Answer
$v=1.3\times10^{8}m/s$.
Work Step by Step
The magnetic force must point toward the center of the Earth and provide the centripetal force.
$$qvB=ma=m\frac{v^2}{r}$$
The radius of the circular path is the Earth’s radius plus 6.0 km.
$$v=\frac{qBr}{m}$$
$$v=\frac{(1.60\times10^{-19}C)(0.50\times10^{-4}T)(6.386\times10^6 m)}{238(1.66\times10^{-27}kg)}$$
$$v=1.3\times10^{8}m/s$$
We may ignore gravity; the magnetic force qvB is over a hundred million times larger than the weight, mg.
