/*! 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 20 In 1956, Frank Lloyd Wright prop... [FREE SOLUTION] | 91Ó°ÊÓ

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In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constructed. Ignoring Earth's rotation, find the change in your weight if you were to ride an elevator from the street level, where you weigh \(600 \mathrm{~N},\) to the top of the building.

Short Answer

Expert verified
The weight changes by approximately 0.3 N.

Step by step solution

01

Understand the Problem

We need to find the change in weight when a person moves from street level to the top of a mile-high building (approximately 1,609 meters high). The weight at street level is given as 600 N, and we are to ignore Earth's rotation.
02

Use the Gravitational Force Formula

Weight is the gravitational force on an object, given by the formula \( F = \frac{G \, M_\text{earth} \, m}{R^2} \), where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( M_\text{earth} \) is Earth's mass, \( m \) is the object's mass, and \( R \) is the distance from the center of the Earth. Since \( F = mg \), we can see that \( g = \frac{G \, M_\text{earth}}{R^2} \).
03

Calculate the New Distance from Earth's Center

The radius of the Earth (\( R_\text{earth} \)) is approximately 6,371,000 meters. At the top of the mile-high building, the total distance from Earth's center (\( R \)) becomes \( R_\text{earth} + 1,609 \, \text{m} \). Thus, \( R \approx 6,371,000 + 1,609 = 6,372,609 \, \text{m} \).
04

Determine the New Gravitational Acceleration

The new gravitational acceleration at the top of the building is \( g' = \frac{G \, M_\text{earth}}{R^2} = \frac{G \, M_\text{earth}}{(6,372,609)^2} \). However, \( g \) at street level is \( \frac{G \, M_\text{earth}}{(6,371,000)^2} \). Since \( F = mg \), the new weight \( F' = mg' = \frac{G \, M_\text{earth} \, m}{(6,372,609)^2} \).
05

Calculate the Ratio of the Change

The ratio of \( \frac{g'}{g} \) is approximately \( \frac{1}{1 + \frac{2\times1609}{6,371,000}} \) which simplifies to about \( 1 - \frac{2 \times 1609}{6,371,000} \).
06

Compute New Weight and Change

The new weight \( F' = 600 \times \left( 1 - \frac{2 \times 1,609}{6,371,000} \right) \approx 600 \left( 1 - 0.000505 \right) = 600 \times 0.999495 \approx 599.7 \, \text{N} \).
07

Determine the Change in Weight

The change in weight is \( 600 \, \text{N} - 599.7 \, \text{N} = 0.3 \, \text{N} \).

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Key Concepts

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

Weight Change
As you ascend to the top of a towering structure, such as a mile-high skyscraper, your weight experiences a subtle change. This change is due to the alteration in your distance from the center of the Earth. At street level, your weight is the force exerted on you by gravity, which depends on how far you are from the Earth's center. Moving higher means increasing that distance.

Therefore, as you take an elevator to the top of such a building, your weight decreases. In this scenario, your weight at street level is 600 N, but once you reach the apex, it shifts to around 599.7 N—a change of just 0.3 N.
  • The higher you go, the weaker the gravitational pull, because the force of gravity decreases with greater distance from the Earth’s core.
  • Thus, while the change may be slight, it demonstrates the power of gravitational forces at different heights.
This concept is a fascinating example of physics in action, showing how our environment affects gravitational forces on our body.
Gravitational Acceleration
Gravitational acceleration, often represented by the symbol \( g \), is a measure of how strong gravity pulls on objects. It is a fundamental constant on the surface of the Earth, approximately \( 9.81 \, \text{m/s}^2 \). However, this value isn't uniform everywhere and varies with distance from the Earth’s center.

As seen in the original exercise, when you ascend to a mile-high point in a building, the acceleration due to gravity decreases slightly. This decrease is calculated using the formula \( g' = \frac{G \, M_\text{earth}}{R^2} \), where \( R \) changes as altitude increases. This becomes evident as the radius of the Earth plus the building's height causes gravity's pull to reduce slightly.
  • The formula \( g' \) helps us understand how your weight changes as you increase altitude, because \( F = mg \).
  • A weaker gravitational pull results in a smaller weight.
This small shift emphasizes the importance of gravitational acceleration in the world of physics.
Physics Problem Solving
Solving physics problems requires a logical approach and a solid understanding of fundamental principles. A structured process can guide through even the toughest challenges. Step-by-step methodology, like the one presented in the original solution, is crucial.

For example, understanding how to analyze a weight change due to altered gravitational acceleration involves interpreting and applying formulas. The steps involved include:
  • Recognizing relevant information, such as given weights and heights.
  • Employing the gravitational force equation to relate weight with gravitational acceleration.
  • Evaluating how changes in height affect gravitational forces and subsequently, weight.
This logical framework aids in breaking down complex problems into more manageable parts, fostering deeper comprehension. By practicing these steps, you can significantly enhance your problem-solving capabilities in physics.

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Most popular questions from this chapter

A projectile is fired vertically from Earth's surface with an initial speed of \(10 \mathrm{~km} / \mathrm{s}\). Neglecting air drag, how far above the surface of Earth will it go?

An object of mass \(m\) is initially held in place at radial distance \(r=3 R_{E}\) from the center of Earth, where \(R_{E}\) is the radius of Earth. Let \(M_{E}\) be the mass of Earth. A force is applied to the object to move it to a radial distance \(r=4 R_{E},\) where it again is held in place. Calculate the work done by the applied force during the move by integrating the force magnitude.

What must the separation be between a \(5.2 \mathrm{~kg}\) particle and a \(2.4 \mathrm{~kg}\) particle for their gravitational attraction to have a magnitude of \(2.3 \times 10^{-12} \mathrm{~N} ?\)

Two Earth satellites, \(A\) and \(B,\) each of mass \(m,\) are to be launched into circular orbits about Earth's center. Satellite \(A\) is to orbit at an altitude of \(6370 \mathrm{~km} .\) Satellite \(B\) is to orbit at an altitude of \(19110 \mathrm{~km} .\) The radius of Earth \(R_{E}\) is \(6370 \mathrm{~km} .\) (a) \(\mathrm{What}\) is the ratio of the potential energy of satellite \(B\) to that of satellite \(A,\) in orbit? (b) What is the ratio of the kinetic energy of satellite \(B\) to that of satellite \(A,\) in orbit? (c) Which satellite has the greater total energy if each has a mass of \(14.6 \mathrm{~kg}\) ? (d) By how much?

We watch two identical astronomical bodies \(A\) and \(B\), each of mass \(m,\) fall toward each other from rest because of the gravitational force on each from the other. Their initial center-to-center separation is \(R_{i}\). Assume that we are in an inertial reference frame that is stationary with respect to the center of mass of this twobody system. Use the principle of conservation of mechanical energy \(\left(K_{f}+U_{f}=K_{i}+U_{i}\right)\) to find the following when the centerto-center separation is \(0.5 R_{i}:\) (a) the total kinetic energy of the system, (b) the kinetic energy of each body, (c) the speed of each body relative to us, and (d) the speed of body \(B\) relative to body \(A\). Next assume that we are in a reference frame attached to body \(A\) (we ride on the body). Now we see body \(B\) fall from rest toward us. From this reference frame, again use \(K_{f}+U_{f}=K_{i}+U_{i}\) to find the following when the center-to-center separation is \(0.5 R_{i}\) : (e) the kinetic energy of body \(B\) and (f) the speed of body \(B\) relative to body \(A\). (g) Why are the answers to (d) and (f) different? Which answer is correct?

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