Question

It follows from Newton's Law of Universal Gravitation that the gravitational force exerted by the Earth on an object above the Earth's surface varies inversely as the square of its distance from the Earth's center. Thus, an object's weight $w(x)$ is related to its distance $x$ from the Earth's center by a formula of the form $$ w(x)=\frac{k}{x^{2}} $$ where $k$ is a constant of proportionality that depends on the mass of the object. (a) Use this fact and the assumption that the Earth is a sphere of radius $4000 \mathrm{mi}$ to obtain the formula for $w(x)$ in Example 4 (b) Find a formula for the weight $w(x)$ of a satellite that is $x \mathrm{mi}$ from the Earth's surface if its weight on Earth is $6000 \mathrm{lb}$ (c) How much work is required to lift the satellite from the surface of the Earth to an orbital position that is 1000 mi high?

          It follows from Newton's Law of Universal Gravitation that the gravitational force exerted by the Earth on an object above the Earth's surface varies inversely as the square of its distance from the Earth's center. Thus, an object's weight $w(x)$ is related to its distance $x$ from the Earth's center by a formula of the form
$$
w(x)=\frac{k}{x^{2}}
$$
where $k$ is a constant of proportionality that depends on the mass of the object.
(a) Use this fact and the assumption that the Earth is a sphere of radius $4000 \mathrm{mi}$ to obtain the formula for $w(x)$ in Example 4
(b) Find a formula for the weight $w(x)$ of a satellite that is $x \mathrm{mi}$ from the Earth's surface if its weight on Earth is $6000 \mathrm{lb}$
(c) How much work is required to lift the satellite from the surface of the Earth to an orbital position that is 1000 mi high?

Show more…

Added by Joseph D.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Supreeta N

07/08/2023

Step 1

When an object is on the Earth's surface, its distance from the Earth's center is 4000 mi. Let $w_0$ be the weight of the object on the Earth's surface. Then, we have $$ w_0 = \frac{k}{(4000)^2}. $$ (b) We are given that the weight of the satellite on Earth is Show more…

Show all steps

lock

Thanks for your feedback!

It follows from Newton's Law of Universal Gravitation that the gravitational force exerted by the Earth on an object above the Earth's surface varies inversely as the square of its distance from the Earth's center. Thus, an object's weight $w(x)$ is related to its distance $x$ from the Earth's center by a formula of the form $$ w(x)=\frac{k}{x^{2}} $$ where $k$ is a constant of proportionality that depends on the mass of the object. (a) Use this fact and the assumption that the Earth is a sphere of radius $4000 \mathrm{mi}$ to obtain the formula for $w(x)$ in Example 4 (b) Find a formula for the weight $w(x)$ of a satellite that is $x \mathrm{mi}$ from the Earth's surface if its weight on Earth is $6000 \mathrm{lb}$ (c) How much work is required to lift the satellite from the surface of the Earth to an orbital position that is 1000 mi high?

Play audio

Feedback

Powered by NumerAI

Supreeta N and 89 other subject Calculus 1 / AB educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 Hello everyone, welcome to this video.

00:02 In the first part of the question we are given with w is equal to k by x square.

00:09 So, now we have to give the formula of w of x using the given information provided in the question.

00:18 So, here we have r is equal to 4000.

00:22 So, for find w of x let us find the integration from 4000 to 4220 k by x square dx.

00:33 So, integrating this we get k into 1 by x 4000 4220.

00:52 So, here on evaluating this we get minus of 1 by 4000 minus 1 by 4220.

01:07 So, evaluating this again we get w is equal to minus k 200 divided by 4000 into 4200.

01:24 So, this will give after evaluating we get w of x is equal to minus k into 11 k 11 k divided by 84400.

01:45 So, this is the answer for the first part of the question.

01:48 Now, let us find the second part.

01:51 So, we have to find work required to lift the satellite.

01:56 So, we are given with w of x is equal to 6000 and we have w of 4000 will be equal to using the formula we can write k by 4000 square.

02:16 So, equating this we get.

02:20 So, k by 4000 square is equal to 6000.

02:27 So, equating this we get k equal to 6000 into 4000 square k will be equal to 6000 into 4000 square.

02:42 So, here our w of x will be equal to k by x square...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later