Phys 7221, Fall 2006: Homework # 5
Gabriela González
October 1, 2006
Prob 3-11: Collapse of an orbital system
Consider two particles falling into each other due to gravitational forces, starting from rest
at a distance a. The system has zero angular momentum, with the energy given by
1
k
k
E = T + V = mṙ2 − = −
2
r
a
where m is the reduced mass of the system, and r is the distance between the masses. Notice
that the value of the energy, −k/a, calculated from the initial condition ṙ = 0, r = a, is
not that of a Kepler’s orbit, −k/2a, because l = 0.
We can derive an equation for r as usual:
r
dr
2√
=
E−V
dt
m
r r
2 k k
−
=
m r
a
r
√
ma rdr
√
dt = −
2k a − r
r
2ma p
=
a − u2 du
k
where we used the substitution u2 = a − r, and used the fact that dr/dt < 0 to add a
negative sign when taking the square root of ṙ2 . We can integrate the equation from the
√
initial time when u = 0, to the collapse time when u = a, obtaining the time of the fall:
r
r
r
Z √a p
2ma
2ma
πa
ma3
t0 =
a − u2 du =
=π
k
k 4
8k
0
p
If the masses were in a circular orbit of radius
a,
the
period
is
τ
=
2π
ma3 /k, so the
√
time of the fall can be expressed as t0 = τ /4 2.
1
Prob 3-21: A modified Kepler’s potential
Consider a central potential of the form V (r) = −k/r + h/r2 . The orbit equation (3.34)
for u(θ) = 1/r(θ) is
d2 u
m d
m d
km 2mh
+u=− 2
V =− 2
(−ku + hu2 ) = 2 − 2 u
2
dθ
l du
l du
l
l
d2 u
km
2mh
u= 2
+ 1+ 2
2
dθ
l
l
The solution to this equation is of the form
u=
km
+ A cos(β(θ − θ0 ))
l2
with β 2 = 1 + 2mh/l2 .
This is the equation of a Kepler orbit (parabola, ellipse or hyperbola) in a coordinate
system where the angular coordinate is θ0 = βθ.
A revolution around the origin sweeps a θ angle equal to 2π. If β 1,there are many
radial oscillations in one revolution; if β ≈ 1, the orbit shows a slow precession. If the
energy is negative and 2mh/l2 1, the orbit is a precessing ellipse. In a cycle of the
periodic motion with period τ , the radial coordinate returns to the original value when
β(θ − θ0 ) = 2π, or
2π
2π
θ − θ0 =
=p
= 2π − Ω̇τ
β
1 + 2mh/l2
The precession speed is then
2π
Ω̇ =
τ
1− p
1
1 + 2mh/l2
!
≈
2πmh
l2 τ
This means orbit precession can be used as a test of Newton’s theory for the gravitational force being derived from a potential −k/r. Using l2 = mka(1 − e2 ), we obtain an
expression for Ω̇ in terms of the perturbation parameter of Kepler’s potential η = h/ka,
and orbital parameters:
Ω̇ ≈
2πmh
2πmh
2πη
=
=
2
2
l τ
mka(1 − e )τ
(1 − e2 )τ
The effect is more pronounced for eccentric and long orbits. The perihelion of Mercury is
observed to precess (after correcting for known planetary perturbations) by 43 arc-seconds
per century:
43 × (2π/360) × (1/3600) rad
Ω̇ =
= 2.1 × 10−6 rad/yr
100yr
2
and thus
(1 − e2 )τ Ω̇
≈ 7.6 × 10−8
2π
This discrepancy is also (and better) explained by General Relativity.
η≈
Prob 3-23: Mass ratio of Sun and Earth
The period and the semi-major axis of elliptical orbits in Kepler’s potential are related
by (τ /2π)2 = a3 µ/k where µ = m1 m2 /(m1 + m2 ) is the reduced mass of the system, and
k = Gm1 m2 . When m1 m2 , we have (τ /2π) ≈ a3 /Gm1 . For the Earth-Sun system,
τ 2
a3
es
= es .
