NEWTON S THEORY OF GRAVITY

Transcription

1 NEWTON S THEOY OF GAVITY 3 Concptual Qustions 3.. Nwton s thid law tlls us that th focs a qual. Thy a also claly qual whn Nwton s law of gavity is xamind: F / = Gm m has th sam valu whth m = Eath and m = sun o vic vsa. Mm s Mm s 3.. Th foc of th sta on plant is F = G. Th foc of th sta on plant is F = G. Sinc m = m, and =, M ( m ) G s ( ) F = = = F Mm G 4 s Mm Mm 3.3. (a) Th foc of th Eath on th fist satllit is F = G whil F = G. Sinc =, F m 000 kg = = =. F m 000 kg (b) With Fnt = ma, F = ma and F = ma, so, a F m 000 kg = = = a F m 000 kg Th f-fall acclation is th sam fo objcts of diffnt mass Th astonauts can b wightlss at any distanc bcaus an objct is said to b wightlss if it is in f fall (as in obit). Fo th gavitational foc to bcom zo, th spaccaft would hav to b an infinit distanc away Th hamm will not fall to th Eath. Th astonaut, shuttl, and hamm a all in f fall aound th Eath (in an obit), so th hamm has th sam acclation as th astonaut and dos not mov away fom him Th acclation du to gavity is M g = G. With g = 0 m/s, M = M, and =, M ( M ) M g = G = G = G = g = 0 m/s ( ) Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist. 3-

2 3- Chapt Th gavitational potntial ngy is ngativ bcaus w choos to plac th zo point of potntial ngy at infinity ( U = 0). With this choic, th gavitational potntial ngy is ngativ bcaus th consvativ foc of at gavity is attactiv. Th two masss will gain kintic ngy as thy appoach ach oth Th scap spd fom Plant X is v scap X GM X = = 0,000 m/s. Plant Y is twic as dns as Plant X but th sam siz so has twic th mass. Thfo v scap Y X X G( MX) = = vscap X = 4,4 m/s 3.9. Th Eath s nw obital piod would b about.9 yas. Fo cicula obits, Kpl s thid law lats obital 4π 3 piod to adius and can b wittn T =. H M is th mass of th body bing obitd (th Sun in this GM poblm.) Th obital piod is indpndnt of th mass of th obiting body Fo a cicula obit, v = GM/ (Equation 3.). As dcass, v incass, so th satllit spds up as th satllit spials inwad. Exciss and Poblms Sction 3.3 Nwton s Law of Gavity 3.. Modl: Modl th sun (s) and th ath () as sphical masss. Du to th lag diffnc btwn you siz and mass and that of ith th sun o th ath, a human body can b tatd as a paticl. GMsM y GMM y Solv: Fs on you = and F on you = s- Dividing ths two quations givs 30 6 s on y s kg m -4 = = = on y s kg m F M F M 3.. Modl: Assum th two lad balls a sphical masss. Gmm on F on ( N m /kg )(0 kg)(0. 00 kg) 9 Solv: (a) F = = = = N (0. 0 m) (b) Th atio of th abov gavitational foc to th gavitational foc on th 00 g ball is N = (0. 00 kg)(9. 8 m/s ) Assss: Th answ in pat (b) shows how small is th gavitational foc btwn two lad balls spaatd by 0 cm compad to th gavitational foc on th 00 g ball Modl: Modl th sun (s), th moon (m), and th ath () as sphical masss. GMsM m GMM m Solv: Fs on m = and F on m = s-m -m 9 Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

3 Nwton s Thoy of Gavity 3-3 Dividing th two quations and using th astonomical data fom Tabl 3., 30 8 s on m s -m kg m 4 on m s-m kg m F M = = =. 8 F M Not that th sun-moon distanc is not noticably diffnt fom th tabulatd sun-ath distanc Solv: F sph on paticl Dividing th two quations, GM M GM M = = s p p and F ath on paticl s-p 6 sph on paticl s 5900 kg m -7 = = =. ath on paticl s-p kg m F M F M 3.5. Modl: Modl th woman (w) and th man (m) as sphical masss o paticls. Solv: F - GM wm m. w on m Fm on w m-w (. 0 m) ( N m /kg )(50 kg)(70 kg) -7 = = = =. 3 0 N 3.6. Modl: Modl th ath () as a sph Th spac shuttl o a.0 kg sph (s) in th spac shuttl is + s = m m = m away fom th cnt of th ath. Solv: (a) 4 GMMs on s 6 ( + s) ( m) F ( N m /kg )( kg)(. 0 kg) = = = 9. 0 N Sction 3.4 Littl g and Big G 3.7. Modl: Modl th sun (s) as a sphical mass. 30 GMs.. gsun sufac 8 s ( m) - 30 GMs ( N m /kg )( kg) sun at ath s- ( m) Solv: (a) (b) g ( N m /kg )( 99 0 kg) = = = 74 m/s -3 = = = m/s 3.8. Modl: Modl th moon (m) and Jupit (J) as sphical masss. Solv: (a) g GM m.. moon sufac 6 m ( m) ( N m /kg )( kg) = = =. 6 m/s Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

4 3-4 Chapt 3 (b) g 7 GM J.. Jupit sufac 7 J ( m) ( N m /kg )( 90 0 kg) = = = 5. 9 m/s 3.9. Modl: Modl th ath () as a sphical mass. Th acclation du to gavity at sa lvl is Tabl 3.) m/s (s Tabl 3.) and = m (s 6 Solv: H g GM GM g = = = = ( ) m/s ath gobsvatoy ( + h) ( + h / ) ( + h / ) ath GM/ = is th acclation du to gavity on a non-otating ath, which is why w v usd th valu m/s. Solving fo h, h= =. 4 km Modl: Modl th ath () as a sphical mass. Solv: Lt th acclation du to gavity b 30g. whn th ath is shunk to a adius of x. Thn, g sufac sufac GM GM = = and 3.0g sufac x 3 GM GM = = = x x. Th ath s adius would nd to b 0.58 tims its psnt valu. 3.. Modl: Modl Plant Z as a sphical mass. Z ( N m /kg ) M Z 4 M 6 Z Z ( m) GM Solv: (a) gz sufac = 8. 0 m/s = = kg (b) Lt h b th hight abov th noth pol. Thus, g GM Z GM Z gz sufac 8. 0 m/s abov N pol 6 Z + h Z + hz + hz = = = = = m/s ( ) ( / ) ( / ) m m Sction 3.5 Gavitational Potntial Engy 3.. Modl: Modl Mas (M) as a sphical mass. Igno ai sistanc. Also consid Mas and th ball as an isolatd systm, so mchanical ngy is consvd. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

