 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,the grain energy isThereforeRelease 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 givesThe 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.