Gravitation: 4 Problems

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1) The following formula represents the period of a pendulum.
T = 2π(l/g)
(a) What would be the period of a 1.8 m long pendulum on the moon's surface? The moon's mass is 7.34 1022 kg, and its radius is 1.74x10^6 m.
The time period T of a simple pendulum of length l is given by

Here g is the acceleration due to gravity at the place of experiment.
Now the acceleration due to gravity at the surface of a planet is given by

Here G is universal gravitation constant, M is the mass of the planet and R is the radius of the planet.
Thus the acceleration due to gravity near the surface of moon will be given by the same formula.
Here mass of the moon M = 7.34*1022 kg
Radius of moon R = 1.74*106 m
And gravitation constant G = 6.67*10-11 N m2/kg2

Thus the acceleration at the surface of moon will be
m/s2
Thus the time period of the pendulum of length l = 1.80 m on the surface of moon is given by

Thus the period of small oscillation of pendulum will be 6.6 s.

2) A force of 40.9 N is ...