 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-316/4/iv/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

Keep the inner-pair Laplace coefficient unexpanded because

α = a_{1} / a_{2}

is close to one. Put

ℓ = L_{1} / L_{2}

and

ϵ_{B} = \frac{M _{3}}{M _{2}} \frac{3 β ^{3}}{α ^{2} b _{3/2}^{(1)} ( α )}, β = \frac{a _{1}}{a _{3}} ≪ 1.

Using the small-argument Laplace coefficient for the distant planet and

n_{2} / n_{1} = α^{3/2}

gives

B_{21} B_{13} B_{23} B_{31} B_{32} = ℓ B_{12}, ≃ ϵ_{B} B_{12}, ≃ α^{- 3/2} ϵ_{B} B_{12}, ≃ \frac{L _{1}}{L _{3}} ϵ_{B} B_{12}, ≃ \frac{L _{2}}{L _{3}} α^{- 3/2} ϵ_{B} B_{12} .

The reverse-coupling expressions follow from the angular momentum identity

L_{j} B_{jk} = L_{k} B_{kj}

. If desired,

L_{1} / L_{2} = (M_{1} / M_{2}) α

L_{1} / L_{3} = (M_{1} / M_{3}) β

, and

L_{2} / L_{3} = (M_{2} / M_{3}) β / α

express all ratios in the supplied quantities. The diagonal entries are

B_{11} = - (B_{12} + B_{13})

B_{22} = - (B_{21} + B_{23})

, and

B_{33} = - (B_{31} + B_{32})

For comparable masses and closely spaced inner orbits,

α ≃ 1

ℓ \sim 1

, and

L_{1} / L_{3}, L_{2} / L_{3} = O (β)

. Thus

B_{12} \sim B_{21} ≫ B_{13} \sim B_{23} ≫ B_{31} \sim B_{32} .

The distant couplings

B_{13}, B_{23}

scale as

β^{3}

relative to a fixed inner architecture, and the reverse couplings have an additional

β

suppression.

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