476Week 11: SoundsxLANAANNm = 1m = 2Figure 141: The pipe closed at both ends isjust like a string fixed at both ends, as long as oneconsiders thedisplacementwave.ork Lm=mπ(1018)form= 1 2 3 . This converts to:, , ...λ m= 2 Lm(1019)andf m= v aλ m=v ma2 L(1020)Them= 1 solution (first harmonic) is called theprinciple harmonicas it was before. The actualtone of a flute pipe with two closed ends will be a superposition of harmonics, usually dominated bythe principle harmonic.11.5.2: Pipe Closed at One Endm = 2sxNALANAm = 1Figure 142: The pipe closed at both ends isjust like a string fixed at one end, as long as oneconsiders thedisplacementwave.In the case of a pipe open at onlyoneend, there is adisplacement nodeat theclosed end,and adisplacement antinodeat theopen end. If one considers the pressure wave, the positionsof nodes and antinodes arereversed. This is just like a string fixed at one end and free at the other.Let’s arbitrarily makex= 0 the closed end. Then:s x, t() =s 0sin(k xm) cos(ω tm )(1021)
Week 11: Sound477has a node atx= 0 for all . To get an antinode at the other end, we require:ksin(k Lm) =± 1(1022)ork Lm= 2m − 22π(1023)form= 1 2 3, , ...(odd half-integral multiples of . As before, you will see different conventions usedπtonamethe harmonics, with some books asserting that only odd harmonics are supported, but Iprefer to make the harmonic index do exactly the same thing for both pipes so it counts the actualnumber of harmonics thataresupported by the pipe. This is much more consistent with what onewill do next semester considering e.g. interference, where one often encounters similar series for aphase angle in terms of odd-half integer multiples of , and makes the second harmonic the lowestπfrequency actually present in the pipe inall three casesof pipes closed at neither, one or both ends.This converts to:λ m=4 L 2m − 1(1024)andf m= v aλ m= v a(2m − 1)4 L(1025)11.5.3: Pipe Open at Both Endsm = 2sxLm = 1AANAANANFigure 143: A pipe open at both ends is the exactoppositeof a pipe (or string) closed (fixed) atboth ends: It hasdisplacement antinodesat the ends. Note well the principle harmonic with a singlenode in the center. Theresonant frequencyseries for the pipe is the same, however, as for a pipeclosed at both ends!This is apanpipe, one of the most primitive (and beautiful) of musical instruments. A panpipeis nothing more than a tube, such as a piece of hollow bamboo, open at both ends. The modes ofthis pipe are driven at resonance by blowing gently across one end, where the random fluctuationsin the airstream are amplified only for the resonant harmonics.To understand the frequencies of those harmonics, we note that there aredisplacement antin-odesatboth ends. This is just like a string free at both ends. The displacement solution mustthus be acosinein order to have a displacement antinode atx= 0:s x, t() =s 0cos(k xm) cos(ω tm )(1026)andcos(k Lm) =± 1(1027)
478Week 11: SoundWe can then writek Lmas a series of suitable multiples ofπand proceed as before to find thewavelengths and frequencies as a function of the mode indexm= 1 2 3 . This is left as a (simple), , ...exercise for you.Alternatively, we couldalsonote that there arepressurenodes at both ends, which makes themlike a string fixed at both ends again as far as thepressurewave is concerned. This gives usexactlythe same result (for frequencies and wavelengths) as the pipe closed at both ends above, althoughthe pipeopenat both ends is probably going to be a bit louder and easier to drive at resonance (howcan you “blow” on a closed pipe to get the waves in there in the first place? How can the sound getout?Either way one will get the same frequencies but thepictureof the displacement waves is differentfrom thepictureof the pressure waves – be sure to draw displacement antinodes at the open endsif you are asked to draw a displacement wave, or vice versa for a pressure wave!You might try drawing the first 2-3 harmonics on a suitable picture like the first two given above,with displacement antinodes at both ends. What does the principle harmonic look like? Show thatthe supported frequencies and wavelengths match those of the string fixed at both ends, or pipeclosed at both ends.11.6: BeatsIf you have ever played around with a guitar, you’ve probably noticed that if two strings are fingeredto be the “same note” but are really slightly out of tune and are struck together, the resulting sound“beats” – it modulates up and down in intensity at a low frequency often in the ballpark of a fewcycles per second.Beats occur because of the superposition principle. We can add any two (or more) solutions tothe wave equation and still get a solution to the wave equation, even if the solutions have differentfrequencies. Recall the identity:sin( ) + sin( ) = 2 sin(ABA + B2) cos(A − B2)(1028)If one adds two waves with different wave numbers/frequencies and uses this rule, one getss x, t()=s 0sin(k x0−ω t0) +s 0sin(k x1−ω t1)(1029)= 2s 0sin(k 0+ k 12x− ω 0+ ω 12t) cos(k 0− k 12x− ω 0− ω 12t )(1030)This describes a wave that has twice the maximum amplitude, theaveragefrequency (the firstterm), and a second term that (at any point ) oscillates like cos(x∆ ωt2).The “frequency” of this second modulating term isf 0− f 1 2, but the earcannot hearthe inversion ofphase that occurs when it is negative and the difference is small. It just hears maximum amplitudein the rapidly oscillatingaveragefrequency part, which goes tozerowhen the slowing varying cosinedoes, twice per cycle. The ear thenhearstwo beats per cycle, making the “beat frequency”:fbeat= ∆ =f|f 0− f 1 |(1031)11.7: Interference and Sound WavesWe will not cover interference and diffraction of harmonic sound waves in this course. Beats are acommon experience in sound as is the doppler shift, but sound wave interference is not so common
Week 11: Sound479an experience (although it can definitely and annoyingly occur if you hook up speakers in your stereoout of phase). Interferencewillbe treated next semester in the context of coherent light waves.Justto give you a head start on that, we’ll indicate the basic ideas underlying interference here.Suppose you have two sources that are at thesamefrequency and have thesameamplitude andphase but are at different locations. One source might be a distancexaway from you and the othera distancex+ ∆ away from you. The waves from these two sources add like:xs x, t()=s 0sin(kx−ωt) +s 0sin( ( + ∆ )k xx−ωt)(1032)=2 s 0sin( ( +k x∆ x2−ωt) cos(k∆ x2)(1033)The sine part describes a wave with twice the amplitude, the same frequency, but shifted slightlyin phase by ∆k x/2. The cosine part istime independentandmodulatesthe first part. For somevalues of ∆ it canxvanish. For others it can have magnitude one.The intensity of the wave is what our ears hear – they are insensitive to the phase (althoughcertain echolocating species such as bats may be sensitive to phase information as well as frequency).The average intensity is proportional to the wave amplitudesquared:I 0= 12ρω s v2 20(1034)With two sources (and a maximum amplitude of two) we get:I=12ρω2(22 2s 0cos (2k∆ x2) v(1035)=4 I 0cos (2k∆ x2)(1036)There are two cases of particular interest in this expression. Whencos (2k∆ x2) = 1(1037)one hasfour timesthe intensity of one source at peak. This occurs when:k∆ x2=nπ(1038)(forn= 0 1 2 ) or, , ...∆ =xnλ(1039)If thepath differencecontains anintegral number of wavelengthsthe waves from the two sourcesarrivein phase, add, and produce sound that has twice the amplitude and four times the intensity.This is calledcomplete constructive interference.On the other hand, whencos (2k∆ x2) = 0(1040)the sound intensityvanishes. This is calleddestructiveinterference. This occurs whenk∆ x2=2 + 1n2π(1041)(forn= 0 1 2 ) or, , ...∆ =x2 + 1n2λ(1042)If the path difference contains ahalf integralnumber of wavelengths, the waves from two sourcesarrive exactly out of phase, andcancel. The sound intensity vanishes.
480Week 11: SoundYou can see why this would make hooking your speakers up out of phase a bad idea. If youhook them up out of phase the wavesstartwith a phase difference ofπ– one speaker is pushing outwhile the other is pulling in. If you sit equidistant from the two speakers and then harmonic waveswith the same frequency from a single source coming from the two speakerscancelas they reach you(usually not perfectly) and the music sounds very odd indeed, because other parts of the music arenot being played equally from the two speakers and don’t cancel.You can also see that there are many other situations where constructive or destructive interfer-ence can occur, both for sound waves and for other waves including water waves, light waves, evenwaves on strings. Our “standing wave solution” can be rederived as the superposition of a left- andright-travelling harmonic wave, for example. You can have interference from more than one source,it doesn’t have to be just two.This leads to some really excellent engineering. Ultrasonic probe arrays, radiotelescope arrays,sonar arrays, diffraction gratings, holograms, are all examples of interference being put to work. Soit is worth it to learn the general idea as early as possible, even if it isn’t assigned.11.8: The EarFigure 144: The anatomy of the human ear.Figure 144 shows a cross-section of the human ear, our basic transduction device for sound. Thisis not a biology course, so we will not dwell uponallof the structure visible in this picture, butrather will concentrate on the parts relevant to the physics.Let’s start with theouter ear. This structure collects sound waves from a larger area than theear canal per se and reflects them down to the ear canal. You can easily experiment with the kindof amplification that results from this by cupping your hands and holding them immediately behindyour ears. You should be able to hear both a qualitative change in the frequencies you are hearingand an effective amplification of the sounds from infrontof you at the expense of sounds originatingbehindyou. Many animals have larger outer ears oriented primarily towards the front, and havemuscles that permit them to further alter the direction of most favorable sound collection withoutturning their heads. Human ears are more nearly omnidirection.Theauditory canal(ear canal in the figure above) acts like a resonant cavity to effectivelyamplify frequencies in the 2.5 kHz range and tune energy deliver to the tympanic membrane or
Week 11: Sound481eardrum. This membrane is a strong, resilient, tightly stretched structure that can vibrate inresponse to driving sound waves. It is connected to a collection of small bones (the ossicles) thatconduct sound from the eardrum to the inner ear and that constitute themiddle eargurefiin the above. The common name of the ossicles are: hammer, anvil and stirrup, the latter so named becausesefiectively ampliffits shape strongly resembles that of the stirrup on a hores saddle. The anvil eoscillations by use of the principle of leverage, as a fulcrum attachment causes the stirrup end tovibrate through a much larger amplitude than the hammer end. The stirrup is directly connectedto theoval window, the gateway into the inner ear.The middle ear is connected to theeustachian tubeto your throat, permitting pressure insideyour middle ear to equalize with ambient air pressure outside. If you pinch your nose, close yourmouth, and try to breath out hard, you can actually blow air out through your ears although this isunpleasant and can be dangerous. This is one way your ears equilbrate by “popping” when you rideerences acrossffy in an airplane. If/when this does not happen, pressure difla car up a hillside or the tympanic membrane reduce its response to ambient sounds reducing auditory acuity.Figure 145: A cross-section of the spiral structure of the cochlea.Sound, amplifed by focal concentration in the outer ear, resonance in the auditory canal, andmechanical leverage in the ossicles, enters thecochlea, a shell-shaped spiral that is the primaryorgan of hearing that transduces sound energy into impulses in our nervous system through the ovalwindow. The cochlea containshair cellsof smoothly varying length lining the narrowing spiral,each of which isresonantto a particular auditory frequency. The arrangement of the cells in agure 145.ficross-section of the cochlea is shown in The nerves stretching from these cells are collected into theauditory nerve bundleand fromwhen they receive sound at the right frequency in to the ffthence carries the impulses they give oauditory cortex(not shown) where it becomes, eventually and through a process still not fullyunderstood, our perception of sound. Our brains take this frequency resolved information – thebiomechanical equivalent of a fourier transform of the sound signal, in a way – and synthesize itback into a detailed perception of sound and music within the general frequency range of 10 Hz to20,000 Hz.As you can see, there are many individual parts that can fail in the human auditory system.er damage to their outer ears through accident or disease. The ear canalffIndividuals can lose or sucan become clogged with cerumen, orearwaxuid that normally cleans and lubricates thefl, a waxy ear canal and eardrum but that can build up and dry out to both load the tympanic membraneso it becomes less responsive and physically occlude part of the canal so less sound energy can getthrough. The eardrum itself is vulnerable to sudden changes in sound pressure or physical contactthat can puncture it. The middle ear, as a closed, warm, damp cavity connected to the throat, is anideal breeding ground for certain bacteria that can cause infections and swelling that both interferewith or damage hearing and that can be quite painful. The ossicles are susceptible to physical
482Week 11: Soundtrauma and infectious damage.Finally, the hair cells of the cochlea itself, which are safely responsive over at least twelve tofourteen orders of magnitude of transient sound intensity (and safely responsive over eight or nineorders of magnitude of sustained sound intensity) are highly vulnerable tobothsudden transientsounds of still higher intensity (e.g. sound levels in the vicinity of 120 to 140 decibels and higherandto sustained excitation at sound levels from roughly 90 decibels and higher. Both disease andmedical conditions such as diabetes (that produces a progressive neuropathy) can further contributeto gradual or acute hearing loss at the neurological level.When hair cells die, they do not regenerate and hearing loss of this sort is thus cumulative overa lifetime. It is therefore a really good idea to wear ear protectors if, for example, you play aninstrument in a marching band or a rock and roll band where your hearing is routinely exposed to100 dB and up sounds. It is also a good reason not to play music too loudly when you are young,however pleasurable it might seem. One is, after all, very probably trading listening to very loudmusic at age seventeen against listening to musicat allat age seventy. Hearing aids do not reallyfix the problem, although they can help restore enough function for somebody to get by.However, it is quite possible that over the next few decades the bright and motivated physicsstudents of today will help create the bioelectronic and/or stem cell replacements of key organs andnervous tissue that will relegate age-related deafness to the past. I would certainly wish, as I sithere typing this with eyes and ears that are gradually failing as I age, that whether or not it comein time for me, it comes in time to help you.
