Chapte Intoducton. Intoducton Mechancs s the theoy of moton of mateal bodes. In ode to dstngush ths theoy fom othe theoes, patculaly quantum mechancs, one also efes to t as Classcal Mechancs. Ths latte specalzaton s also meant to nclude the Specal Theoy of Relatvty. We assume hee famlaty wth fundamental physcal concepts such as space, tme, mass and foce. We begn wth the mechancs of a sngle pontlke mass called a patcle. A couse n Classcal Mechancs s the fst and basc couse n theoetcal physcs. Apat fom coveng the mechancs of mateal bodes, the couse seves also as an essental peequste fo subsequent couses n electodynamcs, quantum mechancs and statstcal mechancs. Although we ecaptulate fst the laws postulated by Newton, we shall then poceed vey dffeently. As n any othe moden text, the bass of the mechancs we develop hee wll be the postulate of extemzaton of the so-called acton ntegal, and we shall convnce ouselves that ths pncple epoduces the totalty of Newtonan mechancs. Moeove, we shall see (to some extent) that the pncple s of much wde genealty, so that ts applcaton can be found n many othe banches of physcs. The Newton equatons of moton ae obtaned fom ths extemzaton pncple as Eule Lagange equatons, and the fomalsm as such s known as the Lagange fomalsm. We shall consde mpotant applcatons. The vtal contnuaton of ths fomalsm leads to the canoncal Hamlton fomalsm, whch seves as a bass fo the tanston to quantum mechancs. It s evdent that fo ths eason alone these fomalsms ae aleady of fundamental sgnfcance. In fact, wth the devaton of the so-called Posson algeba n the context of Hamlton s mechancs, one s appoachng quantum mechancs as closely, as s possble wthn clas-
2 CHAPTER. Intoducton scal mechancs (thee s no devaton, agan one has to stat fom some postulates). Moe complcated poblems ae those nvolvng constants, and we shall be concened wth some hee. A method paallel to that of Posson backets but natually moe complcated can be developed to deal wth constants n quantum mechancs. Thus both, the Lagange and the Hamlton fomalsms, play a fundamental ole not only n mechancs but also n subsequent couses such as electodynamcs and quantum mechancs. In addton they pemt a deep nsght nto the stuctue of mechancs, nto elatonshps between symmety popetes and consevaton laws, and othe basc popetes. In the followng chapte we begn wth a vey bef ecaptulaton of Newton s mechancs manly fo the pupose of makng the dffeence n the postulates undelyng Newton s appoach and those undelyng the moden extemzaton appoach as clea as possble. In ths ecaptulaton of Newton s mechancs, we also emphasze n patcula that ths does not eque a specal postulate fo otatonal moton. In addton, we consde mmedately many patcle systems, and theeby ntoduce also the concept of constants whch s equed fo the subsequent consdeaton of gd bodes and the dynamcs. Ou bef ecaptulaton of Newtonan mechancs seves also ou othe pupose hee, namely to uncove ts connecton wth elatvty,.e. Ensten s mechancs. On the way we make bef and elementay foays nto electodynamcs, goup theoy and statstcal physcs. The lage numbe of woked poblems scatteed thoughout the text s meant to llustate specfc aspects, and moe than that to demonstate the wde spectum of elevance of classcal mechancs n physcs, eachng fom atomc physcs, though phenomena of daly lfe to those of celestal dmensons. Thus we am at a pesentaton hee whch spans the ente doman of mechancs fom Newton to Ensten, whch s one of the most fascnatng domans of the natual scences.
Chapte 2 Recaptulaton of Newtonan Mechancs 2. Intoductoy Remaks Although Newton s appoach to mechancs s today obsolete, t s helpful fo an undestandng of moden easonng to be fully awae of the dffeence n the basc pncples of both appoaches. One can then gan a deepe appecaton of the moe geneal moden appoach leadng also to the neghbouhood of quantum mechancs wthout losng anythng of Newton s mechancs. Thus we fst ecaptulate Newton s laws fo patcles and systems of patcles, and dstngush clealy between lnea and otatonal moton. Snce ths s bascally school mechancs we efan fom pesentng numeous llustatve examples, though t would be wong to conclude that all such examples would be smple o even tval. Many books exst wth hundeds of woked o unwoked examples, of whch some can be qute tcky o ease to handle wth the Lagangan fomalsm. Some examples whch usually eque second thoughts ae, fo nstance, those wth vayng mass, as n the case of ockets, chans slppng off a table and smla cases. Thus we nclude some examples at the end of ths chapte, also fo puposes of llustaton. The powe of the moden Lagangan and Hamltonan methods esults fom the sutablty to teat much moe geneal stuatons as wll become evdent n late chaptes. The eade nteested n a fascnatng dscusson of the development of the concept of spacetme fom Astotle to Galle, Newton and Ensten s ecommended to ead R. Penose [38], Chapte 7, pp. 383 4. Ths book wll be efeed to at vaous ponts of the text. The autho s ndebted to Andew Y.T. Chan, then an Edto of Wold Scentfc, fo bngng ths book to hs attenton. Books wth hundeds of examples (solved and/o unsolved) ae, fo nstance, those of T. W. B. Kbble and F. H. Bekshe [25] and The Physcs Coachng Class [48]. 3
4 CHAPTER 2. Recaptulaton of Newtonan Mechancs 2.2 Recaptulaton of Newton s Laws In Newton s fomulaton of classcal mechancs, hs wellknown thee laws of moton ae the basc postulates o axoms fom whch eveythng else s deduced. We ecall these fst and comment theeafte on cetan tems used:. The law of neta: Evey body (of mass 0) contnues n ts state of est o of unfom moton n a staght lne, unless compelled by some extenal foce to change that state. 2. The equaton of moton: The tme ate of change of moton (.e. of the momentum) s popotonal to the appled foce, and takes place n the decton of ths foce. 3. Acto = eacto: To evey acton thee s an equal and opposte eacton. The fst law may sound tval n the way we all became acquanted wth ths at school. Howeve, the phase state of est o of unfom moton (constant velocty) eveals aleady elatvty,.e. Gallean elatvty: Thee s nothng to dstngush the physcs of the so-called state of est fom that of unfom moton, snce thee s no such noton as a fxed pont n space. The wod neta (fom net meanng sluggsh, slow) means wth nonzeo mass but wthout nheent acton o moton, and hence mples what the fst law spells out. The wod unfom means wth constant velocty, and also that e.g. a bottle wth a cok stuck on moves n ths fom wth the same magntude and decton, and not e.g. sometmes wth a cok flyng somewhee neaby. In the second law the phase tme ate of change means the devatve wth espect to tme t,.e. d/dt, and the wod moton s to be undestood as lnea momentum p. Thus fo a sngle patcle unde the nfluence of an appled foce F the second law mples dp dt = F. (2.) We shall defne late an netal efeence fame as an unacceleated one defned by mass velocty = const.,.e. velocty = const./mass, so that the mass cannot be zeo (hence the name netal fame, and late the name moment of neta meanng, effectvely, moment of mass). The lage the mass s, the sluggshe, moe net s the moton, o the esstance to acceleaton, as R. Penose says [38], p. 392. In Newton s nonelatvstc mechancs such moton s convenently consdeed wth espect to an assumed fxed ogn n space, o coespondngly to the poston of an extemely massve masspont lke that of a fxed sta o the sun. Acceleated systems ae the subject of the Geneal Theoy of Relatvty. Snce acceleaton s thee agued to be ndstngushable fom gavty, netal fames ae those fa away fom any nfluence of gavty. An netal fame of efeence s theefoe also descbed as a non-acceleated fame of efeence, and n what s called fee fall ( fee meanng the ght hand sde of the geodesc equaton of Sec. 3.2 s zeo) all bodes move netally whethe o not gavty s pesent,.e. at least locally (cf. also Secs. 0.2, 0.4).
