Final answer:
The spaceship, launched from Earth at 6 AU/year in the direction of Earth's orbit, will follow a hyperbolic trajectory since its speed exceeds the solar escape velocity at that distance from the Sun.
Step-by-step explanation:
The subsequent orbit of a spaceship launched from Earth with a speed of 6 AU/year, in the direction of Earth's orbit around the Sun, can be determined by comparing this speed to the orbital and escape velocities at Earth's orbit around the Sun. Earth's orbital speed around the Sun is approximately 29.8 km/s, which converts to about 1 AU/year (taking the length of a year and the distance of an AU into account).
Because the spaceship is traveling at 6 times Earth's orbital speed and in the same direction, the spaceship will gain enough velocity to overcome the solar gravitational attraction and move into an open trajectory away from the Sun. This kind of trajectory is a hyperbolic orbit since it is faster than the escape velocity at that distance from the Sun. Therefore, the answer to the question would be c) Hyperbolic.