/*! 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 18 An earth satellite is observed a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 18

An earth satellite is observed at perigee to be \(250 \mathrm{km}\) above the earth's surface and traveling at about \(8500 \mathrm{m} / \mathrm{s}\). Find the eccentricity of its orbit and its height above the earth at apogee. [Hint: The earth's radius is \(R_{e} \approx 6.4 \times 10^{6} \mathrm{m} .\) You will also need to know \(G M_{\mathrm{e}},\) but you can find this if you remember that \(\left.G M_{\mathrm{e}} / R_{\mathrm{e}}^{2}=g .\right]\)

Short Answer

Expert verified
The eccentricity is approximately 0.078 and the height at apogee is about 1292 km.

Step by step solution

01

Understanding the given data

The satellite is initially at perigee, which is the closest point to Earth in its orbit. At perigee, the distance from the center of the Earth to the satellite is the sum of the Earth's radius and the perigee height. Therefore, the perigee distance \( r_p \) is given by: \[ r_p = R_e + 250 \times 10^3 \, \text{m} = 6.4 \times 10^6 \, \text{m} + 250 \times 10^3 \, \text{m} = 6.65 \times 10^6 \, \text{m} \] The velocity at perigee \( v_p \) is given as \( 8500 \, \text{m/s}. \)
02

Applying conservation of specific angular momentum

For elliptical orbits, the specific angular momentum (\( h \)) is conserved and given by the formula: \[ h = r_p \times v_p \] Substituting the known values: \[ h = 6.65 \times 10^6 \, \text{m} \times 8500 \, \text{m/s} = 5.6525 \times 10^{10} \, \text{m}^2/\text{s} \]
03

Using vis-viva equation at perigee

The vis-viva equation gives the orbital equation for any point in orbit:\[ v^2 = GM_e \left(\frac{2}{r} - \frac{1}{a}\right) \]At perigee, substitute for \( v_p \) and \( r_p \):\[ 8500^2 = GM_e \left(\frac{2}{6.65 \times 10^6} - \frac{1}{a}\right) \] Recall \( GM_e = g \times R_e^2 \) where \( g \approx 9.81 \, \text{m/s}^2 \).\[ GM_e = 9.81 \times (6.4 \times 10^6)^2 \] Solve for \( a \), the semi-major axis.
04

Calculating the semi-major axis

First, calculate \( GM_e \):\[ GM_e = 9.81 \times (6.4 \times 10^6)^2 = 3.986 \times 10^{14} \, \text{m}^3/\text{s}^2 \] Substituting back in the vis-viva equation:\[ 8500^2 = 3.986 \times 10^{14} \left(\frac{2}{6.65 \times 10^6} - \frac{1}{a}\right) \] Rearrange to find \( a \):\[ a \approx 7.171 \times 10^6 \, \text{m} \]
05

Calculating the eccentricity of the orbit

The eccentricity \( e \) is given by the formula:\[ e = \frac{r_a - r_p}{r_a + r_p} \] Where \(r_a\) is the apogee distance and \(a\) is the semi-major axis:\[ r_a = 2a - r_p \]\[ r_a = 2 \times 7.171 \times 10^6 - 6.65 \times 10^6 = 7.692 \times 10^6 \, \text{m} \]Now compute \( e \):\[ e = \frac{7.692 \times 10^6 - 6.65 \times 10^6}{7.692 \times 10^6 + 6.65 \times 10^6}\]\[ e \approx 0.078 \]
06

Determining the height at apogee

The height above Earth's surface at apogee is:\[ h_a = r_a - R_e \]\[ h_a = 7.692 \times 10^6 - 6.4 \times 10^6 = 1.292 \times 10^6 \, \text{m} = 1292 \, \text{km} \]

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eccentricity
Eccentricity is a crucial concept when discussing satellite orbits. It measures how an orbit deviates from being circular. In simpler terms, it tells us how "stretched" an orbit is. The eccentricity (\( e \)) of a circle is zero, meaning no deviation from a circular shape. As eccentricity approaches 1, orbits become increasingly elliptical.

When calculating eccentricity in satellite orbits, we often use:
  • the apogee distance (\( r_a \)), the farthest point from Earth in the orbit
  • the perigee distance (\( r_p \)), the closest point to Earth
For example, with the Earth satellite, we derived the eccentricity using measurements of these two points. The formula used is:\[ e = \frac{r_a - r_p}{r_a + r_p} \]In this exercise, \( r_a \) was calculated as \( 7.692 \times 10^6 \text{ m} \), and \( r_p \) was \( 6.65 \times 10^6 \text{ m} \). This gave an eccentricity of approximately 0.078, indicating a slightly elliptical shaped orbit. Understanding this measure is key in predicting satellite motion and ensuring stable orbit trajectories.
Vis-Viva Equation
The vis-viva equation is essential for describing the motion of bodies in an orbit. It allows us to calculate the velocity at any point of a satellite's elliptical orbit. The formula for the vis-viva equation is:\[ v^2 = GM_e \left(\frac{2}{r} - \frac{1}{a}\right) \]where:
  • \( v \)is the orbital speed of the satellite at a specific point.
  • \( G \)is the gravitational constant and \( M_e \)is the Earth鈥檚 mass.
  • \( r \)is the distance from the center of Earth to the satellite.
  • \( a \)is the semi-major axis, the average of the orbit distances at apogee and perigee.
The vis-viva equation connects the kinetic and potential energies of the satellite, allowing us to compute orbital characteristics like the semi-major axis. In the exercise, using the satellite's speed at perigee (\( 8500 \text{ m/s} \)) and the vis-viva equation helped us solve for \( a \). By rearranging the equation and solving for \( a \), we found it to be \( 7.171 \times 10^6 \text{ m} \). This step is vital for further eccentricity and other orbit calculations.
Angular Momentum Conservation
Conservation of angular momentum is a fundamental principle in orbital mechanics. For a satellite, this principle states that the product of its velocity and distance from the center of Earth remains constant at different points in its orbit, assuming no external torque forces.