Week 11: Sound483Homework for Week 11Problem 1.Physics Concepts:Make this week’s physics concepts summary as you work all of the problemsin this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s)they were key to, and include concepts from previous weeks as necessary. Do the work carefullyenough that you can (after it has been handed in and graded) punch it and add it to a three ringbinder for review and study come finals!
484Week 11: SoundProblem 2.aaaaaaaaaa= f oLavaBill and Ted are falling into hell at aconstant speed(terminal velocity), and are screaming atthe frequencyf 0. As they fall, they hear their own voices reflecting back to them from the puddleof molten rock that lies below at a frequency of 2 .f 0How fast are they falling relative to the speed of sound in warm, dry hellish air?
Week 11: Sound485Problem 3.abcABSound waves travel faster in water than they do in air. Light waves travel faster in air than theydo in water. Based on this, which of the three paths pictured above are more likely tominimizethetime required for thea) Sound:b) Light:produced by an underwater explosion to travel from the explosion atAto the pickup atB? Why(explain your answers)?
486Week 11: SoundProblem 4.Fred is standing on the ground and Jane is blowing past him at a closest distance of approach of afew meters at twice the speed of sound in air. Both Fred and Jane are holding a loudspeaker thathas been emitting sound at thefrequencyf 0for some time.a) Who hears the sound produced by theotherperson’s speaker assingle frequency soundwhenthey are approaching one another and what frequency do they hear?b) What does theotherperson hear (when they hear anything at all)?c) What frequenc(ies) do each of them hearafterJane has passed and is receding into thedistance?Problem 5.Discuss and answer the following questions:a) Sunlight reaches the surface of the earth withroughly1000 Watts/meter of intensity. What is2the “sound intensity level” of asoundwave that carries as muchenergy per square meter,indecibels?b) In table 5, what kind of sound sources produce this sort of intensity? Bear in mind that theSun is150 million kilometers awaywhere sound sources capable of reaching the same intensityare typically only a few meters away. The the Sun produces alotof (electromagnetic) energycompared to terrestrial sources of (sound) energy.c) The human body produces energy at the rate of roughly 100 Watts.Estimatethe fraction ofthis energy that goes into my lecture when I am speaking in a loud voice in front of the class(loud enough to be heard as loudly as normal conversation ten meters away).d) Again using table 5, how far away from a jack hammer do you need to stand in order for thesound to (marginally) no longer be dangerous to your hearing?
Week 11: Sound487Problem 6.String one has mass per unit lengthµand is at tensionTand has a travelling harmonic wave on it:y x, t1 () =Asin(kx−ωt)String one is also very long compared to a wavelength:L≫ λ .Identical string two has the superposition of two harmonic travelling waves on it:y x, t2 () =Asin(kx−ωt) + 3 sin(Akx+ωt)If the average energy density (total mechanical energy per unit length) of the first string isE 1 ,what is the average energy density of the second stringE 2intermsofE 1 ?Problem 7.Two identical strings of lengthLhave massµand are fixed at both ends. One string has tensionT. The other has tension 1 21 . When plucked, the first string produces a tone at frequency. Tf 1 ,the second produces a tone at frequencyf 2 .a) What is thebeatfrequency produced if the two strings are is plucked at the same time, interms off 1 ?b) Are the beats likely to be audible iff 1is 500 Hz? How about 50 Hz? Why or why not?Problem 8.You measure the sound intensity level of a single frequency sound wave produced by a loudspeakerwith a calibrated microphone to be 80 dB. At that intensity, thepeakpressure in the sound waveat the microphone isP 0. The loudspeaker’s amplitude is turned up until the intensity level is 100dB. What is the peak pressure of the sound wave now (in terms ofP 0)?Note that you could look this up in the table, but don’t. The point is for you to know how thepeak pressurescaleswith the intensity, as well as how the intensity varies with the sound intensitylevel in decibels.
488Week 11: SoundProblem 9.Resonant Sound WavesTube closed at one endAn organ pipe is made from a brass tube closed at one end as shown. The pipe has lengthLandthe speed of sound isv s. When played, it produces a sound that is a mixture of the first, third andsixth harmonic (mode, countingn= 1 2 3 )., , ...a) What are the frequencies of these modes?b) Qualitatively sketch the wave amplitudes for the first and the third harmonic modes (only) inon the figure, indicating the nodes and antinodes. Be sure to indicate whether the nodes orantinodes drawn are forpressure/densitywaves ordisplacementwaves!c) Evaluate your answers numerically whenL= 3 4 meters long, and.v s= 340 meters/second(as usual).
Week 11: Sound489Problem 10.L 0,10,1 fYou crash land on a strange planet and all your apparatus for determining if the planet’s atmo-sphere is like Earth’s is wrecked. In desperation you decide to measure the speed of sound in theatmosphere before taking off your helmet. You do have a barometer handy and can see that the airpressure outside is approximately one atmosphere and the temperature seems to be about 300 K,◦so if the speed of sound is the same as on Earth the airmightbe breathable.You jury rig a piston and cylinder arrangement like the one shown above (where the cylinder isclosed at both endsbut has a small hole in the side to let sound energy in to resonate) and takeout your two handy tuning forks, one atf 0= 3400 Hz and one atf 1= 6800 Hz.a) Using the 3400 Hz fork as shown, what do youexpect(or rather, hope) to hear as you movethe piston in and out (varyingL 0). In particular, what are the shortest few values ofL 0forwhich you expect to hear a maximum resonant intensity from the tube if the speed of soundin the unknown atmosphere is indeed the same as in air (which you will cleverly note I’mnottelling you as you are supposed to know this number)?b) Using the 6800 Hz fork you hear your first maximum (for the smallest value ofL 1) atL 1= 5cm. Should you sigh with relief and rip off your helmet?c) What is thenextvalue forL 1for which you should hear a maximum (given the measurement inb) and what should the difference between the two equal in terms of the wavelength of the 6800Hz wave in the unknown gas? Draw thedisplacementwave for this case only schematically inon the diagram above (assuming that theLshown is this second-smallest value ofL 1for thef 1tuning fork), and indicate where the nodes and antinodes are.
490Week 12: GravityOptional ProblemsStudy for the final exam!This is the last week of class, and this wraps up both thechapter and the texbook. Students looking for more problems to work on are directed to the onlinereview guide for introductory physics 1 and the online math review, the latter as needed.
Week 12: GravityGravity Summary•Early western (Greek) cosmology was bothgeocentric– simple earth-centered model withfixed stars “lamps” or “windows” on big solid bowl, moon and stars and planets orbiting the(usually flat) Earth “somehow” in between. The simple geocentric models failed to explainretrograde motionof the planets, where for a time they seem to go backwards against thefixed stars in their general orbits. There were also earlyheliocentric– sun centered – models,in particular one by Aristarchus of Samos (270 B.C.E.), who usedparallaxto measure thesize of the earth and the sizes of and distances to the Sun and Moon.•Ptolemy206(140 C.E.) “explained” retrograde motion with ageometric geocentric modelinvolving complexepicycles. Kudos to Ptolemy for inventing geometric modelling in physics!The model was a genuine scientific hypothesis, in principlefalsifiable, and a good startingplace for further research.Sadly, a few hundred years later the state religion of the western world’s largest empire em-braced this geocentric model as beingconsistent with The Book of Genesisin its theisticscriptural mythology (and with many other passages in the old and new testaments) and forover a thousand years alternative explanations were considered heretical and could only bemade at substantial personal risk throughout the Holy Roman Empire.•Copernicus207(1543 C.E.) (re)invented aheliocentric– sun-centered model, explained ret-rograde motion withsimplerplain circular geometry, regular orbits. The work of Copernicus,De Revolutionibus Orbium Coelestium208(On the Revolutions of the Heavenly Spheres) wasforthwith banned by the Catholic Church as heretical at the same time that Galileo was bothpersecuted and prosecuted.•Wealthy Tycho Brahe accumulated data and his paid assistant, Johannes Kepler, fit that datato specific orbits and deducedKepler’s Laws. All Brahe got for his efforts was a lousy mooncrater named afterhim209.•Kepler’s Laws:a) All planets move in elliptical orbits with the sun at one focus.b) A line joining any planet to the sun sweeps out equal areas in equal times (dA/dt=constant).c) The square of the period of any planet is proportional to the cube of the planet’s meandistance from the sun (T 2=CR3). Note that the semimajor or semiminor axis of theellipse will serve as well as the mean, with different contants of proportionality.206Wikipedia: http://www.wikipedia.org/wiki/Ptolemy.207Wikipedia: http://www.wikipedia.org/wiki/Copernicus.208Wikipedia: http://www.wikipedia.org/wiki/De revolutionibus orbium coelestium.209Wikipedia: http://www.wikipedia.org/wiki/Tycho (crater).491
492Week 12: Gravity•Galileo210(1564-1642 C.E.) is known as the Copernican heliocentric model’s most famous earlydefender, not so much because of the quality of his science as for his infamousprosecutionby the Catholic church. In truth, Galileo was a contemporary of Kepler and his workwas nowhere nearly as carefully done or mathematically convincing (or correct!) as Kepler’s,although using atelescopehe made a number of important discoveries that added considerablefurther weight to the argument in favor of heliocentrism in general.•Newton211(1642-1727 C.E.) was the inheritor of the tremendous advances of Brahe, Descartes212(1596-1650 C.E.), Kepler, and Galileo. Applying the analytic geometry invented by Descartesto the empirical laws discovered by Kepler and the kinematics invented by Galileo, he was ableto deduceNewton’s Law of Gravitation:~F= −GMmr 2ˆr(1043)(a simplified form valid when massM≫ m ,~rare coordinates centered on the larger massM, and~Fis the force acting on the smaller mass); we will learn a more precisely statedversion of this law below. This law fully explained at the limit of observational resoution, andcontinues tomostlyexplain, Kepler’s Laws and the motions of the planets, moons, comets,and other visible astronomical objects! Indeed, it allows their orbits to be precisely computedand extrapolated into the distant past or future from a sufficient knowledge of initial state.•In Newton’s Law of Gravitation the constantGis a considered to be aconstant of nature,and was measured by Cavendish213in a famous experiment, thus (as we shall see) “weighingthe planets”. The value ofGwe will use in this class is:G= 6 67.×10− 11N-m2kg2(1044)You are responsible for knowing this number!Like , it is enormously important andguseful as a key to the relative strength of the forces of nature and explanation for why it takesan entire planet to produce a force on your body that is easily opposed by (for example) athin nylon rope.•Thegravitational fieldis a simplification of Newton’s theory of gravitation that emergedover a considerable period of time with no clear author that attempts to resolve the problemNewton first addressed ofaction at a distance– the need for acausefor the gravitationalforce that propagatesfromone objecttothe other. Otherwise it is difficult to understand howone mass “knows” of the mass, direction and distance of its partner in the gravitational force!It is (currently) defined to be thegravitational force per unit massorgravitationalaccelerationproduced at and associated with everypoint in spaceby asinglemassiveobject. This field acts on any mass placed at that point and thereby exerts a force. Thus:~g ~( )r=−GMr 2ˆr(1045)~F m( )~r=m~g ~( ) =r−GMmr 2ˆr(1046)•Important true facts about the gravitational field:–The gravitational field produced by a (thin) spherically symmetric shell of mass ∆Mvanishes inside the shell.210Wikipedia: http://www.wikipedia.org/wiki/Galileo.211Wikipedia: http://www.wikipedia.org/wiki/Newton.212Wikipedia: http://www.wikipedia.org/wiki/Descartes.213Wikipedia: http://www.wikipedia.org/wiki/Cavendish.
