$解：最远点在F到圆O的切线的切点，即点Q$
$∵FQ与圆O相切，∠FQO=90°$
$PF=350\ \mathrm {km}，PO=OQ=6400\ \mathrm {km}，FO=6400+350=6750\ \mathrm {km}$
$∴cosα=\frac {OQ}{OF}=\frac {6400}{6750}$
$∴α≈18.53° \frac {α}2≈9.27°$
$∴PQ=2×6400×sin {\frac {α}2}≈2061.0\ \mathrm {km}$