5 Nwton s Thoy of Gavity 3-5 Solv: A hight of 5 m is vy small in compaison with th adius of ath o Mas. W can us flat-ath gavitational potntial ngy to find th spd with which th astonaut can thow th ball. On ath, with y i = 0 m and v = 0 m/s, ngy consvation givs f mvf mgyf mvi mgy i vi gyf + = + = = (9. 8 m/s )(5 m) = 7. m/s Engy is also consvd on Mas, but th acclation du to gavity is diffnt. g 3 GM M.. Ma s sufac 6 M ( m) ( N m /kg )(6 4 0 kg) = = = m/s On Mas, with y i = 0 m and v f = 0 m/s, ngy consvation is vi (7. m/s) mv f mgyf mv i mgyi yf g m. + = + = = = 39 m (3 77 m/s ) 3.3. Modl: Modl Jupit as a sphical mass and th objct as a point paticl. Th objct and Jupit fom an isolatd systm, so mchanical ngy is consvd. Th minimum launch spd fo scap, which is calld th scap spd, allows an objct to scap to an infinit distanc fom Jupit (o, in gnal, fom its patn objct). Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

6 3-6 Chapt 3 Solv: Th ngy consvation quation K + U = K+ U is wh J and GM m GM m J o J o mv o = mv o J M J a th adius and mass of Jupit. Using th asymptotic condition v = 0 m/s as, 7 GM Jmo GM J ( N m /kg )( kg) 4 mv o v 7 J J m 0 J = = = = m/s Thus, th scap vlocity fom Jupit is 60. km/s Modl: Modl th ath () as a sphical mass. Compad to th ath s siz and mass, th ockt () is modld as a paticl. This is an isolatd systm, so mchanical ngy is consvd. Solv: Th ngy consvation quation K + U = K+ U is In th psnt cas,, so GM m GM m mv = mv GM m GM mv = mv v= v ( N m /kg )( kg) v = (. 5 0 m/s) = 0 km/s m 3.5. Modl: Th pob and th sun fom an isolatd systm, so mchanical ngy is consvd. Th minimum launch spd fo scap, which is calld th scap spd, allows an objct to scap to an infinit distanc fom th sun, wh th objct will hav slowd to zo spd with spct to th sun. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

7 Nwton s Thoy of Gavity 3-7 W dnot by m p th mass of th pob. M s is th sun s mass, and sun and th pob. Solv: Th consvation of ngy quation K + U = K+ U is GM m mv Using th condition v = 0 m/s asymptotically as, s p s p p- = mv p- s-p s-p is th spaation btwn th cnts of th GM m GM m ( N m /kg )( 99 0 kg) mv 50 0 m - 30 s p GMs.. 4 p scap = vscap = = =. s-p s-p. 4 0 m/s 3.6. Modl: Modl th distant plant (p) and th ath () as sphical masss. Bcaus both a isolatd, th mchanical ngy of th objct on both th plant and th ath is consvd. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

8 3-8 Chapt 3 Lt us dnot th mass of th plant by gavity on th sufac of th plant is M p and that of th ath by and th ath, spctivly. Solv: (a) W a givn that M = M and g = g. Sinc g p p g p and on th sufac of th ath is g. P p 4 GM p GM = and g =, w hav GM GM G( M / ) M. You mass is m 0. Also, acclation du to p and p p 6 7 = = p = =. =. p 4 4 (b) Th consvation of ngy quation K + U = K+ U is Using v = 0 m/s as, w hav mv 0 scap GM m = p 0 p GM m mv p 0 p 0 0 = mv 0 p 8 8( m) 80 0 m GM m 4 p ( ) 4( N m /kg )( kg) scap 7 p p m v a th adii of th plant GM G M = = = = 9. 4 km/s Sction 3.6 Satllit Obits and Engis 3.7. Modl: Modl th sun (s) as a sphical mass and th astoid (a) as a point paticl. Th astoid, having mass m a and vlocity v a, obits th sun in a cicl of adius a. Th astoid s tim piod is T a = 50. ath yas = s. 8 Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

9 Nwton s Thoy of Gavity 3-9 Solv: Th gavitational foc btwn th sun (mass = M ) and th astoid povids th cntiptal acclation quid fo cicula motion. Substituting s a a a s π a s a a a a a Ta 4π GM m m v GM GM T = = = G = N m /kg, a = m. Th vlocity of th astoid in its obit will thfo b s 30 M s = kg, and th tim piod of th astoid, w obtain πa π a T 8 a (4.4 0 m) 4 v = = =.7 0 m/s.58 0 s Solv: W giv th answ to two significant figus bcaus th astoid piod is givn to two significant figus Modl: Modl th sun (s) and th ath () as sphical masss. Th ath obits th sun with vlocity v in a cicula path with a adius dnotd by s-. Th sun s and th ath s masss a dnotd by M s and m, spctivly. Solv: Th gavitational foc povids th cntiptal acclation quid fo cicula motion. M Assss: Th tabulatd valu is isn t xactly cicula. s ( π s-) = = s- s- s-t GM m m v m 3 3 4π s- 4 π ( m) s - GT ( N m /kg )( s) = = =. 0 0 kg kg. Th slight diffnc can b ascibd to th fact that th ath s obit 3.9. Fom Kpl s thid law, th obital piod squad is popotional to th obital adius cubd: T y x 3 3 y (4 x) 64 x (8 x) / Thus, at = 4, T = T. Thfo T = 8T. A ya on Plant Y is 8 00 = 600 ath days long Modl: Modl th sta (s) and th plant (p) as sphical masss. Solv: Fom th plant s acclation du to gavity, w find its mass to b M g p g 6 p p (. m/s )( m) 5 p = = =. G N m /kg y GM = (b) Fom th plant s obital piod, w find th mass of th sta Omga to b M p p 4π T = GM s x 5 0 kg 4π 4 π (. 0 m) 30 = = = 5. 0 kg GT ( N m /kg )( s) s 3.. Modl: Modl th plant and satllits as sphical masss. Plas f to Figu EX3.. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

