An astronaut has landed on an asteroid and conducts an experiment to determine the free fall acceleration on that asteroid.He uses a simple pendulum that has a period of oscillation of 2.00 s on Earth and finds that on the asteroid the period is 11.3 s.What is the free fall acceleration on that asteroid?

Question

Quizplus · Accepted Answer

The period of a pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$, where $T$ is the period, $L$ is the length of the pendulum, and $g$ is the acceleration due to gravity. Since the length of the pendulum doesn't change, the ratio of the squares of the periods on Earth and the asteroid will be equal to the inverse ratio of the gravitational accelerations. Thus, $\left(\frac{T_{asteroid}}{T_{Earth}}\right)^2 = \frac{g_{Earth}}{g_{asteroid}}$. Given $T_{Earth} = 2.00s$, $T_{asteroid} = 11.3s$, and $g_{Earth} = 9.81 m/s^2$, solving for $g_{asteroid}$ gives approximately $0.307 m/s^2$.

An Astronaut Has Landed on an Asteroid and Conducts an Experiment