Question

You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel with the clock to the distant planet CornTeen and measure the period of the clock to be 4s. By what factor is the gravitational acceleration constant g different on planet CornTeen compared to g on Earth

Answer 1

`(g_(2))/(g_(1)) = (1)/(4)`

Step-by-step explanation:

The period of the simple pendulum is:

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

Where:

`l` - Cord length, in m.

`g` - Gravity constant, in
`(m)/(s^(2))`.

Given that the same pendulum is test on each planet, the following relation is formed:

`T_(1)^(2)\cdot g_(1) = T_(2)^(2)\cdot g_(2)`

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

`(g_(2))/(g_(1)) = \left((T_(1))/(T_(2)) \right)^(2)`

`(g_(2))/(g_(1)) = \left((2\,s)/(4\,s) \right)^(2)`

`(g_(2))/(g_(1)) = (1)/(4)`