2π
GMs
For the Earth-Moon system,
τ 2
a3
em
= em .
2π
GMe
Taking ratios, we obtain
τes 2 a3em Me
= 3
τem
aem Ms
3 3 2 27.3 2
τem
Ms
aes
1.5 × 108
=
= 3.4 × 105
=
Me
aem
τes
3.8 × 105
365
The measured value, from http://ssd.jpl.nasa.gov/?constants, is 328900.56±0.02;
our estimate is within 3% of this value.
Prob 3-24: Kepler’s equation
The energy in Kepler’s motion is
1
l2
k
E = mṙ2 +
−
2
2
2mr
r
k
For negative energy and elliptical orbits, the energy is E = − 2a
, and the angular momen2
2
tum is l = kma(1 − e ), thus
l2
k
2
2
ṙ =
E−
+
m
2mr2
r
2
k
ka(1 − e2 ) k
=
− −
+
m
2a
2r2
r
k
=
−r2 − a2 (1 − e2 ) + 2ar
2
mar
k
=
a2 e2 − (r − a)2
2
mar
3
Since the period of the motion is τ = 2π/ω, with ω =
p
k/ma3 , we obtain
k
a2 e2 − (r − a)2
2
mar
a2
= ω 2 2 a2 e2 − (r − a)2
r
dr
ap 2 2
= ω
a e − (r − a)2
dt
r
r dr
1
p
dt =
aω a2 e2 − (r − a)2
ṙ2 =
which we can use to integrate t(r). Using the orbit equation r = a(1 − e cos ψ), and
dr = ea sin ψdψ we obtain
ω dt =
=
r dr
1
p
2
2
a a e − (r − a)2
1 a(1 − e cos ψ)ea sin ψdψ
p
a
a2 e2 − (−ae cos ψ)2
= (1 − e cos ψ) dψ
which can be now trivially integrated into Kepler’s equation:
ωt = ψ − e sin ψ
Prob 3-33: A particle in a paraboloid of revolution
A particle with coordinates r = (x, y, z) is constrained to move in a paraboloid f revolution.
i.e., z = r2 /a = (x2 + y 2 )2 /a. We will use generalized polar coordinates r, θ to describe the
motion of the particle, with z = r2 /a. The kinetic energy is
1
1
r2
T = m(ṙ2 + r2 θ̇2 + ż 2 ) = m(ṙ2 + r2 θ̇2 + 4 2 ṙ2 )
2
2
a
The potential energy is
V = −mg · r = +mgz =
mg 2
r
a
The Lagrangian is
1
1
r2
mg 2
L = T − V = mṙ2 + mr2 θ̇2 + 2m 2 ṙ2 −
r
2
2
a
a
We see that the coordinate θ is cyclic, so the z-component of the angular momentum
is conserved (associated with the symmetry of rotation about the z-axis):
mr2 θ̇ = l = constant
4
Lagrange’s equation for the coordinate r is
d
dt
mṙ + 2m
r2
rṙ2 2mg
2
ṙ
−
mr
θ̇
−
4m
+
r = 0
a2
a2
a
l2
2g
r2
1 + 2 2 r̈ − 2 3 + r = 0
a
m r
a
There are solutions for circular orbits, with r4 = r04 = l2 a/(2gm2 ). If the orbit is circular
with radius r0 , the angular momentum is related to the radius and the speed, l = mr0 v.
We can then find the condition between the speed and the radius for circular orbits: v 2 =
2gr02 /a. If the orbit is only approximately circular, we find an approximate equation for
the perturbation δr = r − r0 :
(r0 + δr)2 ¨
1
2g
l2
1+2
− (r0 + δr)
δr
=
3
2
3
2
a
a
m r0 1 + (δr/r0 )
2
r
¨ ≈ − 2g δr
1 + 2 02 δr
a
a
This equation has a periodic solution, with period τ = 2π/ω, with ω 2 = (2g/a)/(1+2r02 /a2 ).
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