10 3-0 Chapt 3 3 / Solv: (a) Th piod of a satllit in a cicula obit is T = [4 π /( GM )] This is indpndnt of th satllit s mass, so w can find th atio of th piods of two satllits a and b: T T a b 3 a = b Satllit has =, so T = T= 50 min. Satllit 3 has 3 = (3/), so T 3/ 3 T = (3/) = 459 min. (b) Th foc on a satllit is F = GMm/. Thus th atio of th focs on th two satllits a and b is Satllit has = and m = m, so m = so F = (/3) () F = 4440 N. 3 m, 3 a b a F m = Fb a mb F F = () () = 0,000 N. Similaly, satllit 3 has 3 = (3/) and (c) Th spd of a satllit in a cicula obit is v = ( GM/), so its kintic ngy is atio of th kintic ngy of two satllits a and b is Ka b ma = Kb a mb Satllit 3 has 3 = (3/) and m 3 = m, so K/ K 3 = (3/)(/) = 3/ =. 50. K = mv = GMm/. Thus th 3.. Modl: Modl th sun (S) as a sphical mass and th satllit (s) as a point paticl. Th satllit, having mass m s and vlocity v s, obits th sun with a mass M S in a cicl of adius s. Solv: Th gavitational foc btwn th sun and th satllit povids th ncssay cntiptal acclation fo cicula motion. Nwton s scond law givs S s msvs = s s GM m Bcaus vs = π s/ Ts wh T s is th piod of th satllit, this quation simplifis to 30 S ( π s) 3 S s ( N m /kg )( kg)( s) 9 = s = = s =. s Ts s 4π 4π GM GM T 9 0 m 3.3. Modl: Modl th ath () as a sphical mass and th satllit (s) as a point paticl. Th satllit has a mass is m s and obits th ath with a vlocity v s. Th adius of th cicula obit is dnotd by s and th mass of th ath by M. Solv: Th satllit xpincs a gavitational foc that povids th cntiptal acclation quid fo cicula motion: 4 s s s.. s s s vs (5500 m/s) 7 πs π(. 3 0 m) 4 Ts = = =. 5 0 s = 4. h vs 5500 m/s GM m m v GM ( N m / kg )( kg) 7 = = = =. 3 0 m 3.4. Modl: Modl Mas (m) as a sphical mass and th satllit (s) as a point paticl. Th gosynchonous satllit whos mass is m s and vlocity is v s obits in a cicl of adius s aound Mas. Lt us dnot mass of Mas by M m. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

11 Nwton s Thoy of Gavity 3- Solv: Th gavitational foc btwn th satllit and Mas povids th cntiptal acclation ndd fo cicula motion: Using G = N m /kg, s 7 = m. Thus, altitud piod, o GM mms msvs ms( π s) GM mt s = = s = s s T s s 4 π 3 M = kg, and T s = (4. 8 hs) = (4. 8)(3600) s = 89,80 s, w obtain s m m 7 = =. 7 0 m. Th spd is th obital cicumfnc dividd by th πs π( m/s) vs = = =. 44 km/s T 89,80 s s 3.5. Solv: W a givn M+ M = 50 kg which mans M= 50 kg M. W also hav 6 6 ( N)(0. 0 m) = N MM = = 4798 kg GM M (0. 0 m) N m /kg Thus, (50 kg M ) M = 4798 kg o M (50 kg) M + (4798 kg ) = 0. Solving this quation givs M = kg and kg. Th two masss a 04 and 46 kg W placd th oigin of th coodinat systm on th 0.0 kg mass ( m ) so that th 5.0 kg mass ( m 3) is on th y-axis and th 0.0 kg mass ( m ) is on th x-axis. 7 /3 Solv: (a) Th focs acting on th 0.0 kg mass ( m ) a Thus th foc is F i i i gmm ˆ ( N m /kg )(0.0 kg)(0.0 kg) ˆ 6 ˆ m on m = = =.33 0 N (0.0 m) gmm 3 ˆ ( N m /kg )(0.0 kg)(5.0 kg) ˆ 7 ˆ m3 on m = = =.67 0 N 3 (0.0 m) 6 Fon m = Fm on m + F ˆ -7 ˆ -6 m on m = i j Fon m = F j j j.33 0 N N.34 0 N 3 ( 6 Fon m ) x.33 0 N θ = tan = tan 83 ( F 7 on m ) =.67 0 N y 6 Fon m =. 3 0 N, 83 cw fom th +y-axis Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

12 3- Chapt 3 (b) Th focs acting on th 5.0 kg mass ( m 3) a 7 F = F =.67 0 ˆj N Thus 3 F m on m m on m 3 3 gmm3 m on m3 3 (0.0 m) + (0.0 m) 8 ˆ 8 m on m = ( N)sin φ ( N) cos 3 ( N m /kg )(0.0 kg)(5.0 kg) 8 = = = N F i φˆj 8 0 cm 8 = ( N) ˆ 0 cm i ( N) ˆj.36 cm.36 cm 8 ˆ 8 = ( N) i ( N) ˆj 8ˆ 7 F = F + F = i N.64 0 ˆj N F 7 m on m m on m m on m on m = ( N) + (.64 0 N) =.84 0 N 3 ( F 8 on m ) x N = tan = tan 7.5 ( F 7 on m ) =.64 0 N 3 y θ 3 F on =. 3 0 N, 7. 5 ccw fom th -y-axis Solv: Th angl θ = tan = Th distanc = = ( m) + (0. 00 m) = m. Th focs 0 on th 0.0 kg mass a Mm F ˆ ˆ = G ( sinθi + cos θ j) Mm F = G (sinθiˆ+ cos θ ˆj) 5 Not m= mand =. Thus, th nt foc on th 0.0 kg mass is Mm F F F G ˆj nt = + = cosθ ( N m /kg )(0. 0 kg)(5. 0 kg)cos(4. 04 ) = ˆ j (0. 06 m) 7 = ˆj N Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

13 Nwton s Thoy of Gavity Solv: Th total gavitational potntial ngy is th sum of th potntial ngis du to th intactions of th pais of masss. U = U + U + U 3 3 mm mm m m = G G G With m m m = 0. 0 kg, = 0. 0 kg, = 5. 0 kg, = 0. 0 m, = 0. 0 m, and 3 = m, 7 U = J Assss: Th gavitational potntial ngy is ngativ bcaus th masss attact ach oth. It is a scala, so th a no vcto calculations quid Solv: Th total gavitational potntial ngy is th sum of th potntial ngis du to th intactions of th pais of masss. U = U + U + U 3 3 mm mm m m = G G G Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

