Radial Motion of Two Mutually Attracting Particles

Size: px

Start display at page:

Download "Radial Motion of Two Mutually Attracting Particles"

Joleen Stone
6 years ago
Views:

1 Radal Moon of Two Muually Aacng Pacles Cal E. Mungan, U.S. Naval Academy, Annapols, MD A pa of masses o oppose-sgn chages eleased fom es wll move decly owad each ohe unde he acon of he nvesedsance-squaed foce of aacon beween hem. An exac expesson fo he sepaaon dsance as a funcon of me can only be found by numecally nveng he soluon of a dffeenal equaon. A smple, appoxmae fomula can be obaned by combnng dmensonal analyss, Keple s hd law, and he famla quadac dependence of dsance on me fo a mass fallng nea Eah s suface. These exac and appoxmae esuls ae appled o seveal neesng examples: he flgh me and maxmum alude aaned by an objec fed sagh upwad fom Eah s suface; he me equed fo an aseod of known sang poson and speed o coss Eah s ob f s beang owad he Sun; and he collson me of wo opposely chaged pacles sang fom es. Poblem Saemen Suppose ha he fs pacle, whose moon s of pmay nees, has mass m and chage q, whle he second pacle has mass M and chage Q. To ensue ha he eleccal foce beween hem s no epulsve, assume ha Qq. (Noe, howeve, ha q can be posve, negave, o zeo. In he las case, he foce of aacon s solely gavaonal.) We wll analyze he poblem n he cene-of-mass fame of efeence and assume ha he wo pacles consue a bound, solaed sysem. In ohe wods, he oal mechancal enegy s negave elave o he usual poenal efeence a nfne sepaaon, and he pacles always have equal and oppose lnea momena. Consequenly, hee s some maxmum dsance beween he wo pacles a whch pon hey ae boh nsananeously a es. Defne he me of ha even as =. I coesponds o he nal sepaaon beween he pacles f hey sa fom es; epesens he unng pons f he pacles sa ou movng away fom one anohe; and epesens a pon n he exapolaed pas f hey ae nally movng owad each ohe. To summaze, whou loss of genealy we suppose ha wo pacles 1 and ae eleased fom es a me = when hey ae a cene-o-cene dsance apa. Defne he coodnae ogn o be a he saonay cene of mass of he sysem. The wo pacles aac and begn o move owad each ohe. The adal sepaaon and elave speed a any lae me (befoe hey collde wh each ohe a me c a he ogn f hey ae of pon sze) ae and u, as descbed n Fg. 1. The aacve foce beween he wo pacles s he sum of he gavaonal and elecosac foces, wh magnude F = g /, whee g kqq + GMm. (Hee k s he Coulomb consan and G s he unvesal gavaonal consan.) In pacce, one of he wo foces wll usually domnae and we can neglec he ohe, bu s jus as easy o be geneal and nclude boh. Newon s second law descbng he magnude of he elave acceleaon a a 1 + a of he pacles owad each ohe s 1 a g (1) m =, 5 DOI: / The Physcs Teache Vol. 47, Novembe 9

2 * Q 1 a c c Fg. 1. Pacle 1 of mass m and chage q, and pacle of mass M and chage Q movng adally owad each ohe. The poson and speed of he fs pacle elave o he cene of mass (ndcaed by he asesk) ae 1 and u 1, whle hose of he second pacle ae and u. Consequenly he sepaaon and elave speed of he wo pacles ae 1 + and u u 1 + u. Fg.. A hghly ellpcal wo-dmensonal ob of a plane of mass m shown a he nal nsan when s locaed a aphelon wh espec o a sun of mass M. whee he educed mass of he sysem s m mm /(m + M), whch smplfes o m < m fo he common case of M >> m. The sgns of he knemac vaables have been chosen so ha, u, and a ae all posve fo < < c. Consequenly u = d/d snce deceases wh dung hs me neval, whle a = du/d = d /d because u nceases. We can now succncly sae he poblem: Fnd he dsance beween he pacles as a funcon of me. An exac soluon fo he nvese elaonshp () has been publshed n seveal places 6 and s evewed n he appendx. The pupose of he pesen pape s o develop an appoxmae fomula usng deas ha would be moe appopae n an noducoy couse. No only s he esulng expesson fo () smple, bu s analycally nveble o gve (), n conas o he exac expesson ha s no. Thee seps ae used o develop he appoxmae fomula. Fs, dmensonal analyss quckly gves a good esmae fo he collson me c of wo pon pacles ha begn a es. Second, Keple s hd law deemnes he mssng numecal pefaco ha dmensonal analyss canno gve. Thd, he famla knemac esul fo an objec fallng n Eah s suface gavy movaes he fnal expesson. Insucos need no follow all hee seps. Fo example, one could skp decly o he second sep o fnd he exac collson me. Alenavely, one could om he second sep and smply sae whou poof he small numecal coecon o he fomula fo he collson me found n he fs sep. A hd possbly would be o skp he hd sep and esc consdeaon o he one-dmensonal collson of wo pon pacles sang fom es. Sep One Appoxmae Soluon fo he Collson Tme Equaon (1) mplcly elaes he nal sepaaon o he collson me c n ems of he quany g /m ha has SI uns of m /s. To conve hs quany no one wh uns of me, we mus dvde by he cube of and ake he ecpocal squae oo. Theefoe, we mmedaely pedc ha µ c, () g assumng boh pacles have neglgble sze elave o he nal sepaaon. (In conas, an example poblem s solved lae n hs acle whee he second pacle s aken o be he Eah and s adus s accouned fo.) As shown below, hs esmae fo c s only 1% smalle han he exac answe and so s an effecve llusaon of he powe of dmensonal easonng. 7 Fo he case of wo unchaged pacles (so ha g = GMm), when he second pacle s much moe massve han he fs (so ha m < m ), hen Eq. () becomes c GM. () Sep Two Numecal Pefaco fo he Collson Tme The ellpcal ob of a plane (of mass m) abou a sun (of mass M >> m) wh sem-majo axs a has a peod squaed of 4p = a (4) GM accodng o Keple s hd law, as can be deved a The Physcs Teache Vol. 47, Novembe 9 5