Week 12: Gravity493–The gravitational field produced by this same shell equals the usual~g ~( ) =r−G M∆r 2ˆr(1047)outsideof the shell. As a consequence the field outside of any spherically symmetricdistribution of mass is just~g ~( ) =r−G M∆r 2ˆr(1048)These two results can be proven by direct integration or by using Gauss’s Law for the gravi-tational field (using methodology developed next semester for the electrostatic field).•The gravitational force is conservative. The gravitational potential energy of massmin thefield of massM is:U m( ) =~r− Z~r∞~F ·d ~ℓ= −GMmr(1049)By convention, the zero of gravitational potential energy is atr 0=∞(in all directions).•Thegravitational potentialis to the potenial energy as the gravitational field is to the force.That is:V( ) =~rU m( )~rm= − Z~r∞~g·d ~ℓ= −GMr(1050)It as a scalar field that depends only on distance, it is the simplest of the ways to describegravitation. Once the potential is known, one can always find the gravitational potentialenergy:U m( ) =~rmV( )~r(1051)or the gravitational field:~g ~( ) =r− ~∇ V( )~r(1052)or the gravitational force:~F m( ) =~r−m V ~∇( ) =~rm~g ~( )r(1053)•Escape velocityis the minimum velocity required to escape from the surface of a planet (orother astronomical body) and coast in free-fall all the way to infinity so that the object “arrivesat infinity at rest”. SinceU (∞) = 0 by definition, theescape energyfor a particle is:Eescape= K (∞ ) +U (∞) = 0 + 0 = 0(1054)Since mechanical energy is conserved moving through the (presumed) vacuum of space, thetotal energy must be zero on the surface of the planet as well, or:12mv2e−GMmR= 0(1055)orv e=r 2GMR(1056)On the earth:v e=r 2GMR=p 2gRe= 11 2.×10 meters/second3(1057)(11.2 kilometers per second). This is also the most reasonable starting estimate for the speedwith which falling astronomical objects, e.g. meteors or asteroids, willstrikethe earth. Alarge falling mass loses basically all of its kinetic energy on impact, so that even a fairly smallasteroid can easily strike with an explosive power greater than that of anuclear bomb, ormany nuclear bombs. It is believed that just such a collision was responsible for at least thefinal Cretaceous extinction event that brought an end to the age of the dinosaurs some sixtymillion years ago, and similar collisions may have caused other great extinctions as well.
494Week 12: Gravity•A (point-like) object in a plane orbit has a kinetic energy that can be written as:K = K rot+ K r=L 22mr2+ 12mv2r(1058)The total mechanical energy of this object is thus:E= K + U= 12mv2r+L 22mr2−GMmr(1059)~Lfor an orbit (in a central force, recall) isconstant, henceL 2is constant in this expression.The total energy and the angular momentum thus become convenient ways to parameterizethe orbit.•Theeffective potential energyis of a massmin an orbit with (magnitude of) angularmomentumLis:U r ′( ) =L 22mr2−GMmr(1060)and the total energy can be written in terms of theradial kinetic energy onlyas:E= 12mv2r+U r ′( )(1061)This is a convenient form to use to makeenergy diagramsand determine theradial turningpoints of an orbit, and permits us to easily classify orbits not only as ellipses but as generalconic sections. The termL / mr 222is called theangular momentum barrierbecauseit’s negative derivative with respect torcan be interpreted as a strongly (radially) repulsivepseudoforce for small .r•The orbit classifications (for a given nonzeroL) are:–Circular: Minimum energy, only one permitted value ofr cin the energy diagram whereE=U r ′( ).c–Elliptical: Negative energy, always have two turning points.–Parabolic: Marginally unbound,E= 0, one radial turning point. This is the “escapeorbit” described above.–Hyperbolic: Unbound,E >0, one radial turning point. This orbit has enough energy toreach infinity while still moving, if you like, although a better way to think of it is thatits asymptotic radial kinetic energy is greater than zero.
Week 12: Gravity49512.1: Cosmological ModelsSSmEmstarsstars(far away)PlanetsepicyclesorbitsEPtolomeic (terricentric epicycles)Copernican (heliocentric orbits)Figure 146: The Ptolemaicgeocentricmodel withepicyclesthat sufficed to explain the observa-tional data of retrograde motion. The Copernicangeocentricmodel also explained the data andwas somewhat simpler. To determine which was correct required the use ofparallaxto determinedistancesas well as angles.Early western (Greek) cosmology was bothgeocentric, with fixed stars “lamps” or “windows”on a big solid bowl, the moon and sun and planets orbiting a fixed, stationary Earth in the center.Plato represented the Earth (approximately correctly) as a sphere and located it at the center ofthe Universe. Astronomical objects were located on transparent “spheres” (or circles) that rotateduniformly around the Earth at differential rates. Euxodus and then Aristotle (both students ofPlato) elaborated on Plato’s original highly idealized description, adding spheres until the model“worked” to some extent, but left a number of phenomena either unexplained or (in the case of e.g.lunar phases) not particularlybelievablyexplained.The principle failure of the Aristotelian geocentric model is that it fails to explainretrogrademotionof the planets, where for a time they seem to go backwards against the fixed stars intheir general orbits. However, in the second century Claudius Ptolemaeus constructed a somewhatsimpler geocentric model that is currently known as thePtolemaic modelthat still involved Plato’scircular orbits with stars embedded on an outer revolving sphere, but added to this the notion ofepicycles– planets orbiting in circlesarounda point that was itself in a circular orbit around theEarth. The model was very complex, but it actuallyexplained the observational data includingretrograde motionwell enough that – for a variety of political, psychological, and religious reasons –it was adopted asthe “official” cosmology of Western Civilization, endorsed and turnedinto canonical dogma by the Catholic Churchas geocentrism agreed (more or less) with thecosmological assertions of the Bible.In this original period – during which the Greeks invented things like mathematics and philosophyand the earliest rudiments of physics – geocentrism was not the only model. The Pythogoreans, forexample, postulated that the earth orbited a “great circle of fire” that was always beneath one’s feetin a flat-earth model, while the sun, stars, moon and so on orbited the whole thing. An “anti-Earth”was supposed to orbit on the far side of the great fire, where we cannot see it. All one can say is gee,they must have had really good recreational/religious hallucinogenic drugs back then...214. Another214Which in fact, they did...
496Week 12: Gravity“out there” model – by the standards of the day – was theheliocentricmodel.The first person known to have proposed a heliocentric system, however, was Aristarchus ofSamos (c. 270 BC). Like Eratosthenes, Aristarchus calculated the size of the Earth, and measuredthe size and distance of the Moon and Sun, in a treatise which has survived. From his estimates,he concluded that the Sun was six to seven times wider than the Earth and thus hundreds of timesmore voluminous. His writings on the heliocentric system are lost, but some information is knownfrom surviving descriptions and critical commentary by his contemporaries, such as Archimedes.Some have suggested that his calculation of the relative size of the Earth and Sun led Aristarchusto conclude that it made more sense for the Earth to be moving than for the huge Sun to be movingaround it.Archimedes was familiar with, and apparently endorsed, this model. This model explained thelack of motion of the stars by putting themvery far awayso that distances to them could noteasily be detected using parallax! This was the first hint that the Universe wasmuch larger thangeocentric models assumed, which eliminated anyneedfor parallax by approximately fixing theearth itself relative to the stars.The heliocentric model explained may things, but it wasn’t clear how it would explain (in par-ticular) retrograde motion. For a variety of reason (mostly political and religious) the platonicgeocentric model was preserved and the heliocentric model officially forgotten and ignored until theearly 1500’s, when a catholic priest and polymath215resurrected it and showed how itexplainedretrograde motion with far less complexity than the Ptolemaic model. Since the work of Aristarchuswas long forgotten, this reborn heliocentric model was called theCopernican model216, and wasperhaps the spark that lit the early Enlightenment217.Initially, the Copernican model, published in 1543 by a Copernicus who was literally on hisdeathbed, attracted little attention. Over the next 70 years, however, it gradually caused more andmore debate, in no little part because it directly contradicted a number of passages in the Christianholy scriptures and thereby strengthened the position of an increasing number of contemporaryphilosophers who challenged the divine inspiration and fidelity of those writings. This drew attentionfrom scholars within the established Catholic church as well as from the new Protestant churchesthat were starting to emerge, as well as from other philosophers.The most important of these philosophers was another polymath by the name of Galileo Galilei218. The first refracting telescopes were built by spectacle makers in the Netherlands in 1608; Galileoheard of the invention in 1609 and immediately built one of his own that had a magnification ofaround 20. With this instrument (and successors also of his own design) he performed an amazingseries of astronomical observations that permitted him toempirically supportthe Copernican modelin preference over the Ptolemaic model.It is important to note well thatbothmodels explain the observations available to the naked eye.Ptolemaeus’ model was somewhat more complex than the Copernican model (which weighs againstit) but one common early complaint against the latter was that it wasn’t provable by observationand all of the sages and holy fathers of the church for nealy 2000 years considered geocentrism to215One who is skilled at many philosophical disciplines. Copernicus made contributions to astronomy, mathematics,medicine, economics, spoke four languages, and had a doctorate in law.216Wikipedia: http://www.wikipedia.org/wiki/Copernican heliocentrism.217Wikipedia: http://www.wikipedia.org/wiki/Age of Enlightenment. The Enlightenment was the philosophicalrevolution that led to the invention of physics and calculus as the core of “natural philosophy” – what we now callscience – as well as economics, democracy, the concept of “human rights” (including racial and sexual equality) withina variety of social models, and to the rejection of scriptural theism as a means to knowledge that had its roots in thediscoveries of Columbus (that the world was not flat), Descartes (who invented analytic geometry), Copernicus (whoproposed that the non-flat Earth was not the center of creation after all), setting the stage in the sixteenth centuryfor radical and rapid change in the seventeenth and eighteenth centuries.218Wikipedia: http://www.wikipedia.org/wiki/Galileo Galilei. It would take too long to recite all of Galileo’s dis-coveries and theories, but Galileo has for good reason been called “The Father of Modern Science”.
