posted by Katy on .
Having landed on a newly discovered planet, an astronaut sets up a simple pedulum of length 1.29 m and finds that it makes 567 oscillations in 1020 s. The amplitude of the oscillations is very small compared to the pendulum's length.
What is the gravitational acceleration on the surface of this planet? (Answer in units of m/s squared)
This is a duplicate post. I answered it yesterday.
Use the formula for the period:
P = 2 pi sqrt (L/g') and solve for g'
The period is 1020/567 s. Convert to decimal.