3 an noducoy level usng consevaon laws and elemenay popees of ellpses. 8 Now consde an ellpcal ob n he lm ha he eccency e 1. Then he sun (locaed a one of he ellpcal foc) moves o an end of he hghly elongaed ob, as llusaed n a lee o he edo 9 by a sequence of obs fo whch e s nceased fom o 1. To make conac wh he pesen pape, le us suppose ha he plane sas a aphelon, whee s adal velocy s zeo. In ha case, as Fg. makes clea, he plane and sun ae nally sepaaed by he majo axs so ha = a. To complee one ob, he plane fs has o move o he sun s poson, whch by defnon akes he collson me c, and hen eun o aphelon n he same me. Thus, he obal peod s = c. Subsung hs esul and a = / no Eq. (4) gves π c =. GM (5) Consequenly he equed numecal pefaco n fon of Eq. () s / p < Lkewse coecng he moe geneal expesson n Eq. () leads o he exac esul fo wo pon pacles, π µ c =. (6) g Nong ha he elave sepaaon beween he wo pacles deceases by n a me of c, poves convenen n he subsequen analyss o ecas he poson and me vaables n dmensonless fom. Defne he nomalzed adal sepaaon as R (7) and he nomalzed me as T. (8) c Noe ha R = 1 a he nal pon T =, and ha pon pacles would each he ogn R = a T = 1. Sep Thee Appoxmae Relaonshp Beween he Sepaaon and Tme Suppose ha a pacle of mass m s dopped nea Eah s suface (so ha s appoxmaely equal o Eah s adus E ) and falls a small dsance D << o a new adal poson = D n a me. Elemenay knemacs ells us ha 1 D = g, (9) whee g = GM/ E = 9.8 N/kg s Eah s suface gavaonal feld sengh (wh M he mass of he Eah). Use Eqs. (7) and (8) o e-expess he fall dsance and me n nomalzed fom as D R = 1 X whee X << 1 and p T = whee c = g c (1) (11) fom Eq. (5). Usng hese vaables, Eq. (9) can be ewen as 1 4 X = T whee n 1.6. n p (1) Mulplyng hs equaon fo X by n and addng and subacng uny gves 1 (1 nx) = T. (1) We wan o exend hs esul o hold fo values of X ha appoach 1. The expesson n paenheses looks lke he fs wo ems n he bnomal expanson of (1 X) n = R n, usng Eq. (1) o elae X o R. Makng hs denfcaon n Eq. (1) leads o he followng hypoheszed expessons elang he nomalzed sepaaon and me, n 1 R T T ( R) 1 R ( T ) 1/ n o R( T ) 1. n (14) Noe ha hese expessons f he wo endpons R = 1 a T =, and R = a T = 1. Equaon (14) s ploed n ed n Fg.. Fo compason, he exac esul fom Eq. () n he appendx s gaphed n blue. The wo cuves ovelap almos pefecly acoss he whole ange! The ageemen can be slghly mpoved by oundng off n o he value 1.6, whch s moe convenen o wok wh han he exac aonal value; n ha case, he mean dscepancy 1 beween he exac and appoxmae values of T(R) s abou a quae of a pecen. 54 The Physcs Teache Vol. 47, Novembe 9