Week 12: Gravity497be true on observational grounds.Galileo’s telescope – which was little more powerful than an ordinary pair of hand-held binocularstoday – was sufficient to provide that proof. Galileo’s instrument clearly revealed that the moonwas aplanetoid object, a truly massive ball of rock that orbited the Earth, so large that it had itsown mountains and “seas”. It revealed that Jupiter had not one, but four similar moons of its ownthat orbiteditin similar manner (moons named “The Galilean Moons” in his honor). He observedthe phases of Venus as it orbited the sun, and correctly interpreted this as positive evidence thatVenus, too, was a huge world orbiting the sun as the Earth orbits the sun while revolving and beingorbited by its own moon. He was one of the first individuals in modern times to observe sunspots(although he incorrectly interpreted them as Mercury in transit across the Sun) and set the stage forcenturies of solar astronomical observations and sunspot counts that date from roughly this time.His (independent) observations on gravity even helped inspire Newton to develop gravity as theuniversal causeof the observed orbital motions.However, the publication of his own observations defending Copernicus corresponded almostexactly with the Church finally taking action to condemn the work of Copernicus and ban hisbook describing the model. In 1600 the Roman Inquisition had found Catholic priest, freethinker,and philosopher Giordano Bruno219guilty of heresy and burned him at the stake, establishing adangerous precedent that put a damper on the development of science everywhere that the Romanchurch held sway.Bruno not only embraced the Copernican theory, he went far beyond it, recognizing that the Sunis a star like other stars, that there were far, far more stars than the human eye could see withouthelp, and he even asserted that many of those stars have planets like the Earth and that those planetswere likely to be inhabited by intelligent beings. While Galileo was aided in his assertions by the useof the telescope, Bruno’s were all the more remarkable because theyprecededthe invention of thetelescope. Note well that the human eye can only make out some3500 stars altogetherunaidedon the darkest, clearest nights. This leap from 3500 to “infinity”, and the other inferences he madeto accompany them, were quite extraordinary. His guess that the stars are effectively numberlesswas validated shortly afterwards by means of the very first telescopes, which revealed more andmore stars in the gaps between the visible stars as the power of the telescopes was systematicallyincreased.We only discovered positive evidence of the first confirmed exoplanet220in1988and arestillin the process of searching for evidence that might yet validate his further hypothesis of life spreadthroughout the Universe, some of it (other than our own) intelligent. Galileo had written a letterto Kepler in 1597, a mere three years before Bruno’s ritualized murder, stating his belief in theCopernican system (which was not, however, the direct cause of Bruno’s conviction for heresy). Thestakes were indeed high, and piled higher still with wood.Against this background, Galileo developed a careful and observationally supported argumentin favor of the Copernican model and began cautiously to publish it within the limited circles ofphilosophical discourse available at the time, proposing it as a “theory” only, but arguing that itdid not contradict the Bible. This finally attracted the attention of the church. Cardinal and SaintRobert Bellarmine wrote a famous letter to Galileo in 1615221explaining the Church’s position onthe matter. This letter should be required reading for all students, and since if you are reading thistextbook you are, in a manner of speaking,mystudent, please indulge me by taking a moment andfollowing the link to read the letter and some of the commentary following.In it Bellarmine makes the following points:219Wikipedia: http://www.wikipedia.org/wiki/Giordano Bruno. Bruno is, sadly, almost unknown as a philosopherand early scientist for all that he was braver and more honest in his martyrdom that Galileo in his capitulation.220Wikipedia: http://www.wikipedia.org/wiki/Extrasolar planet. As of today, some 851 planets in 670 systems havebeen discovered, with more being discovered almost every day using a dazzling array of sophisticated techniques.221http://www.fordham.edu/halsall/mod/1615bellarmine-letter.asp
498Week 12: Gravity•If Copernicus (and Galileo, defending Copernicus and advancing the theory in his own right)are correct, the heliocentric model “is a very dangerous thing, not only by irritating all thephilosophers and scholastic theologians, but also by injuring our holy faith and rendering theHoly Scriptures false.”In other words, if Galileo is correct, the holy scriptures are incorrect. Bellarmine correctlyinfers that this would reduce the degree of belief in the infallibility of the holy scriptures andhence the entire basis of belief in the religion they describe.•Furthermore, Bellarmine continues, Galileo isdisagreeing with established authoritieswith hishypothesis, who “...all agree in explaining literally (ad litteram) that the sun is in the heavensand moves swiftly around the earth, and that the earth is far from the heavens and standsimmobile in the center of the universe. Now consider whether in all prudence the Church couldencourage giving to Scripture a sense contrary to the holy Fathers and all the Latin and Greekcommentators. Nor may it be answered that this is not a matter of faith, for if it is not amatter of faith from the point of view of the subject matter, it is on the part of the ones whohave spoken.”•Finally, Bellarmine concludes that “if there were a true demonstration that the sun was inthe center of the universe and the earth in the third sphere, and that the sun did not travelaround the earth but the earth circled the sun, then it would be necessary to proceed withgreat caution in explaining the passages of Scripture which seemed contrary, and we wouldrather have to say that we did not understand them than to say that something was false whichhas been demonstrated.” He goes on to assert that “the words ’the sun also riseth and the sungoeth down, and hasteneth to the place where he ariseth, etc.’ were those of Solomon, whonot only spoke by divine inspiration but was a man wise above all others and most learned inhuman sciences and in the knowledge of all created things, and his wisdom was from God.”Interested students are invited to playLogical Fallacy Bingo222with the text of the entire doc-ument. Opinion as fact, appeal to consequences, wishful thinking, appeal to tradition, historian’sfallacy, argumentum ad populum, thought-terminating cliche, and more. The argument of Bel-larmine boils down to the following:•If the heliocentric model is true, the Bible is false where that model contradicts it.•If the Bible is falseanywhere, it cannot be trustedeverywhereand Christianity itself canlegitimately be doubted.•The Bible and Christianity are true. Even if they appear to be false they arestilltrue, butdon’t worry, they don’t even appear to be false.•Therefore, while it is all very well to show how a heliocentric model couldmathematically, orhypotheticallyexplain the observational data, itmust be false.In 1633, this same Bellarmine (later made into a saint of the church) prosecuted Galileo in theInquisition. Galileo was found “vehemently suspect of heresy” for precisely the reasons laid out inBellarmine’s original letter to Galileo. He was forced to publicly recant, his book laying out thereasons for believing the Copernican model was added along with the book of Copernicus to thelist of banned books, and he was sentenced to live out his life under house arrest, praying all dayfor forgiveness. He died in 1642 a broken man, his prodigious and productive mind silenced by theactive defenses of the locally dominant religious mythology for almost ten years.I was fortunate enough to be teaching gravitation in the classroom on October 31, 1992, whenPope John Paul II (finally)publicly apologizedfor how the entire Galileo affair was handled. On222http://lifesnow.com/bingo/ http://lifesnow.com/bingo/
Week 12: Gravity499Galileo’s behalf, I accepted the apology, but of course I must also point out thatBellarmine’sargument is essentially correct. The conclusions of modern science have, almost without exception,contradicted the assertions made in the holy scriptures not just of Christianity but of all faiths.They therefore stand as direct evidence that those scriptures arenot, in fact, divinely inspired orperfect truth, at least where we can check them. While this does notprovethat they are incorrectin other claims made elsewhere, it certainly and legitimately makes them less plausible.12.2: Kepler’s LawsGalileo was not, in fact, the person who made the greatest contributions to the rejection of thePtolemaic model as the first step towards first the (better) heliocentric Copernican model, then tothe invention of physics and science as a systematic methodology for successively improving ourbeliefs about the Universe that does not depend on authority or scripture. He wasn’t even one ofthe top two. Let’s put him in the third position and count up to number one.The person in the second position (in my opinion, anyway) wasTycho Brahe223, a wealthy Danishnobleman who in 1571, upon the death of his father, established an observatory and laboratoryequipped with the most modern of contemporary instrumentation in an abbey near his ancestralcastle. He then proceeded to spend a substantial fraction of his life, including countless long Danishwinter nights,making and recording systematic observations of the night sky!His observations bore almost immediate fruit. In 1572 he observed a supernova in the constella-tion Cassiopeia. This one observation refuted a major tenet of Aristotelian and Church philosophy– that the Universe beyond the Moon’s orbit was immutable. A new star had appeared where nonewas observed before. However, his most important contributions were immense tables of very precisemeasurements of the locations of objects visible in the night sky, over time. This was in no smallpart because hisownhybrid model for a mixture of Copernican and Ptolemaic motion proved utterlyincorrect.If you are a wealthy nobleman with a hobby who is generating a huge pile of data but who alsohas no particular mathematical skill, what are you going to do? You hire a lab rat, a flunky, anassistantwho can do the annoying and tedious work of analyzing your data while you continue tohave the pleasure of accumulating still more. And as has been the case many a time, the servantexceeds the master. The number one philosopher who contributed to the Copernican revolution,more important than Brahe, Bruno, Galileo, or indeed any natural philosopher before Newton wasBrahe’s assistant,Johannes Kepler224.Kepler was a brilliant young man who sought geometric order in the motions of the stars andplanets. He was also a protestant living surrounded by Catholics in predominantly Catholic centralEurope and was persecuted for his religious beliefs, which had a distinctly negative impact on hisprofessional career. In 1600 he came to the attention of Tycho Brahe, who was building a newobservatory near Prague. Brahe was impressed with the young man, and gave him access to hisclosely guarded data on the orbit of Mars and attempted to recruit him to work for him. Although hewas was trying hard to be appointed as the mathematician of Archiduke Ferdinand, his religious andpolitical affilations worked against him and he was forced to flee from Graz to Prague in 1601, whereBrahe supported him for a full year until Brahe’s untimely death (either from possibly deliberatemercury poisoning or a bladder that ruptured from enforced continence at a state banquet – it isn’tclear which even today). With Brahe’s support, Kepler was appointed an Imperial mathematicianand “inherited” at least the use of Brahe’s voluminous data. For the next eleven years he put it tovery good use.Although he was largely ignored by contemporaries Galileo and Descartes, Kepler’s work laid, as223Wikipedia: http://www.wikipedia.org/wiki/Tycho Brahe.224Wikipedia: http://www.wikipedia.org/wiki/Johannes Kepler.
