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University Physics Volume 1

13.4 Satellite Orbits and Energy

University Physics Volume 113.4 Satellite Orbits and Energy
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  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

Learning Objectives

By the end of this section, you will be able to:
  • Describe the mechanism for circular orbits
  • Find the orbital periods and speeds of satellites
  • Determine whether objects are gravitationally bound

The Moon orbits Earth. In turn, Earth and the other planets orbit the Sun. The space directly above our atmosphere is filled with artificial satellites in orbit. We examine the simplest of these orbits, the circular orbit, to understand the relationship between the speed and period of planets and satellites in relation to their positions and the bodies that they orbit.

Circular Orbits

As noted at the beginning of this chapter, Nicolaus Copernicus first suggested that Earth and all other planets orbit the Sun in circles. He further noted that orbital periods increased with distance from the Sun. Later analysis by Kepler showed that these orbits are actually ellipses, but the orbits of most planets in the solar system are nearly circular. Earth’s orbital distance from the Sun varies a mere 2%. The exception is the eccentric orbit of Mercury, whose orbital distance varies nearly 40%.

Determining the orbital speed and orbital period of a satellite is much easier for circular orbits, so we make that assumption in the derivation that follows. As we described in the previous section, an object with negative total energy is gravitationally bound and therefore is in orbit. Our computation for the special case of circular orbits will confirm this. We focus on objects orbiting Earth, but our results can be generalized for other cases.

Consider a satellite of mass m in a circular orbit about Earth at distance r from the center of Earth (Figure 13.12). It has centripetal acceleration directed toward the center of Earth. Earth’s gravity is the only force acting, so Newton’s second law gives

GmMEr2=mac=mvorbit2r.GmMEr2=mac=mvorbit2r.
A drawing shows a satellite orbiting the earth at radius r. The orbit is shown as a blue circle centered on the earth. A red arrow at the satellite points toward the center of the earth and is labeled F and a green arrow tangent to the orbit is labeled v.
Figure 13.12 A satellite of mass m orbiting at radius r from the center of Earth. The gravitational force supplies the centripetal acceleration.

We solve for the speed of the orbit, noting that m cancels, to get the orbital speed

vorbit=GMEr.vorbit=GMEr.
(13.7)

Consistent with what we saw in Equation 13.2 and Equation 13.6, m does not appear in Equation 13.7. The value of g, the escape velocity, and orbital velocity depend only upon the distance from the center of the planet, and not upon the mass of the object being acted upon. Notice the similarity in the equations for vorbitvorbit and vescvesc. The escape velocity is exactly 22 times greater, about 40%, than the orbital velocity. This comparison was noted in Example 13.7, and it is true for a satellite at any radius.

To find the period of a circular orbit, we note that the satellite travels the circumference of the orbit 2πr2πr in one period T. Using the definition of speed, we have vorbit=2πr/Tvorbit=2πr/T. We substitute this into Equation 13.7 and rearrange to get

T=2πr3GME.T=2πr3GME.
(13.8)

We see in the next section that this represents Kepler’s third law for the case of circular orbits. It also confirms Copernicus’s observation that the period of a planet increases with increasing distance from the Sun. We need only replace MEME with MSunMSun in Equation 13.8.

We conclude this section by returning to our earlier discussion about astronauts in orbit appearing to be weightless, as if they were free-falling towards Earth. In fact, they are in free fall. Consider the trajectories shown in Figure 13.13. (This figure is based on a drawing by Newton in his Principia and also appeared earlier in Motion in Two and Three Dimensions.) All the trajectories shown that hit the surface of Earth have less than orbital velocity. The astronauts would accelerate toward Earth along the noncircular paths shown and feel weightless. (Astronauts actually train for life in orbit by riding in airplanes that free fall for 30 seconds at a time.) But with the correct orbital velocity, Earth’s surface curves away from them at exactly the same rate as they fall toward Earth. Of course, staying the same distance from the surface is the point of a circular orbit.

The figure shows a drawing of the earth with a tall tower at the north pole and a horizontal arrow labeled v 0 pointing to the right. 5 trajectories that start at the top of the tower are shown. The first reaches the earth near the tower. The second reaches the earth farther from the tower, and the third even farther. The fourth trajectory hits the earth at the equator, and is tangent to the surface at the equator. The fifth trajectory is a circle concentric with the earth.
Figure 13.13 A circular orbit is the result of choosing a tangential velocity such that Earth’s surface curves away at the same rate as the object falls toward Earth.

