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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)

  • Physics - ,

    Use the equation:

    Period = 2 pi sqrt (L/g') = 1020/567 s
    = 1.799 s

    Solve for the acceleration of gravity on that planet, g'. It will be different from the g that exists on Earth.

    L is the pendulum length

    sqrt(L/g') = 1.799/(2 pi) = 0.2863 s

    L/g' = 0.08197 s^2
    g'/L = 12.2 s^-2
    g' = 15.7 m/s^2

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