2.3 Futhe Defntons and Rotatonal Moton 5 Intoducng the poston vecto, the patcle mass m, and the patcle s velocty v, then hee (cf. Fg. 2.) p = mv = m d dt = m lm 2 t 0 t so that Eq. (2.) becomes F = d(mv)/dt. s = m lm t 0 t, (2.2) 2 s O 2 Fg. 2. Path element s. Fo the eason explaned above the mass m hee n Newton s equaton s called netal mass. In Sec. 2.6 we shall encounte the tem gavtatonal mass, and n Chapte 0 we shall see that these ae equal, ths beng (as explaned thee) the obsevaton of Galle. 2.3 Futhe Defntons and Rotatonal Moton The acceleaton of the patcle of mass m s defned as a = d2 dt 2. (2.3) The patcle s angula momentum L aound a pont 0, and t s toque N aound 0 ae espectvely defned as the moments L = p, N = F. (2.4) Thus the angula momentum s the moment of momentum. We also obseve the coespondence p L, F N
6 CHAPTER 2. Recaptulaton of Newtonan Mechancs between lnea moton and otatonal moton. Ths coespondence becomes even moe conspcuous by the followng elaton whch we vefy below: N = dl dt. (2.5) In vefyng ths elaton we gan some expeence n dealng wth vectos and the coss poducts. The mass m s a constant n tme. Hee n Newtonan mechancs we also consde the oentaton of the axes of ou Catesan coodnate system (chaactezed by unt vectos e x,e y,e z ) as not changng n the couse of tme,.e. d dt (e ) = 0, =,2,3 (o x,y,z). The defnton of L theefoe mples dl dt = m d { d }, o dt dt dl dt = m d { d }. (2.6) dt dt Fo a vecto ω we have ( ) ( dω dωx = dt dt e x + dω y dt e y + dω ) z dt e z = d dt (ω). A coss poduct as n the defnton of L can be expessed n tems of the so-called Lev Cvta symbol whch s defned as follows: +, f,j,k a cyclc pemutaton of,2,3, ɛ jk =, f,j,k, an antcyclc pemutaton of,2,3, 0, f two of,j,k ae equal. (2.7) We then have 3 (A B) = ɛ jk A j B k. (2.8) j,k= Thus we have n the above case of Eq. (2.6) dl dt = m d dt = m j,k ( d ) dt ɛ jk [( d dt = m d dt ) ( ) d dt j j,k= k 3 ( ) d ɛ jk () j dt ( d 2 ) + () j dt 2 k k ]. (2.9)
2.4 Consevatve Foces 7 But fo a geneal thee-vecto α (wth half the ognal expesson plus half that wth j,k enamed k,j): 3 j,k= ɛ jk α j α k = 2 3 j,k= α j α k (ɛ jk + ɛ kj ) = 0. (2.0) Ths esult follows fom the antsymmety of the Lev Cvta symbol. Hence the fst tem on the ght hand sde of Eq. (2.9) vanshes, and we obtan: dl dt = m j,k ɛ jk () j ( ) k,.e. dl dt = F = N, (2.) whee F = m. The moment of the foce F s called toque. It s ths whch causes a tunng effect. We thus ave at the analogues of Newton s laws fo otaton (howeve, deved fom these by consdeng effectvely nfntesmal steps and summng these):. If a body s otatng about a gven axs though ts cente of mass, t wll contnue to do so wth constant angula velocty unless compelled by some extenal toque to change that state. 2. The tme ate of change of angula momentum s popotonal to the extenal toque and takes place about the axs about whch the toque s appled and n the same sense. 3. To evey couple thee s an equal and opposte couple (two equal unlke paallel foces consttutng a couple). 2.4 Consevatve Foces We now defne the wok W 2 done by the extenal foce F appled to a patcle n movng ths fom a pont to a pont 2 by the expesson W 2 = 2 F ds. (2.2a) Fo a constant mass m we have: 2 dv F ds = m dt (vdt) = T 2 T, T = 2 mv2. (2.2b) See e.g. E. H. Booth and P. M. Ncol [6], p. 73.