14 3-4 Chapt 3 With m = 0. 0 kg, m = m = 0. 0 kg, = = ( m) + (0. 00 m) = m, and 3 = m, U = J Assss: Th gavitational potntial ngy is ngativ bcaus th masss attact ach oth. It is a scala, so th a no vcto calculations quid Bcaus of th gavitational foc of attaction btwn th lad sphs, th cabls will mak an angl of θ with th vtical. Th distanc btwn th sph cnts is thfo going to b lss than m. Th fbody diagam shows th focs acting on th lad sph. Solv: W can s fom th diagam that th distanc btwn th cnts is d =. 000 m Lsinθ. Each sph is in static quilibium, so Nwton s scond law is F = F Tsinθ = 0 Tsinθ = F x gav F = T cosθ mg = 0 T cosθ = mg Dividing ths two quations to liminat th tnsion T yilds y Fgav Gmm/d Gm sinθ tanθ cosθ = = mg = mg = d g W know that d is going to b only vy, vy slightly lss than.00 m. Th vy slight diffnc is not going to b nough to affct th valu of F gav, th gavitational attaction btwn th two masss, so w ll valuat th ight sid of this quation by using.00 m fo d. This givs ( N m /kg )(00 kg) 0 8 tanθ = = θ = ( ) (. 00 m) (9. 8 m/s ) gav This small angl causs th two sphs to mov clos by th distanc btwn thi cnts is d = m. 7 Lsinθ =. 4 0 m = m. Consquntly, 3.3. W placd th oigin of th coodinat systm on th 0 kg sph ( m ). Th sph ( m ) with a mass of 0 kg is 0 cm away on th x-axis. Th point at which th nt gavitational foc is zo must li btwn th masss m Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

15 Nwton s Thoy of Gavity 3-5 and m. This is bcaus on such a point, th gavitational focs du to m and m a in opposit dictions. As th gavitational foc is dictly popotional to th two masss and invsly popotional to th squa of distanc btwn thm, th mass m must b clos to th 0-kg mass. Th small mass m, if placd ith to th lft of m o to th ight of m, will xpinc gavitational focs fom m and m pointing in th sam diction, thus always lading to a nonzo foc. Solv: mm mm 0 0 Fm on m = Fm on m G = G = x (0. 0 x) x (0. 0 x) 0x 8x = 0 x= m and 0. 7 m Th valu x = cm is unphysical in th cunt situation, sinc this point is not btwn m and m. Thus, th mass should b placd.7 cm to th ight of th lag mass. To two significant figus, this is cm Modl: Modl th ath () as a sphical mass and th satllit (s) as a point paticl. Lt h b th hight fom th sufac of th ath wh th acclation du to gavity of th sufac valu ( g sufac). Solv: (a) Sinc galtitud = (0. 0) gsufac, w hav GM GM = (0. 0) ( + h) ( + h) = h=. 6 h= (. 6)( m) = m =. 4 0 m ( g altitud) is 0% (b) Fo a satllit obiting th ath at a hight h abov th sufac of th ath, th gavitational foc btwn th ath and th satllit povids th cntiptal acclation ncssay fo cicula motion. Fo a satllit obiting with vlocity v s, 4 GMms msvs GM ( N m /kg )( kg) = v s = = = 4. 5 km/s ( ) 6 7 ( + h) + h + h ( m m) Modl: Modl th ath as a sphical mass and th objct (o) as a point paticl. Igno ai sistanc. This is an isolatd systm, so mchanical ngy is consvd. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

16 3-6 Chapt 3 Solv: (a) Th consvation of ngy quation K + Ug = K + Ug is v = GM + y GM m mv GM m o o o = mv o + y 4 = ( N m /kg )( kg) 3 0 km/s 6 6 = m m (b) In th flat-ath appoximation, Ug = mgy. Th ngy consvation quation thus bcoms m v m gy m v m gy o+ o = o+ o 5 v = v + g( y y ) = (9. 8 m/s )( m 0 m) = 3. 3 km/s (c) Th pcnt o in th flat-ath calculation is 330 m/s 300 m/s 3. 6% 300 m/s Modl: Consid th objct as a paticl and tak th plant to b a sphical mass. Solv: Consvation of mchanical ngy of th objct givs Mm Mm G = mv G + h Th objct s mass dops out. Solving fo th spd as it hits th gound, + h GMh v = GM = GM = + h ( + h) ( + h) Assss: Compa this to GMh v = gh =, which is th sult if th potntial U = mgh is usd Modl: Modl th ath and th pojctil as sphical masss. Igno ai sistanc. This is an isolatd systm, so mchanical ngy is consvd. A pictoial psntation of th bfo-and-aft vnts is shown. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

17 Nwton s Thoy of Gavity 3-7 Solv: Aft using v = 0 m/s, th ngy consvation quation K + U = K+ U is Th pojctil mass cancls. Solving fo h, w find GM m GM m 0 J = h p p mv p + v 5 h= = 4. 0 m GM Modl: Modl th ath as a sphical mass and th mtooids as point masss. Solv: (a) Th ngy consvation quation K + U = K+ Ugivs GM m mv = mv v = v + GM m m (000 m/s) ( N m /kg )( kg) m m 4 =. 0 m/s = km/s GM m Th spd dos not dpnd on th mtooid s mass. / (b) This pat diffs in that = km = m. Th shap of th mtooid s tajctoy is not impotant fo using ngy consvation. Thus GM m GM m mv = mv v = v + GM km m km m 3 = m/s = 8. 9 km/s 7 / / Modl: Modl th two stas as sphical masss, and th comt as a point mass. This is an isolatd systm, so mchanical ngy is consvd. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

