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Barycentric orbits of two unequal masses on declared model data: $p_{1} = (m_{2} / M) p$ , $p_{2} = (m_{1} / M) p$ , semiaxes in the ratio $3 : 1$ and areal rates in the ratio $9 : 1$

Physical scope

A numerical check of the barycentric reconstruction on declared model data with unequal masses: the two labels' conic parameters and semimajor axes, the common eccentricity and focus at the centre of mass, the ratio of the areal rates $(m_{2} / m_{1})^{2}$ and the common period about the centre of mass. Every number is declared model data or an exact consequence of the cited items; no observation or measurement is used.

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 $I_{0} = R$ , unequal inertial masses and a declared value of the constant, $m_{1} = 2 \times 10^{24} k g, m_{2} = 6 \times 10^{24} k g, M = 8 \times 10^{24} k g, μ = \frac{m_{1} m_{2}}{M} = \frac{3}{2} \times 10^{24} k g,$ $G_{N} = 10^{- 24} m^{3} k g^{- 1} s^{- 2}, κ = G_{N} m_{1} m_{2} = 1.2 \times 10^{25} k g m^{3} s^{- 2},$ with the couplings identified with the inertial masses, and suppose a motion has relative conic data $J = 6 \times 10^{27} k g m^{2} s^{- 1}, p = \frac{J^{2}}{μ κ} = 2 \times 10^{6} m, e = \frac{1}{2} .$ Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), as in the cited conic and period items. Then the cited items give:

the relative semiaxes $a = p / (1 - e^{2}) = \frac{8}{3} \times 10^{6} m$ and $b = a \sqrt{1 - e^{2}} = \frac{4 \sqrt{3}}{3} \times 10^{6} m$ , energy $E = - κ / (2 a) = - \frac{9}{4} \times 10^{18} J$ , and period $T = 2 π \sqrt{μ a^{3} / κ} = \frac{16 π}{3 \sqrt{3}} \times 10^{9} s$ ;
each label's orbit about the centre of mass is the scaled ellipse with parameters $p_{1} = \frac{m_{2}}{M} p = \frac{3}{2} \times 10^{6} m, p_{2} = \frac{m_{1}}{M} p = \frac{1}{2} \times 10^{6} m,$ the same eccentricity $e = \frac{1}{2}$ , focus at the centre of mass and semimajor axes $a_{1} = \frac{3}{4} a = 2 \times 10^{6} m$ , $a_{2} = \frac{1}{4} a = \frac{2}{3} \times 10^{6} m$ ;
the areal-velocity magnitudes about the centre of mass are in the ratio $(m_{2} / m_{1})^{2} = 9$ : $(\frac{m_{2}}{M})^{2} \frac{J}{2 μ} = 1125 m^{2} s^{- 1}, (\frac{m_{1}}{M})^{2} \frac{J}{2 μ} = 125 m^{2} s^{- 1};$
both labels are periodic about the centre of mass with the common period $T$ .

Facts & Assumptions

Given: The declared data $m_{1} = 2 \times 10^{24}$ , $m_{2} = 6 \times 10^{24}$ , $M = 8 \times 10^{24}$ , $μ = \frac{3}{2} \times 10^{24}$ , $κ = 1.2 \times 10^{25}$ , $J = 6 \times 10^{27}$ , $p = 2 \times 10^{6}$ and $e = \frac{1}{2}$ . Bare numbers in the arithmetic below are numerical values in the SI units declared in the Example; any explicitly unit-bearing equality is an equality of quantities.

[A1]

The Axiom of Countable Choice is assumed (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L1]

Conic data and barycentric scaling (The inverse-square gravitational relative trajectory has the Kepler polar form: $ρ = p / (1 + e \cos (θ - θ_{0}))$ , $E = (μ κ^{2} / 2 J^{2}) (e^{2} - 1)$ , elliptic completeness and barycentric scaled ellipses clauses 1 and 3, 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, 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), Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential): with the identification $m_{i}^{g} = m_{i}$ and $κ = G_{N} m_{1} m_{2}$ , a motion with the stated conic data has $a = p / (1 - e^{2})$ , $b = a \sqrt{1 - e^{2}}$ , and $E = (μ κ^{2} / 2 J^{2}) (e^{2} - 1) = - κ / (2 a)$ ; each label's trajectory about the centre of mass is the scaled conic $x_{i} - X = \pm \frac{m_{j}}{M} r$ with parameter $p_{i} = \frac{m_{j}}{M} p$ , semimajor axis $a_{i} = \frac{m_{j}}{M} a$ and the same eccentricity, the centre of mass being the focus.

[L2]

Period, area law and curves (Kepler's third law for the gravitational two-label model: the relative period is $T = 2 π μ a b / J = 2 π \sqrt{μ / κ} a^{3 / 2}$ , so $T^{2} = 4 π^{2} a^{3} / (G_{N} M)$ under the mass identification, and both labels are periodic relative to the centre of mass clauses 1 and 3, Planarity of the reduced orbit and the area law: the areal-velocity vector is the constant $j / (2 μ)$ , the swept-area rate is $J / (2 μ)$ , and both labels sweep equal areas in equal times about the centre of mass clauses 2 and 4, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): $T = 2 π μ a b / J = 2 π \sqrt{μ / κ} a^{3 / 2}$ ; the relative areal-velocity vector is $j / (2 μ)$ with magnitude $J / (2 μ)$ ; and the label $i$ 's areal-velocity vector about the centre of mass is $(\frac{m_{j}}{M})^{2} ω$ , of magnitude $(\frac{m_{j}}{M})^{2} J / (2 μ)$ , with both labels $T$ -periodic about the centre of mass.