8 CHAPTER 2. Recaptulaton of Newtonan Mechancs Thus the wok W 2 = T 2 T s equal to the change n knetc enegy. The wok can hee, fo nstance, be that pefomed by the foce called weght mg n movng the mass m fom a pont to a pont 2. If the foce o, as one says, the foce feld, s such that the ntegal taken aound a closed path vanshes,.e. F ds = 0, (2.3) the foce (and hence the system) s descbed as consevatve. We emphasze: The ntegal n Eq. (2.2b) does not only depend on the endponts and 2, but also on the path fom pont to pont 2 (ds s an element of ths path). We now ecall Stokes s theoem: F ds = cul F df = ndf cul F, (2.4) F whee n s a unt vecto pependcula out of the plane of f and F the ente aea ntegated ove, so that fo consevatve foces,.e. those satsfyng Eq. (2.3), cul F = 0,.e. F = 0. (2.5) Snce always cul gad of somethng vanshes (as one can vefy by explct evaluaton), t follows that the consevatve foce can be expessed as a gadent,.e. F = V, (2.6) and the scala quantty V () s descbed as potental. If we use fo cul F the epesentaton e x e y e z cul F = x y z, (2.7) F x F y F z the vanshng of culf wth F = V follows fom the theoem, that a detemnant vanshes f the elements of two ows ae dentcal. The potental V () s a scala quantty (fo the defnton see Sec. 5.6) and s also descbed as potental enegy. Thus fo a consevatve system W 2 = 2 F ds = 2 F V s = V V 2. (2.8) Snce also W 2 = T 2 T, cf. Eqs. (2.2a), (2.2b), we have V V 2 = T 2 T, o T + V = T 2 + V 2 = = const. (2.9)
2.5 Mechancs of a System of Patcles 9 Ths elaton expesses the consevaton of the total enegy. We can also evese the agument. Let B A V ds = V B V A (2.20) be a potental dffeence wth V sngle-valued and contnuous. Then ( B A ) V ds = + V ds = (V B V A ) + (V A V B ) = 0, (2.2) A B.e. V ds = 0 f V s sngle-valued and contnuous. Thus fo F = V then F ds = 0. Hence also: F ds Stokes = cul F df = 0, o culf = 0. F Fnally we obseve fom Eq. (2.) the consevaton of the lnea momentum when the total foce F vanshes,.e. p = const. fo F = 0, (2.22) and fom Eq. (2.5) the consevaton of angula momentum when the total toque N vanshes,.e. L = const. fo N = 0. (2.23) 2.5 Mechancs of a System of Patcles In consdeng a system of seveal patcles, we dstngush between extenal foces ognatng fom some souce outsde the patcle system, and ntenal foces, that the patcles exet on each othe. Let F (e) be the extenal foce actng on patcle (supescpt (e) fo extenal ), and F j the (ntenal) foce that patcle j exets on patcle. Newton s second law appled to patcle then mples the equaton: F (e) + j F j = d dt p ṗ, (2.24) whee p s, of couse, the momentum of patcle. Newton s thd law yelds the equatons F j = F j, F = 0. (2.25)
0 CHAPTER 2. Recaptulaton of Newtonan Mechancs It follows that F j = (F j + F j ) = 0. 2,j,j Summng the extenal foces actng on patcle by summng ove ndex, we obtan F (e) = ṗ = m d 2 dt 2. (2.26) We now defne the coodnate of a pont called cente of mass of the system by the vecto R = m m, M = m, (2.27) whee M s the total mass of the system. The cente of mass s a so-called collectve coodnate of the system, snce t descbes the collectve poston of the system. Late we shall use Eq. (2.27) n the fom m Rm = 0. (2.28) It follows that we can ewte Eq. (2.26) as F (e) F (e) = m d 2 R dt 2 = M d2 R dt 2. (2.29) Ths equaton says that the mass cente moves as f the ente extenal foce acts on the total mass of the system. An example s an explodng genade. The mass cente of ts peces moves as f the genade wee stll a sngle pece. Fo the total lnea momentum P of the system we obtan wth Eq. (2.28) P = m d dt = M dr dt. (2.30) It follows that f F (e) = F (e) = 0,.e. Md 2 R/dt 2 = 0, P = M dr dt = const. (2.3) Ths means, the total lnea momentum s constant o, as one says, conseved. Consde next the total toque N about a pont O: N = dl dt = F = F (e) +,j F j. (2.32)
2.5 Mechancs of a System of Patcles Let the extenal toque be the followng and zeo,.e. N (e) = F (e) = 0. (2.33) Then dl dt = F j = [ F j + j F j ] 2,j,j = ( j ) F j = j F j = 0, (2.34) 2 2,j,j snce by defnton the foce F j s paallel to j (cf. Fg. 2.). It follows that f, as hee assumed, the extenal toque N (e) vanshes, the total angula momentum L s conseved,.e. constant n tme. Equaton (2.30) gves the total lnea momentum n tems of cente of mass quanttes. What s the coespondng expesson fo the total angula momentum? We have L = p. (2.35) We set as n Fg. 2.2 and hence = R, and ṙ = v, ṙ = v, (2.36) v = v v, v = Ṙ. (2.37) It follows that we can ewte the expesson (2.35) of the angula momentum as L = m v = m ( + R) (v + v) = m R v + m v + m v + m R v. (2.38a) The sum of the last two tems hee s m ( v + R v ) = m v + R d m = 0 + 0, dt snce m = m m R (2.28) = 0.
2 CHAPTER 2. Recaptulaton of Newtonan Mechancs It follows that L = R Mv + p. (2.38b) Ths means, the total angula momentum of the system s equal to the angula momentum of the system concentated n the cente of mass plus the angula momentum of the moton aound the mass cente. In an analogous way one obtans fo the total knetc enegy of the system T = 2 Mv2 + m v 2. (2.39) 2 cente of mass R O Fg. 2.2 Cente of mass. Consdeng now the wok W 2 pefomed by all foces when the system, ognally n confguaton, has eached a confguaton 2, we have W 2 = 2 F (e) ds + 2 F j ds. (2.40) If the ntenal as well as the extenal foces ae consevatve,.e. f these can be deved fom potentals V j and V espectvely, we have F j = V j = + j V j = F j, F (e) = V, (2.4) j. whee V j = V j ( j ), so that F j = F j, and V j ( j ) = j f, f a scala functon, j j.
2.5 Mechancs of a System of Patcles 3 The second tem on the ght hand sde of Eq. (2.40) can theefoe be wtten: 2 j, F j ds = 2 = 2 Howeve, j, j,, j 2 2 {F j ds + F j ds j } { V j ds j V j ds j }. (2.42) V j = j V j = j V j and ds ds j = d d j = d j, so that 2 j, F j ds = 2 2,j, j j V j d j = 2 2 V j.,j, j Fo the wok W 2 we obtan theefoe 2 W 2 = V V =, V + V j. (2.43) 2,j, j The quantty V defnes the total potental enegy of the system. We see that the total enegy T +V s agan conseved, snce, as we saw above, W 2 s also equal to T 2 T, the dffeence of the knetc enegy n the two onfguatons. + d d j d + j d j = j + d j O j j + d j d j Fg. 2.3 Defnton of d j. The ntenal potental enegy,.e.,j, j V j, emans constant only fo specfc bodes whch ae theefoe known as gd bodes. Thus the ntenal foces of a gd body do not lead to changes n the confguaton of the body,
4 CHAPTER 2. Recaptulaton of Newtonan Mechancs.e. they do not pefom wok and can theefoe be gnoed n the dscusson of the moton of the ente body. We look at ths moe closely,.e. that the wok pefomed by the constanng foces n a gd body vanshes. We have as defnton of d j (see Fg.2.3): d + j d j j + d j o d j = ( j + d j ) j = d d j. (2.44) Thus d j s not paallel to j. If F j = ( j )f = j f, (2.45) (.e. the decton of F j s that of j o of j ), then the wok done by the constanng foce F j s: Ths foce vanshes when ethe F j d j = F j (d d j ). (2.46). d = 0,d j = 0,.e. no dsplacements nsde the gd body, o 2. d d j pependcula to F j, whch also apples n the case of a gd body, snce thee the dstance between two ponts emans constant,.e. ( j ) 2 c 2 j = 0, c j = const., and hence j d j = 0,.e. F j d j = 0. It follows that n the case of gd bodes the (see also below) vtual wok done by the ntenal foces s zeo. 2.6 Newton s Law of Gavtaton The above summay would be ncomplete wthout at least an ntoductoy efeence to Newton s law of unvesal gavtaton whch we shall be concened wth at vaous ponts late; also, we ewte t n vaous dffeent foms as the context demands. Ths law of Newton s states that between any two patcles of matte (note: patcles,.e. mass ponts whch have no spatal extenson) thee acts a foce F of attacton whch s popotonal to the two masses m,m 2 and s nvesely popotonal to the squae of the sepaaton,.e. (cf. Eq. (7.35)) F = G m m 2 2.
