= Solution
At closest approach the <velocity> is tangential. The conserved specific <orbital energy> and <specific angular momentum> are
$$
\varepsilon=\frac{v_p^2}{2}-\frac{GM}{r_p}=-2.1972\times10^8\,\mathrm{m^2s^{-2}},\qquad\ell=r_pv_p.
$$
The negative energy gives an <elliptic orbit>. Its <semi-major axis> is
$$
a=-\frac{GM}{2\varepsilon}=3.0376\times10^{11}\,\mathrm m.
$$
Since $r_p+r_a=2a$,
$$
\boxed{r_a=2a-r_p=5.58\times10^{11}\,\mathrm m.}
$$
This treats the star as fixed, appropriate to $m\ll M$. Equivalently, equate the energy at the two turning points using $v_a=\ell/r_a$. The planet <mass> cancels; a circular-orbit relation would be inappropriate here.
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