Final answer:
If a satellite's orbital velocity is halved, the radius of its orbit must increase by a factor of four to maintain the same angular momentum. This will result in the orbital period of the satellite increasing by a factor of eight.
Step-by-step explanation:
When a satellite in orbit decreases in speed but maintains the same angular momentum, the radius of the orbit must change to compensate. According to the conservation of angular momentum L = mvr, where L is the angular momentum, m is the mass, v is the orbital velocity, and r is the radius of the orbit. Since the mass of the satellite remains constant and the angular momentum must stay the same, if the velocity decreases, the radius r must increase.
Specifically, from Kepler's third law and the conservation of angular momentum, we can derive that if the orbital velocity decreases by a factor of two (reduces to half its original value), the radius must increase by a factor of four to maintain the same angular momentum. This increase in radius will also affect the orbital period, which according to Kepler's third law is proportional to the radius to the power of 3/2. Thus, the orbital period would increase by a factor of 23 or eight times the original period.