= Solution
A non-coplanar encounter may change $I$ while preserving $T_p$, so the projection of the accessible <phase space> onto the $(a,e)$ plane is a region rather than one curve. A pair $(x,e)$ is accessible for some inclination exactly when
$$
\boxed{\left|T_p-\frac1x\right|
\leq2\sqrt{x(1-e^2)}}.
$$
For $T_p=5/2$, the boundary curves are obtained by taking $\cos I=1$ or $-1$; intermediate inclinations fill the region between them. In particular, inclinations at least as large as the value found in part (d) allow the region to meet $e=0$.
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