Note: to obtain the numerical results given in this chapter, the following numerical values of certain physical quantities have been used; kg.1097.5andsm80.9 ,kgmN10673.6 24 E 22211   mgG Use of other tabulated values for these quantities may result in an answer that differs in the third significant figure. 12.1: The ratio will be the product of the ratio of the mass of the sun to the mass of the earth and the square of the ratio of the earth-moon radius to the sun-moon radius. Using the earth-sun radius as an average for the sun-moon radius, the ratio of the forces is .18.2 kg105.97 kg101.99 m101.50 m1084.3 24 30 2 11 8                      12.2: Use of Eq. (12.1) gives N.1067.1 m)106.38m10(7.8 kg)2150)(kg1097.5( )kgmN10673.6( 4 265 24 2211 2 21 g      r mm GF The ratio of this force to the satellite’s weight at the surface of the earth is %.7979.0 )sm80.9)(kg2150( )N1067.1( 2 4   (This numerical result requires keeping one extra significant figure in the intermediate calculation.) The ratio, which is independent of the satellite mass, can be obtained directly as , 2 E 2 E 2 E        r R gr Gm mg rmGm yielding the same result. 12.3: . )( ))(( 12 2 12 21 2 12 21 F r mm G nr nmnm G  12.4: The separation of the centers of the spheres is 2R, so the magnitude of the gravitational attraction is .4)2( 2222 RGMRGM  12.5: a) Denoting the earth-sun separation as R and the distance from the earth as x, the distance for which the forces balance is obtained from , )( 2 E 2 S x mGM xR mGM   which is solved for m.1059.2 1 8 E S    M M R x b) The ship could not be at equilibrium for long, in that the point where the forces balance is moving in a circle, and to move in that circle requires some force. The spaceship could continue toward the sun with a good navigator on board. 12.6: a) Taking force components to be positive to the right, use of Eq. (12.1) twice gives                       11 22 2211 1032.2 , m0.600 kg0.10 m0.400 kg00.5 kg100.0kgm10673.6 g F with the minus sign indicating a net force to the left. b) No, the force found in part (a) is the net force due to the other two spheres. 12.7:         .104.2 m1078.3 kg1035.7kg70 kgm.10673.6 3 28 22 2211      12.8:     4 2 1003.6 500,23 000,333   12.9: Denote the earth-sun separation as 1 r and the earth-moon separation as 2 r . a)   ,1030.6 )( 20 2 2 E 2 21 S M          r m rr m Gm toward the sun. b)The earth-moon distance is sufficiently small compared to the earth- sun distance (r 2 << r 2 ) that the vector from the earth to the moon can be taken to be perpendicular to the vector from the sun to the moon. The components of the gravitational force are then ,1099.1,1034.4 20 2 2 EM 20 2 1 SM  r mGm r mGm and so the force has magnitude  20 1077.4 and is directed 6.24 from the direction toward the sun. c)     ,1037.2 r 20 2 2 E 2 21 S M          r m r m Gm toward the sun. 12.10: square theofcenter the towardN,102.8 m)10.0( kg)800()kgNm1067.6( m)(0.10 45coskg)800)(kgNm1067.6(2 45cos 2 F45cos2 3 2 22211 2 22211 2 AD DA 2 AB BA DBonA             r mGm r mGm FF 12.11: left the toN,106.1 N10043.1 N10668.1 m40.0m;10.0 kg500 3 31 4 2 23 32 3 3 2 12 21 1 2312 321         FFF r mm GF r mm GF rr mmm 12.12: The direction of the force will be toward the larger mass, and the magnitude will be . )(4 )2()2( 2 12 2 1 2 2 d mmGm d mGm d mGm   12.13: For convenience of calculation, recognize that the mass of the small sphere will cancel. The acceleration is then ,sm101.2 0.10 0.6 m)10(10.0 kg)260.0(2 29 22     G directed down. 12.14: Equation (12.4) gives      .sm757.0 m1015.1 kg105.1kgmN10763.6 2 2 6 222211      g 12.15: To decrease the acceleration due to gravity by one-tenth, the distance from the earth must be increased by a factor of ,10 and so the distance above the surface of the earth is   m.1038.1R110 7 E  12.16: a) Using ,sm80.9g 2 E  Eq   4.12 gives ,sm87.8 )905)(.sm80.9( 949. 1 )815(. 