= Solution
The <velocity-addition formula> combines the ship's station-frame <velocity> and the capsule's ship-frame <velocity>:
$$
\boxed{\frac{v_c}{c}=\frac{0.85+0.9}{1+0.85(0.9)}=\frac{350}{353}\simeq0.9915014.}
$$
For the distance convention in <10/c/i>, station-frame arrival takes $D/v_c=(353/350)\,\mathrm{day}=1.008571\,\mathrm{day}$. Transforming its two endpoint events back to the ship frame gives $\gamma_s[D/v_c-0.85D/c]$, agreeing with <10/c/i>. The capsule remains below the <speed of light> in either frame.
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