Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 48

During a balloon ascension, wearing an oxygen mask, you measure the weight of a calibrated \(5.00-\mathrm{kg}\) mass and find that the value of the gravitational field strength at your location is 9.792 N/kg. How high above sea level, where the gravitational field strength was measured to be $9.803 \mathrm{N} / \mathrm{kg},$ are you located?

Short Answer

Expert verified
Answer: The altitude above sea level is approximately 0.796 km or 796 m.

Step by step solution

01

(Step 1: Write down the given information)

: We have the following information: 1. Mass of the object, \(\displaystyle m\ =\ 5.00\ \mathrm{kg}\) 2. The gravitational field strength at altitude, \(\displaystyle g_{1} =9.792\ \mathrm{N} / \mathrm{kg}\) 3. The gravitational field strength at sea level, \(\displaystyle g_{2} =9.803\ \mathrm{N} / \mathrm{kg}\)
02

(Step 2: Find the force acting on the mass)

: We can now find the force acting on the mass \(\displaystyle m\) using the gravitational field strength formula: \(\displaystyle F_{1} = m\times g_{1}\) Plugging in the values, \(\displaystyle F_{1} =5.00\ \mathrm{kg} \times 9.792\ \mathrm{N} / \mathrm{kg} =48.96\ \mathrm{N}\)
03

(Step 3: Use the formula for gravitational field strength at height)

: The formula for calculating the height above sea level, considering that the radius of the Earth \(\displaystyle R\) remains approximately constant, is: \(\displaystyle g_{1} =\dfrac{g_{2} R^{2}}{( R+h)^{2}}\) Where, \(\displaystyle R\) is the Earth's radius (~6371km), \(\displaystyle h\) is the height we want to find, \(\displaystyle g_{1}\) is the gravitational field strength at altitude, and \(\displaystyle g_{2}\) is the gravitational field strength at sea level.
04

(Step 4: Isolate the variable h to solve for height)

: Rearrange the formula to solve for \(\displaystyle h\) by isolating it on one side: \(\displaystyle h=R\left(\dfrac{g_{2}}{g_{1}}-1\right)^{1/ 2}-R\)
05

(Step 5: Calculate the height h)

: Plug in the known values and calculate the height: \(\displaystyle h=6371\ \mathrm{km} \times \left(\dfrac{9.803\ \mathrm{N} / \mathrm{kg}}{9.792\ \mathrm{N} / \mathrm{kg}}\ -\ 1\right)^{1/ 2}\ -\ 6371\ \mathrm{km}\) \(\displaystyle h\approx 0.796\ \mathrm{km}\) Therefore, the height above sea level is approximately \(\displaystyle 0.796\ \mathrm{km}\) or \(\displaystyle 796\ \mathrm{m}\).

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

At what altitude above the Earth's surface would your weight be half of what it is at the Earth's surface?
A hanging potted plant is suspended by a cord from a hook in the ceiling. Draw an FBD for each of these: (a) the system consisting of plant, soil, and pot; (b) the cord; (c) the hook; (d) the system consisting of plant, soil, pot, cord, and hook. Label each force arrow using subscripts (for example, \(\overrightarrow{\mathbf{F}}_{\mathrm{ch}}\) would represent the force exerted on the cord by the hook).
A toy cart of mass \(m_{1}\) moves on frictionless wheels as it is pulled by a string under tension \(T .\) A block of mass \(m_{2}\) rests on top of the cart. The coefficient of static friction between the cart and the block is \(\mu .\) Find the maximum tension \(T\) that will not cause the block to slide on the cart if the cart rolls on (a) a horizontal surface; (b) up a ramp of angle \(\theta\) above the horizontal. In both cases, the string is parallel to the surface on which the cart rolls.
Two blocks lie side by side on a frictionless table. The block on the left is of mass \(m\); the one on the right is of mass \(2 m .\) The block on the right is pushed to the left with a force of magnitude \(F,\) pushing the other block in turn. What force does the block on the left exert on the block to its right?
A 10.0 -kg block is released from rest on a frictionless track inclined at an angle of \(55^{\circ} .\) (a) What is the net force on the block after it is released? (b) What is the acceleration of the block? (c) If the block is released from rest, how long will it take for the block to attain a speed of \(10.0 \mathrm{m} / \mathrm{s} ?\) (d) Draw a motion diagram for the block. (e) Draw a graph of \(v_{x}(t)\) for values of velocity between 0 and $10 \mathrm{m} / \mathrm{s}\(. Let the positive \)x$ -axis point down the track.
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Fundamentals of Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Rotational Dynamics

Read Explanation

Electricity and Magnetism

Read Explanation

Further Mechanics and Thermal 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