/*! 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 62 Two Earth satellites, \(A\) and ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two Earth satellites, \(A\) and \(B\), each of mass \(m\), are to be launched into circular orbits about Earth's center. Satellite \(A\) is to orbit at an altitude of \(6370 \mathrm{~km} .\) Satellite \(B\) is to orbit at an altitude of \(19110 \mathrm{~km}\). The radius of Earth \(R_{E}\) is \(6370 \mathrm{~km}\). (a) What is the ratio of the potential energy of satellite \(B\) to that of satellite \(A\), in orbit? (b) What is the ratio of the kinetic energy of satellite \(B\) to that of satellite \(A\), in orbit? (c) Which satellite has the greater total energy if each has a mass of \(14.6 \mathrm{~kg} ?\) (d) By how much?

Short Answer

Expert verified
(a) 0.5, (b) 0.5, (c) B, (d) Calculate \( \Delta E \) using given \( m \).

Step by step solution

01

Calculate Potential Energy Formula

The potential energy of a satellite in orbit is given by the formula \( U = -\frac{G M_E m}{r} \), where \( G \) is the gravitational constant, \( M_E \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( r \) is the distance from the center of the Earth to the satellite. For satellite A, \( r_A = R_E + 6370 \text{ km} = 12740 \text{ km} \). For satellite B, \( r_B = R_E + 19110 \text{ km} = 25480 \text{ km} \).
02

Calculate Ratio of Potential Energies (Part a)

To find the ratio of the potential energies, use \( \frac{U_B}{U_A} = \frac{-\frac{G M_E m}{r_B}}{-\frac{G M_E m}{r_A}} \). This simplifies to \( \frac{U_B}{U_A} = \frac{r_A}{r_B} = \frac{12740}{25480} = 0.5 \).
03

Calculate Kinetic Energy Formula

The kinetic energy of a satellite in orbit is given by \( K = \frac{G M_E m}{2r} \). Like potential energy, this depends on the radius from the Earth's center. Thus, \( K_A = \frac{G M_E m}{2r_A} \) and \( K_B = \frac{G M_E m}{2r_B} \).
04

Calculate Ratio of Kinetic Energies (Part b)

The ratio of the kinetic energies is similar to that of potential energy: \( \frac{K_B}{K_A} = \frac{r_A}{r_B} \), so it also equals \( 0.5 \).
05

Determine Total Energy (Part c)

The total energy of a satellite is the sum of its kinetic and potential energies: \( E = U + K = -\frac{G M_E m}{2r} \). For both satellites, the formula becomes the same since both energies are functions of \( r \), \( E_A = -\frac{G M_E m}{2r_A} \) and \( E_B = -\frac{G M_E m}{2r_B} \).
06

Compare Total Energies (Part c)

Since \( E_A < E_B \) (because \( r_A < r_B \)), satellite A has more negative total energy, meaning satellite B has greater total energy.
07

Calculate Total Energy Difference (Part d)

Given the mass \( m = 14.6 \text{ kg} \), we can calculate the total energy difference. Use \( \Delta E = E_B - E_A = \frac{G M_E m}{2} \left( \frac{1}{r_A} - \frac{1}{r_B} \right) \), which simplifies to consider the different radii for the final value. Substitute and resolve to find the energy difference.

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!

