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The rockets that launched Voyager 1 and Voyager 2, were designedto escape Earth’s gravity and send them into space to search thesolar system. The Voyager craft have found such things as newmoons orbiting Jupiter, Saturn, Uranus, and Neptune. They havealso discovered volcanoes on Io and rings around Jupiter.However, the majority of satellites are launched into Earth’s orbitand will remain captive in Earth’s gravitational field, destined tocircle the planet year after year and perform tasks of immediateimportance to people on Earth.
Satellites in Earth OrbitSome of these satellites monitor the weather, the growth andhealth of crops, the temperature of the oceans, the presence of ice floes, and the status of the ozone layer. Others actively scan Earth’s surface with radar to enhance our knowledge of thegeography and geology of our planet. These days, many peopleroutinely use satellites to tell them where they are (for example,the Global Positioning System) and to provide them with mobilecommunication and seemingly limitless television entertainment.
Other satellites look outward, monitoring regions of the electromagnetic spectrum that cannot pass easily through Earth’satmosphere. In doing so, they tell us about our own solar system,as well as other solar systems, stars, and galaxies located manylight-years away from us.
The largest artificial satellite in Earth’s orbit is the InternationalSpace Station. During its construction and lifetime, it has beenserviced from other temporary satellites, the space shuttles. Forthese shuttles to rendezvous successfully with the space station,teams of scientists, engineers, and technicians must solve prob-lems involving the orbital motions and energies that are the subject of this section.
Most satellites are in either a circular orbit or a near-circularorbit. In Chapter 3, you learned how the force of gravity acts as acentripetal force, holding each satellite in its own unique orbit.You learned how to calculate the orbital speeds of the satellitesthat orbit at specific radii. In this section, you will focus on theenergies of these satellites.
Energy of Orbiting Satellites6.2
236 MHR • Unit 2 Energy and Momentum
• Analyze the factors affectingthe motion of isolated celestialobjects, and calculate the gravitational potential energyfor such a system.
• Analyze isolated planetary andsatellite motion and describe itin terms of the forms of energyand energy transformations that occur.
• Calculate the kinetic and gravitational potential energy of a satellite that is in a stablecircular orbit around a planet.
• circular orbit
• total orbital energy
T E R M SK E Y
E X P E C T A T I O N SS E C T I O N
Orbital Energies The orbital energy of a satellite consists of two components: itskinetic energy and its gravitational potential energy. To a greatextent, the Earth-satellite system can be treated as an isolated system. Subtle effects, such as the pressure of light, the solarwind, and collisions with the few atmospheric molecules thatexist at that distance from the surface, can change the energy ofthe system. In fact, without the occasional boost from a thruster,the orbits of all satellites will decay. However, this generally takesdecades. These effects are so tiny over the short run that you willneglect them in the following topics.
The kinetic energy of an orbiting satellite of mass m is Ek = GMm2r
.
Ek = GMm2r
� Since 12 mv2 is the kinetic energy
of any object of mass m, you cansubstitute Ek for the expression.
12 mv2 = GMm
2r� Multiply both sides of the equation
by 12 .
mv2rr
= GMmr2 r
mv2 = GMmr
� Multiply both sides of the equationby r.
mv2
r= GMm
r2
� Write the relationship that represents a planet’s gravity providing a centripetal force.
Chapter 6 Energy and Motion in Space • MHR 237
A thoroughunderstanding of orbitalmechanics is necessary fora successful rendezvouswith the space station orwith any other satellite.
Figure 6.2
Gravitational Potential EnergyIn Chapter 5, Conservation of Energy, you demonstrated that thechange in the gravitational potential energy of an object was equalto the work done in raising the object from one height to another.That relationship (W = mg∆h ) was the special case, where anychange in height was very close to Earth’s surface. Since you arenow dealing with objects being launched into space, you cannotuse the special case. You must consider the change in the force ofgravity as the distance from Earth increases. Fortunately, however,you have already developed an expression for the amount of workrequired to lift an object from a distance r1 to a distance r2 fromEarth’s centre. Therefore, the result of your derivation is equal tothe change in the gravitational potential energy between those twopositions.