We can summarize our discussion of orbiting satellites in the following Problem-Solving Strategy.

Problem-Solving Strategy: Orbits and Conservation of Energy

  1. Determine whether the equations for speed, energy, or period are valid for the problem at hand. If not, start with the first principles we used to derive those equations.
  2. To start from first principles, draw a free-body diagram and apply Newton’s law of gravitation and Newton’s second law.
  3. Along with the definitions for speed and energy, apply Newton’s second law of motion to the bodies of interest.

Example 13.9

The International Space Station Determine the orbital speed and period for the International Space Station (ISS).

Strategy Since the ISS orbits 4.00×102km4.00×102km above Earth’s surface, the radius at which it orbits is RE+4.00×102kmRE+4.00×102km. We use Equation 13.7 and Equation 13.8 to find the orbital speed and period, respectively.

Solution Using Equation 13.7, the orbital velocity is

vorbit=GMEr=6.67×10−11N·m2/kg2(5.96×1024kg)(6.36×106+4.00×105m)=7.67×103m/svorbit=GMEr=6.67×10−11N·m2/kg2(5.96×1024kg)(6.36×106+4.00×105m)=7.67×103m/s

which is about 17,000 mph. Using Equation 13.8, the period is

T=2πr3GME=2π(6.37×106+4.00×105m)3(6.67×10−11N·m2/kg2)(5.96×1024kg)=5.55×103sT=2πr3GME=2π(6.37×106+4.00×105m)3(6.67×10−11N·m2/kg2)(5.96×1024kg)=5.55×103s

which is just over 90 minutes.

Significance The ISS is considered to be in low Earth orbit (LEO). Nearly all satellites are in LEO, including most weather satellites. GPS satellites, at about 20,000 km, are considered medium Earth orbit. The higher the orbit, the more energy is required to put it there and the more energy is needed to reach it for repairs. Of particular interest are the satellites in geosynchronous orbit. All fixed satellite dishes on the ground pointing toward the sky, such as TV reception dishes, are pointed toward geosynchronous satellites. These satellites are placed at the exact distance, and just above the equator, such that their period of orbit is 1 day. They remain in a fixed position relative to Earth’s surface.

Check Your Understanding 13.6

By what factor must the radius change to reduce the orbital velocity of a satellite by one-half? By what factor would this change the period?

Example 13.10

Determining the Mass of Earth Determine the mass of Earth from the orbit of the Moon.

Strategy We use Equation 13.8, solve for MEME, and substitute for the period and radius of the orbit. The radius and period of the Moon’s orbit was measured with reasonable accuracy thousands of years ago. From the astronomical data in Appendix D, the period of the Moon is 27.3 days =2.36×106s=2.36×106s, and the average distance between the centers of Earth and the Moon is 384,000 km.

Solution Solving for MEME,

T=2πr3GMEME=4π2r3GΤ2=4π2(3.84×108m)3(6.67×10−11N·m2/kg2)(2.36×106m)2=6.01×1024kg.T=2πr3GMEME=4π2r3GΤ2=4π2(3.84×108m)3(6.67×10−11N·m2/kg2)(2.36×106m)2=6.01×1024kg.

Significance Compare this to the value of 5.96×1024kg5.96×1024kg that we obtained in Example 13.5, using the value of g at the surface of Earth. Although these values are very close (~0.8%), both calculations use average values. The value of g varies from the equator to the poles by approximately 0.5%. But the Moon has an elliptical orbit in which the value of r varies just over 10%. (The apparent size of the full Moon actually varies by about this amount, but it is difficult to notice through casual observation as the time from one extreme to the other is many months.)

Check Your Understanding 13.7

There is another consideration to this last calculation of MEME. We derived Equation 13.8 assuming that the satellite orbits around the center of the astronomical body at the same radius used in the expression for the gravitational force between them. What assumption is made to justify this? Earth is about 81 times more massive than the Moon. Does the Moon orbit about the exact center of Earth?

Example 13.11

Galactic Speed and Period Let’s revisit Example 13.2. Assume that the Milky Way and Andromeda galaxies are in a circular orbit about each other. What would be the velocity of each and how long would their orbital period be? Assume the mass of each is 800 billion solar masses and their centers are separated by 2.5 million light years.

