Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 1

Put the star at the origin, take $\hat{\mathbf{z}}$ along the line of sight toward the observer, and let $\hat{\mathbf{x}}$ point North in the plane of the sky. The $\hat{\mathbf{y}}$ direction is then $90^{\circ }$ anticlockwise from North. Draw the ascending node on the sky plane at position angle $\Omega$, tilt the circular Kepler orbit through the orbital inclination $I$ about that nodal line, and place the planet an angle $f$ along the orbit from the ascending node. The direction of increasing $f$ must cross the sky plane toward $+\hat{\mathbf{z}}$ at $f=0$.

A convenient orthonormal basis in the orbital plane isHere ${\mathbf{e}}_1$ points along the nodal line and ${\mathbf{e}}_2$ fixes the required sense of motion. The planet lies at $\mathbf{r}=a(\cos f{\mathbf{e}}_1+\sin f{\mathbf{e}}_2)$; this relation also specifies all the labels required in the sketch.

Resolving $\mathbf{r}=a(\cos f{\mathbf{e}}_1+\sin f{\mathbf{e}}_2)$ into the sky coordinates givesIn particular, $z=0$ at the orbital nodes, and $dz/df=a\sin I>0$ at $f=0$, so that node is indeed the ascending node.

The projected separation isRotating the projected coordinates through $-\Omega$ gives components $a\cos f$ along the nodal line and $a\cos I\sin f$ perpendicular to it. The position angle is therefore most safely written with the quadrant-preserving functionor, wherever the corresponding tangent is finite,

For the nearly edge-on orbit $I=\pi /2-I^{\prime }$ with $I^{\prime }\ll 1$, the small-angle approximation gives $\cos I=\sin I^{\prime }=I^{\prime }+O(I^{\prime 3})$. HenceAlso,Thus, on a branch where $\cos f>0$ and with separation measured in units of $a$,as stated. The absolute value is required when $R_{\mathrm{sky}}$ denotes the nonnegative separation over the whole orbit.

The oriented normal vectors to the planet and dust-belt planes areTheir inner product is the cosine of the mutual inclination, soIf ${\varphi }_d=\Omega$, this reduces to $\cos I_m=\cos (I-I_d)$; equal inclinations then give $I_m=0$. If both planes are face-on, it likewise gives $I_m=0$, independently of their undefined nodal longitudes. Reversing an observationally unidentified ascending node by $\pi$ produces the familiar orbital-plane orientation degeneracy for an axisymmetric belt.

Let $\Delta =\Omega -{\varphi }_d$. From part (iii), $\varphi -{\varphi }_d=\Delta +(\varphi -\Omega )$ and $\tan (\varphi -\Omega )=\cos I\tan f$. The tangent addition formula therefore givesor, without singular coordinate tangents,For $I=\pi /2-I^{\prime }$ with $I^{\prime }\ll 1$, smooth choice of the angular branch gives

Coplanarity requires both $I=I_d$ and $\Omega ={\varphi }_d$, up to the orbital-plane orientation degeneracy. Under that hypothesis the measured planet angle must obeyFor a nearly edge-on belt this offset is small except close to the projected crossings where $\cos f=0$. Measurements at several orbital phases can therefore test the entire projected ellipse: a systematic offset from the predicted relation implies a nonzero mutual inclination. A single projected location generally cannot establish coplanarity because $f$, the front-back branch, and the nodal orientation can be degenerate.

An exterior misaligned planet exerts an orbit-averaged secular torque on each belt orbit. In the hierarchical limit $a_d\ll a$, its leading quadrupole approximation drives nodal precession about the planet's orbital angular-momentum axis at a rate of orderup to a coefficient of order unity. Because this rate varies across a radially extended belt, differential nodal precession winds an initially flat belt into a warp and eventually a vertically thick, phase-mixed structure. Collisions can damp the free inclinations and make material approach a forced or Laplace plane; at sufficiently large mutual inclination, coupled eccentricity-inclination evolution such as the Kozai–Lidov mechanism may also become important.