Final answer:
An orbit nine times farther from a black hole will have a circular velocity that is one-third (slower) than that of a closer orbit, according to Kepler's third law and the inverse square law of gravity.
Step-by-step explanation:
An orbit that is nine times farther from a black hole will have a circular velocity that is slower than the closer orbit. This is explained by Kepler's third law and the fact that gravitational force decreases with distance. According to this law, the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. This implies that if the orbital radius increases, the orbital velocity decreases because the gravitational pull becomes weaker. In the context of a black hole, the same principle applies.
As an example, Kepler's Laws anticipate that the orbital velocity is inversely proportional to the square root of the distance from the body being orbited. Since the radius increases by a factor of nine, the orbital velocity will decrease by a factor of three, because √9 is 3. Therefore, if you are nine times farther away from the black hole, the orbital velocity is one-third of what it would be at the original distance.