Strategy We cannot use Equation 13.7 and Equation 13.8 directly because they were derived assuming that the object of mass m orbited about the center of a much larger planet of mass M. We determined the gravitational force in Example 13.2 using Newton’s law of universal gravitation. We can use Newton’s second law, applied to the centripetal acceleration of either galaxy, to determine their tangential speed. From that result we can determine the period of the orbit.

Solution In Example 13.2, we found the force between the galaxies to be

F12=Gm1m2r2=(6.67×10−11N·m2/kg2)[(800×109)(2.0×1030kg)]2[(2.5×106)(9.5×1015m)]2=3.0×1029NF12=Gm1m2r2=(6.67×10−11N·m2/kg2)[(800×109)(2.0×1030kg)]2[(2.5×106)(9.5×1015m)]2=3.0×1029N

and that the acceleration of each galaxy is

a=Fm=3.0×1029N(800×109)(2.0×1030kg)=1.9×10−13m/s2.a=Fm=3.0×1029N(800×109)(2.0×1030kg)=1.9×10−13m/s2.

Since the galaxies are in a circular orbit, they have centripetal acceleration. If we ignore the effect of other galaxies, then, as we learned in Linear Momentum and Collisions and Fixed-Axis Rotation, the centers of mass of the two galaxies remain fixed. Hence, the galaxies must orbit about this common center of mass. For equal masses, the center of mass is exactly half way between them. So the radius of the orbit, rorbitrorbit, is not the same as the distance between the galaxies, but one-half that value, or 1.25 million light-years. These two different values are shown in Figure 13.14.

Two galaxies are illustrated, separated by a distance r that is shown on the diagram. The galaxy on the left is larger than the galaxy on the right. A distance from the center of the galaxy on the left to a point between the two galaxies but closer to the left is shown and labeled as r orbit.
Figure 13.14 The distance between two galaxies, which determines the gravitational force between them, is r, and is different from rorbitrorbit, which is the radius of orbit for each. For equal masses, rorbit=1/2rrorbit=1/2r. (credit: modification of work by Marc Van Norden)

Using the expression for centripetal acceleration, we have

ac=vorbit2rorbit1.9×10−13m/s2=vorbit2(1.25×106)(9.5×1015m).ac=vorbit2rorbit1.9×10−13m/s2=vorbit2(1.25×106)(9.5×1015m).

Solving for the orbit velocity, we have vorbit=47km/svorbit=47km/s. Finally, we can determine the period of the orbit directly from T=2πr/vorbitT=2πr/vorbit, to find that the period is T=1.6×1018sT=1.6×1018s, about 50 billion years.

Significance The orbital speed of 47 km/s might seem high at first. But this speed is comparable to the escape speed from the Sun, which we calculated in an earlier example. To give even more perspective, this period is nearly four times longer than the time that the Universe has been in existence.

In fact, the present relative motion of these two galaxies is such that they are expected to collide in about 4 billion years. Although the density of stars in each galaxy makes a direct collision of any two stars unlikely, such a collision will have a dramatic effect on the shape of the galaxies. Examples of such collisions are well known in astronomy.

Check Your Understanding 13.8

Galaxies are not single objects. How does the gravitational force of one galaxy exerted on the “closer” stars of the other galaxy compare to those farther away? What effect would this have on the shape of the galaxies themselves?

Interactive

See the Sloan Digital Sky Survey page for more information on colliding galaxies.

Energy in Circular Orbits

In Gravitational Potential Energy and Total Energy, we argued that objects are gravitationally bound if their total energy is negative. The argument was based on the simple case where the velocity was directly away or toward the planet. We now examine the total energy for a circular orbit and show that indeed, the total energy is negative. As we did earlier, we start with Newton’s second law applied to a circular orbit,

GmMEr2=mac=mv2rGmMEr=mv2.GmMEr2=mac=mv2rGmMEr=mv2.

In the last step, we multiplied by r on each side. The right side is just twice the kinetic energy, so we have

K=12mv2=GmME2r.K=12mv2=GmME2r.

The total energy is the sum of the kinetic and potential energies, so our final result is

E=K+U=GmME2rGmMEr=GmME2r.E=K+U=GmME2rGmMEr=GmME2r.
(13.9)

We can see that the total energy is negative, with the same magnitude as the kinetic energy. For circular orbits, the magnitude of the kinetic energy is exactly one-half the magnitude of the potential energy. Remarkably, this result applies to any two masses in circular orbits about their common center of mass, at a distance r from each other. The proof of this is left as an exercise. We will see in the next section that a very similar expression applies in the case of elliptical orbits.

