A pendulum that on Earth has period 0.52 s is taken to the surface of a large asteroid, where it is measured to have period 1.7 s. What is the value of the gravitational acceleration on the surface of that asteroid?
T₁=2π•sqrt{L/g₁}
T₂=2π•sqrt{L/g₂}
g₂=g₁(T₁/T₂)²
To find the value of the gravitational acceleration on the surface of the asteroid, we can use the formula for the period of a pendulum:
T = 2π√(L/g)
Where T represents the period, L represents the length of the pendulum, and g represents the gravitational acceleration.
Given that the period of the pendulum on Earth is 0.52 s and the period on the asteroid is 1.7 s, we can set up two equations:
For the Earth pendulum:
0.52 = 2π√(L/earth_g)
For the asteroid pendulum:
1.7 = 2π√(L/asteroid_g)
We can divide the two equations to eliminate the length (L) of the pendulum:
0.52/1.7 = √(earth_g/asteroid_g)
To find the gravitational acceleration on the surface of the asteroid, we solve for asteroid_g:
(asteroid_g/earth_g) = (1.7/0.52)^2
asteroid_g = earth_g * (1.7/0.52)^2
Now we know that the gravitational acceleration on Earth is approximately 9.8 m/s^2. We can substitute this value into the equation:
asteroid_g = 9.8 * (1.7/0.52)^2
Using a calculator, we can evaluate the expression:
asteroid_g ≈ 31.42 m/s^2
Therefore, the value of the gravitational acceleration on the surface of the asteroid is approximately 31.42 m/s^2.