2.7 Mscellaneous Examples 5 Hee G s the unvesal gavtatonal constant wth G = 6.7 0 8 g cm 3 s 2. The value of G has been detemned by dvese expemental methods. The adus of the Eath s detemned e.g. astonomcally fom the cuvatue of the Eath, so that the mass of the Eath may be detemned. If m 2 = M = mass of the Eath, the mass m of an object on Eath o theeabouts s called the gavtatonal mass of ths object. As emaked n Sec. 2.2 we shall see n Chapte 0 that ths gavtatonal mass s equal to the netal mass, as obseved by Galle. Fo why should these masses be equal, as one may be nclned to assume at fst sght? Some authos employ the addtnal dstncton between actve gavtatonal mass,.e. one that gves se to a gavtatonal feld, and passve gavtatonal mass, whch s a quantty acted upon by a gavtatonal feld. In Newtonan mechancs the equvalence of these follows fom the equaton fo the gavtatonal potental φ, namely φ = 4πρ(), whee ρ s the densty of netal mass, and F = φ(),φ = Gm m 2 /. Newton s potental wll be deved fom Ensten s equaton n Sec. 5.6. 2.7 Mscellaneous Examples As stated above, we estct ouselves hee to a few mscellaneous examples llustatng a numbe of dectons of applcatons. It wll be seen that these poblems ae not always as tval as mght be supposed. In patcula Examples 2.5 to 2.8 deal wth Newton s law of gavtaton whch wll play a domnant ole n late chaptes. Example 2.: The otatng umbella A vetcally held umbella wth hozontal m s otated about ts axs n tmes n t 0 seconds and theeby scattes dops of wate tangentally away fom ts (hozontal) m. The m s the ccumfeence of a ccle of damete of 2R metes, and t s h metes above the gound. What s the adus of the ccle fomed by the pontlke dops on the gound? Soluton: The angula velocty of the umbella s ω = 2πn/t 0 adan/s. Hence the hozontal velocty of a dop s ωr = 2πnR/t 0 metes/s. The tme t t takes the dop to each the gound s equal to the tme t eques to fall though the dstance of h metes (wth ntal velocty zeo). Hence we have ẍ = g metes/s 2, x = ( ) 2h /2 2 gt2 = h metes, so that t = metes. g In ths nteval of tme the dops move hozontally and tangentally to the m of the umbella though a dstance d = 2πnR ( ) 2h /2 metes. t 0 g The eade s agan ecommended to ead the dscusson of ths topc gven by R. Penose [38], pp. 390 399. C. W. Msne and P. Putnam [32].
6 CHAPTER 2. Recaptulaton of Newtonan Mechancs It follows cf. Fg. 2.4 that the dops fom a ccle on the gound whch has the adus [ R 2 + d 2 = R + 8π2 n 2 ] h /2 gt 2 metes. 0 d R O Fg. 2.4 Vew on the umbella fom above. Example 2.2: The ball hoppng on an nclned plane A pontlke ball falls on an nfntely long, smooth, nclned plane wth nclnaton angle α. Examne the moton of the ball. Afte how many jumps does the ball begn to oll, f at all? Show that the tajectoes of the hoppng moton have the shape of paabolas. y v α α v α α π/2 2α x α Fg. 2.5 The ball hoppng on the nclned plane. Soluton: Accodng to the law of consevaton of momentum the ball wth ncdent angle α to the vetcal to the smooth, nclned plane wth the same angle of nclnaton, s eflected though the same angle, as ndcated n Fg. 2.5. Wth no fcton, the velocty of the ball emans the same as that of ts ncdence, v. We choose the x-axs along the plane as n Fg. 2.5, and the y-axs pependcula to the plane. Then we have fo the components of the velocty v along the x and y dectons n the ntal eflecton, and fo the components of the acceleaton due to gavty g thee (note that g s always vetcally dected) : v x = v sn α, v y = v cos α, g x = g sn α, g y = g cos α.
2.7 Mscellaneous Examples 7 Let x be the dstance tavelled by the ball n t seconds afte ts fst mpact on the plane. Then the veloctes of eflecton ae ẋ = v snα + gtsn α, ẏ = v cos α gtcos α. (2.47) By ntegaton we obtan (wth appopate ntal condtons) x = vt sn α + g t2 2 snα, y = vt cos α g t2 2 cos α. (2.48) At the next (fst) pont of mpact at tme t = t, we have y = 0, and t = 2v g. (2.49) The coespondng dstance along the plane s (nsetng t = t ) x = 2v g v snα + 2 g snα ( 2v g ) 2 = g 4v2 sn α. (2.50) We compute the angle α whch the velocty of mpact v makes wth the vetcal to the nclned plane fom the components of v. Fst, howeve, we have fom Eq. (2.47) wth the negatve of ẏ (fo mpact nstead of eflecton) and ẋ(t ) v x = v snα + g 2v g snα = 3v sn α, ẏ(t ) v y = v cos α + g 2v g cos α = v cos α, (2.5) v = v cos 2 α + 9sn 2 α = v + 8sn 2 α. (2.52) Hence ( ) α = tan vx = tan (3tanα). (2.53) v y The ball s now eflected wth velocty v n the decton of angle α. In ths case we obtan paallel to Eq. (2.47) wth v v, α α but the acceleaton due to gavty g dected vetcally as befoe, so that ts components along and pependcula to the plane eman unchanged: ẋ = v snα + gtsnα, ẏ = v cos α gt cos α (2.54) (sn α = v x /v,cos α = v y /v ). By ntegaton we obtan x = v tsn α + g t2 2 sn α + const. x 2, y = v tcos α g t2 2 cos α + const. y 2. (2.55) Let the pont of mpact (above x, y ) be the ogn x = 0, y = 0 of a new coodnate fame at t=0. Then const. = 0, and x, y hee become x 2, y 2, and fom y 2 = 0 at tme t 2 we obtan t 2 = 2v cos α g cos α = 2v y (2.5) 2v cos α = g cos α g cos α = 2v g = t. (2.56) Thus fo evey hop the ball eques the same length of tme. Moeove: 2v x 2 = v g snα + g ( ) 2v 2 snα = 2v 2 g g (v snα + v snα) = 2v g (v x + v x).
