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
– 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
(2)
• 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
• Put orbit eqtn results r(θ) into (3) & integrate:
– We had [α/r(θ)] = 1 + e cos(θ - θ´)
– With e = [ 1 + {2E2(mk2)}]½
– And 2α = [22(mk)]
(3)
 (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 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 = 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 + {2E2(mk2)}]½
– And 2α = [22(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 + {2E2(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
ψ = cos-1[(1-r/a)/e]
(2´)
(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(ac) - c(ab)

p  L = [mf(r)/r][r(rr) - r2r]
Note that rr = 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  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 || rmin) 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(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 + {2E2(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 + {2E2(mk2)}]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(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: 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 LaplaceRunge-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”).