∆Eg = GMm( 1
r1− 1
r2
)
As you know, you must choose a reference point for all forms of potential energy. Earth’s surface in no longer an appropriate reference, because you are measuring distances from Earth’s centreto deep into space. Physicists have accepted the convention ofassigning the reference or zero point for gravitational potentialenergy as an infinite distance from the centre of the planet or othercelestial body that is exerting the gravitational force on the objectof mass m. This is appropriate because at an infinite distance, thegravitational force goes to zero. You can now state that the gravita-tional potential energy of an object at a distance r2 from Earth’scentre is the amount of work required to move an object from aninfinite distance, r1, to r2.
Eg = GMm( 1
∞ − 1r2
)Eg = GMm
(0 − 1
r2
)Eg = − GMm
r2
Since there is only one distance (r2) in the equation, it is oftenwritten without a subscript. Notice, also, that the value is negative.This is simply a result of the arbitrary choice of an infinite dis-tance for the reference position. You will discover as you workwith the concept that it is a fortunate choice.
238 MHR • Unit 2 Energy and Momentum
It might seem odd that the potential energy is always negative.Since changes in energy are always of interest, however, thesechanges will be the same, regardless of the location of the zerolevel.
To illustrate this concept, consider the houses in the LoireValley in France that are carved out of the face of limestone cliffsas shown in Figure 4.3(B). To the person on the cobblestone street,everyone on floors A, B, and C in Figure 6.3(A) would have posi-tive gravitational energy, due to their height above the street.However, to a person on floor B, those on floor A are at a negativeheight, and so have negative gravitational potential energy relativeto them. At the same time, the person on floor B would consider
Quantity Symbol SI unit
gravitational potentialenergy Eg J (joules)
universal gravitationalconstant G N · m2
kg2 (newton metressquared perkilogramssquared)
mass of the planet orcelestial body M kg (kilograms)
mass of the object m kg (kilograms)
distance from centre ofplanet or celestial body r m (metres)
Unit Analysis
joule =newton · metre2
kilogram2 · kilogram · kilogram
metre
J =N · m2
kg2 · kg · kg
m= N · m = J
Note: Use of this equation implies that the reference or zeroposition is an infinite distance from the planet or celestial body.
Eg = − GMmr
GRAVITATIONAL POTENTIAL ENERGYThe gravitational potential energy of an object is the negativeof the product of the universal gravitational constant, the massof the planet or celestial body, and the mass of the object,divided by the distance from the centre of the planet or celestial body.
Chapter 6 Energy and Motion in Space • MHR 239
240 MHR • Unit 2 Energy and Momentum
that people on floor C would have a positive gravitational potentialenergy because they are higher up the cliff.
Naturally, the person standing on the roof beside the chimneywould consider that everyone in the house had negative gravita-tional potential energy. All of the residents would agree, however,on the amount of work that it took to carry a chair up from floor Ato floor C, so the energy change would remain the same, regardlessof the observer’s level. At the same time, a book dropped from awindow in floor B would hit the ground with the same kineticenergy, regardless of the location of the zero level for gravitationalpotential energy.
Figure 6.4 is a graph of the gravitational potential energy of a 1.0 kg object as it moves away from Earth’s surface. Since workmust be done on that object to increase the separation, the objectis often referred to as being in a gravitational potential energy“well.”
Since work must be done on the 1.0 kg object to move it away from Earth, although the gravitational potential energy is always negative, it is increasing (becoming less negative) as it retreats farther and farther from Earth.
Figure 6.4
Earth
approaching zeroat infinite separation
−2
−4
−6
−8Gra
vita
tion
al p
oten
tial
ener
gy (
J ×
102 )
2 4 6 8 10 12 14 16 18
Distance from Earth’s centre (Earth’s radii)
C
B
A
Some houses in the Loire Valley are carved out of limestone cliffs.