500Week 12: Gravitywe shall see, the foundation upon which one Isaac Newton built his physics. That foundation canbe summarized inKepler’s Lawsdescribing the motion of the orbiting objects of the solar system.They wereobservationallaws, propounded on the basis of careful analysis of the Brahe data andfurther observations to verify them. Newton was able toderivetrajectories that rather preciselyagreed with Kepler’s Laws on the basis of his physics and law of gravitation.The laws themselves are surprisingly simple and geometric:a) Planets move around the Sun inelliptical orbitswith the Sun at one focus (see next sectionfor a review of ellipses).b) Planets sweep outequal areas in equal timesas they orbit the Sun.c) Themean radiusof a planetary orbit (in particular, the semimajor axis of the ellipse)cubedis directly proportional to theperiodof the planetary orbitsquared, with the same constantof proportionality for all of the planets.The first law can be proven directly from Newton’s Law of Gravitation (although we will notprove it in this course, as the proof is mathematically involved). Instead we will content ourselveswith the observation that a circular orbit is certainly consistent, and by using energy diagramswe will see that elliptical orbits are at least rather plausible. The second law will turn out to beequivalent to theconservation of angular momentumof the orbits, because gravitation is acentral forceand exerts no torque. The third, again, is difficult to formally prove for ellipticalorbits but straightforward to verify for circular orbits.Since most planets havenearlycircular orbits, we will not go far astray by idealizing and re-stricting our analysis of orbits to the circular case. After all, not even elliptical orbits are preciselycorrect, because Kepler’s results and Newton’s demonstrationignore the influence of the planets oneach otheras they orbit the Sun, which constantly perturb even elliptical orbits so that they are atbest a not-quite-constant approximation. The best one can do is directly and numerically integratethe equations of motion for the entire solar system (which can now be done to quite high precision)but even that eventually fails as small errors from ignored factors accumulate in time.Nevertheless, the path from Ptolemy to Copernicus, Galileo and Kepler to Newton stands outas a great triumph in the intellectual and philosophical development of the human species. It is forthat reason that we study it.12.2.1: Ellipses and Conic SectionsThe following is a short review of the properties of ellipses (and, to a lesser extent, the other conicsections). Recall that a conic section is the intersection of a plane with a right circular cone alignedwith (say) the -axis, where the intersecting plane can intercept at any value ofzzand parallel,perpendicular, or at an angle to the - plane.x yA circle is the intersection of the cone with a plane parallel to the - plane. An ellipse is thex yintersection of the cone with a plane tipped at an anglelessthan the angle of the cone with thecone. A parabola is the intersection of the cone with a planeat the same angleas that of the cone.A hyperbola is the intersection of the cone with a plane tipped at agreaterangle than that of thecone, so that it produces two disjoint curves and hasasymptotes. An example of each is drawn infigure 147, the hyperbola for the special case where the intersecting plane is parallel to the -axis.zProperly speaking, gravitational two-body orbits are conic sections: hyperbolas, parabolas, el-lipses, or circles, not just ellipses per se. However,bound planetary orbitsare elliptical, so we willconcentrate on that.
Week 12: Gravity501ellipsecircleparabolahyperbolaFigure 147: The various conic sections. Note that a circle is really just a special case of the ellipse.abffPFigure 148:Figure 148 illustrates the general geometry of the ellipse in the - plane drawn such that itsx ymajor axis is aligned with thexaxis. In this simple case the equation of the ellipse can be written:x 2a 2+ y 2b 2= 1(1062)There are certain terms you should recall that describe the ellipse. The major axis is the longest“diameter”, the one that contains both foci and the center of the ellipse. The minor axis is theshortest diameter and is at right angles to the major axis. The semimajor axis is the long-direction“radius” (half the major axis); the semiminor axis is the short-direction “radius” (half the minoraxis).In the equation and figure above,ais the semimajor axis andbis the semiminor axis.Not all ellipses have major/minor axes that can be easily chosen to bexandycoordinates.Another general parameterization of an ellipse that is useful to us is a parametric cartesian repre-sentation:x t( )=x 0+ cos(aωt+ φ x )(1063)y t( )=y 0+ cos(bωt+ φ y )(1064)This equation will describeanyellipse centered on (x , y00) by varyingωtfrom 0 to 2 . Adjustingπthe phase anglesφ xandφ yand amplitudesaandbvary the orientation and eccentricity of theellipse from a straight line at arbitrary angle to a circle.
502Week 12: GravityThefociof an ellipse are defined by the property that the sum of the distances from the focito every point on an ellipse is a constant (so an ellipse can be drawn with a loop of string and twothumbtacks at the foci). Iffis the distance of the foci from the origin, then the sum of the distancesmust be 2 = ( + ) + (dfaa− f) = 2 (from the pointax= ,a y= 0. Also,a 2= f 2+ b 2(from thepointx= 0,y= ). Sobf= √ a 2− b 2where by conventiona≥ b .This is all you need to know (really more than you need to know) about ellipses in order tounderstand Kepler’s First and Third Laws. The key things to understand are the meanings of theterms “focus of an ellipse” (because the Sun is located at one of the foci of an elliptical orbit)and “semimajor axis” as a measure of the “average radius” of a periodic elliptical orbit. As notedabove, we will concentrate inthiscourse on circular orbits because they are easy to solve for andunderstand, but in future, more advanced physics courses students will actually solve the equationsof motion in 2 dimensions (the third being irrelevant) for planetary motion using Newton’s Lawof Gravitation as the force and prove that the solutions are parametrically described ellipses. Insome versions of eventhiscourse, students might use a tool such as octave, mathematica, or matlabto solve the equations of motion numerically and graph the resulting orbits for a variety of initialconditions.12.3: Newton’s Law of GravitationIn spite of the church’s opposition, the early seventeenth century saw the formal development of theheliocentric hypothesis, supported by Kepler’s empirical laws. Instrumentation improved, and thegeometric methods involvingparallaxto determine distance produced a systematically improvingpicture of the solar system that was not only heliocentric but verified Kepler’s Laws in detail foradditional planetary bodies. The debate with the geocentric/ptolemaic model supporters continued,but in countries far away from Rome where its influence waned, a consensus was gradually formingthat the geocentric hypothesis was incorrect. The observations of Brahe and Galileo and analysis ofKepler was compelling.However, thecauseof heliocentric motion was a mystery. There was clearly substantial geometryand order in the motion of the planets, although it was not precisely the geometry proposed by Platoand advanced by Aristotle and Ptolemaeus and others. This geometry was subtle, and best describedwithin the confines of the new Analytic Geometry invented by Descartes225where ellipses (as wecan see above) were not “just” conic sections or objects visualized in a solid geometry: They couldbe represented byequations.Descartes was another advocate of the heliocentric theory, but when, in 1633, he heard thatGalileo had been condemned for his advocacy of Copernicus and arguments against the Ptolemaicgeocentric model, he abruptly changed his mind about publishing a work to that effect! As notedabove, these were dangerous times for freethinking philosophers who were literally forbidden by therulers of the predominant religion under threat of torture and murder from speculating in ways thatcontradicted the scriptures of that religion. A powerful voice was thus silenced and the geocentricmodel persisted without anyopenchallenge for fifty more years.So things remained until one of the most brilliant and revered men of all time came along: IsaacNewton. Born on December 25, 1642, Newton was only 8 in 1650 when Descartes died, but he wastaught Descartes’ geometry at Cambridge (before it closed in the midst of a bout of the plagueso that he was sent home for a while) and by the age of 24 had transformed it into a theory of“fluxions” – the first rudimentary description of calculus. Calculus, or the mathematics of related225Wikipedia: http://www.wikipedia.org/wiki/Rene Descartes. Descartes was another of the “renaissance man”polymaths of the age. He was brilliant and led a most interesting life, making contributions to mathematics (where“Cartesian Coordinates” are named in his honor), physics, and philosophy. He reportedly liked to sleep late, neverrising before 11 a.m., and when an opportunity to become a court mathematician and tutor arose that forced him tochange his habits and arise at 5 a.m. every day, he sickened and died (in 1650) a short while thereafter!
Week 12: Gravity503rates of change established on top of a coordinatized geometry, was the missing ingredient, the keypiece needed to transform thestrictly geometricobservations of philosophers from Plato throughKepler into ananalytic descriptionof both the causes and effects of motion.Even so, Newton workedthirteen more yearsproducing and presenting advances in mathematics,optics, and alchemy before (in 1679), having recently completed a speculative theory of optics, heturned his attention wholly towards the problem of celestial mechanics and Kepler’s Laws. In thishe was reportedly inspired by the intuition that the force ofgravity– the same force that makes theproverbial apple fall from the tree – was responsible for holding the moon in its orbit around theEarth.Initially he corresponded heavily withRobert Hooke226, known to us throughHooke’s Lawinthe text above, who had been appointed secretary of the brand newRoyal Society227, the world’sfirst “official” scientific organization, devoted to an eclectic mix of mathematics, philosophy, andthe brand new “natural philosophy” (the correct and common termin for “science” almost to theend of the nineteenth century). Hooke later claimed (quite possibly correctly) that he suggestedthe inverse-square force law to Newton, but what Hooke did not do that Newton did is to take thepostulated inverse square force law, add to it a set of axioms (Newton’s Laws) thatdefinedforce ina particular mathematical way, and then show that the equations of motion that followed from aninverse square force law, evaluated through the use of calculus,completely predicted and explainedKepler’s Laws and moreby means of explicit functional solutions built on top of Descartes’ analyticgeometry, where the “more” was the apparent non-elliptical orbits of other celestial bodies, notablycomets.It is difficult to properly explain how revolutionary, how world-shattering this combination ofinvention and discovery was. Initially it was communicated privately to the Royal Society itself in1684; three years later it was formally published as thePhilosophiae Naturalis Principia Mathemat-ica228, or “The Mathematical Principles of Natural Philosophy”. This book changed everything.It utterly destroyed, forever, any possibility that the geocentric hypothesis was correct. The readermust determine for themselves if it initiated the very process anticipated and feared by Robert Bel-larmine – as the consequences of Newton’s work unfolded, they have proven the Bible and all of theother religious mythologies and scriptures of the worldliterallyfalse time and again.As we have seen from a full semester of work with its core principles, Newton’s Laws and a smallset of actual force laws permit the nearly full description and prediction of virtually all everydaymechanical phenomena, and itsideas(in some cases extended far beyond what Newton originallyanticipated) survive to some extent even in its eventual replacement, quantum mechanics.PrincipiaMathematicalaid down a template for theprocessof scientific endeavor – a mix of accumulationand analysis of experimental data, formal axiomatic mathematics, and analytic reasoning leading toa detailed description of the visible Universe of ever-improving consistency. It was truly asystemof the world, the basis of thescientific worldview. It was a radically different worldview thanthe one based on faith, authority, and the threat of violence divine or mundane to any that daredchallenge it that preceded it.Let us take a look at the force law invented or discovered (as you please) by Newton and seehow it works to explain Kepler’s Laws, at least for simple cases we can readily solve without muchcalculus.Here are Newton’s axioms, the essential individual assumptions that are assembled compactlyinto the law of gravitation. Note that these assumptions were initially applied to objects like theSun and the planets and moons that are spherically symmetric to a close approximation; the alsoapply to “particles” of mass or chunks of mass small enough to be treated as particles. Followingalong with figure 149 above:226Wikipedia: http://www.wikipedia.org/wiki/Robert Hooke.227Wikipedia: http://www.wikipedia.org/wiki/Royal Society.228Wikipedia: http://www.wikipedia.org/wiki/Philosophiae Naturalis Principia Mathematica.