18 3-8 Chapt 3 In th initial stat, th comt is fa away fom th two stas and thus it has nith kintic ngy no potntial ngy. In th final stat, as th comt passs though th midpoint conncting th two stas, it posssss both kintic ngy and potntial ngy. Solv: Th consvation of ngy quation Kf + Uf = Ki + Ui givs mv f f f 30 f f. GMm GMm = 0 J + 0 J 4GM 4( N m /kg )( kg) v = = = 3,600 m/s = 33 km/s m Assss: Not that th final vlocity of 33 km/s dos not dpnd on th mass of th comt Modl: Modl th astoid as a sphical mass and youslf as a point mass. This is an isolatd systm, so mchanical ngy is consvd. Th adius of th astoid is M a and its mass is a. Solv: Th consvation of ngy quation Kf + Uf = Ki + Ui fo th astoid givs GM m GMam mv = a f mvi a + Th minimum spd fo scap is th on that will caus you to stop only whn th spaation btwn you and th astoid bcoms vy lag. Noting that vf 0 m/s as, w hav a i a v 4 GM ( N m /kg )(. 0 0 kg) = = v 3 i =. 58 m/s. 0 0 m That is, you nd a spd of.58 m/s to scap fom th astoid. W can now calculat you jumping spd on th ath. Th consvation of ngy quation givs 0 J mvi mg(0 50 m) vi (9 8 m/s )(0 50 m) 3 3 m/s =. =.. =. This mans you can scap fom th astoid Modl: Th pojctil is a paticl. Th ath and moon a sphical masss. Solv: Th pojctil is attactd to both th moon and ath. Its final vlocity and potntial ngy a zo. Sinc th pojctil is fid fom th fa sid of th moon, its initial distanc fom th cnt of th ath is th ath-moon a Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

19 Nwton s Thoy of Gavity 3-9 distanc -m plus th adius of th moon m. Lt th pojctil hav mass m. Th consvation of mchanical ngy quation is Fom Tabl 3., givs scap v =. 78 km/s. Mm M m mv -G - G = 0 J + 0 J ( ) m scap -m + m m M M v G m scap = + -m + m m m m = m, = m, M = kg, and Assss: Th scap vlocity dos not dpnd on th mass of th objct which is tying to scap. m M = kg, which Modl: Th two astoids mak an isolatd systm, so mchanical ngy is consvd. W will also us th law of consvation of momntum fo ou systm. Solv: Th consvation of momntum quation pfx = pix is M( v ) + M( v ) = 0 kg m/s ( v ) = ( v ) fx fx fx fx Th quation fo mchanical ngy consvation Kf + Uf = Ki + Ui is GM ( )( M) GM ( )( M) 0. 8GM M( vfx) + ( M)( vfx) = [( vfx) ] + ( vfx) = 0 GM ( vfx) = ( vfx) = ( vfx) =. 03 / Th havi astoid has a spd of 0. 56( GM/ ) and th light on a spd of. 03( GM/ ) Modl: Gavity is a consvativ foc, so w can us consvation of ngy. GM / Th plants will b pulld togth by gavity and ach will hav spd v as thy cash and th spaation btwn thi cnts will b. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

20 3-0 Chapt 3 Solv: Th plants bgin with only gavitational potntial ngy. Whn thy cash, thy hav both potntial and kintic ngy. Thus, Bcaus th plant is Jupit-siz, w ll us GMM K U Mv Mv K U + = + = + = 0 J v = GM 7 M = M Jupit =. 9 0 kg and GMM = =. 4 0 m. Insting Jupit ths valus into th xpssion abov givs th cash spd of ach plant as v = m/s. Assss: Not that th foc is not constant, bcaus it vais with distanc, so th motion is not constant acclation motion. Th fomulas fom constant-acclation kinmatics do not wok fo poblms such as this Modl: Modl th distant plant (P) as a sphical mass. Solv: Th acclation at th sufac of th plant and at th altitud h a GM M GM g g g G ( + h) P P P sufac = and altitud = sufac = ( + h) = + h= h= ( ) = That is, th staship is obiting at an altitud of Modl: Th stas a sphical masss. 4 8 Solv: Th stats a idntical, so thi final spds v f a th sam. Thy collid whn thi cnts a apat. Fom ngy consvation, M M 3 G = 3 Mv 9 f 3 G (5 0 0 m). vf = GM m v = m/s f Modl: Th stas a sphical masss. Thy ach otat about th systm s cnt of mass. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

21 Nwton s Thoy of Gavity 3- Solv: (a) Th stas otat about th systm s cnt of mass with th sam piod: T = T = T. W can locat th cnt of mass by ltting th oigin b at th small-mass sta. Thn (. 0 0 kg)(0 m) + ( kg)(. 0 0 m) = cm = =. 5 0 m kg kg Mass m undgos unifom cicula motion with adius = m du to th gavitational foc of mass m at distanc =. 0 0 m. Th gavitational foc is sponsibl fo th cntiptal acclation, so Gmm mv m p 4p m Fgav = = ma cntiptal = = = T T / / p 8 30 ( N m /kg )(.0 0 kg) 4p 4 (0.5 0 m)(0.5 0 m) T= = = s = 4 yas Gm (b) Th spd of ach sta is v= ( π )/ T. Thus π π. 8 ( 5 0 m) v = = =. 3 km/s T s π π. 8 (0 5 0 m) v = = = 4. km/s T s Modl: Modl th moon (m) as a sphical mass and th land (l) as a paticl. This is an isolatd systm, so mchanical ngy is consvd. Th initial position of th luna land (mass = m ) is at a distanc = m + 50 km fom th cnt of th moon. Th final position of th luna land is th obit whos distanc fom th cnt of th moon is = km. m Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

22 3- Chapt 3 Solv: Th xtnal wok don by th thusts is xt = mch = g W E U wh w usd E = U fo a cicula obit. Th chang in potntial ngy is fom th initial obit at mch g i = m + 50 km to th final obit f = m km. Thus W xt GM mm GM mm GM mm = = f i i f ( N m /kg )( kg)(4000 kg) = m m 8 = J Modl: Modl th ath () as a sphical mass and th spac shuttl (s) as a point paticl. This is an isolatd systm, so th mchanical ngy is consvd. Th spac shuttl (mass = m ) is at a distanc of = + 50 km. s Solv: Th xtnal wok don by th thusts is xt = mch = g W E U wh w usd E = U fo a cicula obit. Th chang in potntial ngy is fom th initial obit at mch g = + 50 km to th final obit f = + 60 km. Thus W xt GMm GMm GMm = = f i i f 4 ( N m /kg )( kg)(75,000 kg) = m m =. 0 J This much ngy must b supplid by buning th on-boad ful Modl: Assum a sphical astoid and a point mass modl fo th satllit. This is an isolatd systm, so mchanical ngy is consvd. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