4 snce T = / c, whee c = p E / g accodng o Eq. (5). Fo compason, Eq. (14) pedcs a me of E 1.6 = p ( 1 ), (17) g usng he ounded-off value of n. The numecal values of Eqs. (16) and (17) dsagee by less han 1 s. Fg.. Appoxmae (ed) and exac (blue) plos of he sepaaon of he wo pacles as a funcon of me n dmensonless fom. Thee Example Poblems 1. Suppose a pacle s launched upwad fom he suface of he Eah wh a speed u ob equal o ha of a low-eah-ob saelle. Wha s he maxmum hegh ha eaches above Eah s suface and how long does ake o ge hee? 11 The speed of ob of a saelle of mass m jus above he suface of he Eah (gnong s amosphee and opogaphy) s obaned fom Newon s second law as ob E GMm u = m f E uob = ge = 7.9 km/s, (15) whee M and E ae Eah s mass and adus, especvely. Snce he velocy on he way up s equal and oppose o he subsequen velocy back down fom he peak hegh, we can subsue = E, u = u ob, and g /m = GM = g E no Eq. (5) o oban he maxmum adal poson = E,.e., he pacle aans a maxmum hegh of one Eah adus (67 km) above he suface. I hen falls back o he gound, a whch pon R = E / = 1/. Subsung hs value no Eq. (), one hen fnds ha he fall (o equvalenly he se) me s E p = 1 + = 4.5 mn g (16). A new aseod s dscoveed ha s headng sagh fo he Sun. I s nally obseved a a dsance fom he Sun of AU wh a speed of half of he speed u esc would need o escape fom he Sun a ha pon n s ob. How much me wll elapse unl he aseod cosses Eah s ob? Le = m and f = m be he nal and fnal adal posons of he aseod, especvely, and le denoe he maxmum dsance fom he Sun ha he aseod would aan f wee avelng decly away fom he Sun wh an nal speed of u esc /. Also le R / and R f f /. The nal speed of he aseod (of mass m) s obaned fom enegy consevaon as 1 GMm muesc = 1 GM u uesc = = 1.1 km/s, whee M s he mass of he Sun. Nex we use Eq. (5) wh g /m = GM o fnd he ecpocal of he aseod s adal unng pon, 1 1 u = =, GM 4 wh he help of Eq. (18). Consequenly, R = /4 and R f = 1/4. Thus Eq. () gves (18) (19) D f = cos cos GM 4 4 () p f = = 17 days, GM usng Eqs. (5) and (8). In compason, Eq. (14) mples ha f D = p GM (1) The Physcs Teache Vol. 47, Novembe 9 55

5 and he quany n squae backes s nealy equal o 1/, n ageemen wh he esul of Eq. (). If such an aseod eally wee dscoveed omoow, we would heefoe have abou half a yea o pepae f he Moon o Eah wee o be mpaced.. An elecon and a poson sa ou a es a dsance = 1 nm apa. How much me wll ake fo hem o annhlae each ohe? The wo pacles have equal and oppose chages of magnude q = Q = e, whee e = C. Fuhemoe hey have equal masses m = M = kg so ha he educed mass s m = m/. The elecosac foce compleely ovewhelms he gavaonal foce beween hem and hus g = ke. Equaon (6) now becomes p m c = = 49 fs. 4e k They dsappea n less han a blnk of an eye! () Appendx: Exac Relaonshp Beween he Sepaaon and Tme Subsung a = d /d no Eq. (1) esuls n a second-ode dffeenal equaon fo () ha can be solved n wo seps. Fs we fnd he elave speed u of he pacles as a funcon of he sepaaon ; hen we negae ha o deemne he me T equed fo he pacles o appoach each ohe o whn some dsance R n dmensonless fom. I. Relave Speed as a Funcon of he Sepaaon In he cene-of-mass efeence fame, he wo pacles have equal and oppose lnea momena of magnude 1 mu 1 = Mu = mu p, so ha he oal knec enegy s p p m u = + = mu. m M m M () The poenal enegy coespondng o he foce of neacon n Eq. (1) s g/. Consevaon of mechancal enegy fo he sysem of wo pacles beween he nal and an abay fnal sae heefoe becomes 1 g g mu =, whch can be eaanged as g 1 1 υ = µ (4). (5) The squae oo s always eal because. Noe ha he speed becomes nfne as, bu n he eal wold ha would be pevened by he fne szes of he wo pacles. II. Relaonshp Beween he Sepaaon and Tme Subsung u = d/d no Eq. (5), he vaables and can be sepaaed and ewen n dmensonless fom usng Eqs. (6) hough (8) o ge T p dr dt = 1/. R 1 R 1 The lowe lms coespond o he nal condon R = 1 a T =. The me negal on he lef-hand sde s equal o T. One can solve he adal negal by makng a change of vaable. By al and eo, one s led o he gonomec subsuon R = cos q, whose vue s ha lends self o he Pyhagoean deny. (6) sec q 1 = an q. (7) Subsung hs deny and he dffeenal dr = cos q sn qdq no Eq. (6) leads o q p cos ( 1 cos ), T = q d q = + q d q q (8) usng he cosne double-angle fomula and nong ha q = when R = 1. Inegang gves p sn q T = q + = q + cosq snq (9) = q + cosq 1 cos q wh he help of he sne double-angle fomula and anohe Pyhagoean deny. Reunng o he og- 56 The Physcs Teache Vol. 47, Novembe 9