504Week 12: GravityrFMm2112Figure 149:a) The force of gravity is atwo body forceand does not change if three or more bodies arepresent.b) The force of gravity isaction at a distanceand does not require the two objects to “touch”in order to act.c) The force of gravity acts along (in the direction of) a linejoining centers of sphericallysymmetric masses, in this case along .~rd) The force of gravity isattractive.e) The force of gravity isproportional to each mass.f) The force of gravity isinversely proportional to the distance between the centers ofthe masses.We will add to this list the assumption that one of the two masses ismuch larger than the othersothat the center of mass and the center of coordinates can both be placed at the center of the largermass. This isnot at all necessaryand proper treatments dating all the way back to Newton accountfor motion around a more general center of mass, but for us it will greatly simplify our pictures andtreatments if we idealize in this way and in the case of systems like the Earth and the moon, or theSun and the Earth, it isn’t a terrible idealization. The Sun’s mass is a thousand times larger thaneven that of Jupiter!These axioms are rather prolix in words, but in the form of analgebraic equationthey are ratherbeautiful:~F 21= −G M m12r 2ˆr(1065)whereG= 6 67 10.×− 11N-m /kg is the textbfuniversal gravitational constant, added as the constant22of proportionality that establishes the connections between all of the differentunitsin question. Notethat we continue to use the convention that~F 21stands for the force acting on mass 2 due to mass1; the force~F 12= − ~F 21both from Newton’s third law and because the force isattractivefor bothmasses.Kepler’s first law follow from solving Newton’s laws and the equations of motion in three di-mensions for this particular force law. Even though one dimension turns out to be irrelevant (themotion is strictly in a plane), even though the motion turns out to have two constants of the motionthat permits it to be further simplified (the energy and the angular momentum) the actual solutionof the resulting differential equations is a bit difficult and beyond the scope of this course. We willinstead show that circular orbits are one special solution that easily satisfy Kepler’s First and ThirdLaws, while Kepler’s Second Law is a trivial consequence of conservation of angular momentum.Let us begin with Kepler’s Second Law, as it stands alone (the other two proofs are related).It is proven by observing that the force isradial, and hence exertsno torque. Thus the angularmomentum of a planetary orbit is constant!
Week 12: Gravity505dA = | r x vdt |v tr∆Figure 150: The area swept out in an elliptical orbit in time ∆ is shaded in the ellipse above.tWe start by noting that the area enclosed by an parallelogram formed out of two vectors is themagnitude of the the cross product of those vectors. Hence the area in the shaded triangle in figure150 is half of that:dA=12| × ~r~vdt|= 12| ||~ ~vrdt|sinθ(1066)=12m| × ~rm dt~v|(1067)If we divide the ∆ over to the other side we get the area per unit time being swept out by thetorbit:dAdt=12m| × | ~r~ p=12m| | ~L= a constant(1068)because angular momentum is conserved for a central force (see the chapter/week on torque andangular momentum if you have forgotten this argument) and Kepler’s second law is proved for thisforce.That was pretty easy! Let’s reiterate the point of this demonstration:Kepler’s Second Law is equivalent to the Law of Conservation of AngularMomentum and is true for any central force (not just gravitation)!The proofs of Kepler’s First and the Third lawsfor circular orbitsrely on a common algebraicargument, so we group them together. They key formula is, as one might expect the fact thatifan orbiting mass moves in a circular orbit,thenthe gravitational force has to be equal to the masstimes the centripetal acceleration:G M mspr 2=m ap r= m pv 2r(1069)whereM sis the mass of the central attracting body (which we implicitly assume is much larger thanthe mass of the orbiting body so that its center of mass is more or less at the center of mass of thesystem),m pis the mass of the planet,vis its speed in its circular orbit of radius . This situationris illustrated in figure 151.This equation in and of itself “proves” that Newton’s Laws plus Newton’s Law of Gravitationhave a solution consisting of a circular orbit, where a circle is a special case of an ellipse. Thisproof isn’t very exciting, however, asanyattractive radial force law we might attempt would havea similarly consistent circular solution. The kinematic radial acceleration of a particle moving inuniform circular motion is independent of the particular force law that produces it!What is a lot more interesting is the demonstration that the circular orbit satisfies Kepler’sThirdLaw, as this law quite specifically defines the relationship between the radius of the orbitand its period. We can easily see that onlyoneradial force law will lead to consistency with theobservational data for circular orbits.
506Week 12: Gravityr2F = GMm= mv2rSEMmrFv = 2 rT πFigure 151: The geometry used to prove Kepler’s and Third Laws for a circular (approximately)orbit like that of the Earth around the Sun.We start by cancelling the mass of the planet and one of the factors of :rv 2=G Msr(1070)But,vis related torand the periodTby:v= 2πrT(1071)so thatv 2= 4π r2 2T 2=G Msr(1072)Finally, we isolate the powers of :rr 3=G Ms4 π 2T 2(1073)and Kepler’s third law is proved for circular orbits.Since there is nothing unique about circular orbits andallclosed elliptical orbits around thesamecentral attracting body have to have the same constant of proportionality, we have both proven thatNewton’s Law of Gravitation has circular solutions that satisfy Kepler’s Third Lawandwe haveevaluated theuniversalconstant of proportionality, valid for all of the planets in the solar system!We can then write the law more compactly:R 3sm=G Ms4 π 2T 2(1074)where nowR smis the semimajor axis of the elliptical orbit, which happens to berfor a circularorbit.Note well that this constant is easily measured! In fact we can evaluate it from our knowledge ofthe semimajor axis of Earth’s nearly circular orbit –R E≈1 5 10.×11meters (150 million kilometers)plus our knowledge of its period –T= 3 153.×10 seconds (1 year, in seconds). These two numbers7are well worth remembering – the first is called anastronomical unitand is one of the fundamentallengths upon which our knowledge of the distances to the nearer stars is based; the second physicists
Week 12: Gravity507tend to remember as “ten million timesπseconds per year” because that is accurate to well withinone percent and easier to remember than 3.153.Combining the two we get:G Ms4 π 2=T 2R 3sm=π 2×10143 375.×1033≈ 3×10− 19(1075)where we usedanotherphysics geek cheat:π 2≈10, and then approximated 10 3 375/ .≈3 as well.That way we can get an answer, good to within a couple of percent, without using a calculator orlooking anything up!Note well! If only we knewG, we’d know the mass of the Sun! If we use the same logic todetermine thesameconstant for objects orbiting theEarth(where we might use the semimajoraxis of the moon’s orbit, 384,000 kilometers, and the period of the moon’s orbit, 27.3 days, to getGM / πE42) we would also be able to determine the mass of the Earth!Of course wedoknowGnow, but when Newton proposed his theory, it wasn’t so easy to figureout! This is becausegravitation is the weakest of the forces of nature, by far!It is so weak that it isremarkably difficult to measure the direct gravitational force between two objects of known massesseparated by a known distance in the laboratory, so that all of the quantities in Newton’s Law ofGravitation were measuredbutG .In fact, it took over a century for Henry Cavendish229to build a clever apparatus that wassufficiently sensitive that it could measureGfrom these three known quantities. This experimentwas said to “weigh the Earth” not because it actually did so – far from it – but because onceGwasknown experiments that had long since been done instantly gave us the mass of the Sun, the massof the Earth, the mass of Jupiter and Saturn and Mars (any planet where we can remotely observethe semimajor axis and period of a moon) and much more.These in turn gave us some serious conundrums! The Sun turns out to be 1.4million kilometersin diameter, and to have a mass of 2×1030kilograms! With a surface temperature of some 6000 K,what mechanism keeps it so hot? Any sort of chemical fire would soon burn out!Laboratory experiments plus astronomical observations based on the use of parallax with theentire diameter of the Earth’s orbit used as a triangle base and with exquisitely sensitive measure-ments of the angles between the lines of sight to the nearer stars (which allowed us to determinethe distance to these stars) all analyzed by means of Newton’s Laws (including gravitation), allowedastronomers to rapidly infer a startling series of facts about the Solar system, our local galaxy (theMilky Way), and the Earth.Not only was thegeocentrichypothesis wrong, so was theheliocentrichypothesis. The Earthturned out to be a mostly unremarkable planet, a relatively small one of a rather large numberorbiting an entirely unremarkable star that itself was orbiting in a huge collection of stars, that wasonly one of a truly staggering number of similar collections of stars, where every new generation oftelescopes revealed still more of everything, still further away. At the moment, there appear to beon the order of a hundredbilliongalaxies, containing somewhere in the ballpark of 1023stars, in thevisible Universe, which is (allowing for its original inflation, 13.7 billion years ago) around 46 billionlight years in radius. At least one method of estimation has claimed to establish a radius aroundtwice this large as alower boundfor its size (so that all of these estimates are probably low by anorder of magnitude) – and there is no upper bound.Exoplanets are being discovered at a rate that suggests that planetary systems around those starsarecommon, not rare (especially so given that we can only “see” or infer the existence of extremelylarge planets so far – we would find it almost impossible to detect a planet as small as the Earth).Bruno’s original assertion that the Universe is infinite, contains and infinite number of stars, with229Wikipedia: http://www.wikipedia.org/wiki/Cavendish Experiment.
