Suppose that a pendulum clock keeps time exactly correctly on Earth. Now suppose that you take this clock to planet X and find that the clock’s hour hand makes α revolutions every hour. Find a formula for the magnitude of the acceleration due to gravity near the surface of planet X. What would α be on the moon?

The gravity at planet X is  

The value of on the moon is  

Explanation:

  From the question we are told that

        The clocks hour hand makes revolution every 1 hour which is 3600 sec

this implies that the time peroid for 1 revolution would be  

   The peroid  for a pendulum is mathematically represented as

Where L is the pendulum length

            g is the acceleration due to gravity

Let assume that we have a pendulum that counts in second on earth

   This implies that its peroid would be = 2 second

   i.e one second to swimg forward and one second to swing back to its original position

  Now the length of this pendulum on earth is

                  [Making L the subject in above equation]

 Substituting values

When the same pendulum is taken to planet X  the peroid would be

Recall this value was obtained above for 1 revolution (from start point to end point back to start point)

      So the acceleration due to gravity on this planet would be mathematically represented as

                  [making g the subject in the above equation]

     substituting values

On moon the acceleration due to gravity has a constant value of

The period of this pendulum  on the moon can be mathematically evaluated as

 substituting value

 given that  

         {Note 1 revolution takes a peroid }

Making the subject of the formula

no

sodium is solid at rt