Solution

Substitute

r = 1/ u

and

\overset{r}{˙} = - h u^{'}

into the conserved radial energy. Differentiation with respect to

ϕ

gives

h^{2} u^{'} u^{''} - m u^{'} + h^{2} u u^{'} - 3 m h^{2} u^{2} u^{'} = 0.

After division by

h^{2} u^{'}

and continuity through turning points,

u^{''} + u = \frac{m}{h ^{2}} + 3 m u^{2} .

Put

q = m / h^{2}

and substitute

u = q (1 + α) + A cos [(1 + β) ϕ] .

Keeping the leading nontrivial powers of

m / h

and terms through first order in

A

, the constant and oscillatory terms give

q (1 + α) = q + 3 m q^{2}, - 2 β A = 6 m q A .

Therefore

α = \frac{3 m ^{2}}{h ^{2}}, β = - \frac{3 m ^{2}}{h ^{2}} .

The radial oscillation has angular period

2 π / (1 + β)

, so successive periapses advance by

Δ ϕ = - 2 π β = \frac{6 π m ^{2}}{h ^{2}}

Solved by gpt-5.6-sol high.

 Ancestors (11)