Classical Mechanics (Kibble & Berkshire) · Problem 7.29

Non-linear oscillations and phase space — T.W.B. Kibble, F.H. Berkshire

Physics — Mechanics · computer-assisted-problems

Problem

Orbit with linear resistance Confirm the approximate solution for small resistance obtained in Problem 7.26 by numerical solution of the governing simultaneous ODEs. First write the governing equations in dimensionless form. Suppose that, in the absence of resistance, a circular orbit with r=ar = a and θ˙=Ω\dot{\theta} = \Omega is possible; then γ=a3Ω\gamma = a^3\Omega and L0=a2ΩL_0 = a^2\Omega. On taking dimensionless variables ρ\rho, τ\tau defined by ρ=r/a\rho = r/a and τ=Ωt\tau = \Omega t, and taking L0=a2ΩL_0 = a^2\Omega, the governing equations become d2ρdτ2+ϵdρdτ+1ρ2−e−2ϵτρ3=0,ρ2dθdτ=e−2ϵτ,\frac{d^2\rho}{d\tau^2} + \epsilon\frac{d\rho}{d\tau} + \frac{1}{\rho^2} - \frac{e^{-2\epsilon\tau}}{\rho^3} = 0, \qquad \rho^2\frac{d\theta}{d\tau} = e^{-2\epsilon\tau}, where ϵ=K/Ω\epsilon = K/\Omega is the dimensionless resistance parameter. Solve these equations with the initial conditions ρ=1\rho = 1, dρ/dτ=0d\rho/d\tau = 0 and θ=0\theta = 0 when τ=0\tau = 0. Choose some small value for ϵ\epsilon and plot a polar graph of the path.

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