10 m, so the distance from the Sun to the Moon during a solar eclipse is. The mass of the Sun, Earth, and Moon are = =
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1 Chpte 1 nivesl Gvittion 11 *P1. () The un-th distnce is 1.4 nd the th-moon 8 distnce is.84, so the distnce fom the un to the Moon duing sol eclipse is = 1.4 The mss of the un, th, nd Moon e nd M = kg M = M kg M = kg We hve F M 11 0 (.7 10 )( )( ) ( ) Gm1m = = = N (b) F M 11 4 (.7 10 N m kg )( )( ) 8 ( ) = = N (c) F (.7 10 N m kg )( )( ) 11 ( ) = = N (d) The foce exeted by the un on the Moon is much stonge thn the foce of the th on the Moon. In sense, the Moon obits the un moe thn it obits the th. The Moon s pth is eveywhee concve towd the un. Only by subtcting out the sol obitl motion of the th-moon system do we see the Moon obiting the cente of mss of this system. P1. MG.80 m s = = = 0.1 m s towd the th. ( R ) P1.11 If ( πr ) Gρ 4 / 4 g= = = πgρr R R gm 1 4 πgρmrm/ = = g 4 πgρ R / ρm M R then = = ( 4) ρ g 1 = g R M
2 P1.14 By Keple s Thid Lw, T = k ( = semi-mjo xis) Fo ny object obiting the un, with T in yes nd in A.., k = Theefoe, fo Comet Hlley y = ( 75.) ( 1.00) The fthest distnce the comet gets fom the un is y = ( 75.) = 5. A.. (out ound the obit of Pluto). FIG. P1.14 P1.15 4π T = (Keple s thid lw with m M << ) 8 4 ( 4. ) (.7 10 N m kg )( s) 4π π M GT 11 = = = kg (ppoximtely 1 th msses) Gm P1.1 ˆ Gm ˆ Gm g= i+ j+ ( cos45.0 ˆ i+ sin 45.0 ĵ ) l l l 1 so g= + ˆ ˆ ( i+ ) l 1 j o y m l m l Gm 1 g = + towd the opposite cone l O m x FIG. P1.1 P1. () 11 0 (.7 10 N m kg ) 100( kg)( 10 kg ) ( m ) m F = = = 4 m m (b) Δ F = font bck ΔF Δ g = = m Δ g = ( bck font ) font bck N (.7 10 ) 100( ) ( 1.01 ) ( 1.00 ) 4 4 ( 1.00 ) ( 1.01 ) Δ g = N kg FIG. P1.
3 P1.4 () m 11 4 (.7 10 N m kg )( kg )( 100 kg ) = = = (b), (c) Plnet nd stellite exet foces of equl mgnitude on ech othe, diected downwd on the stellite nd upwd on the plnet (.7 10 N m kg )( kg)( 100 kg) ( 8.7 ) m F = = = 5 N P1.5 () (b) (c) 0 ( ) ( ) M kg ρ = = = kg m 4 π 4π (.7 10 N m kg )( kg ) (.7 ) g = = =.7 m ( N m kg )( kg)( 1.00 kg) 1 g = = = P1.1 mvi + m = mv f f i v f = = 1. *P1. tot m = M J v + 0 = v i f R o vf = v1 R nd v f = v m kg 1 1 Δ = = i f Δ = = Both in the oiginl obit nd in the finl obit, the totl enegy is negtive, with n bsolute vlue equl to the positive kinetic enegy. The potentil enegy is negtive nd twice s lge s the totl enegy. As the stellite is lifted fom the lowe to the highe obit, the gvittionl enegy inceses, the kinetic enegy deceses, nd the totl R 1
4 enegy inceses. The vlue of ech becomes close to zeo. Numeiclly, the gvittionl enegy inceses by 8 MJ, the kinetic enegy deceses by 4 MJ, nd the totl enegy inceses by 4 MJ. P1. To obtin the obitl velocity, we use mm G mv F = = R R MG o v = R We cn obtin the escpe velocity 1 mmg fom mvesc = R MG o vesc = = v R *P1.4 Gvittionl sceening does not exist. The pesence of the stellite hs no effect on the foce the plnet exets on the ocket. The ocket is in potentil well t Gnymede s sufce with enegy m (.4 ) 11 Gm1m.7 10 N m kg 1 = = =.78 m s 1 The potentil well fom Jupite t the distnce of Gnymede is kg m ( ) 11 7 Gm1m.7 10 N m kg = = = 1.18 m s 8 To escpe fom both equies 1 8 mv esc =+ ( ) m m s 8 v = 1. = 15. km s esc P1.41 Let m epesent the mss of the spcecft, the dius of the th s obit, nd x the distnce fom th to the spcecft. The un exets on the spcecft dil inwd foce of kg s F = m s ( x) while the th exets on it dil outwd foce of F = m x The net foce on the spcecft must poduce the coect centipetl cceletion fo it to hve n obitl peiod of ye.
