Answer
Calculate g at the approximate height of the shuttle.
$$g= G \frac{M_{Earth}}{d^{2}}$$
$$g= (6.67 \times 10^{-11} \frac {N \cdot m^{2}}{kg^{2}}) \frac{(6.0 \times 10^{24} kg)}{((6380 + 200) \times 10^{3} m))^{2}} = 9.24 \frac{m}{s^{2}} $$
This is not zero. If Earth's gravity at that height were zero, an unpowered space shuttle would feel no force, and move in a straight line at constant velocity.
