An astronaut lands on a distant planet and places a mathematical pendulum with a length of 50 cm. He measures that the pendulum makes 100 oscillations in 136 sec. What is the gravity acceleration on that planet.
T = 136s/100cycles. = 1.36 s./cycle.
T^2 = 39.48*L/g = (1.36)^2.
L/g = 0.0468, g = 0.50/0.0468 = 10.7 m/s^2.
To find the gravity acceleration on the distant planet, we can use the formula for the period of a simple pendulum:
T = 2 * π * √(L / g)
Where:
T is the period of the pendulum,
L is the length of the pendulum, and
g is the acceleration due to gravity.
We are given the length of the pendulum (L = 50 cm) and the number of oscillations in a given time (100 oscillations in 136 seconds). Let's calculate the period of the pendulum first:
Period (T) = Total time / Number of oscillations
T = 136 sec / 100 oscillations
Now, we can substitute the values of T and L into the formula and solve for g:
136 sec / 100 oscillations = 2 * π * √(50 cm / g)
Simplifying the equation:
1.36 sec/oscillation = 2 * π * √(50 cm / g)
Now, let's isolate g:
√(50 cm / g) = 1.36 sec/oscillation / (2 * π)
Square both sides of the equation:
50 cm / g = (1.36 sec/oscillation / (2 * π))^2
Solving for g:
g = 50 cm / ((1.36 sec/oscillation / (2 * π))^2)
Evaluating the expression:
g ≈ 9.81 cm/sec^2
Therefore, the gravity acceleration on that distant planet is approximately 9.81 cm/sec^2.