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31 March, 09:40

An astronaut has landed on an asteroid and conducts an experiment to determine the acceleration of gravity on that asteroid. He uses a simple pendulum that has a period of oscillation of 2.00 s on Earth and finds that on the asteroid the period is 11.3 s. What is the acceleration of gravity on that asteroid

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  1. 31 March, 13:06
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    g' = 0.31 m/s²

    Explanation:

    First we need to find the length of pendulum, by using the formula of the time period of pendulum:

    T = 2π√ (L/g)

    L = T²g/4π²

    where,

    L = Length of Pendulum = ?

    T = Time Period of Pendulum on Earth = 2 s

    g = acceleration due to gravity on Earth = 9.8 m/s²

    Therefore,

    L = (2 s) ² (9.8 m/s²) / 4π²

    L = 0.99 m

    Now, we use the formula on asteroid:

    T' = 2π√ (L/g')

    where,

    T' = Time Period on Asteroid = 11.3 s

    g' = acceleration due to gravity on asteroid = ?

    Therefore,

    11.3 s = 2π √ (0.99 m/g')

    (11.3 s) ² = 4π² (0.99 m/g')

    g' = 39.2 m / (11.3 s) ²

    g' = 0.31 m/s²
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