Question

Imagine you are an astronaut who has landed on another planet and wants to determine the free-fall acceleration on that planet. In one of the experiments you decide to conduct, you use a pendulum \(0.50 \mathrm{~m}\) long and find that the period of oscillation for this pendulum is \(1.50 \mathrm{~s}\). What is the acceleration due to gravity on that planet?

Short answer

Answer: The acceleration due to gravity on this planet is approximately \(6.076 \mathrm{~m/s^2}\).

Step-by-step solution

Write down the known values

We know the length of the pendulum, L = 0.50 m and the period of oscillation, T = 1.50 s.

Write the formula for the period of oscillation

The formula for the period of oscillation of a pendulum is given by \(T=2\pi\sqrt{\frac{L}{g}}\).

Plug the known values into the formula

We know the values of T and L, so our equation becomes: \(1.50 = 2\pi\sqrt{\frac{0.50}{g}}\)

Solve for the acceleration due to gravity, g

To solve for g, we need to isolate g in the equation. First, we can divide both sides of the equation by 2\(\pi\), which results in: \(\frac{1.50}{2\pi}=\sqrt{\frac{0.50}{g}}\) Next, we square both sides of the equation to eliminate the square root: \(\left(\frac{1.50}{2\pi}\right)^2 = \frac{0.50}{g}\) Now, we can cross multiply to isolate g: \(g = \frac{0.50}{\left(\frac{1.50}{2\pi}\right)^2}\)

Compute the value of g

Compute the value of g using a calculator: \(g = \frac{0.50}{\left(\frac{1.50}{2\pi}\right)^2} \approx 6.076 \mathrm{~m/s^2}\)

State the final answer

The acceleration due to gravity on this planet is approximately \(6.076 \mathrm{~m/s^2}\).

Key concepts

Acceleration Due to Gravity

The acceleration due to gravity, usually denoted as \( g \), describes how fast an object speeds up as it falls freely toward the surface of a celestial body, like a planet. On Earth, the standard acceleration due to gravity is about \( 9.81 \, \text{m/s}^2 \), but this value varies on other planets. It's essential for understanding the dynamics of objects on any given planet.

When you're landing on another planet, as in the example of the astronaut's experiment, determining this value helps in various scientific studies and engineering applications. You use a simple pendulum to help with the calculations because it provides a straightforward method to measure \( g \).

In this specific exercise, by swinging the pendulum and measuring its period of oscillation, this process allows for the determination of \( g \) on the new planet using the equation \( T = 2\pi\sqrt{\frac{L}{g}} \), which is derived from physics principles governing pendular motion.

Period of Oscillation

The period of oscillation, represented as \( T \), is the time it takes for a pendulum to complete one full back-and-forth swing. This property can be affected by various factors, but in this instance, for a simple pendulum, it mainly depends on the length of the pendulum \( L \) and the local gravitational acceleration \( g \).

The period is calculated using the formula \( T = 2\pi\sqrt{\frac{L}{g}} \). This shows a direct relationship between the length of the pendulum and gravity. As the length \( L \) increases, the period \( T \) also increases, given constant \( g \). Conversely, if the gravitational acceleration \( g \) increases, the period decreases for a constant \( L \).

Understanding the period of a pendulum is not only crucial in mechanics but also in applications such as timekeeping, where pendulums are used in earlier clock designs for their consistent periodical motion.

Free-Fall Acceleration

Free-fall acceleration is the acceleration of an object when it is moving only under the force of gravity, with no other forces acting on it. In any context outside of Earth, determining the free-fall acceleration is crucial, since it directly influences all motion dynamics experienced on the planet or celestial body.

In the astronaut's experiment with the pendulum, discovering the free-fall acceleration of the unknown planet helps us understand how objects would move or fall if released from a height. The formula \( g = \frac{0.50}{\left(\frac{1.50}{2\pi}\right)^2} \) used in the exercise, calculated this value for the planet to be approximately \( 6.076 \, \text{m/s}^2 \).

Experiments like this are fundamental for space missions, where knowing the exact free-fall acceleration ensures the designs and strategies for missions are safe and effective. It also influences activities like walking, driving, or any action involving motion or balance.