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A gravitational circular relative orbit checked on declared model data: r0=2⋅106 m, μr0ω2=u′(r0)=3⋅1012 N, E=−3⋅1018 J and ωlin=ω=10−9 s−1

Physical scope

A numerical check of the circular-orbit criterion and the linearised radial frequency on declared model data for the gravitational relative problem: the critical radius, the force-balance identity μr0ω2=u′(r0), the energy at the minimum of the effective potential, the equality ωlin=ω for the inverse-square potential, and the two labels' circular motions about the centre of mass. Every number is declared model data or an exact consequence of the cited items; no observation is used and no nonlinear stability is claimed.

Example

Declare the gravitational two-label model of Adopted two-label gravitational model: the inverse-square pair potential with coupling parameters, the empirical mass identification, and the measured value and uncertainty of the Newtonian constant with time window I0=R and the same data as the barycentric check (Barycentric orbits of two unequal masses on declared model data: p1=(m2/M)p, p2=(m1/M)p, semiaxes in the ratio 3:1 and areal rates in the ratio 9:1): m1=2×1024 kg,m2=6×1024 kg,μ=32×1024 kg,GN=10−24 m3 kg−1 s−2, κ=GNm1m2=1.2×1025 kg m3 s−2,J=6×1027 kg m2 s−1, with the couplings identified with the inertial masses. Then Circular relative orbits of the central model: the critical-point criterion u′(r0)=J2/(μr03), uniform rotation of both labels about the centre of mass, and the linearised radial frequency gives:

  1. the critical radius and angular velocity r0=J2μκ=2×106 m,ω=Jμr02=10−9 s−1, and the criterion holds in all three equivalent forms (UJeff)′(r0)=0, u′(r0)=J2/(μr03)=3×1012 N, μr0ω2=3×1012 N;
  2. the orbital energy E=UJeff(r0)=−μκ2/(2J2)=−3×1018 J, the minimum of the effective potential;
  3. the linearised radial frequency ωlin=(UJeff)′′(r0)/μ=κ/(μr03)=10−9 s−1, equal to the orbital angular velocity ω for this inverse-square potential;
  4. both labels move uniformly on circles about the centre of mass with angular velocity ω=10−9 s−1 and radii m2Mr0=1.5×106 m and m1Mr0=5×105 m, each with period 2π/ω=2π×109 s.

The stated period is relative to the centre of mass, for the coordinates xi−X. An absolute label trajectory has this period only if the centre of mass velocity V vanishes; otherwise the centre of mass drifts uniformly.

Facts & Assumptions

Given: The declared data μ=32×1024, κ=1.2×1025, J=6×1027, m1=2×1024, m2=6×1024, M=8×1024.

[L1]

Circular-orbit theorem (Circular relative orbits of the central model: the critical-point criterion u′(r0)=J2/(μr03), uniform rotation of both labels about the centre of mass, and the linearised radial frequency clauses 1-4, Adopted central-force model class: static central pair potentials, the reduced one-label problem, the fixed-centre idealization, reading conventions and exclusions clauses (i)-(ii), Adopted two-label gravitational model: the inverse-square pair potential with coupling parameters, the empirical mass identification, and the measured value and uncertainty of the Newtonian constant clauses (i)-(ii), Two-body reduction of a static central pair-potential model: the relative coordinate is a central-force trajectory for the reduced mass, the centre of mass moves uniformly, and the reconstruction is a bijection clause 4, Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential): the equivalences of clause 1 hold with UJeff(ρ)=u(ρ)+J2/(2μρ2) and u(ρ)=−κ/ρ; under the condition, the relative motion is the uniform rotation at radius r0 with angular velocity ω=J/(μr02), energy E=UJeff(r0) and angular momentum Je3; the linearised radial equation at r0 has frequency ωlin=(UJeff)′′(r0)/μ when (UJeff)′′(r0)>0; and the two labels move uniformly on circles about the centre of mass with radii mjMr0 and angular velocity ω.

[L2]

Arithmetic: r0=3.6×1055/(1.8×1049)=2×106 m; ω=6×1027/(1.5×1024⋅4×1012)=10−9 s−1; u′(r0)=κ/r02=1.2×1025/4×1012=3×1012 N; J2/(μr03)=3.6×1055/(1.5×1024⋅8×1018)=3×1012 N; μr0ω2=1.5×1024⋅2×106⋅10−18=3×1012 N; and, using J2=μκr0, (UJeff)′′(r0)=κ/r03=1.5×106 (Intervals of R: the nine order-convex forms, nondegeneracy, and length, Dimensions and SI unit conventions of the non-relativistic classical mechanics framework).

Verification

technique · direct
1.1givenL1L2algebra

Critical radius and criterion: substituting the data gives r0=J2/(μκ)=(36×1054)/(18×1048)=2×106 m, which lies in J0=(0,∞), and (UJeff)′(r0)=u′(r0)−J2/(μr03)=3×1012−3×1012=0 by the three arithmetic evaluations of [L2]; since μr0ω2=3×1012 as well, all three forms of the criterion of Circular relative orbits of the central model: the critical-point criterion u′(r0)=J2/(μr03), uniform rotation of both labels about the centre of mass, and the linearised radial frequency clause 1 hold simultaneously.

2.1step 1.1L1L2algebra

Angular velocity, energy and the labels: ω=J/(μr02)=10−9 s−1 by [L2]; the energy is E=UJeff(r0)=−κ/r0+J2/(2μr02)=−6×1018+3×1018=−3×1018 J, which also equals −μκ2/(2J2)=(32×1024⋅144×1048)/(2⋅36×1054)⋅(−1)=−3×1018 J; by [L1] the labels' circles have radii 34r0=32×106 m and 14r0=5×105 m with angular velocity ω, each with period 2π×109 s.

3.1step 1.1step 2.1L1L2algebra∎

Linearised frequency: (UJeff)′′(r0)=−2κ/r03+3J2/(μr04); with J2=μκr0 the two terms are −2κ/r03 and +3κ/r03, giving (UJeff)′′(r0)=κ/r03=32×106, so ωlin=32×106/(32×1024)=10−18=10−9 s−1=ω. This is the equality of the linearised radial and orbital frequencies for the inverse-square potential.

Remarks

Depends on

Dependency tree · 2 levels

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