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Phys 7221, Fall 2006: Homework # 5
Gabriela Gonz´alez
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
E = T + V =
2 m^ r˙
(^2) − k r =^ −^
k 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 ˙r = 0, r = a, is not that of a Kepler’s orbit, −k/ 2 a, because l = 0. We can derive an equation for r as usual:
dr dt =
m
E − V
m
k r
− k a dt = −
ma 2 k
√rdr √ a − r
=
2 ma k
a − u^2 du
where we used the substitution u^2 = a − r, and used the fact that dr/dt < 0 to add a negative sign when taking the square root of ˙r^2. 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:
t 0 =
2 ma k
∫ √a
0
a − u^2 du =
2 ma k
πa 4 =^ π
ma^3 8 k
If the masses were in a circular orbit of radius a, the period is τ = 2π
ma^3 /k, so the time of the fall can be expressed as t 0 = τ / 4
Prob 3-21: A modified Kepler’s potential
Consider a central potential of the form V (r) = −k/r + h/r^2. The orbit equation (3.34) for u(θ) = 1/r(θ) is
d^2 u dθ^2 +^ u^ =^ −^
m l^2
d du V^ =^ −^
m l^2
d du (−ku^ +^ hu
(^2) ) = km l^2 −^
2 mh l^2 u d^2 u dθ^2 +
2 mh l^2
u =
km l^2
The solution to this equation is of the form
u = km l^2
with β^2 = 1 + 2mh/l^2. This is the equation of a Kepler orbit (parabola, ellipse or hyperbola) in a coordinate system where the angular coordinate is θ′^ = 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/l^2 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 − θ 0 ) = 2π, or
θ − θ 0 =
2 π β =^
√^2 π 1 + 2mh/l^2
= 2π − Ω˙τ
The precession speed is then
Ω = ˙^2 π τ
1 − √^1
1 + 2mh/l^2
≈ 2 πmh l^2 τ
This means orbit precession can be used as a test of Newton’s theory for the gravita- tional force being derived from a potential −k/r. Using l^2 = mka(1 − e^2 ), we obtain an expression for Ω in terms of the perturbation parameter of Kepler’s potential˙ η = h/ka, and orbital parameters:
Ω˙ ≈ 2 πmh l^2 τ =^
2 πmh mka(1 − e^2 )τ =^
2 πη (1 − e^2 )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:
Ω = ˙^43 ×^ (2π/360)^ ×^ (1/3600) rad 100yr
= 2. 1 × 10 −^6 rad/yr
Since the period of the motion is τ = 2π/ω, with ω =
k/ma^3 , we obtain
r˙^2 =
k mar^2
a^2 e^2 − (r − a)^2
= ω^2
a^2 r^2
a^2 e^2 − (r − a)^2
dr dt
= ω a r
a^2 e^2 − (r − a)^2
dt =
aω
√ r dr a^2 e^2 − (r − a)^2
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 = (^1) a √ r dr a^2 e^2 − (r − a)^2
=
a
a √(1 − e cos ψ)ea sin ψdψ a^2 e^2 − (−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 = r^2 /a = (x^2 + y^2 )^2 /a. We will use generalized polar coordinates r, θ to describe the motion of the particle, with z = r^2 /a. The kinetic energy is
T =
2 m( ˙r
(^2) + r (^2) θ˙ (^2) + ˙z (^2) ) =^1 2 m( ˙r
(^2) + r (^2) θ˙ (^2) + 4 r^2 a^2 r˙
The potential energy is V = −mg · r = +mgz =
mg a r
2
The Lagrangian is
L = T − V =^1 2
m r˙^2 +^1 2
mr^2 θ˙^2 + 2m r
2 a^2
r˙^2 − mg a
r^2
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):
mr^2 θ˙ = l = constant
Lagrange’s equation for the coordinate r is
d dt
m r˙ + 2m r
2 a^2 r˙
− mr θ˙^2 − 4 m r^ r˙
2 a^2 +
2 mg a r^ =^0 ( 1 + 2
r^2 a^2
r ¨ −
l^2 m^2 r^3 +
2 g a r^ =^0
There are solutions for circular orbits, with r^4 = r^40 = l^2 a/(2gm^2 ). If the orbit is circular with radius r 0 , the angular momentum is related to the radius and the speed, l = mr 0 v. We can then find the condition between the speed and the radius for circular orbits: v^2 = 2 gr^20 /a. If the orbit is only approximately circular, we find an approximate equation for the perturbation δr = r − r 0 : ( 1 + 2 (r^0 +^ δr)
2 a^2
δr^ ¨ = l
2 m^2 r^30
1 + (δr/r 0 )^3 −^
2 g a (r^0 +^ δr) ( 1 + 2 r
(^20) a^2
δr^ ¨ ≈ − 2 g a
δr
This equation has a periodic solution, with period τ = 2π/ω, with ω^2 = (2g/a)/(1+2r^20 /a^2 ).