508Week 12: Gravityan infinite number of planets, an infinite number of which have some sort of intelligent or otherwiselifemay be impossible to verify or refute, but infinite or not the Universe isenormouscompared tothe scale of the Solar system, which ishugecompared to the scale of the Earth, and contains many,many stars with many, many planetary systems.In fact, the only thing about the Earth that is remarkable may turn out to be – us!12.4: The Gravitational FieldAs noted above, Newton proposed the gravitational force as thecauseof the observed orbital motionsof the celestial objects. However, this force wasaction at a distance– it exists between two objectsthat are not touching and that indeed are separated bynothing: a vacuum! What then, causes thegravitational force itself? Let us suggest that there must besomethingthat is produced by oneplanet acting as asourcethat is present at the location of the other planet that is the proximatecause of the force that planet experiences. We define thegravitational fieldto be thiscauseofthe gravitational force, the thing that is present at all points in space surrounding a masswhetheror not some other mass is present there to be acted on!We define the gravitational field conveniently to be the force per unit mass, a quantity that hasthe units ofacceleration:~g ~( ) =r−G Mr 2ˆr=~Fm(1076)The magnitude of the gravitational field at the surface of the earth is thus:g= (g RE) =Fm =G MER 2E(1077)and we see that the quantity that we have been calling the gravitationalaccelerationis in fact moreproperly called the near-Earth gravitationalfield.This is a very useful equation. It can be used to find any one of ,g RE ,M E, orG, from aknowledge of any of the other three, depending on which ones you think you know best.gis easy;students typically measuregin physics labs at some point or another several different ways!R Eis actually also easy to measure independently and some classical methods were used to do solongbefore Columbus.M E, however is hard! This is because it always appears in the company ofG, so that knowinggandR Eonly gives you their product. This turns out to be the case nearly everywhere – any ordinarymeasurement you might make turns out to tell youGMEtogether, not either one separately.What aboutG ?To measureGin the laboratory, one needs a very sensitive apparatus for measuring forces. Sincewe know already thatGis on the order of 10− 10N-m /kg , we can see that gravitational forces22between kilogram-scale masses separated by ten centimeters or so are on the order of a fewbillionthsof a Newton.Henry Cavendish made the first direct measurement ofGusing a torsional pendulum – basicallya barbell suspended by a very thin, strong thread – and some really massive balls whose relativeposition could be smoothly adjusted to bring them closer to and farter from the barbell balls. Asyou can imagine, it takes very little torque to twist a long thread from its equilibrium angle to a newone, so this apparatus has – when utilized by someone with a great deal of patience, using a lightsource and a mirror to further amplify the resolution of the twist angle – proven to be sufficientlysensitive to measure the tiny forces required to determineG, even to some reasonable precision.Using this apparatus, he was able to findGand hence to “weigh the earth” (findM E). By mea-suring ∆ as a function of the distanceθrmeasured between the centers of the balls, and calibrating
Week 12: Gravity509equilibrium position whenattracted by mass MMMMMmmtorsional pivotequilibrium position (no mass M)∆θFigure 152: The apparatus associated with theCavendish experiment, which established the firstaccurate estimates forGand thereby “weighed the Earth”, the Sun, and many of the other objectswe could see in the sky.the torsional response of the string using known forces, he managed to get 6.754 (vs 6.673 currentlyaccepted)× 10− 11N-m /kg . This is within just about one percent. Not bad!2212.4.1: Spheres, Shells, General Mass DistributionsSo far, our empirically founded expression for gravitational force (and by inheritance, field) appliesonly tospherically symmetric mass distributions– planets and stars, which are generallyalmost perfectly roundbecauseof the gravitational field – or particles small enough that they canbe treated like spheres. Our pathway towards the gravitational field of more general distributionsof mass starts by formulating the field of a single point-like chunk of mass in such a distribution:d ~g= −G dm0| − ~r~r 0 |3( ~r− ~r 0 )(1078)This equation can be integrated as usual over an arbitrary mass distribution using the usualconnection: The mass of each chunk is the mass per unit volume times the volume of the chunk, ordm=ρdV0 .~g= − ZG ρ dV0| − ~r~r 0 |3( ~r− ~r 0 )(1079)where for exampledV0=dx dy dz000(Cartesian) ordV0= r 20sin( )θ dθ dφ dr0000(Sphereical Polar)etc. This integral is not always easy, but it can generally be done very accurately, if necessarynumerically. In simple cases we can actually do the calculus and evaluate the integral.Inthispart ofthiscourse, we will avoid doing the integral, although we will tackle many examplesof doing it in simple cases next semester. We will content ourselves with learning the followingTrueFactsabout the gravitational field:•The gravitational field produced by a (thin) spherically symmetric shell of mass ∆Mvanishesinside the shell.
510Week 12: Gravity•The gravitational field produced by this same shell equals the usual~g ~( ) =r−G M∆r 2ˆr(1080)outsideof the shell. As a consequence the field outside of any spherically symmetric distributionof mass is just~g ~( ) =r−G M∆r 2ˆr(1081)These two results can be proven by direct integration or by using Gauss’s Law for the gravitationalfield (using methodology developed next semester for the electrostatic field). The latter is so easythat it is hardly worth the time to learn the former for this special case.Note well the most important consequence for our purposes in the homework of this rule is thatwhen we descend a tunnel into a uniformly dense planet, the gravity willdiminishas we are onlypulled down by the massinsideour radius. This means that the gravitational field we experience is:~g ~( ) =r−G M r∆( )r 2ˆr(1082)whereM r( ) = 4ρ πr /33 for a uniform density, something more complicated in cases where the densityitself changes with . You will use this expression in several homework problems.r12.5: Gravitational Potential EnergyW12W t= 0r2r1W12W t= 0ABFigure 153: A crude illustration of how one can show the gravitational force to be conservative (sothat the work done by the force is independent of the path taken between two points), permittingthe evaluation of a potential energy function.If you examine figure 153 above, and note that the force is always “down” along , it is easy~rto conclude that gravity must be a conservative force. Gravity produced by some (sphericallysymmetric or point-like) mass does work on another mass only when that mass is moved in orout along~rconnecting them; moving at right angles to this along a surface of constant radiusrinvolves no gravitational work. Any path between two points near the source can be broken up intoapproximating segments parallel to~rand perpendicular to~rat each point, and one can make theapproximation as good as you like by choosing small enough segments.This permits us to easily compute the gravitational potential energy as the negative work done
Week 12: Gravity511moving a massmfrom a reference position~r 0nal position :fito a ~rU r( )=− Zrr 0~F ·d ~r(1083)=− Zrr 0−GMmr 2dr(1084)=− (GMmr−GMmr 0)(1085)=−GMmr+GMmr 0(1086)Note that the potential energy function depends only on thescalar magnitudeof~r 0and , and~rthatr 0ne the potential energy to be zero.fiis in the end the radius of an arbitrary point where we deBy convention, unless there is a good reason to choose otherwise, we require the zero of thegravitational potential energy function to be atr 0=∞. Thus:U r( ) =−GMmr(1087)Note that since energy in some sense is more fundamental than force (the latter is the negativederivative of the former) we could just as easily have learned Newton’s Law of Gravitation directlyas thisscalarpotential energy function and then evaluated the force by taking its negative gradient(multidimensional derivative).The most important thing to note about this function is that it isalways negative. Recall thatthe force points in the direction that the potential energydecreases most stronglyin. SinceU r( ) isnegative and gets larger in magnitude for smaller , gravitation (correctly) pointsrdownto smallerrwhere the potential energy is “smaller” (more negative).The potential energy function will be very useful to us when we wish to consider things like escapevelocity/energy, killer asteroids, energy diagrams, and orbits. Let’s start with energy diagrams andorbits.12.6: Energy Diagrams and Orbitseld in a clever way that isolatesfiLet’s write the total energy of a particle moving in a gravitational theradial kinetic energy:E tot=12mv2−GMmr(1088)=12mv2r+ 12mv2t−GMmr(1089)=12mv2r+12mr2(mv rt) 2−GMmr(1090)=12mv2r+L 22mr2−GMmr(1091)=12mv2rff+ U e( )r(1092)In this equation,12mv2ris the radial kinetic energy, andffU e( ) =rL 22mr2−GMmr(1093)is the radial potential energyplusthe rotational kinetic energy of the orbiting particle, formed outof the transverse velocityv tasK rot= 12mv2t=L / mr 222ective potential (and itsff. If we plot the egure 154.fipieces) we get a one-dimensional radial energy plot as illustrated in
512Week 12: Gravity____ L 222mr______− GMmr____ − GMm L 222mr______reff ,Utot EUeffr= Figure 154: A typical energy diagram illustrating theeffective potential energy, which is basicallythe sum of the radial potential energy and the angular kinetic energy of the orbiting object.By drawing a constant total energy on this plot, the difference betweenE totandU eff( ) is therradial kinetic energy, which must be positive. We can determine lots of interesting things from thisdiagram.In figure 155, we show orbits with a given fixed angular momentum~L 6= 0 and four generic totalenergiesE tot. These orbits have the following characteristics and names:a)E tot>0. This is ahyperbolicorbit.b)E tot= 0. This is aparabolicorbit. This orbit definesescape velocityas we shall see later.c)E tot<0. This is generally anellipticalorbit (consistent with Kepler’s First Law).d)E tot= Ueff min,. This is a circular orbit. This is a special case of an elliptical orbit, but deservesspecial mention.Note well that all of the orbits areconic sections. This interesting geometric connection between1/r2forces and conic section orbits was a tremendous motivation for important mathematical worktwo or three hundred years ago.12.7: Escape Velocity, Escape EnergyAs we noted in the previous section, a particle has “escape energy” if and only if its total energyis greater than or equal to zero, provided that we set the zero of potential energy at infinity in thefirst place. We define theescape velocity(a misnomer!) of the particle as the minimumspeed(!)that it must have to escape from its current gravitational field – typically that of a moon, or planet,or star. Thus:E tot= 0 =12mv2escape−GMmr(1094)so thatvescape=r 2GMr=p 2gr(1095)
Week 12: Gravity513E , UtoteffEtotEtotEtotEtotEkEkE < 0 (forbidden)krmaxr0rminr1342Figure 155: A radial total energy diagram illustrating the four distinct named orbits in terms oftheir total energy: 1) is ahyperbolicorbit. 2) is aparabolicorbit. 3) is anellipticalorbit. 4)is acircularorbit. Note that all of these orbits are conic sections, and that the classical ellipticorbits have tworadialturning points at the apogee and perigee along the major axis of the ellipse.where in the last formg=GMr 2(the magnitude of the gravitational field – see next item).To escape from theEarth’s surface, one needs to start with a speed of:vescape=r 2GMER E=p 2gRE= 11 2.km/sec(1096)Note: Recall the form derived by equating Newton’s Law of Gravitation andmv /r2in an earliersection for the velocity of a massmin a circular orbit around a larger massM :v 2circ=GMr(1097)from which we see thatvescape= √ 2 vcirc.)It is often interesting to contemplate this reasoning in reverse. If we drop a rock onto the earthfrom a state of rest “far away” (much farther than the radius of the earth, far enough away to beconsidered “infinity”), it will REACH the earth with escape (kinetic) energy and a total energy closeto zero. Since the earth is likely to be much larger than the rock, it will undergo aninelasticcollisionand release nearlyall its kinetic energy as heat. If the rock is small, this is not necessarily aproblem. If it is large – say, 1 km and up – it releases alotof energy.Example 12.7.1: How to Cause an Extinction EventHow much energy? Time to do an estimate, and in the process become just a tiny bit scared of avery, very unlikely event that could conceivably cause the extinction ofus.Let’s take a “typical” rocky asteroid that might at any time decide to “drop in” for a one-wayvisit. While the asteroid might well have any shape – that of a potato, orpikachu230, we’ll follow230Wikipedia: http://www.wikipedia.org/wiki/Pikachu. If you don’t already know, don’t ask...
514Week 12: Gravitythe usual lazy physicist route and assume that it is a simple spherical ball of rock with a radius .rIn this case we can estimate its total mass as a function of its size as:M = 4πρ3r 3(1098)Of course, now we need to estimate its density, . Here it helps to know two numbers: Theρdensity of water, or ice, is around 10 kg/m (a metric ton per cubic meter), and the33specificgravity or rockis highly variable, but in the ballpark of 2 to 10 (depending on how much of whatkinds of metals the rock might contain, for example), say around 5.If we then letr≈1000 meters (a bit over a mile in diameter), this works out toM ≈1 67.×1012kg, or around 2 billion metric tons of rock, about the mass of a small mountain.This mass will land on earth withescape velocity, 11.2 km/sec, if it falls in “from rest” from faraway. Or more, of course – it may have started with velocity and energy from some other source –this is pretty much a minimum. As an exercise, compute the number of Joules this collision wouldrelease to toast the dinosaurs – or us! As a further exercise, convert the answer to “tons of TNT”(a unit often used to describe nuclear-grade explosions – the original nuclear fission bombs had anexplosive power of around 20,000 tons of TNT, and the largest nuclear fusion bombs built duringthe height of the cold war had an explosive power on the order of 1 to 15 million tons of TNT.The conversion factor is 4.184 gigajoules per ton of TNT. You can easily do this by hand, althoughthe internet now boasts of calculators that will do the entire conversion for you. I get ballpark oftento the twentiethjoules or 25 gigatons – that isbillions of tons– of TNT. In contrast, wikipediacurrently lists the combined explosive power of all of the world’s 30,000 or so extant nuclear weaponsto be around 5 gigatons. The explosion of Tambora (see last chapter) was estimated to be around 1gigaton. The asteroid that might have caused the K-T extinction event that ended the Cretaceousand wiped out the dinosaurs and created the 180 kilometer in diameterChicxulub crater231hada diameter estimated at around 10 km and would have released around 1000 times as much energy,between 25 and 100teratonsof TNT, the equivalent of some 25,000 Tambora’s happening all atonce.Such impacts are geologically rare, but obviously can have enormous effects on the climate andenvironment. On a smaller scale, they are one very good reason to oppose the military exploitationof space – it is all too easy to attack any point on Earth by dropping rocks on it, where the asteroidbelt could provide a virtually unlimited supply of rocks.12.8: Bridging the Gap: Coulomb’s Law and ElectrostaticsThis concludes our treatment of basic mechanics. Gravitation is our first actuallaw of nature– aforce or energy law that describes the way we think the Universe actually works at a fundamentallevel.Gravity is, as we have seen, important in the sense that we live gravitationally bound to the outersurface of a planet that is itself gravitationally bound to a star that is gravitationally compressedat its core to the extent that thermonuclear fusion keeps the entire star white hot over billions ofyears, providing us with our primary source of usable energy. It isunimportantin the sense that itisvery weak, the weakest of all of the known forces.Next, in the second volume of this book, you will study one of thestrongestof the forces, theone that dominates almost every aspect of your daily life. It is the force that binds atoms andmolecules together, mediates chemistry, permits the exchange of energy we call light, and indeed isthe fundamental source ofnearly everyof the “forces” we treated in this semester in collective form:The electromagnetic interaction.231Wikipedia: http://www.wikipedia.org/wiki/Chicxulum Crater.
