University:
Amherst College
Course:
PHYS 343 | Dynamics
Academic year:

2005
Views:

92

Pages:

4
Author:

Janessa Figueroa

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

of 4

Description

Free up your schedule!

Take 5 seconds to unlock