/*! 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 49 At what altitude above the Ear... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 49

At what altitude above the Earth's surface would your weight be half of what it is at the Earth's surface?

Short Answer

Expert verified
Explain your answer. Answer: No, it is not possible for someone to have half their weight at any altitude above Earth's surface. The solution resulted in a negative altitude, which indicates that the gravitational force does not decrease fast enough with increasing altitude to achieve half the weight before reaching the Earth's surface.

Step by step solution

01

Choose a weight for the person at Earth's surface

Let's label a person's weight (gravitational force) on the Earth's surface as \(F_1\) and the weight at the desired altitude as \(F_2\). We know that: \(F_2 = 0.5 F_1\) Let's choose a weight for the person on the Earth's surface. This doesn't matter, as it will cancel out in the final calculations. For example, let's say the person's weight is \(80 N\) on Earth's surface (\(F_1 = 80 N\)), and thus the weight (\(F_2\)) at the desired altitude should be \(40 N\).
02

Write down the equations for the weights at Earth's surface and desired altitude

Now we have the following equations for the gravitational force at Earth's surface and the desired altitude: \(F_1 = \cfrac{GMm}{R^2}\) \(F_2 = \cfrac{GMm}{(R+h)^2}\)
03

Find the relationship between the two equations

Divide the equation for \(F_2\) by the equation for \(F_1\) to find the relationship between the two equations: \(\cfrac{F_2}{F_1} = \cfrac{\cfrac{GMm}{(R+h)^2}}{\cfrac{GMm}{R^2}}\) We already know that \(F_2 = 0.5 F_1\), so: \(0.5 = \cfrac{(R+h)^2}{R^2}\)
04

Solve for the altitude (h)

Now, we need to solve for h in the equation: \(0.5 = \cfrac{(R+h)^2}{R^2}\) First, multiply both sides by \(R^2\): \(0.5R^2 = (R+h)^2\) Find the square root of both sides: \(\sqrt{0.5}R = R + h\) Now, solve for h: \(h = (\sqrt{0.5} - 1)R\) Plug in the Earth's radius (R) value: \(h = (6.371\times10^6m)(\sqrt{0.5} - 1)\) Calculate the altitude: \(h \approx -2.26\times10^6m\) This is a negative result, which means that it is not possible to have half of the weight at any altitude above the Earth's surface. This is because the gravitational force decreases with increasing altitude, but not at a fast enough rate to achieve half the weight before reaching the Earth's surface.

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!

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

Mechanical advantage is the ratio of the force required without the use of a simple machine to that needed when using the simple machine. Compare the force to lift an object with that needed to slide the same object up a frictionless incline and show that the mechanical advantage of the inclined plane is the length of the incline divided by the height of the incline \((d / h\) in Fig. \(4.25).\)
A crate full of artichokes rests on a ramp that is inclined \(10.0^{\circ}\) above the horizontal. Give the direction of the normal force and the friction force acting on the crate in each of these situations. (a) The crate is at rest. (b) The crate is being pushed and is sliding up the ramp. (c) The crate is being pushed and is sliding down the ramp.
A crow perches on a clothesline midway between two poles. Each end of the rope makes an angle of \(\theta\) below the horizontal where it connects to the pole. If the weight of the crow is \(W,\) what is the tension in the rope? Ignore the weight of the rope.
Spring scale \(A\) is attached to the ceiling. A 10.0 -kg mass is suspended from the scale. A second spring scale, \(\mathrm{B}\), is hanging from a hook at the bottom of the 10.0 -kg mass and a 4.0 -kg mass hangs from the second spring scale. (a) What are the readings of the two scales if the masses of the scales are negligible? (b) What are the readings if each scale has a mass of $1.0 \mathrm{kg} ?$
(a) What is the magnitude of the gravitational force that the Earth exerts on the Moon? (b) What is the magnitude of the gravitational force that the Moon exerts on the Earth? See the inside front and back covers for necessary information.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Modern Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Kinematics Physics

Read Explanation

Electromagnetism

Read Explanation

Dynamics

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.