Question

Astronauts on the first trip to Mars take along a pendulum that has a period on earth of 1.26 s. The period on the other planet turns out to be 3.82 s. What is the free-fall acceleration on that other planet?

Answer 1

Answer:

1.066 m/s^2

Explanation:

On the earth and on Mars respectively, the time periods of the pendulums are expressed as:

Tearth = 2pi * sqrt(L/gearth)

Tmars = 2pi * sqrt(L/gMars)

Divide the two equations above:

Tearth/Tmars = sqrt(gMars/gearth)

gMars = gearth(Tearth/Tmars)^2 = (9.80m/s^2)(1.26s/3.82s)^2 = 1.066 m/s^2