/*! 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 34 A spy satellite is in circular o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 34

A spy satellite is in circular orbit around Earth. It makes one revolution in \(6.00 \mathrm{h}\). (a) How high above Earth's surface is the satellite? (b) What is the satellite's acceleration?

Short Answer

Expert verified
Answer: The altitude of the satellite above Earth's surface is 3.594 x 10^7 meters.

Step by step solution

01

1. Identify relevant physical quantities for the satellite

First, list the given quantities and constants that are relevant to this problem. - Time taken for one revolution (period): \(T = 6.00 \hspace{1mm} h = 21600 \hspace{1mm} s\) - Earth's mass: \(M_{e} = 5.97 \times 10^{24} \hspace{1mm} kg\) - Earth's radius: \(R_{e} = 6.38 \times 10^{6} \hspace{1mm} m\) - Gravitational constant: \(G = 6.67408 \times 10^{-11}\hspace{1mm}m^3鈰卥g^{鈭�1}鈰卻^{鈭�2}\)
02

2. Calculate the satellite's angular velocity

We can calculate the angular velocity of the satellite by using the formula \(\omega = \frac{2\pi}{T}\), where \(\omega\) is the angular velocity and \(T\) is the period. $$\omega = \frac{2\pi}{T} = \frac{2\pi}{21600} = 2.908 \times 10^{-4} \hspace{1mm} rad/s$$
03

3. Determine the satellite's orbital radius

Use the centripetal force equation, which is equal to the gravitational force acting on the satellite: \(m_s \omega^2 r = \frac{G M_e m_s}{r^2}\). The mass of the satellite, \(m_s\), cancels out when we solve for \(r\) (the satellite's orbital radius): $$r = \sqrt[\leftroot{-2}\uproot{2}]{\frac{G M_e}{\omega^2}}$$ Now, plug in the values: $$r = \sqrt[\leftroot{-2}\uproot{2}]{\frac{6.67408 \times 10^{-11} m^{3}鈰卥g^{鈭�1}鈰卻^{鈭�2} \times 5.97 \times 10^{24} kg}{(2.908 \times 10^{-4} rad/s)^2}} = 4.23 \times 10^{7} m$$
04

4. Calculate the satellite's altitude above the Earth

To get the satellite's altitude above the Earth's surface, subtract the Earth's radius from the satellite's orbital radius: $$h = r - R_e = (4.23 \times 10^7 m) - (6.38 \times 10^6 m) = 3.594 \times 10^7 m$$ This is the satellite's altitude above the Earth.

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

Earth's orbit around the Sun is nearly circular. The period is 1 yr \(=365.25\) d. (a) In an elapsed time of 1 d what is Earth's angular displacement? (b) What is the change in Earth's velocity, \(\Delta \overrightarrow{\mathbf{v}} ?\) (c) What is Earth's average acceleration during 1 d? (d) Compare your answer for (c) to the magnitude of Earth's instantaneous radial acceleration. Explain.

The rotor is an amusement park ride where people stand against the inside of a cylinder. Once the cylinder is spinning fast enough, the floor drops out. (a) What force keeps the people from falling out the bottom of the cylinder? (b) If the coefficient of friction is 0.40 and the cylinder has a radius of \(2.5 \mathrm{m},\) what is the minimum angular speed of the cylinder so that the people don't fall out? (Normally the operator runs it considerably faster as a safety measure.) (IMAGE CAN'T COPY)

In an amusement park rocket ride, cars are suspended from 4.25 -m cables attached to rotating arms at a distance of \(6.00 \mathrm{m}\) from the axis of rotation. The cables swing out at an angle of \(45.0^{\circ}\) when the ride is operating. What is the angular speed of rotation?

A biologist is studying growth in space. He wants to simulate Earth's gravitational field, so he positions the plants on a rotating platform in the spaceship. The distance of each plant from the central axis of rotation is \(r=0.20 \mathrm{m} .\) What angular speed is required?
During normal operation, a computer's hard disk spins at 7200 rpm. If it takes the hard disk 4.0 s to reach this angular velocity starting from rest, what is the average angular acceleration of the hard disk in $\mathrm{rad} / \mathrm{s}^{2} ?$
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Translational Dynamics

Read Explanation

Fields in Physics

Read Explanation

Modern Physics

Read Explanation

Magnetism

Read Explanation

Thermodynamics

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.