23 Nwton s Thoy of Gavity 3-3 Th obital adius of th satllit is Solv: (a) Th spd of a satllit in a cicula obit is = a + h= 8,800 m + 5,000 m = 3,800 m 6 GM ( N m /kg )(. 0 0 kg) v = = = 7. 0 m/s 3,800 m (b) Th minimum launch spd fo scap ( v i ) will caus th satllit to stop asymptotically ( v f = 0 m/s) as. Using th ngy consvation quation K + U = K + U, w gt f GM m GM m GM mv mv v v a s a s a s f = s i 0 J 0 J = scap f a a scap a / 6 GMa ( N m /kg )(. 0 0 kg) = = = m/s 8800 m Modl: Modl th moon as a sphical mass and th satllit as a point mass. Th otational piod of th satllit is th sam as th otational piod of th moon aound its own axis. This tim happns to b 7.3 days. Solv: Th gavitational foc btwn th moon and th satllit povids th cntiptal acclation ncssay fo cicula motion aound th moon. Thfo, Sinc = m + h, thn GM mm ω m m T π = = 3 GM mt ( N m /kg )( kg)( s) = = 4π 4π 7 = m h= m = m m = m Modl: Modl th ath as a sphical mass and th satllit as a point mass. Th satllit is dictly ov a point on th quato onc vy two days. Thus, s = s. T = T = 4 Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

24 3-4 Chapt 3 Solv: Fom Kpl s laws applid to a cicula obit: 4π GMT ( N m /kg )( kg)( ) = = 4π 4π T = GM 7 = m Assss: Th adius of th obit is lag than th gosynchonous obit Plas f to Figu P3.50. Solv: Th gavitational foc on on of th masss is du to th sta and th oth plant. Thus Mm Gmm mv m π GM Gm 4π G + = = + = 4 ( ) T T 3 G m 4π 4π M + = T 4 = T G ( M + m/4) 3.5. Solv: (a) Taking th logaithm of both sids of v But p q = Cu givs p q q logc [log( v ) = plog v] = [log( Cu ) = logc + qlog u] logv = logu + p p x= logu and y = log v, so x and y a latd by q logc y = x+ p p (b) Th pvious sult shows th is a lina lationship btwn x and y, so th is a lina lationship btwn log u and log v. Th gaph of a lina lationship is a staight lin, so th gaph of log v-vsus-log u will b a staight lin. (c) Th slop of th staight lin psntd by th quation y= ( qp / ) x+ log Cp / is qp. / Thus, th slop of th log v- vsus-log u gaph will b qp. / (d) Th pdictd y-intcpt of th gaph is log C/p, and th xpimntally dtmind valu is Equating ths, w can solv fo M. Bcaus th plants all obit th sun, th mass w a finding is M = M. 4π 4π logc = log = = 0 = GMsun GMsun 0 M sun Th tabulatd valu, to th significant figus, is wigh th sun! 4π = (0 ) = kg G 3.5. Solv: (a) Dividing th cicumfnc of th obit by th piod givs (b) Using th fomula fo th acclation at th sufac, g sum 30 / M = kg. W hav usd th obits of th plants to πs π(. 0 0 m) 4 v = = = m/s T. 0 s 30 GMs.. sufac 4 s (. 0 0 m) 4 ( N m /kg )( 99 0 kg) = = =. 3 0 m/s sun Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

25 Nwton s Thoy of Gavity 3-5 (c) Th mass of an objct on th ath will b th sam as its mass on th sta. Th gavitational foc is (d) Th adius of th obit of th satllit is T G sta sufac ( F ) = mg =. 3 0 N = 0 m m =. 0 m. Th piod is π 4 π (. 0 m) 4 = = T =. GM 30 s ( N m /kg )( kg) This mans th a 589 volutions p scond o obits p minut. () Applying Equation 3.5 fo a gosynchonous obit, s 30 3 GMs ( N m /kg )( kg)(. 0 s) 6 = = =. T 4π 4π 5 0 m Modl: Assum th sola systm is a point paticl. Solv: (a) Th adius of th obit of th sola systm in th galaxy is 5,000 light yas. This mans = 5,000 light yas = 500( m/s)(365 d/y)(4 h/d)(3600 s/h)( y) = m π π( m) 5 8 T = = = s =. 0 yas v m/s yas (b) Th numb of obits = = 4 obits yas (c) Applying Nwton s scond law yilds GM m mv v G g cnt ss ss ( m/s) ( m) 4 = M g cnt = = =. (d) Th numb of stas in th cnt of th galaxy is kg = kg Modl: Assum th th stas a sphical masss N m /kg kg Th stas otat about th cnt of mass, which is th cnt of th tiangl and qual distanc fom all th stas. Th gavitational foc btwn any two stas is th sam. On a givn sta th two focs fom th oth stas mak an angl of 60. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

26 3-6 Chapt 3 Solv: Th valu of can b found as follows: Th gavitational foc btwn any two stas is L/ L. 0 0 m = cos(30 ) = = = m cos(30 ) cos(30 ) 30 GM ( N m /kg )( kg) 6 Fg = = = N L (. 0 0 m) Th componnt of this foc towad th cnt is c g 6 6 F = F cos(30 ) = ( N)cos(30 ) =. 9 0 N Th nt foc on a sta towad th cnt is twic this foc, and that foc quals 6 π. 9 0 N = Mω = M T 30 Mω. This mans 4π M 4 π ( kg)( m) 8 T = = = s = 0 yas N N Modl: Angula momntum is consvd fo a paticl following a tajctoy of any shap. Fo a paticl in an lliptical obit, th paticl s angula momntum is L = mvt = mvsin β, wh v is th vlocity tangnt to th tajctoy and β is th angl btwn and v. Solv: At th distanc of closst appoach ( min ) and also at th most distant point, β = 90. Sinc th is no tangntial foc (th only foc bing th adial foc) th is no toqu, so angula momntum must b consvd: min max m v = m v Pluto min Pluto max 3 v = v ( / ) = (6. 0 m/s)(( m)/( m) = 3. 7 km/s Modl: Angula momntum is consvd fo a paticl following a tajctoy of any shap. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

27 Nwton s Thoy of Gavity 3-7 Fo a paticl in an lliptical obit, th paticl s angula momntum is L = mvt = mvsin β, wh v is th vlocity tangnt to th tajctoy, and β is th angl btwn and v. Solv: At th distanc of closst appoach ( min ) and also at th most distant point, β = 90. Sinc th is no tangntial foc (th only foc bing th adial foc) th is no toqu, so angula momntum must b consvd: m v = m v Mcuy min 0 Mcuy max ( m)(38. 8 km/s) 0 min = maxv/ v = = m km/s Modl: Fo th sun + comt systm, th mchanical ngy is consvd. Solv: Th consvation of ngy quation Kf + Uf = Ki + Ui givs Using G = Nm /kg, w gt v = km/s. GM M Mv GM M s c s c c = Mv c 30 M s = kg, 0 = m, = m, and v = km/s, Modl: Modl th plant (p) as a sphical mass and th spacship (s) as a point mass. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

