8-14-1
=
8.14*** Consider a particle of reduced mass orbiting in a central force with Ukr" where kn > 0. (a) Explain what the condition kn 0 tells us about the force. Sketch the effective potential energy Uefr for the cases that n=2,-1, and-3. (b) Find the radius at which the particle (with given angular momentum ) can orbit at a fixed radius. For what values of n is this circular orbit stable? Do your sketches confirm this conclusion? (e) For the stable case, show that the period of small oscillations about the circular orbit is Tose Torb/n+2. Argue that if n + 2 is a rational number, these orbits are closed. Sketch them for the cases that n = 2, -1, and 7. U=kr kn>0
@what dues en indicate abut force?
F=
ar
-knn-1
→
kn>0
F(r) 0
U(r)
dominates for smaller r
dominates for larger r
2μ 2
Jeff =
(r) + l
Zur
2
2μг
radid mi
ar
→want values ofr for which
1=const. For the se values oft, will have =0
solins to
ar
Гаго
Neff
ar
knr-le
0
2
2
3
ہر
ra wat
krnje
[
→For what valves of on is this stable? (assme kn>0)
Orbit is stable if
>0.
kn
ORM 33101
(2) Ueff = Ur) + l = tr + 2a
radial sign: - mi -
2μeff 2μr
→ want values of r for which Ueff will have r=0
r=const. For these values of r, will have =0
sols to
ar
Гаго
Neff
ar
knr-le
0
2
2
3
ہر
ra wat
krnje
[
→ for what values of n is this stable? (assume kn>0)
Orbit is stable if
>0.
kn
Ueff = 0: kn(nr-1)r - 2μr = 0
kn(nr-1)r - 2μ = 0
nr-1 - 2μ/knr = 0
nr - 1 - 2μ/knr = 0
r = [1 + √(1 + 8μ/kn)] / 2n
8-14-3
at r=ro:
d²Ueff/dr² = kn(n-1)r₀ⁿ⁻² + 3l²/μr₀⁴ > 0
multiply by r₀²(n-1) to:
kn(n-1)r₀ⁿ + 3l²/μ > 0
(n+2) > 0
n > -2
This is consistent w/sketches:
for n=2,-1 it's a minire of U at r=ro
for n=-3 it's a local max at r=ro.
Show that period of small oscillations about circular orbit is:
τosc = τorb/√(n+2)
radial equation: μr̈ = -∂Ueff(r)/∂r
For small perturbations about ro
r → r₀ + ε
Ueff → Ueff(r₀) + ε∂Ueff(r₀)/∂r + ε²/2 ∂²Ueff(r₀)/∂r²
eqn of motion is: με̈ + ε∂²Ueff(r₀)/∂r² = 0
∂²Ueff(r₀)/∂r² = kn(n-1)r₀ⁿ⁻² + 3l²/μr₀⁴ = (kn(n-1)r₀ⁿ + 3l²/μ)/r₀²
ε̈ + ε[kn(n-1)r₀ⁿ + 3l²/μ]/μr₀² = 0
ε̈ + ω₀²ε = 0, ω₀² = [kn(n-1)r₀ⁿ + 3l²/μ]/μr₀² = l²/μ²r₀⁴(n+2)
and τ = 2π/ω₀ = 2πμ/l r₀²/√(n+2)
8-14-4
Period of angular motion?
θ = Δθ/Δt = ωt for Δθ = 2π, t = T_ang, ω = 2π/T_ang
using r₀ = [l²/(knμ)]^(1/(n+2)),
we get T_ang = 2πμr₀²/lz
So: T_orb = T_ang/(n+2) = (1.0, T_ang/3, T_ang/2)
If n+2 is rational,
call it p/q, we have
T_orb = pT_ang/q. After a time pT_ang and q*T_ang = T,
the angular position and the radial position are the
same as the initial positions and the orbit closes.
n=2: T_osc = T_orb/2 --> T_orb = 2*T_osc
n=-1: T_osc = T_orb
n=7: T_osc/T_orb = 1/3 --> T_orb = 3*T_osc
n=2: 2 osc. per orbit (oscillating path crosses circular orbit 4 times per orbit)
n=-1: 1 osc. per orbit (osc. path crosses circular orbit twice per orbit)
n=7: 3 osc. per orbit (osc. path crosses circular orbit 6 times per orbit)
Physics 43 F05 Exam 2
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