[L3]

Arithmetic: $κ / (2 a) = (1.2 \times 10^{25}) / (\frac{16}{3} \times 10^{6}) = \frac{9}{4} \times 10^{18}$ ; $μ / κ = \frac{3 / 2}{1.2} \times 10^{- 1} s^{2} m^{- 3} = \frac{1}{8} s^{2} m^{- 3}$ ; $μ a^{3} / κ = \frac{3 / 2}{1.2} \times 10^{- 1} \cdot \frac{512}{27} \times 10^{18} s^{2} = \frac{64}{27} \times 10^{18} s^{2}$ ; $2 π \sqrt{64 / 27} \times 10^{9} = \frac{16 π}{3 \sqrt{3}} \times 10^{9}$ ; $(m_{2} / M)^{2} = 9 / 16$ , $(m_{1} / M)^{2} = 1 / 16$ and $J / (2 μ) = 2000 m^{2} s^{- 1}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Verification

technique · direct

1.1givenA1L1L3algebra

Relative conic data: $1 - e^{2} = \frac{3}{4}$ , so $a = p / (1 - e^{2}) = 2 \times 10^{6} \cdot \frac{4}{3} = \frac{8}{3} \times 10^{6}$ m and $b = \frac{8}{3} \times 10^{6} \cdot \frac{\sqrt{3}}{2} = \frac{4 \sqrt{3}}{3} \times 10^{6}$ m; the energy identity of [L1] gives $E = - κ / (2 a) = - (\frac{6}{5} \times 10^{25}) / (\frac{16}{3} \times 10^{6}) = - \frac{9}{4} \times 10^{18}$ J; and the period is $T = 2 π \sqrt{μ a^{3} / κ} = 2 π \sqrt{\frac{64}{27} \times 10^{18}} = \frac{16 π}{3 \sqrt{3}} \times 10^{9}$ s.

2.1step 1.1L1L3algebra

Barycentric scaling: $m_{2} / M = \frac{3}{4}$ and $m_{1} / M = \frac{1}{4}$ , so [L1] gives $p_{1} = \frac{3}{4} \cdot 2 \times 10^{6} = \frac{3}{2} \times 10^{6}$ m, $p_{2} = \frac{1}{4} \cdot 2 \times 10^{6} = \frac{1}{2} \times 10^{6}$ m, $a_{1} = \frac{3}{4} \cdot \frac{8}{3} \times 10^{6} = 2 \times 10^{6}$ m and $a_{2} = \frac{1}{4} \cdot \frac{8}{3} \times 10^{6} = \frac{2}{3} \times 10^{6}$ m, with eccentricity $e = \frac{1}{2}$ and the focus at the centre of mass for both.

3.1step 2.1L2L3algebra

Areal rates: by [L2] the label-1 areal-velocity magnitude is $(\frac{3}{4})^{2} J / (2 μ) = \frac{9}{16} \cdot 2000 = 1125 m^{2} s^{- 1}$ and that of label 2 is $(\frac{1}{4})^{2} \cdot 2000 = 125 m^{2} s^{- 1}$ ; their ratio is $9 = (m_{2} / m_{1})^{2}$ .

4.1step 1.1step 2.1L2L3algebra∎

Periods: by [L2] both labels are $T$ -periodic about the centre of mass with the common period $T = \frac{16 π}{3 \sqrt{3}} \times 10^{9}$ s from step 1.1; equivalently $T^{2} = 4 π^{2} μ a^{3} / κ = \frac{256 π^{2}}{27} \times 10^{18} s^{2}$ with the arithmetic of [L3].

Remarks

The focus is the centre of mass, not the heavier body. Both barycentric conics have the centre of mass at a focus; the lighter label 1 has the larger orbit, in the ratio $m_{2} : m_{1} = 3 : 1$ of the parameters and semiaxes.
Declared data only. The masses, $J$ , $p$ , $e$ and the declared $G_{N}$ are model data chosen for exact arithmetic; no measured constant or observation is used, and no real system is modelled.

Barycentric orbits of two unequal masses on declared model data: $p_{1} = (m_{2} / M) p$ , $p_{2} = (m_{1} / M) p$ , semiaxes in the ratio $3 : 1$ and areal rates in the ratio $9 : 1$

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

Barycentric orbits of two unequal masses on declared model data: p1=(m2/M)pp1​=(m2​/M)p, p2=(m1/M)pp2​=(m1​/M)p, semiaxes in the ratio 3:13:1 and areal rates in the ratio 9:19:1

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

Barycentric orbits of two unequal masses on declared model data: $p_{1} = (m_{2} / M) p$ , $p_{2} = (m_{1} / M) p$ , semiaxes in the ratio $3 : 1$ and areal rates in the ratio $9 : 1$