Astronauts on the first trip to mars take along a pendulum that has a period on earth of 1.09 s. the period on the other planet turns out to be 3.23 s. what is the free-fall acceleration on that other planet?

Respuesta :

Not sure exactly. I am not sure what this
T=2pi* sqrt(l/g)
1.09=2*3.14*sqrt(l/9.8)
=> l=0.295
3.23=2*3.14* sqrt(0.295/gg)
=> gg=1.116