Week 12: Gravity515Just to whet your interest (and explain why we have spent so long on gravity when it is weak andmostly irrelevant outside of its near-Earth form in everyday affairs) let is take note ofCoulomb’sLaw, the force that governs the all-important electrostatic interaction that binds electrons to atomicnuclei to make atoms, and binds atoms together to make molecules. It is the force that exists betweentwocharges, and can be written as:~F 12=k q qe12r 212ˆr12(1099)Hmmm, this equation looks rather familiar! It isalmost identicalto Newton’s Law of Gravitation,only it seems to involve thecharge( ) of the particles involved, not their mass, and anqelectrostaticconstantk einstead of the gravitational constantG .In fact, it issosimilar that you instantly “know” lots of things about electrostatics from this oneequation, plus your knowledge of gravitation. You will, for example, learn about the electrostaticfield, the electrostatic potential energy and potential, you will analyze circular orbits, you will analyzetrajectories of charged particles in uniform fields – all pretty much the same idea (and algebra, andcalculus) as their gravitational counterparts.The one reallyinterestingthing you will learn in the first couple of weeks is how toproperlydescribe the geometry of 1/r2force laws and their underlying fields – a result calledGauss’s Law.This law and the other Maxwell Equations will turn out to govern nearly everything you experience.In some very fundamental sense, youareelectromagnetism.Good luck!Homework for Week 12Problem 1.Physics Concepts:Make this week’s physics concepts summary as you work all of the problemsin this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s)they were key to, and include concepts from previous weeks as necessary. Do the work carefullyenough that you can (after it has been handed in and graded) punch it and add it to a three ringbinder for review and study come finals!Problem 2.It is a horriblemisconceptionthat astronauts in orbit around the Earth areweightless, whereweight (recall) is a measure of the actual gravitational force exerted on an object. Suppose you arein a space shuttle orbiting the Earth at a distance of two times the Earth’s radius (R e= 6 4.×106meters) from its center.a) What is your weightrelativeto your weight on the Earth’s surface?b) Does your weight depend on whether or not you are moving at a constant speed? Does itdepend on whether or not you are accelerating?c) Why would youfeelweightless inside an orbiting shuttle?d) Can youfeelas “weightless” as an astronaut on the space shuttle (however briefly) in yourown dorm room? How?
516Week 12: GravityProblem 3.Physicists are working to understand “dark matter”, a phenomenological hypothesis invented toexplain the fact that things such as the orbital periods around the centers ofgalaxiescannot beexplained on the basis of estimates of Newton’s Law of Gravitation using the totalvisiblematter inthe galaxy (which works well for the mass we can see in planetary or stellar context). By addingmass we cannot see until the orbital rates are explained, Newton’s Law of Gravitation is preserved(and so are its general relativistic equivalents).However, there are alternative hypotheses, one of which is that Newton’s Law of Gravitationiswrong, deviating from a 1/r2force law at very large distances (but remaining a central force).The orbits produced by such a 1/rnforce law (withn 6= 2) would not be elliptical any more, andr 3 6=CT2– but would they still sweep out equal areas in equal times? Explain.
Week 12: Gravity517Problem 4.rRMωThis problem will help you learn required concepts such as:•Newton’s Law of Gravitation•Circular Orbits•Centripetal Acceleration•Kepler’s Lawsso please review them before you begin.A straight, smooth (frictionless) transit tunnel is dug through a spherical asteroid of radiusRand massMthat has been converted into Darth Vader’sdeath star. The tunnel is in the equatorialplane and passes through the center of the death star. The death star moves about in a hard vacuum,of course, and the tunnel is open so there are no drag forces acting on masses moving through it.a) Find the force acting on a car of massma distancer < Rfrom the center of the death star.b) You are commanded to find the precise rotational frequency of the death starωsuch thatobjects in the tunnel will orbitatthat frequency and hence will appear toremain at restrelative to the tunnel at any point along it. That way Darth can Use the Dark Side to movehimself along it almost without straining his midichlorians. In the meantime, he is reachinghis crooked fingers towards you and you feel a choking sensation, so better start to work.c) Which of Kepler’s laws does your orbit satisfy, and why?
518Week 12: GravityProblem 5.r = R0 ρ0SNmThis problem will help you learn required concepts such as:•Newton’s Second Law.•Newton’s Law of Gravitation•Gravitational Field/Force Inside a Spherical Shell or Solid Sphere.•Harmonic Oscillation Given Linear Restoring Forces.•Definitions and Relations InvolvingωandT .so please review them before you begin.A straight, smooth (frictionless) transit tunnel is dug through a planet of radiusRwhose massdensityρ 0is constant. The tunnel passes through the center of the planet and is lined up with itsaxis of rotation (so that the planet’s rotation isirrelevantto this problem). All the air is evacuatedfrom the tunnel to eliminate drag forces.a) Find the force acting on a car of massma distancer < Rfrom the center of the planet.b) Write Newton’s second law for the car, and extract the differential equation of motion. Fromthis find ( ) for the car, assuming that itr tstarts at restatr 0= Ron the North Pole at timet= 0.c) How long does it take the car to get to the South Pole starting from rest at the North Pole?How long does it take to get back to the North Pole? Compare this (second answer) to theperiod of a circular orbit inside the death staryou found (disguised as ) in the previousωproblem.d) A final thought question: Suppose it is released at rest from an initial positionr 0=R/2(halfway to the center) instead of fromr 0. How long does it take for the mass to get back tothis pointnow(compare the periods)?All answers should be given in terms ofG ρ,0 ,Randm .
Week 12: Gravity519Problem 6.EE01E2E3reff UTheeffective radial potentialof a planetary object of massmin an orbit around a star of massM is:U eff( ) =rL 22mr2−GMmr(a form you already explored in a previous homework problem). The total energy of four orbitsare drawn as dashed lines on the figure above for some given value ofL. Name the kind of orbit(circular, elliptical, parabolic, hyperbolic) each energy represents and mark its turning point(s).
520Week 12: GravityProblem 7.In a few lines prove Kepler’s third law forcircularorbits around a planet or star of massM :r 3=CT2and determine the constantCand then answer the following questions:a) Jupiter has a mean radius of orbit around the sun equal to 5.2 times the radius of Earth’sorbit. How long does it take Jupiter to go around the sun (what is its orbital period or “year”T J)?b) Given the distance to the Moon of 3 84.×10 meters and its (sidereal) orbital period of 27.38days, find the mass of the EarthM e .c) Using the mass you just evaluated and your knowledge ofgon the surface, estimate the radiusof the EarthR e .Check your answers using google/wikipedia. Think for just one short moment how much of thephysics you have learned this semester is verified by the correspondance. Remember, I don’t wantyou to believe anything I am teaching you because of myauthorityas a teacher but because itworks.
Week 12: Gravity521Problem 8.It is very costly (in energy) to lift a payload from the surface of the earth into a circular orbit, butonce you are there, it only costs you that same amount of energy again to get from that circularorbit to anywhere you like – if you are willing to wait a long time to get there. Science Fictionauthor Robert A. Heinlein succinctly stated this as: “By the time you are in orbit, you’re halfwayto anywhere.”Prove this by comparing the total energy of a mass:a) On the ground. Neglect its kinetic energy due to the rotation of the Earth.b) In a (very low) circular orbit with at radiusR ≈ R E– assume that it is still more or less thesame distance from the center of the Earth as it was when it was on the ground.c) The orbit with minimal escape energy (that will arrive, at rest, “at infinity” after an infiniteamount of time).Problem 9.mxyM = 80mD = 5ddThe large mass above is the Earth, the smaller mass the Moon. Find thevector gravitationalfieldacting on the spaceship on its way from Earth to Mars (swinging past the Moon at the instantdrawn) in the picture above.
522Week 12: GravityProblem 10.M eR eM aThis problem will help you learn required concepts such as:•Gravitational Energy•Fully Inelastic Collisionsso please review them before you begin.A bitter day comes: a roughlysphericalasteroid of radiusR aand densityρis discovered thatis falling in fromfar awayso that it will strike the Earth. Ignore the gravity of the Sun in thisproblem. Determine:a) If it strikes the Earth (aninelastic collisionif there ever was one) how much energy willbe liberated as heat? Express your answer in terms ofR eand eithergorGandM eas youprefer. It is probably very safe to say thatM a≪ M e...b) Chuck Norris lands on the surface of the asteroid to save the Earth, but instead of screwingaround with drills and nuclear bombs Chuck jumps up from the surface of the asteroid at aspeed ofvcnto deliver a roundhouse kick that would surely break the asteroid in half andcause it to miss the earth – if it knows what’s good for it (thisisChuck Norris, after all).However, if you jump up too fast on an asteroid, you don’t come down again!DoesChuckever fall down onto the asteroid after his jump?c) Evaluate your answers to a-b above for the following data:ρ=6×10 kilograms/meter33R a=10 meters4R e=6 4.×10 meters6g=10 meters/second2M e=6×1024kilogramsvcn=5 meters/second(1100)Express your answer to a) both in joules and in “tons” (of TNT) where 1 ton-of-TNT =4 2.×10 joules. Compare the answer to (say) 30 Gigatons as a safe upper bound for the total9combined explosive power of every weapon (including all thenuclearweapons) on earth.d) Find the size of an asteroid that (when it hits) liberates only the energy of a typical thermonu-clear bomb, 1 megaton of TNT.
Week 12: Gravity523Problem 11.There is an old physics joke involving cows, and you will need to use its punchline to solve thisproblem.A cow is standing in the middle of an open, flat field. A plumb bob with a mass of 1 kg issuspended via an unstretchable string 10 meters long so that it is hanging down roughly 2 metersaway from the center of mass of the cow. Making any reasonable assumptions you like or need to,estimatethe angle of deflection of the plumb bob from vertical due to the gravitational field of thecow.
524Week 12: GravityOptional ProblemsThe following problems arenot required or to be handed in, but are provided to give yousome extra things to work on or test yourself withaftermastering the required problems and conceptsabove and to prepare for quizzes and exams.Continue Studying for Finalsusing problems from the online review!
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