Worked example 11.5: Gravity on a new planet

Question: Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length $0.6 {\rm m}$, and finds that it makes 51 complete oscillations in 1 minute. The amplitude of the oscillations is small compared to the length of the pendulum. What is the surface gravitational acceleration on the planet?

Answer: The frequency of the oscillations is

$$f = \frac{51}{60} = 0.85 {\rm Hz}.$$

Hence, the angular frequency is

$$\omega = 2 \pi f = 2\times \pi\times 1.833 = 5.341 {\rm rad./s}.$$

Now, $\omega = \sqrt{g/l}$ for small amplitude oscillations of a simple pendulum. Rearrangement off this formula gives

$$g = \omega^2 l = 5.341\times 5.341\times 0.6 = 17.11 {\rm m/s^2}.$$

Hence, the surface gravitational acceleration is $17.11 {\rm m/s^2}$.