Figure 6.3
Chapter 6 Energy and Motion in Space • MHR 241
In summary, the energies of orbiting objects can be expressed as
Eg = − GMmr
Ek = GMm2r
Adding, you obtain
Etotal = Ek + Eg
Etotal = GMm2r
+(
− GMmr
)
Etotal = − GMm2r
The last equation, the total orbital energy, involves only themechanical energies — gravitational potential energy and kineticenergy. Other forms of energy, such as thermal energy, are not considered unless the satellite comes down in flames through the atmosphere.
You can obtain key information by determining whether thetotal orbital energy of an object is positive, zero, or negative. First,consider the conditions under which an object would have zerototal orbital energy around a central object, such as a planet orstar.
If an object is so far from Earth that gravity cannot pull it back,its gravitational potential energy is zero. If the object is motionlessat that point, its kinetic energy is also zero, which gives a totalenergy of zero. The total orbital energy could also be zero if themagnitude of the kinetic and potential energies were equal. Underthese conditions, the kinetic energy would be just great enough tocarry the object to a distance at which gravity could no longer pullit back. It would then have no kinetic energy left and it would bemotionless. By a similar analysis, you could draw all of the following conclusions.
� If the total of the kinetic and gravitational potential energies ofan object is zero, it can just escape from the central object.
� If the total of the kinetic and gravitational potential energies ofan object is greater than zero, it can escape from the centralobject and keep on going.
� If the total of the kinetic and gravitational potential energies ofan object is less than zero, it cannot escape from the centralobject. It is said to be bound to the object.
The extra energy needed to free the object is called the bindingenergy. Since the object will be free with a total energy of zero, the binding energy is always the negative of the total energy:Ebinding = −Etotal .
Fire a cannon ball into space? In the early 1960s, Project HARP (HighAltitude Research Project) did justthat. Scientists at McGill University in Montréal welded two U.S. Navycannon barrels together into a “super-gun” that fired 91 kg instrumentationpackages to a height of more than 145 km from a launch site on theCaribbean island of Barbados.
HARP cannonPhoto courtesy from the web sites:http://www.phy6.org/stargaze/Smartlet.htmand http://www-istp.gsfc.nasa.gov/stargaze/Smartlet.htm taken by Peter Millman.
TECHNOLOGY LINK
The orbital energy equations also have some informative simplerelationships among themselves, as listed below.
By comparing the last two equations, you can see that the satellite in a circular orbit close to the planet already has half ofthe energy it needs to completely escape from that planet. The following problems will help you to develop a deeper understand-ing of orbital energies.
Ebinding = GMm2r
� At the planet’s surface, theenergy needed to break freewas seen in Section 6.1 to be
Ek = GMm2r
� If a satellite is in an orbit closeto the planet, the radius of theorbit is essentially the same asthe radius of the planet:rorbit ≅ rplanet ≅ r .
|Eg| = 2|Ek| = 2|Et| = 2|Ebinding|
� The magnitude of the gravita-tional potential energy is twicethat of the other energies.
|Ek| = |Et| = |Ebinding|
� The magnitudes of the kineticenergy, the total energy, and thebinding energy of an orbitingobject are the same
242 MHR • Unit 2 Energy and Momentum
Space Problems1. On March 6, 2001, the Mir space station was deliberately
crashed into Earth. At the time, its mass was 1.39 × 103 kg andits altitude was 220 km.
(a) Prior to the crash, what was its binding energy to Earth?
(b) How much energy was released in the crash? Assume that its orbit was circular.
Conceptualize the Problem� When Mir was in Earth orbit, it had kinetic and gravitational
potential energy, both of which are determined by its orbitalradius.
� Mir’s binding energy is the negative of its total energy.
� After the crash, Mir had zero kinetic energy.
SAMPLE PROBLEMS
� The law of conservation of energy applies; therefore, the energyreleased in the crash is the difference between the total energy inorbit and the total energy when resting on Earth’s surface.
Identify the GoalsThe binding energy, Ebinding, of the Mir space station to Earth
The energy released during the crash of the Mir space station
Identify the Variables and ConstantsKnown Implied Unknownm = 1.39 × 103 kg (Mir)
h = 2.20 × 105 m
G = 6.673 × 10−11 N · m2
kg2
M = 5.978 × 1024 kg (Earth)
rEarth = 6.378 × 106 m
Etotal in orbit
Ebinding in orbit
Eg on ground
∆Etotal
Develop a Strategy
(a) The binding energy of the Mir space station in orbit was 4.20 × 1010 J.
