Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 62

A pendulum clock has a period of 0.650 s on Earth. It is taken to another planet and found to have a period of \(0.862 \mathrm{s}\). The change in the pendulum's length is negligible. (a) Is the gravitational field strength on the other planet greater than or less than that on Earth? (b) Find the gravitational field strength on the other planet.

Short Answer

Expert verified
Answer: The gravitational field strength on the other planet is approximately 5.57 m/s².

Step by step solution

01

Find the ratio of the periods

Divide the period of the pendulum on the other planet \(T_2\) by its period on Earth \(T_1\). $$\frac{T_2}{T_1} = \frac{0.862 \mathrm{s}}{0.650 \mathrm{s}}$$ Calculate the value of the ratio: $$\frac{T_2}{T_1} = 1.326$$
02

Square the ratio of the periods

Square the ratio \(\frac{T_2}{T_1}\) to obtain the ratio of \(\frac{l_2}{g_2}\) to \(\frac{l_1}{g_1}\). $$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{1.326}{1}\right)^2 = 1.76$$
03

Determine the relationship between the gravitational field strengths

Since the lengths of the pendulum are negligibly different (the same), we can write the following equation for the ratio of gravitational field strengths: $$\frac{l_2}{g_2} = \frac{l_1}{g_1}$$ Since the ratio obtained in Step 2 is greater than 1, we conclude that the gravitational field strength on the other planet is less than that on Earth: $$(l_1 = l_2) \Rightarrow g_2 < g_1$$
04

Find the gravitational field strength on the other planet

As we know the ratio of \(g_1\) to \(g_2\) after obtaining the ratio of periods in Step 2, we can find the value of \(g_2\) using Earth's gravitational field strength, \(g_1 = 9.81 \mathrm{m/s}^2\). $$g_2 = \frac{g_1}{1.76}$$ $$g_2 = \frac{9.81 \mathrm{m/s^2}}{1.76}$$ Calculate the gravitational field strength on the other planet: $$g_2 = 5.57 \mathrm{m/s^2}$$ So the gravitational field strength on the other planet is approximately 5.57 m/s².

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A wire of length \(5.00 \mathrm{m}\) with a cross-sectional area of $0.100 \mathrm{cm}^{2}\( stretches by \)6.50 \mathrm{mm}\( when a load of \)1.00 \mathrm{kN}$ is hung from it. What is the Young's modulus for this wire?
A pendulum is made from a uniform rod of mass \(m_{1}\) and a small block of mass \(m_{2}\) attached at the lower end. (a) If the length of the pendulum is \(L\) and the oscillations are small, find the period of the oscillations in terms of \(m_{1}, m_{2}, L,\) and \(g .\) (b) Check your answer to part (a) in the two special cases \(m_{1} \gg m_{2}\) and \(m_{1}<<m_{2}\).
A bungee jumper leaps from a bridge and undergoes a series of oscillations. Assume \(g=9.78 \mathrm{m} / \mathrm{s}^{2} .\) (a) If a \(60.0-\mathrm{kg}\) jumper uses a bungee cord that has an unstretched length of \(33.0 \mathrm{m}\) and she jumps from a height of \(50.0 \mathrm{m}\) above a river, coming to rest just a few centimeters above the water surface on the first downward descent, what is the period of the oscillations? Assume the bungee cord follows Hooke's law. (b) The next jumper in line has a mass of \(80.0 \mathrm{kg} .\) Should he jump using the same cord? Explain.
In Problem \(8.41,\) we found that the force of the tibia (shinbone) on the ankle joint for a person (of weight \(750 \mathrm{N})\) standing on the ball of one foot was \(2800 \mathrm{N}\). The ankle joint therefore pushes upward on the bottom of the tibia with a force of \(2800 \mathrm{N},\) while the top end of the tibia must feel a net downward force of approximately \(2800 \mathrm{N}\) (ignoring the weight of the tibia itself). The tibia has a length of $0.40 \mathrm{m},\( an average inner diameter of \)1.3 \mathrm{cm},$ and an average outer diameter of \(2.5 \mathrm{cm} .\) (The central core of the bone contains marrow that has negligible compressive strength.) (a) Find the average crosssectional area of the tibia. (b) Find the compressive stress in the tibia. (c) Find the change in length for the tibia due to the compressive forces.
The maximum height of a cylindrical column is limited by the compressive strength of the material; if the compressive stress at the bottom were to exceed the compressive strength of the material, the column would be crushed under its own weight. (a) For a cylindrical column of height \(h\) and radius \(r,\) made of material of density \(\rho,\) calculate the compressive stress at the bottom of the column. (b) since the answer to part (a) is independent of the radius \(r,\) there is an absolute limit to the height of a cylindrical column, regardless of how wide it is. For marble, which has a density of $2.7 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\( and a compressive strength of \)2.0 \times 10^{8} \mathrm{Pa},$ find the maximum height of a cylindrical column. (c) Is this limit a practical concern in the construction of marble columns? Might it limit the height of a beanstalk?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Quantum Physics

Read Explanation

Space Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Nuclear Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free