= Solution
The transfer ellipse has <semi-major axis> and <semi-minor axis>
$$
a=\frac{r_1+r_2}{2},
\qquad
b=a\sqrt{1-e^2}=\sqrt{r_1r_2}.
$$
Conservation of angular momentum gives constant <areal velocity> $h/2$. The satellite sweeps half the ellipse, whose area is $\pi ab/2$, so
$$
t_2-t_1=\frac{\pi ab/2}{h/2}=\frac{\pi ab}{h}.
$$
Using part (i),
$$
\boxed{t_2-t_1
=\frac{\pi(r_1+r_2)^{3/2}}{2\sqrt{2k}}}
=\pi\sqrt{\frac{a^3}{k}}.
$$
This is half the orbital period from <Kepler third law>[Kepler's third law].
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