Posted by dallas on .
Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.29 m and finds that it makes 407 complete oscillations in 698 s. The amplitude of the oscillations is very small compared to the pendulumâ€™s length.
What is the gravitational acceleration on the surface of this planet?
Answer in units of m/s2

Physics 
drwls,
Use the smallamplitude formula:
Period = 2*pi*sqrt(L/g)
and solve for g.
L = 1.29 m
Period P = 698/407 = 1.715 seconds
g/L = 4*pi^2/P^2 = 13.42 s^2
g = 17.31 m/s^2