables to x = tan(½θ)

Gives: t(θ) = [(3)/(2mk2)]∫dx (1+x2) (5´´)(Limits 0 tan (½θ))

t(θ) = [(3)/(2mk2)][tan(½θ) + (⅓)tan3(½θ)] (6)

• t(θ) = [(3)/(2mk2)][tan(½θ) + (⅓)tan3(½θ)] (6)(-π < θ < π) To understand what the particle is doing atdifferent 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 = rmin

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(m2r2)}]-½

Time: Elliptic Orbit• Elliptic orbit: [α/r(θ)] = 1 + e cosθ (θ´ = 0) (1)

– With e = [1 + {2E2(mk2)}]½ – And 2α = [22(mk)]

• Rewrite (2): r = [a(1- e2)]/[1 + e cosθ] (2) a Semimajor axis (α)/[1 - e2] = (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, rmin occurs at ψ = θ = 0.

– Aphelion, rmax occurs at ψ = θ = π.

• Back to time for the elliptic orbit:t(r) = (m/2)½∫dr[E + (k/r) - {2(m2r2)}]-½ (3)– Limits r0 r, r0 rmin = perihelion distance– Rewrite (3) using eccentricity e [1 + {2E2(mk2)}]½ a Semimajor axis (α)/[1 - e2] = (k)/(2|E|), α [2(mk)]

(3) becomes: t(r) = -(m/2k)½∫rdr[r - (r2/2a) – (½)a(1-e2) ]-½ (3´)

By definition: r a(1 - e cosψ) (2´)Change integration variables from r to ψ

(3´) is: t(ψ) = (ma3/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) = (ma3/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/ma3)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 - cos2ψ]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/ma3)½

– If integrate over full range, ψ = 0 to 2π t τ = period of orbit.

• This gives, τ = 2π(m/k)½a3/2 2π/ω τ2 = [(4π2m)/(k)] a3

Same as earlier, of course! Kepler’s 3rd 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/ma3)½ (4´) Kepler’s Equation• Recall also: r a(1 - e cosψ) [a(1- e2)]/[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 = 0.2056) & Pluto (e = 0.2484) all planet’s have e < 0.1. Several have e < 0.05. 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/ma3)½ Kepler’s Equation

• Also: r a(1 - e cosψ) [a(1- e2)]/[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)]½tan(½ψ)

• Use this to get θ once ψ is known.

tan(½θ) = [(1-e)/(1+e)]½tan(½ψ) Use this to get θ once ψ is known.• Solving this & Kepler’s Equation

ωt(ψ) = ψ - e sinψ, ω (k/ma3)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 (1st integrals of motion) in the Central

Force Problem (& so in Kepler r-2 force problem):

Total mechanical energy:E = (½)m(r2 + r2θ2) + V(r) = const = (½)mr2 + [2(2mr2)] + V(r) Total angular momentum: L = r p = const (magnitude & direction!)

3 components or 2 components + magnitude: Constant magnitude pθ mr2θ = const.

• Can show: There is also another conserved vector quantity Laplace-Runge-Lenz Vector, A

• Newton’s 2nd 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(ac) - c(ab) p L = [mf(r)/r][r(rr) - r2r]Note that rr = rr. L = const p L = d(p L)/dt Combining these gives:

d(p L)/dt = - [mf(r)r2][(r/r) - (rr/r2)]

Or: d(p L)/dt - [mf(r)r2][d(r/r)/dt] (2)• Valid for a general central force!

• General central force: d(p L)/dt = - [mf(r)r2][d(r/r)/dt] (2)• Look at (2) in case of r-2 force:

f(r) = -(k/r2) - mf(r)r2 = 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 1st 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 AL = 0 (L is to p L; r is L = r p)

A = fixed (direction & magnitude) in the orbit plane.

A (p L) - (mkr/r) AL = 0 A = fixed in orbit planeθ angle between r & fixed A direction Ar = Ar cos θ = r(p L) - mkr

Identity: r(p L) = L(r p) = LL 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 || rmin) see diagram. (1) Another way to derive that, for the Kepler Problem, orbit eqtn is a conic section!

• The Laplace-Runge-Lenz VectorA = (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 + {2E2(mk2)}]½

• Comparison of (1) & (2) gives relation between A and the eccentricity e (& thus between A, energy E, & angular momentum ): A mke = mk[ 1 + {2E2(mk2)}]½

• Physical interpretation of Laplace-Runge-Lenz Vector, A = (p L) - (mkr/r)

• Direction of A is the same as the perihelion direction: (A || rmin).

• Magnitude of A: A mke = mk[ 1 + {2E2(mk2)}]½ (3)• For the Kepler problem, we’ve found 7 conserved quantities:

3 components of vector angular momentum, L3 components of Laplace-Runge-Lenz Vector, A

1 scalar energy E

A mke = mk[ 1 + {2E2(mk2)}]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 = rmin) 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: AL = 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”).