Two stars of masses and orbit their common centre of mass under their mutual…

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Gravitation and central-force motion · Lagrangian mechanics

Action and Symmetry, Ch The Kepler problem redone: reduction, quadrature and the Runge-Lenz vector

Problem

Two stars of masses m1=2.0×1030 kgm_1 = 2.0\times10^{30}\,\mathrm{kg} and m2=1.0×1030 kgm_2 = 1.0\times10^{30}\,\mathrm{kg} orbit their common centre of mass under their mutual gravitational attraction, V(r)=−GNm1m2/rV(r) = -G_{\rm N}m_1m_2/r, on a bound elliptical orbit of total energy (for the reduced one-body problem) E=−3.0×1033 JE = -3.0\times10^{33}\,\mathrm{J}. All other bodies and all dissipation are ignored. ((a)) Remove the cyclic centre-of-mass coordinate to reduce the pair to one body of reduced mass μ\mu moving in V(r)V(r) about a fixed centre. State μ\mu numerically. ((b)) Mechanical similarity applied to a potential V∝rnV\propto r^n relates the long-time averages ⟨T⟩\langle T\rangle and ⟨V⟩\langle V\rangle of a bound orbit through the exponent nn alone. State that relation for n=−1n=-1, and say what the scaling argument does and does not establish about why such an average exists. ((c)) The heavy symmetric top's Lagrangian has two cyclic angles, precession ϕ\phi and spin ψ\psi, and removing them leaves a single effective potential for the nutation angle θ\theta. The two-body reduction removes the three components of the cyclic Rcm\mathbf R_{\rm cm}, uses the conserved direction of L\mathbf L to fix the plane of the orbit, and then removes the cyclic polar angle φ\varphi. Count the coordinates left after each step, and say which of the steps has a counterpart in the top's reduction and which does not. ((d)) Derive the relation of part (b) directly from the virial theorem: define D=p⋅rD = \mathbf p\cdot\mathbf r for the reduced one-body problem, find D˙\dot D, state the condition on the orbit that makes ⟨D˙⟩→0\langle\dot D\rangle\to0 over a long time, and confirm the bound orbit meets it. Then use E=⟨T⟩+⟨V⟩E=\langle T\rangle+\langle V\rangle to find ⟨T⟩\langle T\rangle and ⟨V⟩\langle V\rangle numerically.

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