The oriented <normal vectors> to the planet and dust-belt planes are
$$
\widehat{\mathbf h}
=(\sin\Omega\sin I,-\cos\Omega\sin I,\cos I),
$$
$$
\widehat{\mathbf h}_d
=(\sin\phi_d\sin I_d,-\cos\phi_d\sin I_d,\cos I_d).
$$
Their <inner product> is the cosine of the <mutual inclination>, so
$$
\boxed{\cos I_m
=\cos I\cos I_d+\sin I\sin I_d\cos(\Omega-\phi_d)}.
$$
If $\phi_d=\Omega$, 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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-316/1/v/solution