Ek = 0 J
Eg = − GMmr
Eg = −
(6.673 × 10−11 N · m2
kg2
)(5.978 × 1024 kg
)(1.39 × 103 kg
)6.378 × 106 m
Eg = −8.6938 × 1010 J
Etotal = Ek + Eg
Etotal = 0 J − 8.6938 × 1010 J
Etotal = −8.6938 × 1010 J
Calculate the mechanical energyafter the crash. Note that whenMir was on Earth’s surface, itskinetic energy was zero.
Ebinding = +4.2019 × 1010 J
Ebinding ≅ +4.20 × 1010 J
Binding energy is the negativeof the total energy.
Et = − GMm2r
Et = −
(6.673 × 10−11 N · m2
kg2
)(5.978 × 1024 kg
)(1.39 × 103 kg
)2(6.598 × 106 m
)Et = −4.2019 × 1010 J
Calculate the total orbital energy before the crash.
rorbit = rearth + hfrom Earth
rorbit = 6.378 × 106 m + 2.20 × 105 m
rorbit = 6.598 × 106 m
Determine the orbital radius.
Chapter 6 Energy and Motion in Space • MHR 243
continued
(b) When Mir crashed, 4.49 × 1010 J of energy were released into the environment.
Validate the SolutionThe final answer is negative, which indicates a decrease in the energyof the system or a loss of energy to the environment.
2. A 4025 kg spacecraft (including the astronauts) is in a circularorbit 256 km above the lunar surface. Determine
(a) the kinetic energy of the spacecraft
(b) the total orbital energy of the spacecraft
(c) the binding energy of the spacecraft
(d) the speed required for escape
Conceptualize the Problem� The spacecraft is in a circular orbit around the Moon, so the
Moon is the central body.
� The spacecraft is moving, so it has kinetic energy.
� The spacecraft is in orbit, so it has gravitational potential energy.
� Binding energy is the amount of energy necessary to escape thegravitational pull of the central body.
� To escape a central body, a spacecraft must increase its kinetic energy until the total energy is zero.
� If you know the kinetic energy, you can find speed.
Identify the Goals(a) The kinetic energy of the spacecraft, Ek
(b) The total orbital energy of the spacecraft, Et
(c) The binding energy of the spacecraft, Ebinding
(d) The speed required for escape, vescape
Identify the Variables and ConstantsKnown Implied Unknownm = 4025 kg G = 6.673 × 10−11 N · m2
kg2
rMoon = 1.738 × 106 m
MMoon = 7.36 × 1022 kg
rorbit
h = 256 km Ek
Et
Ebinding
vescape
∆E = E′total − Etotal
∆E = −8.6938 × 1010 J − (−4.2019 × 1010 J)
∆E = −4.4919 × 1010 J
∆E ≅ −4.49 × 1010 J
Determine the difference intotal energy before and after the crash.
244 MHR • Unit 2 Energy and Momentum
continued from previous page
Chapter 6 Energy and Motion in Space • MHR 245
Develop a Strategy
(a) The kinetic energy of the spacecraft is 4.96 × 109 J.
(b) The total energy of the spacecraft is −4.96 × 109 J.
(c) The binding energy of the spacecraft is 4.96 × 109 J.
(d) The escape speed for the spacecraft is 2.22 × 103 m/s.
12 mv2 = Ek
v =√
2Ekm
v =
√2(9.9138 × 109 J)
4025 kg
v = 2.2195 × 103 ms
v ≅ 2.22 × 103 ms
Find the speed from the kineticenergy.
E′k = Ek + Ebinding
E′k = 4.9569 × 109 J + 4.9569 × 109 J
E′k = 9.9138 × 109 J
The binding energy must comethrough additional kinetic energy
Ebinding = −Et
Ebinding = −(−4.952 × 109 J)
Ebinding = 4.96 × 109 J
The binding energy is the negative of the total energy
Et = −Ek
Et = −4.96 × 109 J
Total orbital energy is the negative of the kinetic energy.
