/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Astronauts on the first trip to ... [FREE SOLUTION] | 91Ó°ÊÓ

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Astronauts on the first trip to Mars take along a pendulum that has a period on earth of 1.50 s. The period on Mars turns out to be 2.45 s. Use this data to calculate the Martian free-fall acceleration.

Short Answer

Expert verified
The Martian free-fall acceleration is approximately \(3.71 m/s^2\).

Step by step solution

01

Start with the Formula for the Period of a Pendulum

First, we start with the formula for the period of a pendulum: \(T = 2\pi \sqrt{\frac{l}{g}}\). It is important to note that, since the pendulum's length does not change, the ratio of the periods is equal to inversely proportional square root of the ratio of the gravitational accelerations.
02

Set Up the Equations

Next, we set up the equations. Since the ratio of the periods is equal to inversely proportional square root of the ratio of the gravitational accelerations, we can write the equation as: \(\frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}}\) where T1, g1 are the period and gravitational acceleration on earth, and T2, g2 are the period and gravitational acceleration on Mars.
03

Solve for Martian Gravitational Acceleration

Substitute the given values into the equation \(\frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}}\). We know \(T_1=1.5s, T_2=2.45s, g_1=9.8 m/s^2\). We solve for g2, which represents Martian gravitational acceleration. We get \(g_2 = g_1 \times \left(\frac{T_1}{T_2}\right)^2\). Putting in the values, we calculate \(g_2 = 9.8 m/s^2 \times \left(\frac{1.5s}{2.45s}\right)^2 = 3.71 m/s^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pendulum Period
The pendulum period refers to the time it takes for a pendulum to complete one full swing, back and forth. On earth, variables that affect the pendulum's period include the length of the pendulum and the local gravitational acceleration, represented by the formula:\[ T = 2\pi \sqrt{\frac{l}{g}} \]where:- \(T\) is the period,- \(l\) is the pendulum's length,- \(g\) is the acceleration due to gravity.The length remains constant whether on Earth or Mars, making the period largely dependent on the gravitational acceleration. If the gravity changes, as it does when you compare Earth and Mars, the period will likewise change. This is why a pendulum that takes 1.50 seconds on Earth swings with a different period, 2.45 seconds, on Mars.
Gravity on Mars
Gravity on Mars is significantly weaker than on Earth. While Earth's gravitational acceleration is approximately 9.8 m/s², Mars exhibits only about 3.71 m/s². This affects not only the behavior of pendulums but also everything from human movement to the way vehicles operate on the Martian surface. Understanding gravity on Mars is crucial for activities like landing spacecraft or operating rovers. Differences in gravity influence: - Walking and jumping - The design of buildings and machinery - The fuel needed for rockets to take off The lower gravity means that objects feel lighter, and you could jump higher than on Earth.
Free-Fall Acceleration
Free-fall acceleration refers to the acceleration of an object solely under the influence of gravity. It excludes any effects from air resistance or other forces. On Earth, this is standardized as 9.8 m/s²; on Mars, it’s lower at approximately 3.71 m/s². This acceleration affects how fast and how an object falls, with the speed increasing linearly over time. The acceleration of free-fall is consistent for all objects, regardless of their weight, which means that all objects fall at the same rate in a vacuum. This concept is vital to understanding how objects would behave when dropped on Mars compared to Earth.
Gravitational Acceleration Calculation
The calculation of gravitational acceleration on Mars using a pendulum involves understanding the inverse relationship between periods and gravitational forces. For a pendulum swinging on Mars, the relationship can be expressed as:\[ \frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} \]where:- \(T_1\) and \(T_2\) are the pendulum periods on Earth and Mars,- \(g_1\) and \(g_2\) are the gravitational accelerations on Earth and Mars.We rearrange for \(g_2\), leading to:\[ g_2 = g_1 \times \left(\frac{T_1}{T_2}\right)^2 \]Using the values \(T_1 = 1.5\) s, \(T_2 = 2.45\) s, and \(g_1 = 9.8\, \text{m/s}^2\), we find \(g_2 = 3.71\, \text{m/s}^2\). This gives us the Martian free-fall acceleration, which describes the weaker gravity compared to Earth. Understanding this formula helps in calculating how different conditions affect movement and energy transfer on Mars.

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Most popular questions from this chapter

An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a period of 0.50 s. His older sister pulls the spring a bit more than intended. She pulls the animal \(30 \mathrm{cm}\) below its equilibrium position, then lets go. The animal flies upward and detaches from the spring right at the animal's equilibrium position. If the animal does not hit anything on the way up, how far above its equilibrium position will it go?

In taking your pulse, you count 75 heartbeats in 1 min. What are the period (in s) and frequency (in \(\mathrm{Hz}\) ) of your heart's oscillations?

Vision is blurred if the head is vibrated at \(29 \mathrm{Hz}\) because the vibrations are resonant with the natural frequency of the eyeball held by the musculature in its socket. If the mass of the eyeball is 7.5 g, a typical value, what is the effective spring constant of the musculature attached to the eyeball?

A block with a mass of \(0.28 \mathrm{kg}\) is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of \(1.0 \mathrm{N}\) on the block. When the block is released, it oscillates with a frequency of \(1.2 \mathrm{Hz} .\) How far was the block pulled back before being released?

An air-track glider attached to a spring oscillates between the \(10 \mathrm{cm}\) mark and the \(60 \mathrm{cm}\) mark on the track. The glider completes 10 oscillations in 33 s. What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?

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