8 CHAPTER 2. Recaptulaton of Newtonan Mechancs Does the heght whch the ball falls though ncease o decease? Snce (usng Eq. (2.5)) v snα = v x = 3v snα, we have x 2 = 2v g (v sn α + v sn α) = 8v2 g snα > 4v2 snα = x. g Thus the heght the ball has fallen though fom the fst to the second mpact, x 2 sn α, s lage than the heght fallen though fom ts ntal to the fst mpact, x sn α. We obtan the angle α 2, whch the velocty of mpact v 2 makes wth the pependcula to the plane, fom the components of v 2 : It follows that (2.54) ẋ(t 2 ) v 2x = (2.54) {}}{ ẏ(t 2 ) v 2y = v cos α +g 2v g v 2 = = v x {}}{ v sn α +g 2v (2.5) snα = 3v snα + 2v snα = 5v sn α, g v y cos α (2.5) = v cos α + 2v cos α = v cos α, v 2 2x + v2 2y = cos 2 α + 25sn 2 α = v + 24sn 2 α. ( ) α 2 = tan v2x = tan (5 tanα),.e. tanα 2 = 5tanα = 5 v 2y 3 tanα,.e. α 2 > α. We obseve that the decton of eflecton becomes nceasngly flatte (α 2 > α ). Ths effect contnues ndefntely on the nfntely long, smooth, nclned plane. The y-component of the velocty of mpact s the constant value v cos α; howeve, ts x-component nceases. Hence tanα also nceases. We obtan the angle of eflecton afte n eflectons, α n, fom the elatons: tanα = 3tanα, tanα 2 = 5 tanα, tanα 3 = 7tanα,..., so that tanα n = (2n + )tanα. In ode that y n+ = v ntcos α n g t2 2 cos α = 0, we have always t = 2v g. The ball contnues to jump untl α n = π/2, so that cos α n = 0. Afte that y n+ s always zeo o negatve. We now show that the tajectoy of evey hop s a paabola. We acheve ths by elmnatng t fom x and y. Fom Eqs. (2.48) we obtan t = v snα ± v 2 sn 2 α + (4gx snα)/2, t = v cos α ± v 2 cos 2 α (4gx cos α)/2, (2g sn α)/2 (2g cos α)/2 2v ± v 2 + 2gx snα = ± v 2 2gy cos α. Squang both sdes of the last equaton and then a second tme, we obtan [ ( x 2v 2 +g snα + y )] 2 ( = 4v 2 v 2 + 2gx ) ( x, g 2 cos α snα snα + y ) 2 ( +4gv 2 x cos α snα + y ) = 0, cos α whch assumes the standad fom of the equaton of a paabola wth the tansfomaton X = x sn α + y cos α, Y = x snα y cos α.e. X 2 = 4v2 g Y.
2.7 Mscellaneous Examples 9 Example 2.3: The velocty of escape fom the Eath Calculate the smallest velocty, known as the escape velocty, wth whch a patcle has to be shot vetcally upwads n ode to escape fom the Eath. Ignoe fcton of the atmosphee and effects esultng fom the otaton of the Eath and the moon. (Ths s a wellknown and smple poblem. Howeve, the esult wll e-appea n the Schwazschld soluton of Ensten s equaton, as we shall see n Chapte 6, e.g. Eq. (6.58b), and n the tck calculaton of Sec..3). Soluton: Let M be the mass of the Eath and m that of the patcle. Let R be the adus of the Eath and v the velocty of the patcle at the heght x above ts pont of launchng nto the a. Then accodng to both of Newton s laws: mẍ = m d ( ) 2ẋ2 = G mm dx (x + R) 2, (2.57) whee G s Newton s gavtatonal constant G = 6.67 0 8 cm 3 g s 2. Integatng the equaton we obtan 2ẋ2 dx = GM (x + R) 2 = GM x + R + const. Wth ẋ = 0 at x =, the constant vanshes. Hence the velocty v at x = 0 s gven by v 2 = 2GM/R. Fo a patcle of unt mass at the suface of the Eath we have g = G ( M)/R 2, so that GM/R = gr and v 2 = 2gR. Wth R = 6370 km, g = 980 cm s 2, we obtan v 2 = 2 980 6370 0 5 cm 2 s 2, v.2 km s. The velocty of escape v esc = 2GM/R obtaned hee s a chaactestc speed n Newton s theoy of gavty, and s elated to the esult of elatvty that lght fom M cannot each a dstant obseve when v esc > c, whee c s the velocty of lght. Example 2.4: The ocket fed vetcally upwads Show that the equaton of moton of a ocket of ntal mass m 0 (no constant mass!), whch s fed vetcally upwads n a unfom gavtatonal feld of neglgble a esstance s m dv dt = v dm dt mg, (2.58) whee m s the mass of the ocket at tme t and v the fowad velocty of the expelled gas elatve to that of the ocket (v > 0, v < 0). Integate ths equaton and detemne v as a functon of m, assumng that the loss of mass s popotonal to tme. Also show that fo a ocket statng fom est wth v = 2070 m s and a loss of mass of /60 of the ntal mass pe second, and whch s to attan the velocty of escape fom the Eath, the ato of loss of mass of fuel elatve to that of the empty ocket must be almost 300. Soluton: In the tme nteval dt the change of momentum p of the ocket s dp = (m + dm)(v + dv) mv dm(v + v) = mdv v dm,.e. we have (obseve that dm s negatve, m m 0 dm = m m 0 < 0) dp dt = mdv dt v dm dt. (2.59) See also D. Rane and E. Thomas [40], p.. We emphasze: v s defned as elatve to v. If dm had an absolute fowad velocty v, Eq. (2.59) would be (ths s the equaton gven e.g. by K. E. Bullen [8], p. 79) dp dt = mdv dt (v v) dm dt = d(mv) v dm dt dt. A smla poblem s gven n The Physcs Coachng Class [48], poblem 38, p. 228.