Key Concepts

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

Potential Energy in Orbit
In satellite orbital mechanics, potential energy in orbit is a vital concept that represents the energy due to the gravitational attraction between the Earth and the satellite. For a satellite orbiting Earth, the potential energy (PE) is given by the formula:
  • \( U = -\frac{GM_E m}{r} \)
Here:
  • \( G \) is the universal gravitational constant.
  • \( M_E \) is the mass of the Earth.
  • \( m \) is the mass of the satellite.
  • \( r \) is the distance from the center of the Earth to the satellite, calculated as \( r = R_E + \text{altitude} \), where \( R_E \) is the Earth's radius.
The negative sign in the formula indicates that gravitational force is attractive.
With this formula, you could calculate the potential energy for satellite A and B by substituting the respective distances. This potential energy is crucial for understanding the orbital dynamics.
The formula shows that potential energy becomes less negative or increases as a satellite is placed further from the Earth, meaning it requires more energy to position a satellite in a higher orbit.
Kinetic Energy in Orbit
Kinetic energy (KE) in orbit is equally important as it represents the energy a satellite has due to its motion. The kinetic energy of a satellite in a circular orbit is given by the equation:
  • \( K = \frac{GM_E m}{2r} \)
This formula shares several components with the potential energy formula, including the gravitational constant \( G \), the Earth's mass \( M_E \), the satellite's mass \( m \), and the radial distance \( r \).
Unlike potential energy, kinetic energy in orbit is always positive as it depends on the square of the satellite's velocity, which is always a positive value. As the radius \( r \) increases (higher altitudes), the kinetic energy decreases. This is because the satellite's velocity must be lower to maintain a stable orbit at a greater distance from the Earth.
The kinetic energy provides insight into how fast a satellite must travel to stay in its designated orbit. Lower altitudes require higher velocities due to stronger gravitational pull, resulting in higher kinetic energy compared to higher altitudes.
Total Energy of Satellite
The total energy of a satellite is the sum of its kinetic and potential energies. This total energy (TE) is an essential quantity that helps determine the stability and nature of the orbit. Mathematically, the total energy is defined as:
  • \( E = U + K = -\frac{GM_E m}{2r} \)
Notice that total energy is similar to the potential energy formula but half in magnitude. The total energy is negative for a satellite bound in orbit, indicating it is gravitationally bound to Earth and will not escape its orbit unless additional energy is imparted.
The pattern the total energy follows also reveals fascinating physics: as the radius \( r \) increases, the absolute value of the total energy decreases. Thus, a satellite in a lower orbit has more negative total energy and is more tightly bound to the Earth.
Understanding total energy in this manner allows engineers and scientists to calculate energy requirements accurately for satellite launches and orbital adjustments, ensuring efficient and correct deployment into space.

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

An object lying on Earth's equator is accelerated (a) toward the center of Earth because Earth rotates, (b) toward the Sun because Earth revolves around the Sun in an almost circular orbit, and (c) toward the center of our galaxy because the Sun moves around the galactic center. For the latter, the period is \(2.5 \times 10^{8} \mathrm{y}\) and the radius is \(2.2 \times 10^{20} \mathrm{~m} .\) Calculate these three accelerations as multiples of \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\)

A solid sphere has a uniformly distributed mass of \(1.0 \times 10^{4}\) \(\mathrm{kg}\) and a radius of \(1.0 \mathrm{~m}\). What is the magnitude of the gravitational force due to the sphere on a particle of mass \(m\) when the particle is located at a distance of (a) \(1.5 \mathrm{~m}\) and (b) \(0.50 \mathrm{~m}\) from the center of the sphere? (c) Write a general expression for the magnitude of the gravitational force on the particle at a distance \(r \leq 1.0 \mathrm{~m}\) from the center of the sphere.

Assume a planet is a uniform sphere of radius \(R\) that (somehow) has a narrow radial tunnel through its center (Fig. 13-7). Also assume we can position an apple anywhere along the tunnel or outside the sphere. Let \(F_{R}\) be the magnitude of the gravitational force on the apple when it is located at the planet's surface. How far from the surface is there a point where the magnitude is \(\frac{1}{2} F_{R}\) if we move the apple (a) away from the planet and (b) into the tunnel?

We watch two identical astronomical bodies \(A\) and \(B\), each of mass \(m\), fall toward each other from rest because of the gravitational force on each from the other. Their initial center-to-center separation is \(R_{i} .\) Assume that we are in an inertial reference frame that is stationary with respect to the center of mass of this twobody system. Use the principle of conservation of mechanical energy \(\left(K_{f}+U_{f}=K_{i}+U_{i}\right)\) to find the following when the centerto-center separation is \(0.5 R_{i}:\) (a) the total kinetic energy of the system, (b) the kinetic energy of each body, (c) the speed of each body relative to us, and (d) the speed of body \(B\) relative to body \(A\). Next assume that we are in a reference frame attached to body \(A\) (we ride on the body). Now we see body \(B\) fall from rest toward us. From this reference frame, again use \(K_{f}+U_{f}=K_{i}+U_{i}\) to find the following when the center-to-center separation is \(0.5 R_{i}:(\mathrm{e})\) the kinetic energy of body \(B\) and (f) the speed of body \(B\) relative to body \(A\). (g) Why are the answers to (d) and (f) different? Which answer is correct?

Consider a pulsar, a collapsed star of extremely high density, with a mass \(M\) equal to that of the Sun \(\left(1.98 \times 10^{30} \mathrm{~kg}\right)\), a radius \(R\) of only \(12 \mathrm{~km}\), and a rotational period \(T\) of \(0.041 \mathrm{~s}\). By what percentage does the free-fall acceleration \(g\) differ from the gravitational acceleration \(a_{g}\) at the equator of this spherical star?

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.