For the nearly <edge-on orbit> $I=\pi/2-I'$ with $I'\ll1$, the <small-angle approximation> gives $\cos I=\sin I'=I'+O(I'^3)$. Hence
$$
\boxed{\tan(\phi-\Omega)=I'\tan f+O(I'^3)}.
$$
Also,
$$
R_{\rm sky}=a|\cos f|\sqrt{1+\cos^2I\tan^2f}
=a|\cos f|\left[1+\frac12(I'\tan f)^2+O(I'^4)\right].
$$
Thus, on a branch where $\cos f\gt0$ and with separation measured in units of $a$,
$$
\boxed{R_{\rm sky}\simeq\left[1+\frac12(I'\tan f)^2\right]\cos f},
$$
as stated. The absolute value is required when $R_{\rm sky}$ denotes the nonnegative separation over the whole orbit.

Solution

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

 Solution

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