20 CHAPTER 2. Recaptulaton of Newtonan Mechancs Newton s equaton of moton,.e. dp/dt = mg, theefoe mples m dv dt = v dm dt mg, o dv = v dm m gdt, (2.60) whee also m = m(t). Integatng wth v 0, m 0 the ntal values of v, m, we obtan: v = v 0 + v ln m m 0 gt. (2.6) Snce the loss of mass s popotonal to tme t, we have m 0 m = µt, µ = const., and the equaton becomes v = v 0 + v ln m + g m m 0. (2.62) m 0 µ We ae gven: v 0 = 0, v = 2070 m s, v = velocty of escape = 200 m s (see Example 2.3), g = 9.8 ms 2, µ = m 0 /60. In Eq. (2.62) we have m m 0, f we assume that the ntal mass of the ocket s ovewhelmngly that of the fuel. Then m m 0 m 0, and wth the numecal values: 200 = 2070ln m/m 0 9.8 60 m/s Fom ths we obtan: m 0 /m = e 5.7 = 299 300. Example 2.5: Equlbum at and appoxmate lmt of atmosphee A body of mass m and ognally at est falls fom a vetcal heght h 0 on the suface of the Eath. Ignong atmosphec effects calculate the velocty of the body takng nto account the dffeence n the acceleaton due to gavty at the heght and at the suface of the Eath, assumng the latte s a sphee of adus. How does the velocty dffe n the two cases h 0 / and h 0 /? The otaton of the Eath also mposes on the body a centfugal foce. The angula velocty of the Eath s 0.00007292 adan/s and = 6370284 metes at the equato, and g = 980 cm/s 2. What s the dstance h above the equato at whch the body s n a state of equlbum? h s Fg. 2.6 The patcle at heght h above the suface of the Eath. Soluton: Let g be the acceleaton due to gavty at the suface of the Eath, and g that at a heght h above, as ndcated n Fg. 2.6. Then, accodng to Newton s law of gavtaton, and wth G the gavtatonal constant and M the mass of the Eath (and dvdng out m fom Newton s equaton of moton and usng s as n Fg. 2.6): g = G M 2, g = G M ( + h) 2, so that g ( + h)2 = g 2 s2 2, g = g2 s 2. But ( ) g = d2 s dt 2 = d ds 2, d2 s 2 ds dt dt 2 = g2 s 2 = 2 d ds ( ) ds 2, and dt ( ) ds 2 = 2g2 + const. dt s
2.7 Mscellaneous Examples 2 At tme t = 0 we have (ds/dt) s=s0 = 0. Hence const. = 2g 2 /s 0,.e. v 2 = ( ) ds 2 ( = 2g 2 dt s ) ( = 2g 2 s 0 + h ). s 0 At the suface of the Eath, whee h = 0, the velocty s v 0 and s gven by ( v0 2 = 2g2 ) (, o v0 2 s = 2g2 0 ), + h 0 whee s 0 = + h 0. Fo h 0 the latte expesson can be appoxmated by v 2 0 = 2gh 0, and fo h 0 we have v 2 = 2g. Ths velocty v can also be ntepeted as that patcula velocty wth whch a body must be shot vetcally upwads n ode not to etun to the Eath (n othe wods, n ode to tavel an nfntely long dstance, h 0 ). Ths velocty s the velocty of escape, also consdeed n Example 2.3. Fo the body of mass m the centfugal foce of the Eath s at the suface (note that ω 2 = 3.4cm/s 2, as shown late, see Eq. (8.28)): mω 2 = ω2 mg mg = g 289 ad2 kg mete s 2. At the dstance h = s above the suface, the outwadly dected centfugal foce s F = (mg/289)(s/), whee the nwadly dected attactve foce of the Eath s F = mg, g = g 2 /s 2. Thus fo equlbum we have mg s 289 = mg2 s 2, s3 = 289 3, s = 6.6. The equested dstance s theefoe oughly 5.6. Vey oughly ths s the top lmt of the atmosphee. Example 2.6: The sphecal ol feld (a) A lage sphee of adus R 0 conssts of a mateal of constant densty ρ 0 except fo a sphecal nset of adus R and densty ρ < ρ 0 whose cente s located at a depth t beneath the suface of the lage sphee. Calculate the dependence of the vetcal and hozontal components of the foce of gavty at a pont P on the suface of the lage sphee as a functon of the angle φ ndcated n Fg. 2.7. (b) Pecson measuements of odnay gavmetes ae based on the deflecton of a hozontally adjusted quatz fbe. These gavmetes have the senstvty to detect a gavtatonal acceleaton of appoxmately 0 4 cm/s 2. Consdeng the hghly dealzed case of a sphecal ol feld at a depth of t = 500 metes n the cust of the Eath and denstes ρ 0 = 5.5, ρ = g/cm 3, calculate what the damete of the sphecal ol feld would have to be n ode to be gavmetcally obsevable. Soluton: (a) We assume the geomety as shown n Fg. 2.7. Thee O s the ogn of the small sphecal nset, and s = O P, as ndcated thee. Then fom tangles O OP, O NP: s 2 = R 2 0 + (R 0 t) 2 2R 0 (R 0 t) cos φ, cos θ = R ( ) 0 (R 0 t)cos φ π = sn s 2 θ, snθ = (R 0 t) snφ. (2.63) s Fom Newton s law of gavtaton, F = Gm m 2 / 3, fo the attactve foce F between two masses m, m 2 at a sepaaton, we obtan fo the foce actng on a unt mass at a pont P as ndcated n Fg. 2.7: F unt mass G = ρ 0(4/3)πR 3 0 y R 2 0 y + (ρ ρ 0 )(4/3)πR 3 s s 2 s,
