Question

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?

Answer 1

The gravity at planet X is
`g= 3.05*10^(-6) \alpha^2 m/s^2`

The value of
`\alpha` on the moon is
`\alpha = 730.38 \ revolutions`

Step-by-step explanation:

From the question we are told that

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

this implies that the time peroid for 1 revolution would be
`= (3600)/(\alpha )sec`

The peroid for a pendulum is mathematically represented as

`T = 2 \pi\sqrt{(L)/(g) }`

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

`L = (gT^2)/(4 \pi^2)` [Making L the subject in above equation]

Substituting values

`L = (9.8 * (2)^2)/(4 * (3.142)^2)`

`= 1`

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

`T = (3600)/(\alpha )`

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

`g = (4 \pi L )/(T^2)` [making g the subject in the above equation]

substituting values

`g = (4 * 3.142^2 * 1)/([ (3600)/(\alpha ) ]^2)`

`g= 3.05*10^(-6) \alpha^2 m/s^2`

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

`g = 1.625 m/s^2`

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

`T = 2\pi \sqrt{(L)/(g) }`

substituting value

`T = 2 *3.142 \sqrt{(1)/(1.625) }`

`= 4.929s`

given that

`1 \ revolution ----> 4.929s\\ \\ \alpha \ revolution -------> 3600 \ s` {Note 1 revolution takes a peroid }

Making
`\alpha` the subject of the formula

`\alpha =(3600)/(4.929)`

`\alpha = 730.38 \ revolutions`