1 mg 2 2 2 E 2 v E E v 2 E E 2 E 2 v E E E v 2 v v v                                                    g R R m m R Gm RR R m m G R Gm where the subscripts v refer to the quantities pertinent to Venus. b) kg)00.5()sm87.8( 2 N.3.44 12.17: a) See Exercise 12.16;   .sm369.0 1700 8 )sm80.9( 2 2 2 Titania           g b) 3 T 3 E E T E T . r r m m ρ ρ  , or rearranging and solving for density,  3 E 3 E E )81( 1700)1( E r r m m T E ρρ   ,mkg1656)mkg5500( 3 1700 512 3  or about E.0.39  12.18: kg1044.2 21 2  G gR M and   .mKg101.30 33 34 3  Rπ M ρ 12.19: 2 E r mm GF  E 3 m10600 Rr  so N610F At the surface of the earth, N.735g  mw The gravity force is not zero in orbit. The satellite and the astronaut have the same acceleration so the astronaut’s apparent weight is zero. 12.20: Get g on the neutron star 2 ns ns 2 ns ns R GM g R GmM mg   Your weight would be 2 ns nsns R mGM mgw  24 302211 2 m)10( kg)1099.1)(kgNm10(6.67 sm8.9 N675           N101.9 13  12.21: From eq. (12.1), ; (12.4), Eq.fromand,G 2 E 2 1 2 RGmgmmFr E  combining and solving for E R , kg.1098.5 24 2 2 E21 E  Fr Rmgm m 12.22: a) From Example 12.4 the mass of the lander is 4000 kg. Assuming Phobos to be spherical, its mass in terms of its density ,)34( isradiusand 3 ρRRρ  and so the gravitational force is N.27)m1012)(mkg2000)(kg4000)(34( )kg4000)(34( 33 2 3    G R RG b) The force calculated in part (a) is much less than the force exerted by Mars in Example 12.4. 12.23: )m700()kg106.3()kgmN10673.6(22 122211   RGM s.m83.0 One could certainly walk that fast. 12.24: a) so ,and E 2 rmGmUrmGmF E  the altitude above the surface of the earth is m.1036.9 5 E  R F U b) Either of Eq. (12.1) or Eq. (12.9) can be used with the result of part (a ) to find m, or noting that , 22 2 22 rmMGU E  E 2 FGMUm  kg.1055.2 3  12.25: The escape speed, from the results of Example 12.5, is .2 RGM a) .sm1002.5)m1040.3()kg1042.6()kgmN10673.6(2 36232211   b) s.m1006.6)m1091.6()kg1090.1()kgmN10673.6(2 47272211   c) Both the kinetic energy and the gravitational potential energy are proportional to the mass. 12.26: a) The kinetic energy is ,)sm1033.3)(kg629(or , 23 2 1 2 2 1  KmvK J.103.49or 9 KE b) , m1087.2 )kg629)(kg1097.5)(kgmN10673.6(GM 9 242211     r m U or J.1073.8 7 U 12.27: a) Eliminating the orbit radius r between Equations (12.12) and (12.14) gives      min.1751005.1 sm6200 kg1097.5kgmN10673.622 4 3 242211 3     s v Gm T E  b) .sm71.3 2 2  T πv 12.28: Substitution into Eq. (12.14) gives s,1096.6 3 T or 116 minutes. 12.29: Using Eq. (12.12),      s.m1046.7 m1080.7m1038.6 kg1097.5kgmN10673.6 3 56 242211      v 12.30: Applying Newton’s second law to the Earth   kg1001.2 )]()3.365[()kgNm1067.6( m)1050.1(4 4 2 and ; 30 2 1064.8 2211 3112 2 E 32 2 2 Earth 2 2 2 4           d s T r s s E sE d GT r G r m T r v G rv m r v m r mGm maF E     12.31: baseball.for the c maF  The net force is the gravity force exerted on the baseball by Deimos, so sm7.4m)100.6(kg)100.2()kgmN1067.6( 3152211 2 2    DD DD D RGmv R v m R mm G A world-class sprinter runs 100 m in 10 s so have sm4.7s;m10  vv for a thrown baseball is very achieveable. 12.32: Apply Newton’s second law to Vulcan.   days9.47 s400,86 d1 s1014.4 kg)1099.1)(kgNm1067.6( m)1079.5(4 4 2 2 : 6 302211 310 3 2 2 s 32 2 s 2 v 2 vs                           Gm r T T r r Gm T r v r v m r mGm maF 12.33: a) s.m1027.8 ))11.0m)(10((1.50kg)1099.185.0)(kgmN10673.6( 4 11302211     rGmv b) weeks).(about twos1025.12 6 vr  12.34: From either Eq. (12.14) or Eq. (12.19), kg.1098.1 d))s1064.8d)(7.224(()kgmN10673.6( m)1008.1(44 30 242211 3112 2 32 S       GT r m 12.35: a) The result follows directly from Fig. 12.18. b) m)1092.5)(248.01( 12  m,1045.4 12  y.248c) m.104.55m)1050.4)(010.01( 1212  T [...].. .12. 36: a) r Gm1m2  7.07  1010 m F b) From Eq (12. 19), using the result of part (a), 2 (7.07  1010 m)3 2 T  1.05  107 s  121 days 2 2 30 11 (6.673  10 N  m kg )(1.90  10 kg) c) From Eq (12. 