Lecture 10 Outline - Kepler
•Diff. Eqn. for the orbit (Section 3.5)
•Bertrand’s Theorem (Section 3.6)
•The Kepler Problem (Sections 3.7, 3.8)
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Orbit Stability and Kepler's Problem: Circular Orbits, Power Laws, and Bertrand's Theorem , Study notes of Physics
The stability of circular orbits in celestial mechanics through the analysis of kepler's problem. Topics include the differential equation for the orbit, bertrand's theorem, and power law stability. The document also discusses the relationship between circular orbit stability and power laws, as well as deviating circles and harmonic motion.
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Lecture 10 Outline - Kepler
Diff. Eqn. for the orbit (Section 3.5)
Bertrand’s Theorem (Section 3.6)
The Kepler Problem (Sections 3.7, 3.8)
Circular Orbit Stability
Which orbits retrace their steps?
Obviously circles (
dt^ dr
) at radius where
V
ef f
is at
minima given
`
and
E
V
ef f
r 0 ) =
V
r 0 ) +
`
2 / 2 mr
(^02)
f ef f
r 0 ) =
∂r ∂V
r 0
= 0
thus
f (^) ( r 0 ) =
`
2
mr
(^03)
Two cases depending on whether minima or maxima:
Deviate Circles
harmonic motionDeviating slightly from circular motion gives rise to
u = u 0 + α
(^) cos(
βθ
/r
OR
r = r 0 + α
(^) cos(
βθ
Here
α
depends on deviation of energy
E
and
β
from first-order Taylor expansion of
f (^) ( r )
at
r
0
Bertrand’s Theorem
Anyways, can
somehow
show that for
r = r 0 + α
(^) cos(
βθ
Condition for a closed orbital depends on
β
β
r 0
f 0
∂r^ ∂f
r 0
andcanalsoshow
f (^) ( r ) =
k
r 3 − β 2
If rational number
β
p/q
with integers
p, q
After
q
revolutions
cos(
pθ/q
) = cos(
πp
Second-order can show that only
β
are stable
f (^) ( r ) =
k
r 3 −
β 2
either
f (^) ( r ) =
r k 2
OR
f (^) ( r ) =
kr
inverse-square and Hooke’s law”.have closed orbitals for all bound particles areBertrand’s Theorem (1873) ”only central forces that
Inverse Square Law
f (^) ( r ) =
r k 2
and thus
V
r ) =
k/r
Several ways to integrate the eqn for orbit, eg.
f (^) ( r ) =
r k 2
`
2
r 2
d
dθ
`
2
mr
2
dθ dr
`
2
mr
3
however, Goldstein
knows
that the indefinite integral
dx
α
βx
γx
2
γ
cos
−
1
(
β
γx
β
2
−
αγ
θ = θ ′ − ∫
du
2 ` mE 2
2 mku ` 2 − u 2 = θ ′ −
cos
−
1
` 2 u
mk
2 E`
2
mk
2
Note
θ
′
is a constant of integration. Thus rearranging
Equation of
r 2
Orbit
Contains
E, `, θ
′
three constants of integration
Note
θ ′
is also the turning angle!
r 1
mk ` 2
E`
2
mk
2
cos(
θ − θ ′ ) )
The (missing) fourth specifies the location on orbit
θ 0
...
which we need to find
r ( t ) , θ
t )
again we can use
mr
2 dθ
`dt
Semimajor Axis of Ellipse (Relation)
So the eccentricity
e
=
2 E`
2
k 2
Motion in time (for
F
r
k/r
Fairly simple, using conservation
`dt
mr
2 dθ
again:
where we throw in
r 1
=
mk ` 2
e
cos(
θ
−
θ ′ ))
t
=
m 2
r
r 0
dr
r k
−
` 2
2 mr
2 + E = ` 3
mk
2
∫
θ
θ 0
dθ
e
cos(
θ
−
θ ′ ))
2
BUT either integrals to give
r
( t ) , θ
t )
are tough!