Answer:
The gravity at planet X is [tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]
The value of [tex]\alpha[/tex] on the moon is [tex]\alpha = 730.38 \ revolutions[/tex]
Explanation:
From the question we are told that
The clocks hour hand makes [tex]\alpha[/tex] revolution every 1 hour which is 3600 sec
this implies that the time peroid for 1 revolution would be [tex]= \frac{3600}{\alpha }sec[/tex]
The peroid for a pendulum is mathematically represented as
[tex]T = 2 \pi\sqrt{\frac{L}{g} }[/tex]
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
[tex]L = \frac{gT^2}{4 \pi^2}[/tex] [Making L the subject in above equation]
Substituting values
[tex]L = \frac{9.8 * (2)^2}{4 * (3.142)^2}[/tex]
[tex]= 1[/tex]
When the same pendulum is taken to planet X the peroid would be
[tex]T = \frac{3600}{\alpha }[/tex]
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
[tex]g = \frac{4 \pi L }{T^2}[/tex] [making g the subject in the above equation]
substituting values
[tex]g = \frac{4 * 3.142^2 * 1}{[ \frac{3600}{\alpha } ]^2}[/tex]
[tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]
On moon the acceleration due to gravity has a constant value of
[tex]g = 1.625 m/s^2[/tex]
The period of this pendulum on the moon can be mathematically evaluated as
[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]
substituting value
[tex]T = 2 *3.142 \sqrt{\frac{1}{1.625} }[/tex]
[tex]= 4.929s[/tex]
given that
[tex]1 \ revolution ----> 4.929s\\ \\ \alpha \ revolution -------> 3600 \ s[/tex] {Note 1 revolution takes a peroid }
Making [tex]\alpha[/tex] the subject of the formula
[tex]\alpha =\frac{3600}{4.929}[/tex]
[tex]\alpha = 730.38 \ revolutions[/tex]