Mathematically, this is expressed as:\[ h = r \times v \]Where:
  • \( h \)is the specific angular momentum, a constant value for the satellite's orbit.
  • \( r \)is the distance from Earth's center to the satellite.
  • \( v \)is the satellite's velocity.
In the exercise, the initial calculation of the specific angular momentum (\( h = 5.6525 \times 10^{10} \text{ m}^2/ ext{s} \)) used data at perigee. This value ensures we understand how the orbit conserves momentum, informing velocity and distance variations at different orbital points effectively. This conservation concept not only applies to satellites but also to several natural phenomena and is key for reliable space mission calculations.

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

Consider two particles interacting by a Hooke's law potential energy, \(U=\frac{1}{2} k r^{2},\) where \(\mathbf{r}\) is their relative position \(\mathbf{r}=\mathbf{r}_{1}-\mathbf{r}_{2},\) and subject to no external forces. Show that \(\mathbf{r}(t)\) describes an ellipse. Hence show that both particles move on similar ellipses around their common CM. [This is surprisingly awkward. Perhaps the simplest procedure is to choose the \(x y\) plane as the plane of the orbit and then solve the equation of motion (8.15) for \(x\) and \(y .\) Your solution will have the form \(x=A \cos \omega t+B \sin \omega t,\) with a similar expression for \(y .\) If you solve these for sin \(\omega t\) and \(\cos \omega t\) and remember that \(\sin ^{2}+\cos ^{2}=1,\) you can put the orbital equation in the form \(a x^{2}+2 b x y+c y^{2}=k\) where \(k\) is a positive constant. Now invoke the standard result that if \(a\) and \(c\) are positive and \(a c>b^{2}\) this equation defines an ellipse.]

Two particles of equal masses \(m_{1}=m_{2}\) move on a frictionless horizontal surface in the vicinity of a fixed force center, with potential energies \(U_{1}=\frac{1}{2} k r_{1}^{2}\) and \(U_{2}=\frac{1}{2} k r_{2}^{2} .\) In addition, they interact with each other via a potential energy \(U_{12}=\frac{1}{2} \alpha k r^{2},\) where \(r\) is the distance between them and \(\alpha\) and \(k\) are positive constants. (a) Find the Lagrangian in terms of the CM position \(\mathbf{R}\) and the relative position \(\mathbf{r}=\mathbf{r}_{1}-\mathbf{r}_{2} \cdot(\mathbf{b})\) Write down and solve the Lagrange equations for the \(\mathrm{CM}\) and relative coordinates \(X, Y\) and \(x, y .\) Describe the motion.

Consider two particles of equal masses, \(m_{1}=m_{2},\) attached to each other by a light straight spring (force constant \(k\), natural length \(L\) ) and free to slide over a frictionless horizontal table. (a) Write down the Lagrangian in terms of the coordinates \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\), and rewrite it in terms of the \(\mathrm{CM}\) and relative positions, \(\mathbf{R}\) and \(\mathbf{r},\) using polar coordinates \((r, \phi)\) for \(\mathbf{r} .\) (b) Write down and solve the Lagrange equations for the CM coordinates \(X, Y\). (c) Write down the Lagrange equations for \(r\) and A. Solve these for the two special cases that \(r\) remains constant and that \(\phi\) remains constant. Describe the corresponding motions. In particular, show that the frequency of oscillations in the second case is \(\omega=\sqrt{2 k / m_{1}}\).

Prove that for circular orbits around a given gravitational force center (such as the sun) the speed of the orbiting body is inversely proportional to the square root of the orbital radius.

Consider two particles of equal masses, \(m_{1}=m_{2},\) attached to each other by a light straight spring (force constant \(k\), natural length \(L\) ) and free to slide over a frictionless horizontal table. (a) Write down the Lagrangian in terms of the coordinates \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\), and rewrite it in terms of the \(\mathrm{CM}\) and relative positions, \(\mathbf{R}\) and \(\mathbf{r},\) using polar coordinates \((r, \phi)\) for \(\mathbf{r} .\) (b) Write down and solve the Lagrange equations for the CM coordinates \(X, Y\). (c) Write down the Lagrange equations for \(r\) and A. Solve these for the two special cases that \(r\) remains constant and that \(\phi\) remains constant. Describe (i) the contant the corresponding motions. In particular, show that the frequency of oscillations in the second case is \(\omega=\sqrt{2 k / m_{1}}.\)
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Astrophysics

Read Explanation

Radiation

Read Explanation

Dynamics

Read Explanation

Work Energy and Power

Read Explanation

Classical Mechanics

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.