22 CHAPTER 2. Recaptulaton of Newtonan Mechancs so that (s x = (R 0 t) snφ, s y = R 0 (R 0 t) cos φ wth x and y axes as at P n Fg. 2.7) F x G = (ρ ρ 0 )(4/3)πR 3 s 2 (R 0 t) sn φ, s F y G = ρ(4/3)πr3 0 R 2 + (ρ ρ 0 )(4/3)πR 3 0 s 2 R 0 (R 0 t)cos φ. (2.64) s t O R s θ P θ y φ φ N O O R 0 x Wth the expessons of Eq. (2.63) these become Fg. 2.7 The sphecal ol feld n the Eath. F x G = (ρ ρ 0 )(4/3)πR 3 (R 0 t) snφ {R 2 0 + (R 0 t) 2 2R 0 (R 0 t) cos φ} 3/2, F y G = ρ 0(4/3)πR 3 0 R 2 0 Fo R 0 t we have appoxmately: F x G = (ρ ρ 0 )(4/3)πR 3 R 0 sn φ R 3 0 23/2 ( cos φ) 3/2, + (ρ ρ 0 )(4/3)πR 3 {R 0 (R 0 t) cos φ} {R 2 0 + (R 0 t) 2. (2.65) 2R 0 (R 0 t)cos φ} 3/2 F y G = ρ 0(4/3)πR 3 0 R 2 + (ρ ρ 0 )(4/3)πR 3 R 0( cos φ) 0 R 3 0 23/2 ( cos φ) 3/2. (2.66) One should note that only n ths appoxmaton (but not exactly) s F y/g at φ = 0 equal to ρ 0 (4/3)πR 0. (b) We go to φ = 0,.e. dectly to the ol feld. Then, accodng to Eq. (2.66) F x = 0, and fom Eq. (2.65) we obtan F y G = 4 3 πr 0ρ 0 + (ρ ρ 0 )(4/3)πR 3 t 2. (2.67) Fo F y g y unt mass we have g y = G 4 3 πr 0ρ 0 + G(ρ ρ 0 )(4/3)πR 3 t 2. The second tem ognates fom the dffeence ρ ρ 0 ; hence t s esponsble fo the devaton fom the fst tem. Hence g y = G(ρ ρ 0 )(4/3)πR 3 t 2. (2.68)
2.7 Mscellaneous Examples 23 Fo g y = 0 4 cm/s 2, G 2 = 6.6732 0 8 cm 3 g s 2, ρ 0 = 5.5gcm 3, ρ =.0gcm 3, R 0 = 6370 km, t = 500 metes, we obtan R 3 = 0.2 08 mete 3, R = 58.35 metes. z P l dm=σ dv O x y Fg. 2.8 The potental of the nonsphecal Eath. Example 2.7: Gavtatonal potental of the nonsphecal, oblate Eath Consde a sold oblate ellpsod lke the Eath of mass M E and of constant mass densty σ, whch s otatonally symmetc about the z-axs, the equatoal plane beng the (x, y)-plane, wth the ogn at the geometcal cente, and R the adus of the ccula plane at the equato. Establsh the gavtatonal potental U() at a pont P wth coodnates = (x, y, z) outsde the ellpsod and show that fo R ths can be expessed n the followng fom (whee µ = GM E ): U() = µ J 2 = σ M [ ( ) R 2 ] P 2 (sn θ)j 2 +, ( ellpsod R ) 2 P 2 (sn θ )d, P 2 (x) = 2 ( 3x2 ). (2.69) What s the physcal sgnfcance of J 2 (e-expessed n tems of Catesan coodnates)? Fnally show that the equpotental sufaces U() = const. have the shape of otatonal ellpsods. The Catesan and pola equatons of an ellpsod, the latte wth pole at the cente and coodnates ρ, φ, θ (see tansfomaton below), ae gven by = x2 a 2 + y2 b 2 + z2 c 2, and ρ 2 = cos2 θ cos 2 φ a 2 + cos2 θ sn 2 φ b 2 + sn2 θ c 2. In the case of an ellpsod whch s otatonally symmetc about the z-axs (.e. whose equpotentals ae ndependent of φ) as depcted n Fg. 2.9 we can choose φ = 0 and obtan as n Fg. 2.9 the coss sectonal ellpse ρ 2 = cos2 θ a 2 + sn2 θ c 2. Soluton: We can choose the sphecal pola coodnates of a pont P (cf. Fg. 2.8) as x = cos θ cos φ, y = cos θ sn φ, z = snθ, 0 φ 2π, 0 θ π. The potental U() at a pont outsde the ellpsod due to an nfntesmal mass element dm located at s U() = G ellpsod dm, dm = σdv = σ 2 d cos θ dθ dφ, l 2 = 2 + 2 2 cos χ. (2.70) l
24 CHAPTER 2. Recaptulaton of Newtonan Mechancs We also have (cf. x, y, z) cos χ = = cos θ cos θ cos φ cos φ + cos θ cos θ snφ snφ + snθ snθ = snθ snθ + cos θ cos θ cos(φ φ ). (2.7) z equpotental P π/2 θ x y φ Fg. 2.9 Rotatonal symmety about the z-axs. Wth the followng expessons whch ae known as Legende polynomals, P (cos χ) = cos χ, P 2 (cos χ) = 2 ( 3cos2 χ), P 3 (cos χ) = 2 cos χ(3 5cos2 χ), etc., we have (2.72) l = ( 2 + 2 2 cos χ) /2 = ) /2 ( + 2 2 2 cos χ = [ ( 2 ) 2 2 2 cos χ + 3 ( 2 ) 2 ] 8 2 2 cos χ = [ + cos χ 2 ] 2 2 ( 3cos2 χ) + = ] [ + P + 2 2 P 2 +. (2.73) Hee dm = M E = mass of the Eath. We assume that the Eath s symmetc about ts otatonal axs. In Eq. (2.73) we consde tems up to and ncludng the tem n P 2. Then we have to evaluate some ntegals. We begn wth the ntegal contanng P and use the esults: π π [ sn sn θ cos θ dθ = sn θ d(sn θ 2 θ ] π ) = = 0, 0 0 2 0 2π cos(φ φ )dφ = [sn(φ φ )] 2π 0 = sn(φ 2π) sn φ = 0. 0
2.7 Mscellaneous Examples 25 Then: dmp (cos χ) = σ dmp 2 (cos χ) = [ 3 dm 2 cos2 χ ] = 2 2 d cos θ dθ dφ [snθ snθ + cos θ cos θ cos(φ φ )] = 0, [ 3 dm + 3 4 sn2θ sn2θ cos(φ φ ) 2 2 sn2 θ sn 2 θ + 3 2 cos2 θ cos 2 θ cos 2 (φ φ ) ]. In the last expesson we eplace cos 2 (φ φ ) by [ + cos 2(φ φ )]/2. The tems n cos n(φ φ ) do not contbute because 2π [ snn(φ φ cos n(φ φ )dφ ] ) 2π = 0 n 0 = [snnφ sn nφ] = 0. n z U 0 π/2 P x y 0 We now have Fg. 2.0 The equpotental and ts envelope n the dpole appoxmaton. dmp 2 (cos χ) 2 = = = = [ 3 dm 2 sn2 θ sn 2 θ + 3 4 cos2 θ cos 2 θ ] 2 2 [ 3 dm 2 sn2 θ sn 2 θ + 3 4 ( sn2 θ)( sn 2 θ ) 2 [ 9 2 dm 4 sn2 θ sn 2 θ 3 4 (sn2 θ + sn 2 θ ) + ] 4 ( 3 2 dm 2 sn2 θ )( 3 2 2 sn2 θ ). 2 ] 2
