/*! 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 41 A satellite is placed in a nonci... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A satellite is placed in a noncircular orbit about the Earth. The farthest point of its orbit (apogee) is 4 Earth radii from the center of the Earth, while its nearest point (perigee) is 2 Earth radii from the Earth's center. If we define the gravitational potential energy \(U\) to be zero for an infinite separation of Earth and satellite, find the ratio $U_{\text {perigce }} / U_{\text {apogec }}$

Short Answer

Expert verified
Answer: The ratio of the gravitational potential energy at the perigee point to the apogee point is 2:1.

Step by step solution

01

Define the gravitational potential energy formula

The gravitational potential energy is given by the formula: \(U = -\frac{G * m_{1} * m_{2}}{r}\) where: - \(U\) is the gravitational potential energy - \(G\) is the gravitational constant, approximately \(6.674 \times 10^{-11} \ Nm^2/kg^2\) - \(m_1\) is the mass of the satellite - \(m_2\) is the mass of the Earth, approximately \(5.972 \times 10^{24} \ kg\) - \(r\) is the distance between the centers of the two masses
02

Use the potential energy formula to find the energy at the perigee point

At the perigee point, the satellite is \(2\) Earth radii away from the center of the Earth. Since the radius of the Earth is approximately \(6.371 \times 10^6 \ m\), the perigee distance is \(2 * 6.371 \times 10^6 \ m\). Now we can find the potential energy at the perigee point: \(U_{perigee} = -\frac{G * m_{1} * m_{2}}{2*6.371 \times 10^6} \)
03

Use the potential energy formula to find the energy at the apogee point

At the apogee point, the satellite is \(4\) Earth radii away from the center of the Earth. So, the apogee distance is \(4 * 6.371 \times 10^6 \ m\). Now we can find the potential energy at the apogee point: \(U_{apogee} = -\frac{G * m_{1} * m_{2}}{4*6.371 \times 10^6} \)
04

Compute the ratio of the potential energy at the perigee to the apogee points

We need to find the ratio \(U_{perigee} / U_{apogee} \). We can divide \(U_{perigee}\) by \(U_{apogee}\) as follows: \(\frac{U_{perigee}}{U_{apogee}} = \frac{-\frac{G * m_{1} * m_{2}}{2*6.371 \times 10^6}}{-\frac{G * m_{1} * m_{2}}{4*6.371 \times 10^6}}\) The masses and the gravitational constant will cancel out: \(\frac{U_{perigee}}{U_{apogee}} = \frac{-\frac{1}{2*6.371 \times 10^6}}{-\frac{1}{4*6.371 \times 10^6}} = \frac{4*6.371 \times 10^6}{2*6.371 \times 10^6}\) Finally, the ratio is: \(\frac{U_{perigee}}{U_{apogee}} = 4 / 2 = 2\) So, the ratio of the gravitational potential energy at the perigee point to the apogee point is 2:1.

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 91Ó°ÊÓ!

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

Jennifer lifts a 2.5 -kg carton of cat litter from the floor to a height of \(0.75 \mathrm{m} .\) (a) How much total work is done on the carton during this operation? Jennifer then pours \(1.2 \mathrm{kg}\) of the litter into the cat's litter box on the floor. (b) How much work is done by gravity on the $1.2 \mathrm{kg}$ of litter as it falls into the litter box?
A planet with a radius of \(6.00 \times 10^{7} \mathrm{m}\) has a gravitational field of magnitude \(30.0 \mathrm{m} / \mathrm{s}^{2}\) at the surface. What is the escape speed from the planet?
(a) How much work does a major-league pitcher do on the baseball when he throws a \(90.0 \mathrm{mi} / \mathrm{h}(40.2 \mathrm{m} / \mathrm{s})\) fastball? The mass of a baseball is 153 g. (b) How many fastballs would a pitcher have to throw to "burn off' a 1520 -Calorie meal? (1 Calorie \(=1000\) cal \(=1\) kcal.)Assume that \(80.0 \%\) of the chemical energy in the food is converted to thermal energy and only \(20.0 \%\) becomes the kinetic energy of the fastballs.
A record company executive is on his way to a TV interview and is carrying a promotional CD in his briefcase. The mass of the briefcase and its contents is \(5.00 \mathrm{kg}\) The executive realizes that he is going to be late. Starting from rest, he starts to run, reaching a speed of $2.50 \mathrm{m} / \mathrm{s} .$ What is the work done by the executive on the briefcase during this time? Ignore air resistance.
The maximum speed of a child on a swing is \(4.9 \mathrm{m} / \mathrm{s}\) The child's height above the ground is \(0.70 \mathrm{m}\) at the lowest point in his motion. How high above the ground is he at his highest point?
See all solutions

Recommended explanations on Physics Textbooks

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.