Answer: Part A:
The period of a simple pendulum is given by the formula: T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. If the mass of the bob is doubled, the period of the pendulum will change. To see how much it changes, we can use the fact that the period depends only on the length of the pendulum and the acceleration due to gravity, and not on the mass of the bob. Therefore, the new period of the pendulum will be the same as the old period: T.
Part B:
If the pendulum is brought to the moon, where the acceleration due to gravity is about g/6, the new period of the pendulum can be found using the same formula as before: T = 2π√(L/g). However, now we need to use the value of the acceleration due to gravity on the moon, which is g/6. Therefore, the new period of the pendulum is T' = 2π√(L/(g/6)) = 2π√(6L/g) = √6T.
Therefore, the answer is: T/√6.
Part C:
If the pendulum is taken into an orbiting space station, the bob will continue to oscillate in a vertical plane with the same period as it did on the surface of the Earth. This is because the period of the pendulum depends only on the length of the pendulum and the acceleration due to gravity, and not on the location of the pendulum. In the space station, both the pendulum and the point to which it is attached are in free fall, but they are falling together and maintaining their relative positions, so the pendulum will continue to oscillate as before.