= Solution
Eliminate $v_p,v_a$ using the <angular momentum> expressions from part (b):
$$
\frac{J^2}{2mr_p^2}-\frac{GMm}{r_p}
=\frac{J^2}{2mr_a^2}-\frac{GMm}{r_a}.
$$
Hence
$$
\frac{J^2}{2m}\left(\frac1{r_p^2}-\frac1{r_a^2}\right)
=GMm\left(\frac1{r_p}-\frac1{r_a}\right).
$$
For distinct turning radii, factor the difference of squares and cancel the common factor to obtain
$$
\boxed{J^2=\frac{2GMm^2}{1/r_p+1/r_a}
=\frac{2GMm^2r_pr_a}{r_p+r_a}.}
$$
The <circular Kepler orbit> case is recovered continuously by setting $r_p=r_a=r$, giving $J^2=GMm^2r$. Combining this with the energy expression also gives $E=-GMm/(r_p+r_a)$ for the transfer <ellipse>.
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