6 nal vaable R gves he fnal esul p 1 cos (1 ). () T = R + R R Ths equaon coecly pedcs ha T = a he nal pon R = 1, and ha T = 1 a R =. We canno analycally nve o fnd R(T), bu by compung a able of values (as ploed n Fg. ) we can fnd any desed esul numecally, whch pons ou he advanage of wokng wh a unvesal (.e., dmensonally nomalzed) fom of he equaon. One can show ha Eq. () becomes Eq. (1) o lowes ode n X f Eq. (1) holds. Ths esul follows by povng ha each of he wo ems on he ghhand sde of Eq. () s appoxmaely equal o X f R = 1 X, whee X << 1. Fo he fs em, one demonsaes he nvese esul 1 X < cos X by Taylo expandng boh sdes up o lnea powes n X. Noe added n poof: S.K. Foong [ Fom Moon-fall o moons unde nvese squae laws, Eu. J. Phys. 9, (Sep. 8)] has obaned ohe appoxmae expessons fo R(T ). 8. E. D. Noll, Keple s hd law fo ellpcal obs, Phys. Teach. 4, 4 4 (Jan. 1996). A pape of he same le wh essenally he same devaon has been publshed by R. Wensock n Am. J. Phys., (Nov. 196). 9. A. J. Mallnckod, Fallng no he sun, Phys. Teach. 6, 4 (Sep. 1998). 1. Le he faconal dscepancy beween he exac and appoxmae values of T be f (R) (T exac T appox )/ T exac. Then I defne he mean dscepancy o be he squae oo of he negal of f fom R = o R = 1, n accodance wh he usual sascal dea of vaance. 11. Poblem adaped fom B. Kosunsky, Physcs Challenge fo Teaches and Sudens: The dll eam ocks! Phys. Teach. 45, 568 (Dec. 7). PACS codes: b, 45.. Cal Mungan s a fequen conbuo o The Physcs Teache. Ths acle was nsped by a queson by hs colleague John Fonanella abou he vecal moon of an objec n Eah s gavy when s change n alude s lage. Physcs Depamen, U.S. Naval Academy, Annapols, MD 14; mungan@usna.edu Refeences 1. By elmnang beween he wo equaons 1 + and m 1 = M, one can quckly show ha m = m 1. Dffeenang boh sdes wh espec o me poves ha mu = mu 1. Dffeenang agan gves ma = ma 1. Then Eq. (1) s equvalen o Newon s second law fo pacle 1. Alenavely one can esablsh ha mu = Mu and ma = Ma, denfyng Eq. (1) as Newon s second law fo pacle. Noe ha ma 1 = Ma n ageemen wh Newon s hd law.. S. Van Wyk, Poblem: The fallng Eah and Soluon, Am. J. Phys. 54, 91 & 945 (Oc. 1986).. A. J. Mallmann, J. L. Hock, and K. M. Ogden, Supsng facs abou gavaonal foces, Phys. Teach., (Nov. 1994). 4. M. B. Sewa, Fallng and obng, Phys. Teach. 6, 1 15 (Feb. 1998). 5. J. Gallan and J. Calson, Long-dsance fee fall, Phys. Teach. 7, (Mach 1999) and Eaa & Commens on pages 61 6 of he May 1999 ssue. 6. M. McCall, Gavaonal obs n one dmenson, Am. J. Phys. 74, (Dec. 6). 7. C. F. Bohen, Dmensonal analyss, fallng bodes, and he fne a of no solvng dffeenal equaons, Am. J. Phys. 7, (Apl 4). The Physcs Teache Vol. 47, Novembe 9 57

5-1. We apply Newton s second law (specifically, Eq. 5-2). F = ma = ma sin 20.0 = 1.0 kg 2.00 m/s sin 20.0 = 0.684N. ( ) ( )

5-1. We apply Newton s second law (specifically, Eq. 5-2). F = ma = ma sin 20.0 = 1.0 kg 2.00 m/s sin 20.0 = 0.684N. ( ) ( ) 5-1. We apply Newon s second law (specfcally, Eq. 5-). (a) We fnd he componen of he foce s ( ) ( ) F = ma = ma cos 0.0 = 1.00kg.00m/s cos 0.0 = 1.88N. (b) The y componen of he foce s ( ) ( ) F = ma = ma