= Solution
First adopt the uniform-angular-motion approximation implicit in the requested formula. Set the line of sight to longitude zero, the $b$ <exoplanet transit> to $t=0$, and the next $c$ <exoplanet transit> to $t=\tau$, with $0\leq\tau<P_c$. Write $n_b=2\pi/P_b$, $n_c=2\pi/P_c$, with $P_c>P_b$. The next <conjunction> satisfies
$$
(n_b-n_c)t+n_c\tau=2\pi.
$$
The common longitude is $n_bt$ modulo $2\pi$. Subtracting one $2\pi$ gives the <conjunction longitude from a transit time lag>
$$
\Lambda_{bc}=\frac{n_c(2\pi-n_b\tau)}{n_b-n_c}
=2\pi\frac{1-\tau/P_b}{P_c/P_b-1}.
$$
Substitution of the period ratio yields
$$
\boxed{\Lambda_{bc}=2\pi\left(1-\frac{\tau}{P_b}\right)\frac pq
\left[1+\delta\frac{p+q}{q}\right]^{-1}\pmod{2\pi}}.
$$
It is exact in this uniform-angle model; no first-order expansion in $\delta$ is necessary here. For other choices of which <exoplanet transit> is used, choose the appropriate whole-turn branch.
\b[For the finite-eccentricity <orbit> specified earlier, this is an approximation]. Let $f_L$ and $M_L$ be the <true anomaly> and <mean anomaly> at the line of sight. The eccentric <planet> has $M_c(t)=M_L+n_c(t-\tau)$ and true angular advance $n_c(t-\tau)+\Delta_e(t)$, where
$$
\Delta_e(t)=f_c(t)-M_c(t)-(f_L-M_L).
$$
True <conjunction> therefore requires $(n_b-n_c)t+n_c\tau-\Delta_e(t)=2\pi$. The PDF omits this term, which is generally $O(e_c)$.
For a concrete counterexample take $p=q=1$, $\delta=0$, $e_c=1/5$, $\tau=P_b/4$, and a line of sight along $c$'s <periapsis>. The uniform model predicts $t=7P_b/4$ and longitude $3\pi/2$. At that time $M_c=3\pi/2$, but <Kepler's equation> gives $f_c\ne3\pi/2$, so the <planets> are not truly aligned. Accurate finite-eccentricity <conjunctions> must be found from the corrected equation.
Back to article page
