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For two nearly edge-on orbits with small orbital inclination difference $\Delta I$ and small longitude of ascending node difference $\Delta \Omega$, expanding the dot product of their normals gives $I_m^2=(\Delta I{)}^2+(\Delta \Omega {)}^2+O({\delta }^4)$, with angles in radians and all offsets from edge-on of order $\delta$. Thus equality of observed sky-plane inclinations alone does not imply aligned orbital planes.

In a right-handed reference frame, the rotation matrix from an orbital plane frame whose first axis points along the ascending node is $R_z(\Omega )R_x(I)$. Its columns are the orbital frame's orthonormal basis expressed in reference coordinates. The unit vector normal is $(\sin \Omega \sin I,-\cos \Omega \sin I,\cos I)$; its dot product with another such normal gives the cosine of their mutual inclination.

Take the sky frame and disk frame to be right-handed orthonormal bases, with positive angular displacement measured from $\hat{X}$ toward $\hat{Y}$. In particular, positive $Z$ is toward the observer, so material crossing the sky plane with positive $\dot{Z}$ is at the ascending node. Write $c_d=\cos {\Omega }_d$, $s_d=\sin {\Omega }_d$, $C_d=\cos I_d$, and $S_d=\sin I_d$. The disk frame's first unit vector lies along its ascending node, and its second has positive $Z$ component. The orbital-frame rotation from inclination and node therefore givesEach column is a disk-frame unit vector in sky coordinates, so $\mathbf{X}=T\mathbf{x}$ and the inverse is $\mathbf{x}=T^{\mathsf{T}}\mathbf{X}$. At the disk's ascending node, the $Z$ component of its tangential velocity is positive, confirming the sign of the tilt.

Let $T_p=R_Z({\Omega }_p)R_X(I_p)$ define a provisional planet frame whose first axis is the planet's sky-plane ascending node. Its normal is ${\hat{z}}_p=(\sin {\Omega }_p\sin I_p,-\cos {\Omega }_p\sin I_p,\cos I_p)$. Expressing that normal in the disk orthonormal basis, with $\Delta \Omega ={\Omega }_p-{\Omega }_d$, givesThe last component is the dot product of the two orbital normals, hence the cosine of the mutual inclination. The mutual ascending node points along ${\hat{z}}_d\times {\hat{z}}_p$: motion there has positive disk-normal component. SetThen $\hat{x}\prime =(A\hat{x}+B\hat{y})/\sqrt{A^2+B^2}$, and the quadrant-correct longitude of ascending node is ${\Omega }_m=atan2(B,A)$. ThusThe tangent alone cannot fix the quadrant, and ${\Omega }_m$ is undefined for exactly coincident orbital planes.

For the two rotation routes, define $\psi$ as the oriented angle, within the planet's orbital plane, from its sky-plane ascending node to its disk-plane ascending node. If ${\hat{x}}_p,{\hat{y}}_p$ are the first two columns of $T_p$, take $\psi =atan2(\hat{x}\prime \cdot {\hat{y}}_p,\hat{x}\prime \cdot {\hat{x}}_p)$. Both routes describe the same planet orthonormal basis:The left route goes through the disk frame; the right goes through the sky-node planet frame. Comparing their third columns gives the two boxed relations, while comparison of first columns fixes the extra in-plane angle.

Put $I_d=\pi /2+{\delta }_d$ and $I_p=\pi /2+{\delta }_p$. A second-order Taylor expansion gives $\cos I_m=1-[({\delta }_p-{\delta }_d{)}^2+\Delta {\Omega }^2]/2+O({\delta }^4)$, and therefore the mutual inclination of nearly edge-on orbits satisfiesAll expansion angles are in radians. For plotting in degrees this is $I_m\simeq \sqrt{(I_p-87^{\circ }{)}^2+(4^{\circ }{)}^2}$. The given planet has exact $I_m\simeq 4.119^{\circ }$ and ${\Omega }_m\simeq 76.044^{\circ }$. The exact minimum occurs at $I_p=atan2(\sin I_d\cos \Delta \Omega ,\cos I_d)\simeq 86.993^{\circ }$, with $I_{m,\min }\simeq 3.995^{\circ }$. At $I_p=85^{\circ }$ and $95^{\circ }$, the second-order values are ${\sqrt{20}}^{\circ }\simeq 4.472^{\circ }$ and ${\sqrt{80}}^{\circ }\simeq 8.944^{\circ }$.

For the disk evolution, neglect disk self-gravity, collisions that align the planes, and back-reaction on the planet. In the initial disk plane, use the complex inclination $w=I{e}^{i\Omega }$ for a ring and $w_p=I_m{e}^{i{\Omega }_m}$ for the planet. Linear Laplace-Lagrange secular theory has forced inclination $w_p$ and solution $w(a,t)=w_p[1-e^{-i\nu (a)t}]$ for initially flat rings. For an inner circular perturber, the Laplace coefficient formula is $\nu =n(M_p/M_{\star })\alpha b_{3/2}^{(1)}(\alpha )/4$, with $\alpha =a_p/a$. Its quadrupole approximation uses $b_{3/2}^{(1)}\simeq 3\alpha$, yieldingIn the $(I\cos \Omega ,I\sin \Omega )$ plot referenced to the initial disk, every ring runs clockwise on the same circle centered at $w_p$, starting at the origin. Inner rings advance more quickly. If the axes are instead referenced to the planet's plane, the relative tilt is $w-w_p=-w_p{e}^{-i\nu t}$: these circles are centered at the origin, and the mutual inclination is constant. Both conventions describe the same nodal precession; they should not be mixed.