Example 13.12

Energy Required to Orbit In Example 13.8, we calculated the energy required to simply lift the 9000-kg Soyuz vehicle from Earth’s surface to the height of the ISS, 400 km above the surface. In other words, we found its change in potential energy. We now ask, what total energy change in the Soyuz vehicle is required to take it from Earth’s surface and put it in orbit with the ISS for a rendezvous (Figure 13.15)? How much of that total energy is kinetic energy?

A demonstration of the ISS and Soyuz in parallel orbits about the earth is shown.
Figure 13.15 The Soyuz in a rendezvous with the ISS. Note that this diagram is not to scale; the Soyuz is very small compared to the ISS and its orbit is much closer to Earth. (credit: modification of works by NASA)

Strategy The energy required is the difference in the Soyuz’s total energy in orbit and that at Earth’s surface. We can use Equation 13.9 to find the total energy of the Soyuz at the ISS orbit. But the total energy at the surface is simply the potential energy, since it starts from rest. [Note that we do not use Equation 13.9 at the surface, since we are not in orbit at the surface.] The kinetic energy can then be found from the difference in the total energy change and the change in potential energy found in Example 13.8. Alternatively, we can use Equation 13.7 to find vorbitvorbit and calculate the kinetic energy directly from that. The total energy required is then the kinetic energy plus the change in potential energy found in Example 13.8.

Solution From Equation 13.9, the total energy of the Soyuz in the same orbit as the ISS is

Eorbit=Korbit+Uorbit=GmME2r=(6.67×10−11N·m2/kg2)(9000kg)(5.96×1024kg)2(6.36×106+4.00×105m)=−2.65×1011J.Eorbit=Korbit+Uorbit=GmME2r=(6.67×10−11N·m2/kg2)(9000kg)(5.96×1024kg)2(6.36×106+4.00×105m)=−2.65×1011J.

The total energy at Earth’s surface is

Esurface=Ksurface+Usurface=0GmMEr=(6.67×10−11N·m2/kg2)(9000kg)(5.96×1024kg)(6.36×106m)=−5.63×1011J.Esurface=Ksurface+Usurface=0GmMEr=(6.67×10−11N·m2/kg2)(9000kg)(5.96×1024kg)(6.36×106m)=−5.63×1011J.

The change in energy is ΔE=EorbitEsurface=2.98×1011JΔE=EorbitEsurface=2.98×1011J. To get the kinetic energy, we subtract the change in potential energy from Example 13.6, ΔU=3.32×1010JΔU=3.32×1010J. That gives us Korbit=2.98×10113.32×1010=2.65×1011JKorbit=2.98×10113.32×1010=2.65×1011J. As stated earlier, the kinetic energy of a circular orbit is always one-half the magnitude of the potential energy, and the same as the magnitude of the total energy. Our result confirms this.

The second approach is to use Equation 13.7 to find the orbital speed of the Soyuz, which we did for the ISS in Example 13.9.

vorbit=GMEr=(6.67×10−11N·m2/kg2)(5.96×1024kg)(6.36×106+4.00×105m)=7.67×103m/s.vorbit=GMEr=(6.67×10−11N·m2/kg2)(5.96×1024kg)(6.36×106+4.00×105m)=7.67×103m/s.

So the kinetic energy of the Soyuz in orbit is

Korbit=12mvorbit2=12(9000kg)(7.67×103m/s)2=2.65×1011J,Korbit=12mvorbit2=12(9000kg)(7.67×103m/s)2=2.65×1011J,

the same as in the previous method. The total energy is just

Eorbit=Korbit+ΔU=2.65×1011+3.32×1010=2.95×1011J.Eorbit=Korbit+ΔU=2.65×1011+3.32×1010=2.95×1011J.

Significance The kinetic energy of the Soyuz is nearly eight times the change in its potential energy, or 90% of the total energy needed for the rendezvous with the ISS. And it is important to remember that this energy represents only the energy that must be given to the Soyuz. With our present rocket technology, the mass of the propulsion system (the rocket fuel, its container and combustion system) far exceeds that of the payload, and a tremendous amount of kinetic energy must be given to that mass. So the actual cost in energy is many times that of the change in energy of the payload itself.

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