26 CHAPTER 2. Recaptulaton of Newtonan Mechancs It follows that, wth GM E = µ, U() = µ = µ [ + M E [ ( R whee, wth R the equatoal adus of the Eath, 2 ( 3 2 2 sn2 θ )( 3 2 2 sn2 θ ) ] dm + 2 ) 2 ] P 2 (sn θ)j 2 +, (2.74) J 2 = ( ) 2 P 2 (sn θ )dm. (2.75) M E Eath R The esult (2.74) contans a devaton of the equpotental sufaces U = const. fom pue sphecal symmety, wheeas J 2 descbes the nonsphecal defomaton of the Eath; ths tem, popotonal to / 3, s known as the dpole contbuton, and s moe famla n electodynamcs. We also note that the potental s ndependent of the angle φ. The eason fo ths s the otatonal symmety of the potental about the z-axs, as llustated n Fg. 2.0. We now estct ouselves to the dpole appoxmaton and expess J 2 n tems of Catesan coodnates. We have: 2 = x 2 + y 2 + z 2, sn θ = z, so that J 2 = = 2M E R 2 2M E R 2 2 ( 3sn 2 θ )dm = (x 2 + y 2 2z 2 )dm = 2M E R 2 ( 2 3z 2 )dm 2(C A), (2.76) 2M E R2 whee C = dm(x 2 + y 2 ), A = dm(x 2 + z 2 ) = dm(y 2 + z 2 ). The last equalty esults fom the otatonal symmety of the Eath. The quantty C s, of couse, the pncpal moment of neta of the Eath (cf. Chapte 8) and A the smallest moment of neta wth espect to any abtay equatoal damete. Thus the quantty J 2 s a measue of the devaton of the shape of the Eath fom exact sphecal symmety. The pola equaton (pola coodnates ρ, θ) of an ellpse wth espect to the cente as pole s, as we saw (these aspects ae consdeed n Chapte 7 n much moe detal), ρ 2 = cos2 θ a 2 + sn2 θ c 2, o [ γ 2 a2 ρ 2 = + a2 c 2 ] c 2 sn 2 θ. (2.77) Equpotental sufaces ae sufaces of the same constant potental,.e. U() = U 0 = const. Insetng ths nto Eq. (2.74), we obtan fo these sufaces (note that the facto 3 ognates fom P 2 (snθ)): U 0 = µ + β 3 ( 3sn2 θ), β = 2 µj 2R 2. (2.78) In ode to be able to compae ths equaton wth Eq. (2.77), we have to ewte t n a fom whch has the facto ( + 3sn 2 θ) on the ght hand sde. Thus we obtan µ + 2β 3 + U 0 = β 3 ( + 3sn2 θ), o 2 + 2 µ + U 0 = ( + 3sn 2 θ). (2.79) β See e.g. H. J. W. Mülle Ksten [34], pp. 78 82. Fo an analogous case see Example 0.3 on the tdal effect.
2.7 Mscellaneous Examples 27 Compang wth Eq. (2.77) we obtan γ 2 a2 ρ 2 = 2 + µ + U 2 0, a 2 = 4c 2. (2.80) β We can see the behavou of the equpotentals wth espect to the ccle U 0 = µ/ by lookng at the ponts θ = 0, π and θ = ±π/2. In these cases we have: 0, θ = 0, π : U 0 = µ + β 0 0 3, π/2, θ = ± π 2 : U 0 = µ π/2 2β 3, π/2 0 > π/2. (2.8) Thus 0 > π/2 as ndcated n Fg. 2.0. Ths example demonstates that the potental of the nonsphecal Eath s gven by a sees of the followng fom: U() = µ [ n=2 ( ) R n ] P n(snθ)j n. (2.82) Example 2.8: A patcle gong back and foth n a tunnel unde gavty A chod lke tunnel of length AB = 2R cos θ s dven though the Eath of adus R fom a pont A on the suface to a pont B on the suface. The patcle of mass m s nseted at A wth zeo velocty. Detemne the peod of oscllaton of the patcle between ponts A and B. A θ R C P R B Fg. 2. The tunnel AB though the Eath of adus R. Soluton: The potental of the patcle at pont P of the tunnel n Fg. 2., a dstance fom the cente of the Eath and dstance d fom A, s V () = G mm E, 2 = R 2 + d 2 2Rd cos θ, (2.83) whee M E s the mass of the Eath, and θ s the angle shown n Fg. 2.. Hence V () = GmM E [R 2 + d 2 2Rd cos θ] /2 G mm [ ] E d2 2Rd cos θ R 2R 2. (2.84) Fo a somewhat smla poblem see The Physcs Coachng Class [48], poblem 49, p. 245. Fo an analogous poblem see The Physcs Coachng Class [48], poblem 030, p. 38.
28 CHAPTER 2. Recaptulaton of Newtonan Mechancs Takng the potental at the suface of the Eath as zeo, the potental enegy of the patcle s V () = G mm E 2R 3 [d2 2Rd cos θ] = G mm E 2R 3 [2 R 2 ]. (2.85) Thus at the suface wth = R ths potental enegy vanshes. The total enegy of the patcle whch stats off fom est at pont A s theefoe E = 2 mv2 + G mm E 2R 3 [2 R 2 ] = 0. (2.86) Settng v = dx/dt, x = AP n Fg. 2., we have 2 = R 2 + x 2 2xR cos θ, so that v 2 = G M E R 3 [2xR cos θ x2 ], (2.87) and theefoe we obtan fo the tme t takes the patcle to tavel fom A to B: B AB R t AB = dt = dx 3 A x=0 GM. (2.88) E 2xR cos θ x 2 Wth D = 2R cos θ and f = R 3 /GM E and x y + D/2, ths s D t AB = f x=0 = f dx D/2 = f Dx x 2 [ sn ( y D/2 dy D 2 /4 y 2 y= D/2 )] D/2 = πf. (2.89) D/2 Thus the peod T = 2t AB s 2π R R 2 /GM E = 2π R/ g, whee g s the acceleaton due to gavty.