Final answer:
The seconds pendulum on the Moon would need to have a length 1/6 times that on Earth to maintain its status as a seconds pendulum. This is due to the gravitational acceleration on the Moon being 1/6 that of Earth's gravity.
Step-by-step explanation:
The length of a seconds pendulum on the moon compared to Earth would be 1/6 times. This is because the period of a pendulum is directly proportional to the square root of its length. The formula for the period T of a pendulum is T = 2π√(L/g), where L is the length and g is the acceleration due to gravity. Since gravity on the Moon is 1/6th that of Earth, to keep the period the same (to remain a seconds pendulum), we need to adjust the length accordingly.
On Earth, where the acceleration due to gravity is g, the period is given by T = 2π√(L/g). On the Moon, with 1/6th the acceleration due to gravity, the period is T = 2π√(L_moon/(g/6)). To find the new length L_moon that results in the same period, we set the two expressions equal to each other and solve for L_moon:
L/g = L_moon/(g/6) → L_moon = L/6
Therefore, the length of the seconds pendulum on the Moon is 1/6 times the length on Earth, which corresponds to option C. The correct answer is C. 1/6 times.