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

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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