/*! 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 14 Empire State Building You weigh ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 14

Empire State Building You weigh \(530 \mathrm{~N}\) at sidewalk level outside the Empire State Building in New York City. Suppose that you ride from this level to the 102 nd floor tower, a height of \(373 \mathrm{~m}\). Ig- noring Earth's rotation, how much less would you weigh there (because you are slightly farther from the center of Earth)?

Short Answer

Expert verified
0.82 N

Step by step solution

01

Identify the Initial and Final Distances

Determine the initial distance from the Earth's center to the sidewalk level (approximately the Earth's radius, denoted as R) and the final distance after going up to the 102nd floor. Initial distance: \[ R = 6.371 \times 10^6 \text{ meters} \] Final distance: \[ R + h = 6.371 \times 10^6 \text{ meters} + 373 \text{ meters} \]
02

Calculate Gravitational Force at Sidewalk Level

The weight at sidewalk level can be calculated by the formula \[ W = mg \] where W is the weight (530 N) and g is the gravitational acceleration (approximately 9.8 m/s虏). Using this formula, find the mass: \[ m = \frac{W}{g} = \frac{530 \text{ N}}{9.8 \text{ m/s}^2} = 54.08 \text{ kg} \]
03

Calculate Gravitational Acceleration at Height

Gravitational acceleration at a height h is given by: \[ g_h = g \left( \frac{R}{R + h} \right)^2 \] Substituting values, \[ g_h = 9.8 \left( \frac{6.371 \times 10^6}{6.371 \times 10^6 + 373} \right)^2 \]
04

Simplify the Gravitational Acceleration Equation

Simplify the right-hand side of the equation to find the new gravitational acceleration, \(g_h\): \[ g_h \approx 9.799 \text{ m/s}^2 \]
05

Calculate New Weight

New weight at the height \( h \) is: \[ W_h = mg_h \] Substituting the values: \[ W_h = 54.08 \text{ kg} \times 9.799 \text{ m/s}^2 = 529.18 \text{ N} \]
06

Determine Weight Difference

Calculate the difference between the initial weight and the weight at height \( h \): \[ \Delta W = W - W_h = 530 \text{ N} - 529.18 \text{ N} = 0.82 \text{ N} \]

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

gravitational acceleration
Gravitational acceleration is a fundamental concept in physics. It denotes how fast an object speeds up when moving under the influence of Earth's gravity. On Earth's surface, the average gravitational acceleration is approximately 9.8 m/s虏.
When you go higher above the Earth's surface, the gravitational acceleration decreases slightly. This is because gravitational force weakens with distance from the Earth's center.
The formula to calculate gravitational acceleration at a height h is: \[ g_h = g \left( \frac{R}{R + h} \right)^2 \]
Here, \( g_h \) is the gravitational acceleration at height h, \( g \) is the gravitational acceleration at sea level, and \( R \) is the Earth's radius.
At sidewalk level outside the Empire State Building, the gravitational acceleration is 9.8 m/s虏. When you ascend to the 102nd floor, the gravitational acceleration decreases slightly, but the change is minimal.
weight calculation
Weight is a measure of the gravitational force acting on an object. It's calculated using the formula: \[ W = mg \]

Here, \( W \) is the weight, \( m \) is the mass, and \( g \) is the gravitational acceleration. On the Earth's surface, using a gravitational acceleration of 9.8 m/s虏, we can calculate the weight. To find the mass of a person weighing 530 N, we use: \[ m = \frac{W}{g} = \frac{530 \text{ N}}{9.8 \text{ m/s}^2} = 54.08 \text{ kg} \]
This means the person has a mass of approximately 54.08 kg. The actual weight can change depending on the gravitational acceleration, which varies with altitude.
height and gravity relationship
As you move higher above the Earth's surface, the force of gravity decreases. This is due to the inverse-square law, which states that gravitational force decreases with the square of the distance from the center of the Earth. When you go up to the 102nd floor of the Empire State Building, you are 373 meters higher than at sidewalk level. To find how this affects your weight, we use the formula for gravitational acceleration at height: \[ g_h = g \left( \frac{R}{R + h} \right)^2 \]
Substituting the values, we get: \[ g_h = 9.8 \left( \frac{6.371 \times 10^6}{6.371 \times 10^6 + 373} \right)^2 \approx 9.799 \text{ m/s}^2 \]
Now, to find the new weight: \[ W_h = mg_h = 54.08 \text{ kg} \times 9.799 \text{ m/s}^2 = 529.18 \text{ N} \]
So, at the 102nd floor, you weigh slightly less, around 529.18 N, compared to 530 N at the sidewalk level. This minor change illustrates how gravitational force decreases with height.

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

In the Shuttle When we see the astronauts in orbit in the space shuttle on TV, they seem to float. If they let go of something, it just stays where they put it. It doesn't fall. What happens to gravity for objects in orbit? Does gravity stop at the Earth's atmosphere? Explain what's happening in terms of the physics you have learned.

Zero, a hypothetical planet, has a mass of \(5.0 \times 10^{23} \mathrm{~kg}\), a radius of \(3.0 \times 10^{6} \mathrm{~m}\), and no atmosphere. A \(10 \mathrm{~kg}\) space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial energy of \(5.0 \times 10^{7} \mathrm{~J}\), what will be its kinetic energy when it is \(4.0 \times 10^{6} \mathrm{~m}\) from the center of Zero? (b) If the probe is to achieve a maximum distance of \(8.0 \times 10^{6} \mathrm{~m}\) from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

The masses and coordinates of three spheres are as follows: \(20 \mathrm{~kg}, x=0.50 \mathrm{~m}, y=1.0 \mathrm{~m} ; 40 \mathrm{~kg}\), \(x=-1.0 \mathrm{~m}, y=-1.0 \mathrm{~m} ; 60 \mathrm{~kg}, x=0 \mathrm{~m}, y=-0.50 \mathrm{~m} .\) What is the magnitude of the gravitational force on a \(20 \mathrm{~kg}\) sphere located at the origin due to the other spheres?

One model for a certain planet has a core of radius \(R\) and mass \(M\) surrounded by an outer shell of inner radius \(R\), outer radius \(2 R\), and mass \(4 M\). If \(M=4.1 \times 10^{24} \mathrm{~kg}\) and \(R=6.0 \times 10^{6} \mathrm{~m}\), what is the gravitational acceleration of a particle at points (a) \(R\) and (b) \(3 R\) from the center of the planet?

(A) What is the escape speed on a spherical asteroid whose radius is \(500 \mathrm{~km}\) and whose gravitational acceleration at the surface is \(3.0 \mathrm{~m} / \mathrm{s}^{2} ?\) (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of \(1000 \mathrm{~m} / \mathrm{s} ?(\mathrm{c})\) With what speed will an object hit the asteroid if it is dropped from \(1000 \mathrm{~km}\) above the surface?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Oscillations

Read Explanation

Thermodynamics

Read Explanation

Electromagnetism

Read Explanation

Rotational Dynamics

Read Explanation

Electricity and Magnetism

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.