Ek = GMm2r
Ek =
(6.673 × 10−11 N · m2
kg2
)(7.36 × 1022 kg
)(4025 kg
)2(1.994 × 106 m
)Ek = 4.9569 × 109 J
Ek ≅ 4.96 × 109 J
Calculate the orbital kinetic energy.
rorbit = rMoon + h
rorbit = 1.738 × 106 m + 0.256 × 106 m
rorbit = 1.994 × 106 m
Determine the orbital radius of thespacecraft.
continued
246 MHR • Unit 2 Energy and Momentum
Validate the SolutionThe escape speed for the spacecraft could have been determined from
the equation vescape =√
2GMr
. Carrying out this calculation yields the
same escape speed. Since it agrees with the results of the energy calculations, it acts as a check on the first three answers.
4. A 55 kg satellite is in a circular orbit aroundEarth with an orbital radius of 7.4 × 106 m.Determine the satellite’s
(a) kinetic energy
(b) gravitational potential energy
(c) total energy
(d) binding energy
5. A 125 kg satellite in a circular orbit aroundEarth has a potential energy of −6.64 × 109 J.Determine the satellite’s
(a) kinetic energy
(b) orbital speed
(c) orbital radius
6. A 562 kg satellite is in a circular orbit around Mars. Data: rMars = 3.375 × 106 m;rorbit = 4.000 × 106 m; MMars = 6.420 × 1023 kg
(a) If the satellite is allowed to crash on Mars,how much energy will be released to theMartian environment?
(b) List several of the forms that the releasedenergy might take.
7. From the orbital kinetic energy of the lunarspacecraft in the second sample problem,determine its orbital speed. What increasebeyond that speed was required for escapefrom the Moon?
8. A 60.0 kg space probe is in a circular orbitaround Europa, a moon of Jupiter. If theorbital radius is 2.00 × 106 m and the mass of Europa is 4.87 × 1022 kg, determine the
(a) kinetic energy of the probe and its orbitalspeed
(b) gravitational potential energy of the probe
(c) total orbital energy of the probe
(d) binding energy of the probe
(e) additional speed that the probe must gainin order to break free of Europa
9. A 1.00 × 102 kg space probe is in a circularorbit, 25 km above the surface of Titan, amoon of Saturn. If the radius of Titan is 2575 km and its mass is 1.346 × 1023 kg,determine the
(a) orbital kinetic energy and speed of thespace probe
(b) gravitational potential energy of the spaceprobe
(c) total orbital energy of the space probe
(d) binding energy of the space probe
(e) additional speed required for the spaceprobe to break free from Titan
10. Material has been observed in a circular orbitaround a black hole some five thousandlight-years away from Earth. Spectroscopicanalysis of the material indicates that it isorbiting with a speed of 3.1 × 107 m/s. If theradius of the orbit is 9.8 × 105 m, determinethe mass of the black hole.
PRACTICE PROBLEMS
continued from previous page
Chapter 6 Energy and Motion in Space • MHR 247
6.2 Section Review
1. Explain why the determination of orbitalspeed does not require knowledge of thesatellite’s mass, while determination oforbital energies does require knowledge of the satellite’s mass.
2. Explain why the binding energy of asatellite is the negative of its total orbitalenergy. Why does this relationship notdepend on the satellite being in a circularorbit?
3. Draw a concept organizer to show thelinks between the general equations for workand energy and the orbital energy equations.Indicate in the organizer which equations arejoined together to produce the new equation.
4. A satellite with an orbital speed of vorbit
is in a circular orbit around a planet. Provethat the speed for a satellite to escape fromorbit and completely leave the planet isgiven by vescape =
√2(vorbit).
5. The magnitude of the attractive forcebetween an electron and a proton is given by
F = kqeqp
r2 , where qe is the magnitude of the
charge on the electron, qp is the magnitude ofthe charge on the proton, r is the separationbetween them, and k is a constant that playsthe same role as G. If the mass of the electronis represented by me, derive an equation forthe orbital speed of the electron.
MC
I
C
K/U
C