5 Thus, ( x) ( ) ( ) ( x) m m mv m F = = = F x x x T π 4 which educes to π x = x x T (1) Cleed of fctions, this eqution would contin powes of x nging fom the fifth to the zeoth. We do not solve it lgebiclly. We my test the ssetion tht x is between 1.47 nd 1.48 by substituting both of these s til solutions, long with the following dt: M = kg, M = kg, = 1.4, nd 7 T = y = s. With x = 1.47 substituted into eqution (1), we obtin o m s With x = 1.48 substituted into the sme eqution, the esult is o m s ince the fist til solution mkes the left-hnd side of eqution (1) slightly less thn the ight hnd side, nd the second til solution does the opposite, the tue solution is detemined s between the til vlues. To thee-digit pecision, it is As n eqution of fifth degee, eqution (1) hs five oots. The un-th system hs five Lgnge points, ll evolving ound the un synchonously with the th. The OHO nd AC stellites e t one. Anothe is beyond the f side of the un. Anothe is beyond the night side of the th. Two moe e on the th s obit, hed of the plnet nd behind it by 0. Plns e unde wy to gin pespective on the un by plcing spcecft t one of these two co-obitl Lgnge points. The Geek nd Tojn steoids e t the co-obitl Lgnge points of the Jupite-un system. P1.47 Fom the wlk, π = 5000 m. Thus, the dius of the plnet is m.8 10 m = π = Fom the dop: 1 1 Δ y = gt = g(. s) = 1.40 m so, (. s) 1.40 m.8 10 MG g = = m s = M = kg
6 P1.51 () At infinite seption = 0 nd t est K = 0. ince enegy of the two-plnet system is conseved we hve, 1 1 Gm1m 0 = mv mv d (1) The initil momentum of the system is zeo nd momentum is conseved. Theefoe, 0 = mv 1 1 mv () Combine equtions (1) nd (): G v 1 = m dm m 1 ( + ) G v = m 1 dm m 1 ( + ) nd Reltive velocity v = v v = 1 ( + m) Gm 1 d (b) ubstitute given numeicl vlues into the eqution found fo nd v in pt () to find v1 4 v 1 = 1.0 nd v =.58 Theefoe, 1 K1 = m1v1 = K = mv =.7 nd P1.5 () Fom the dt bout peigee, the enegy of the stellite-th system is p p ( )( )( 1.0) 1 m 1 = mv = ( 1.0 )( ) (b) (c) o 7 = kg m s L = mv sin θ = mv sin 0.0 = 1.0 kg p p = ince both the enegy of the stellite-th system nd the ngul momentum of the th e conseved, t pogee we must hve 1 m mv =
7 nd mv sin 0.0 = L Thus, ( )( )( 1.0) 1 7 ( 1.0 ) v = kg v =.4 10 kg m s nd 10 olving simultneously, ( )( )( 1.0)( 1.0) 1 v ( 1.0 ) v = 10 which educes to so v 11 04v = 0 v ± = This gives v = 8 0 m s o m s. The smlle nswe efes to the velocity t the pogee while the lge efes to peigee. Thus, kg m s L = = = mv ( 1.0 kg)( 5.58 ) (d) The mjo xis is = p +, so the semi-mjo xis is (e) 4π T = = 1 7 ( ) = + = m 4π ( 8. ) 11 4 (.7 10 N m kg )( kg ) T = 8 00 s = 14 min *P1.5 The gvittionl foces the pticles exet on ech othe e in the x diection. They do not ffect the velocity of the cente of mss. negy is conseved fo the pi of pticles in efeence fme costing long with thei cente of mss, nd momentum consevtion mens tht the identicl pticles move towd ech othe with equl speeds in this fme: gi + K i + K i = gf + K f + K f
8 Gm m Gm m + 0 = + mv i f (.7 10 N m / kg )(1000 kg) (.7 10 N m / kg )(1000 kg) 1 = + (1000 kg) v 0 m m 1/ kg = v = mv / s Then thei vecto velocities e ( ) î m/s nd ( ) î m/s. fo the tiling pticle nd the leding pticle, espectively
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