28 3-8 Chapt 3 Solv: (a) Fo th cicula motion of th spacship aound th plant, GM m GM p s mv0 p = v 0 = Immdiatly aft th ockts w fid v = v 0 / and = 0. Thfo, v = (b) Th spacship s maximum distanc is max = 0. Its minimum distanc occus at th oth nd of th llips. Th ngy at th fiing point is qual to th ngy at th oth nd of th lliptical tajctoy. That is, GM GM m mv p s p s s = mv s Sinc th angula momntum at ths two nds is consvd, w hav 0 p GM m mv = mv v = v ( / ) With this xpssion fo v, th ngy quation simplifis to Using = 0 and v GM p = v / =, 0 0 GM v v GM p p = ( / ) GM GM GM GM p p p 0 p 0 = = = 0 + = Th solutions a = 0 (th initial distanc) and = 0 /7. Thus th minimum distanc is min = 0 / Solv: (a) At what distanc fom th cnt of Satun is th acclation du to gavity th sam as on th sufac of th ath? Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

29 Nwton s Thoy of Gavity 3-9 (b) (c) Th distanc is m. This is. 06Satun Solv: (a) A 000 kg satllit obits th ath with a spd of 997 m/s. What is th adius of th obit? (b) (c) Th adius of th obit is payload 4 GM E ( N m /kg )( kg) 8 = = = m ( v ) (997 m/s) 3.6. Solv: (a) A 00 kg objct is lasd fom st at an altitud abov th moon qual to th moon s adius. At what spd dos it impact th moon s sufac? (b) Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

30 3-30 Chapt 3 (c) Th spd is v = 680 m/s Solv: (a) Kpl s thid law fo cicula obits is 4π 3 4π T = T = GM GM 4π 3 Ltting a =, th fist satllit obys T = a. Fo th scond satllit, which obits th sam mass, GM 3 3 T + T = a( + ) = a + n Sinc, w can us th appoximation ( ± x) ± nx. Thus 3 3 T + T a + Subtacting th quation 3 T a = fo th fist satllit fom this, 3 3 T = a Dividing this by th quation fo th fist satllit, T 3 = T (b) Th satllits obit th ath. Th factional diffnc in thi piods is T 3 km = = T 6700 km Aft = 4467 piods thy will mt again. Fo th inn satllit, 4π 6 T = ( m) 4 G( kg) = 5456 s =. 5 hs So th satllits will mt again in hs = 6770 hs = 8 days. Assss: A communications satllit has an obital piod of aound.5 h. Th supising lngth of tim btwn th two satllits mting is du to th small diffncs in thi piods Modl: Modl th ath and sun as sphical masss, th satllit as a point mass. Assum th satllit s distanc fom th ath is vy small compad to th ath s (and satllit s) distanc to th sun Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

31 Nwton s Thoy of Gavity 3-3 Solv: Th nt foc on th satllit is th sum of th gavitational foc towad th sun and th gavitational foc towad th ath. This nt foc is sponsibl fo cicula motion aound th sun. W want to chos th distanc d to mak th piod T match th piod T with which th ath obits th sun. Th ath s obital piod is givn by 3 (4 π / s) T = GM. Thus Using s p 4p s cntiptal T 3 GM m GM m mv m GM Fnt = = ma = = = m = m d T = d and cancling th Gm tm givs M M = M ( d ) ( d) d s s 3 This quation can t b solvd xactly, but w can mak us of th fact that d to us th binomial appoximation. Facto th out of th xpssions d to gt M M M = ( d / ) d d s s ( / ) If w think of d / = x, w can simplify th fist tm by using ( ± x) ± nx. H n =, so w gt Thus d = [ M /(3 M )] = m. Assss: M M M M M /3 9 s d / = 0. 00, so ou assumption that s s s [ ( ) d / ] = ( d / ) = 3 d d d / is justifid. n Modl: Th ath and sun a sphical masss. Th ath is in a cicula obit aound th sun. Th pojctil is a paticl. Th ffct of th ath s otation on th pojctil s vlocity is ignod. Th ath and sun a so massiv compad to th ockt that thy a unaffctd by th ockt s motion (no coil). Th ockt scaps fom th influnc of th ath s gavitation in a distanc small in compaison with th distanc fom th ath to th sun, so that th chang in sola potntial of th ockt as it scaps th ath is ngligibl. Solv: To scap th sun s gavitational pull at th ath s distanc fom th sun, a pojctil must hav spd v Th ngy consvation quation fo th pojctil is scap. mv Mm - = 0 J s scap G -s 30 Ms.. scap -s ( m) - ( N m /kg )( 99 0 kg) 4 v = G = = m/s Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

32 3-3 Chapt 3 Th ath s spd in its obit is found by considing Equation 3.: GMs 4 v = = m/s -s Th pojctil must also scap fom th ath s gavitational pull. Fa fom th ath, th pojctil s total spd is th sum of its spd lativ to th ath fa fom th influnc of ath s gavity, v p, and th ath s spd. v = v + v scap p vp = vscap v = m/s m/s =. 3 0 m/s. To obtain th spd v p fa fom th ath, th pojctil must b launchd fom th ath with spd v launch. Using ngy consvation, Mm M mv G mv v G v launch = p launch = + p 4 4 ( N m /kg )( kg) = + (. 3 0 m/s) 6 ( m) 4 = m/s Assss: Th pojctil must attain a spd of 6.6 km/s if launchd in th diction of ath s motion. This is about a facto of 7 tims lss than what is quid fom st. Not that th spd of th ath in its obit about th sun v diffs fom th scap spd fom th sun vscap by a facto of Modl: Modl th 400 kg satllit and th 00 kg satllit as point masss and modl th ath as a sphical mass. Momntum is consvd duing th inlastic collision of th two satllits. 6 6 Solv: Fo th givn obit, 0 = + 0 m = m. Th spd of a satllit in this obit is GM v0 = = 7357 m/s 0 Th two satllits collid, stick togth, and mov with vlocity v. Th quation fo momntum consvation fo th pfctly inlastic collision givs (400 kg + 00 kg) v = (400 kg)(7357 m/s) (00 kg)(7357 m/s) v = 444 m/s Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

