= Solution
Immediately after release, the grain has the comet's position and velocity, but its effective stellar gravitational parameter is $\mu_d=(1-\beta)\mu$. Using the comet's <specific orbital energy>,
$$
\frac{v^2}{2}-\frac\mu r=-\frac\mu{2a},
$$
the grain energy is
$$
\frac{v^2}{2}-\frac{(1-\beta)\mu}{r}
=-\frac\mu{2a}+\frac{\beta\mu}{r}
=-\frac{(1-\beta)\mu}{2a_d}.
$$
Therefore
$$
\boxed{a_d=\frac{a(1-\beta)}{1-2\beta a/r}}.
$$
Release with zero relative velocity preserves the <specific angular momentum> $h^2=\mu a(1-e^2)$. Applying $h^2=\mu_d a_d(1-e_d^2)$ then gives
$$
\boxed{e_d^2=
\frac{e^2-2\beta+\beta^2+2\beta a(1-e^2)/r}{(1-\beta)^2}}.
$$
The paper instead prints $e^2-2e\beta+\beta^2$ in the first three terms. That expression is incompatible with both its printed $a_d$ and conservation of $h$, except at $e=1$. The next part is the result obtained from the printed eccentricity, so both consequences are recorded below.
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