= Solution
At infinity, the meteorite's incoming asymptote has <speed> $V$ and perpendicular distance $b$ from the Earth’s centre. The magnitude of its <specific angular momentum> is the magnitude of $\mathbf r\times\mathbf v$. On the asymptote, this is $bV$, so <conservation of angular momentum> gives $\boxed{h=bV}$. Its <mechanical energy> at infinity is $E=mV^2/2$. Comparing with part (ii) gives
$$
\boxed{e=\sqrt{1+\frac{b^2V^4}{G^2M^2}}}.
$$
For $b>0$ this <Kepler orbit> is a <hyperbola>. The smallest $r=1/u$ occurs at <periapsis>, where $\cos(\theta-\theta_0)=1$. Thus
$$
\boxed{r_{\min}=\frac{b^2V^2}{GM(1+e)}=\frac{GM}{V^2}(e-1)
=\frac{\sqrt{G^2M^2+b^2V^4}-GM}{V^2}.}
$$
The middle equality follows by substituting $e^2-1=b^2V^4/(G^2M^2)$. A trajectory entering the Earth satisfies $r_{\min}<R$, equivalently
$$
e<1+\frac{RV^2}{GM}\iff b^2<R^2+\frac{2GMR}{V^2}.
$$
Therefore the sufficient collision condition is
$$
\boxed{b<\sqrt{R^2+\frac{2GMR}{V^2}}.}
$$
This is <gravitational focusing>: <Newtonian gravity> increases the collision radius measured at infinity. Equality is the ideal grazing trajectory, which touches the surface; a strict inequality penetrates it. For $b=0$ the incoming motion is radial and collides, with $r_{\min}=0$ as the limiting value of the above expression.
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