33 Nwton s Thoy of Gavity 3-33 Th nw satllit s adius immdiatly aft th collision is still = 0 = m. Now it is moving in an lliptical obit. W nd to dtmin if th minimum distanc is lag o small than th ath s adius = m. 6 Th combind satllits will continu moving in an lliptical obit. Th momntum of th combind satllit is L = mv sin β (s Equation 3.6) and is consvd in a tajctoy of any shap. Th angl β is 90 whn v = 444 m/s and whn th satllit is at its closst appoach to th ath. Fom consvation of angula momntum, w hav 0 0 v = v = (444 m/s) = m /s m /s = v Using th consvation of ngy quation at positions and, GM(500 kg) GM(500 kg) (500 kg)(444 m/s) = (500 kg) v m Using th abov xpssion fo, w can simplify th ngy quation to A vlocity of 0,07 m/s fo v yilds v 4 7 ( m/s) v + ( m /s ) = 0 m /s v = 0,07 m/s and 444 m/s m /s 6 = =. 6 0 m 0,07 m/s 6 Sinc 6 < = m, th combind mass of th two satllits will cash into th ath Modl: Plant Physics is a sphical mass. Th cuis ship is in a cicula obit. Solv: (a) At th sufac, th f-fall acclation is 0 g = GM/. Fom kinmatics, y = y + v t g( t) 0 m = 0 m + ( m/s)(. 5 s) g(. 5 s) f i g =. 0 m/s Th piod of th cuis ship s obit is = 3,800 s. Fo th cicula obit of th cuis ship, Th mass is thus M= ( / Gg ) = kg. (b) Fom pat (a), 4π 3 T T = ( ) = = GM 3π GM g (. 0 m/s )(3,800 s) 6 = = m. 3π 6 =. 3 0 m Modl: Th moon is a sphical mass. Th moon land is oiginally in a cicula obit. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

34 3-34 Chapt 3 Solv: Engy and momntum a consvd btwn points and in th lliptical obit. Also, at both points GM β = 90, so L = mvsin β = mv. Lt h = 000 km. Th oiginal spd of th land is m v = m/s + h =. v m + h Consvation of angula momntum quis m v = mv = =. Th ngy consvation qua- v tion givs m + h Substituting v = v, m m m + M m Mmm mv G = mv G h Mm m + h Mm = m h m m v G v G + m h v + = GM m m m + h m 6 = m = m m 39 0 m /s ( ) =. m m + h [( m + h)/ m] ( m + h) m + h v GM h GM v = 80 m/s Th factional chang in spd quid to just gaz th moon at point is 338 m/s 80 m/s =. 8% 338 m/s Assss: A duction in spd by almost % is asonabl Modl: Modl th ath as a sphical mass and th satllit as a point mass. This is an isolatd systm, so mchanical ngy is consvd. Also, th angula momntum of th satllit is consvd. Plas f to Figu CP3.68. Solv: (a) Angula momntum is L = mvsin β. Th angl β = 90 at points and, so consvation of angula momntum quis Th ngy consvation quation is m v m v v v = = GMm GMm mv ( ) = mv ( ) ( v ) ( v ) = GM m m m Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

35 Nwton s Thoy of Gavity 3-35 Using th angula momntum sult fo v givs (b) Fo th cicula obit, Fo th lliptical obit, = = ( v ) ( v ) GM ( v ) GM GM ( / ) GM ( / ) v = and v = v = GM ( N m /kg )( kg) v = = = 7730 m/s ( m 3 0 m) = km = m m = m = + 35,900 km = m m = m GM ( / ) v = v = 0, 60 m/s 0 (c) Fom th wok-kintic ngy thom, W = K = mv mv =. 7 0 J GM ( (d) / ) v = + Using th sam valus of and as in (b), v = 600 m/s. Fo th cicula obit, 9 () W = mv mv = J 0 GM v = = 3070 m/s (f) Th total wok don is. 5 0 J. This is th sam as in Exampl 3.6, but h w v land how th wok has to b dividd btwn th two buns Modl: Th od is thin and unifom. Solv: (a) Th od is not sphical so must b dividd into thin sctions ach d wid and having mass dm. Sinc th od is unifom, dm d M = dm = d M L L Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

36 3-36 Chapt 3 Th width of th od is small nough so that all of dm is distanc away fom m. Th gavitational potntial ngy of dm and m is m dm m( M/L) d du = G = G Th total potntial ngy of th od and mass m is found by adding th contibutions du fom vy point along th od in an intgal: Not is incasing with th limits chosn as thy a. (b) Th foc on m whn at x is L x+ GMm d GMm x + L/ U = du = ln L = L x L/ L x du GMm d L L GMm F = = ln x ln x dx L dx + = L x + L/ x L/ 4 L = GMm, x 4x L Assss: Th diction of th foc is towad th x diction, as xpctd. Th foc magnitud appoachs as th mass m appoachs th nd of th od, but gos to zo lik away th od looks lik a point mass. x as x gts lag. This is xpctd sinc fom fa Modl: Th ing is unifom and is so thin that vy point on it may b considd to b th sam distanc fom m. Solv: (a) W must dtmin th gavitational potntial (du) btwn m and an abitay pat of th ing dm, thn add using an intgal all th contibutions to U. Sinc th ing is unifom, Th distanc fom m to dm is Th total gavitational potntial is dm dl M = dm = dl M π π = x +. Th gavitational potntial btwn m and dm is m dm mm dl du = G = G π x + GmM U = du = dl π x + ing Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

37 Nwton s Thoy of Gavity 3-37 Not that x and do not chang fo any location of dm. Th intgal is just th lngth of th ing. ing dl = π Thus GmM U = x + (b) Th foc on m whn at x is du d 3 Fx = = GmM [( x + ) ] = GmM ( x + ) ( x) dx dx x = GmM 3 ( x + ) Thus th magnitud of th foc is x F = GmM ( x + ) Assss: Th foc is zo at th cnt of th ing. Elswh its diction is towad th oigin. As x gts lag, th foc dcass lik x This is xpctd sinc fom fa away th ing looks lik a point mass. 3. Copyight 03 Pason Education, Inc. All ights svd. This matial is potctd und all copyight laws as thy cuntly xist.

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