At an intermediate epoch, rings with $\nu t\gtrsim 1$ have tilted substantially while outer rings with $\nu t\ll 1$ remain near the original disk plane. This radial variation is a planet-induced debris-disk warp. The characteristic affected radius grows like $t^{2/7}$. Conservative differential nodal precession preserves each ring's tilt relative to the planet, so it does not by itself align every orbit; an unresolved inner region can acquire a mean plane near the planetary plane through phase mixing. The secular interpretation is also described in Wyatt's planetary dynamics lectures.

Treat $a,e$ as instantaneous osculating orbital elements of the fixed-central-parameter Kepler orbit. The perturbing work changes the specific orbital energy according toThe specific angular momentum and conic equation giveSubstituting proves the Gauss planetary equation for the semi-major axis:Here $f$ denotes true anomaly, not the fractional luminosity used in Question 1. The equation follows from work alone; a normal perturbing acceleration has no instantaneous contribution because the velocity lies in the orbital plane. For slowly averaged evolution, the perturbation must be weak over one orbital period.

Let ${\mu }_p=G{M}_p$, $A=\beta G{M}_{\star }/c$ and $\kappa =A/a_p^2$. A tightly bound circumplanetary orbit lies well inside the Hill sphere, so $a/a_p\ll 1$ and its orbital period is short compared with the planet's year. Treat the stellar flux and stellar direction $\hat{\mathbf{R}}$ as constant during one dust orbit. With stellar-frame velocity ${\mathbf{V}}_p+\mathbf{u}$, the velocity-dependent acceleration isThe terms independent of $\mathbf{u}$, including the leading static radiation pressure, do no net work on an unperturbed closed circular orbit. HenceFor a coplanar circular orbit, $u^2={\mu }_p/a$ and the component along the stellar direction has mean square $u^2/2$. Thus $\dot{a}=2a^2{\dot{\epsilon }}_p/{\mu }_p$ yields the Poynting–Robertson decay of a circumplanetary orbitThe coefficient three assumes coplanarity, which the PDF does not state. For orbital normal $\hat{\mathbf{k}}$, the general circular covariance is $⟨u_i{u}_j⟩=(u^2/2)({\delta }_{ij}-k_i{k}_j)$, givingAn orbital plane normal to the stellar radial direction has coefficient two during that orbit, a counterexample to an orientation-independent coefficient three. Averaging also over the planet's circular stellar orbit, with fixed dust-plane orbital inclination $I$ to it, gives coefficient $3-\frac{1}{2}{\sin }^{2}I$. The conservative forces must be weak enough for the assumed approximately circular, planet-bound orbit to persist; the small planet-to-star mass ratio alone does not ensure this for arbitrary grain $\beta$.

A simultaneous transit in a resonant chain requires both conjunction patterns to admit the observer's longitude at the same time. In the leading mean-conjunction geometry put $x=\Lambda -{\varpi }_c$. The necessary phase compatibility at symmetric centres isLet $d=\gcd (q,n)$. Eliminating $x$, and using that $q\ell -nk$ ranges over multiples of $d$, gives the exact modular compatibility criterionHere $\gcd$ denotes the greatest common divisor. For the orders $q,n\in \{1,2,3\}$ above, the two phase sets intersect only whenThe common directions are the two quadratures relative to periapsis. In reduced resonances this requires $p$ and $m$ odd; otherwise the apparent second-order ratios reduce to different resonance orders. There is no additional restriction on those odd values from angular compatibility alone. A shared direction and suitable temporal phase can be chosen, and the rational period ratios allow its recurrence.

This is a potential low-amplitude configuration, not a guarantee of an observed triple exoplanet transit. The observer must lie in the common orbital plane and near an allowed direction. Finite stellar radii, nonzero libration amplitudes of a resonant argument, apsidal motion and asymmetric centres broaden or change the possibilities. At finite $e_c$, true simultaneous alignment obeys the more precise equations $(p+q)y-px={\varphi }_{bc}$ and $(m+n)x-my={\varphi }_{cd}$, with $y=M_c(x)$. Thus the order-only conclusion is not a universal necessary condition for every eccentric resonant chain.

An inclined inner planet causes exterior debris disk rings to undergo nodal precession at different rates. In the quadrupole approximation, $\nu \simeq (3/4)n(M_p/M_{\star })(a_p/a{)}^2\propto a^{-7/2}$. Rings that have precessed appreciably tilt away from their original orbital plane, whereas distant rings remain near it, making an evolving warp. Conservative secular perturbations alone do not damp every ring into the planet's plane; unresolved phase mixing can instead produce a thick inner disk whose mean normal is near the forced plane.