The period of a pendulum (for short swings) is about 2 PI (L/g)1/2. The gravity on the moon is less than that on Earth by a factor of six, so the period of the pendulum on the moon would be greater, i.e. slower, by about a factor of 2.5.
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As the force of gravity increases the period would decrease. So shortest period on the sun (if you can keep it intact), then sea level, then mountain top and then moon.
The period is not likely to be charged. However, it would change due to the weaker gravitational force on the moon. Since the surface gravity of the moon is 0.165 that of the earth, the period would increase by a multiple of 1/sqrt(0.165) = 2.462 approx.
... dependent on the length of the pendulum. ... longer than the period of the same pendulum on Earth. Both of these are correct ways of finishing that sentence.
Yes. The period of the pendulum (the time it takes it swing back and forth once) depends on the length of the pendulum, and also on how strong gravity is. The moon is much smaller and less massive than the earth, and as a result, gravity is considerably weaker. This would make the period of a pendulum longer on the moon than the period of the same pendulum would be on earth.
The period of a simple pendulum swinging at a small angle is approximately 2*pi*Sqrt(L/g), where L is the length of the pendulum, and g is acceleration due to gravity. Since gravity on the moon is approximately 1/6 of Earth's gravity, the period of a pendulum on the moon with the same length will be approximately 2.45 times of the same pendulum on the Earth (that's square root of 6).