14) the radius is (8) 2 3  four times that of the large planet’s orbit, or 2.83  1011 m 12. 37: a) For a circular orbit, Eq (12. 12) predicts a speed of (6.673  1011... 2 12. 43:   a) From Eq (12. 12),    Rv 2 7.5 ly 9.461  1015 m ly 200  103 m s M  G 6.673  1011 N  m 2 kg 2   2  4.3  1037 kg  2.1  107 M S b) It would seem not RS  c) 2GM 2v 2 R  2  6.32  1010 m, 2 c c which does fit 12. 44: Using the mass of the sun for M in Eq (12. 32) gives RS    m s 2 6.673  10 11 N  m 2 kg 2 1.99  1030 kg 3.00  10 8 2   2.95 km That is, Eq (12. 32)... 0.4 R, and x1  32 m  1.6 R 12. 51: Solving Eq (12. 14) for r,  T  R  GmE    2  2 3  (27.3 d)(86,400 s d)   (6.673  10 N  m kg )(5.97  10 kg)   2    5.614  10 25 m 3 , from which r  3.83  10 8 m 11 12. 52:  g g  2 2 ( 6.67310 11 N  m 2 kg 2 )( 20.0 kg) center of the sphere (1.50 m) 2 2 24  5.93  1010 N kg, directed toward the 12. 53: a) From Eq (12. 14), 2 T   86,164 s... 6057 km h 2 12. 64: Combining Equations (12. 13) and (3.28) and setting arad  9.80 m s 2 so that ω  0 in Eq 12. 30 , T  2 R  5.07  10 3 s, a rad which is 84.5 min, or about an hour and a half 12. 65: The change in gravitational potential energy is GmE m GmE m h   GmE m , RE  h RE RE RR  h  so the speed of the hammer is, from the work-energy theorem, U  2GmE h RE  h RE 12. 66: a) The... calculating g 0 from Eq (12. 4) gives ma rad  2.40 N 12. 41: a) GmN m R 2  10.7 m s 2  5.00 kg   53.5 N, or 54 N to two figures    4 2 2.5  10 7 m  10.7 m s 2    52.0 N b) m g 0  arad   5.00 kg   16 h 3600 s h 2    12. 42:   GMm RSc 2 2 mc 2 RS   r2 r2 2r 2 a) 5.00 kg 3.00  108 m s 2 1.4  10 2 m   350 N b)  2 3.00  10 6 m  2 c) Solving Eq (12. 32) for M ,  ... Fx  Gm M x 2 , as expected The derivative may also be taken by expressing  L ln1    ln( x  L)  ln x x  at the cost of a little more algebra 12. 39: a) Refer to the derivation of Eq (12. 26) and Fig (12. 22) In this case, the red ring in Fig (12. 22) has mass M and the common distance s is x 2  a 2 Then, U   GMm x 2  a 2 b) When x >> a, the term in the square root approaches x 2 and U ... Starting with T  2GM (Eq. (12. 14)), T  2 r v , and v  GM r (Eq. (12. 12)) find the velocity and period of the initial orbit: (6.673  1011 N  m 2 kg 2 )(5.97  1024 kg)  7.672  103 m s, and 6.776  106 m T  2 r v  5549 s  92.5 min We then can use the two derived equations to approximate the 3 100   T and  v,  T  3 πv r and v  πT r T  7.672(10 3m) s  0 .122 8 s, and v m v  π r... 7.492  10 22 m 3 , 2 3 and so h  r  RE  3.58  107 m Note that the period to use for the earth’s rotation is the siderial day, not the solar day (see Section 12. 7) b) For these observers, the satellite is below the horizon 12. 54: Equation 12. 14 in the text will give us the planet’s mass: 2r 3 2 T GM P T2  4 2 r 3 GM P MP  4 2 r 3 4 2 (5.75  105 m  4.80  106 m)3  GT 2 (6.673  1011 N ... 6 2  677 N 12. 55: In terms of the density  , the ratio M R is 4 3 R 2 , and so the escape speed is v 8 3 6.673  1011 N  m 2   kg 2 2500 kg m 3 150  103 m  2  177 m s 12. 56: a) Following the hint, use as the escape velocity v  2gh, where h is the height one can jump from the surface of the earth Equating this to the expression for the escape speed found in Problem 12. 55, 2 gh... rotating If the angular speed of the satellite is ωs and the angular speed of the earth is ωE , the angular speed ωrel of the satellite relative to you is ωrel  ωs  ωE 1 ωrel  1 rev  12 h   12 rev h 1 ωE  12 rev h ωs  ωrel  ωE  1 rev h  2.18  10-4 rad s 8   mm v2  F  ma says G 2 E  m r r GmE Gm v2  and with v  r this gives r 3  2E ; r  2.03  107 m r ω This is the radius . Fig. 12. 18. b) m)1092.5)(248.01( 12  m,1045.4 12  y.248c) m.104.55m)1050.4)(010.01( 121 2  T 12. 36: a) m.1007.7 10 21  F mGm r b) From Eq. (12. 19),. T E  b) .sm71.3 2 2  T πv 12. 28: Substitution into Eq. (12. 14) gives s,1096.6 3 T or 116 minutes. 12. 29: Using Eq. (12. 12),      s.m1046.7 m1080.7m1038.6