Physics exercises_solution: Chapter 12

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Physics exercises_solution: Chapter 12

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 12

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Nội dung Text: Physics exercises_solution: Chapter 12

1. Note: to obtain the numerical results given in this chapter, the following numerical values of certain physical quantities have been used; G  6.673  1011 N  m 2 kg 2 , g  9.80 m s 2 and mE  5.97  1024 kg. 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 2  3.84  10 8 m   1.99  10 30 kg    1.50  1011 m     5.97  10 24 kg   2.18.      12.2: Use of Eq. (12.1) gives m1m2 11 (5.97  10 24 kg)(2150 kg) Fg  G 2  (6.673  10 N  m kg ) 2 2  1.67  10 4 N. r (7.8  10 m  6.38  10 m) 5 6 2 The ratio of this force to the satellite’s weight at the surface of the earth is (1.67  10 4 N)  0.79  79%. (2150 kg)(9.80 m s 2 ) (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 GmE m r 2 GmE  RE   2   , mg r g  r  yielding the same result. (nm1 )(nm2 ) mm 12.3: G 2  G 1 2 2  F12 . (nr12 ) r12 12.4: The separation of the centers of the spheres is 2R, so the magnitude of the gravitational attraction is GM 2 (2 R) 2  GM 2 4 R 2 .
2. 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 GM S m GM E m  , ( R  x) 2 x2 which is solved for R x  2.59  108 m. MS 1 ME 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  5.00 kg  10.0 kg  ,   Fg  6.673  1011   m 2 kg 2 0.100 kg    2  0.400 m  0.600 m   2  2.32  1011  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: 6.673  10 11 . m kg 2 2  70kg  7.35  1022 kg   2.4  103 . 3.78  10 m  8 2 12.8: 333, 000  6.03  10 4 23, 5002
3. 12.9: Denote the earth-sun separation as r1 and the earth-moon separation as r2 .  mS mE  a) GmM    2   6.30  1020 ,  (r1  r2 ) 2 r2  toward the sun. b)The earth-moon distance is sufficiently small compared to the earth- sun distance (r2
4. 12.11: m1  m2  m3  500 kg r12  0.10 m; r23  0.40 m m1m2 F1  G 2  1.668  10 3 N r12 m2 m3 F3  G 2  1.043  10 4 N r23 F  F1  F3  1.6  10 3 N, to the left 12.12: The direction of the force will be toward the larger mass, and the magnitude will be Gm2 m Gm1m 4Gm(m2  m1 )   . (d 2) 2 (d 2) 2 d2 12.13: For convenience of calculation, recognize that the mass of the small sphere will cancel. The acceleration is then 2G (0.260 kg) 6.0 2   2.1  109 m s 2 , (10.0  10 m) 10.0 2 directed down. 12.14: Equation (12.4) gives g 6.763  10 11  N  m 2 kg 2 1.5  10 22 kg   0.757 m s 2 . 1.15  10 m 6 2
5. 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  10  1 R E  1.38  107 m. 12.16: a) Using g E  9.80 m s 2 , Eq 12.4 gives 2 Gm m  R   1  g v  2 v  G v  m E  E  m  R   2 R  Rv  E  v  E 2 2 Gm  m   R   1   2E  v   E   g E (.815) m  R   RE  E   v   .949   (9.80 m s 2 )(.905)  8.87 m s 2 , where the subscripts v refer to the quantities pertinent to Venus. b) (8.87 m s 2 ) (5.00 kg)  44.3N. 12.17: a) See Exercise 12.16;  82  g Titania  (9.80 m s 2 )    1700   0.369 m s . 2   3 3 b) ρT  mT r (1 1700) m r ρE mE . rE3 , or rearranging and solving for density, ρT  ρE . mE E . (1 8E r )3  T E (5500 kg m 3 ) 1700   1656 kg m 3 , or about 0.39 E. 512 gR 2 12.18: M G  2.44  1021 kg and ρ  M 4 π 3  R 3  1.30  103 Kg m3 .
6. mmE 12.19: F G r2 r  600  10 3 m  RE so F  610 N At the surface of the earth, w  mg  735 N. 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 GmM ns mg ns  R2 GM ns g ns  R2 Your weight would be mGM ns wns  mg ns  R2  675 N  (6.67  1011 Nm 2 kg 2 )(1.99  10 30 kg)   9.8 m s 2     (10 4 m) 2  9.1  1013 N 12.21: From eq. (12.1), G  Fr 2 m1m 2 , and from Eq. (12.4), g  GmE RE ; combining 2 and solving for RE , 2 gm1m2 RE mE  2  5.98  1024 kg. Fr
7. 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 ρ and radius R is (4 3) ρR3 , and so the gravitational force is G (4 3)(4000 kg) R 3 2  G (4 3)(4000 kg)(2000 kg m 3 )(12  10 3 m)  27 N. R b) The force calculated in part (a) is much less than the force exerted by Mars in Example 12.4. 12.23: 2 GM R  2(6.673  1011 N  m 2 kg 2 ) (3.6  1012 kg) (700 m)  0.83 m s. One could certainly walk that fast. 12.24: a) F  GmE m r 2 and U  GmE m r , so the altitude above the surface of the U earth is F  RE  9.36  105 m. b) Either of Eq. (12.1) or Eq. (12.9) can be used with the 2 result of part (a ) to find m, or noting that U 2  G 2 M E m 2 r 2 , m  U 2 FGM E  2.55  103 kg. 12.25: The escape speed, from the results of Example 12.5, is 2GM R. a) 2(6.673  10 11 N  m 2 kg 2 ) (6.42  10 23 kg) (3.40  10 6 m)  5.02  103 m s . b) 2(6.673  10 11 N  m 2 kg 2 ) (1.90  10 27 kg) (6.91  107 m)  6.06  10 4 m s. c) Both the kinetic energy and the gravitational potential energy are proportional to the mass.
8. 12.26: a) The kinetic energy is K  1 mv 2 , or K  1 (629 kg )(3.33  103 m s) 2 , 2 2 or KE  3.49  109 J. GMm (6.673  1011 N  m 2 kg 2 )(5.97  1024 kg)(629 kg) b) U    , r 2.87  109 m or U  8.73  10 7 J. 12.27: a) Eliminating the orbit radius r between Equations (12.12) and (12.14) gives T   2GmE 2 6.673  1011 N  m 2 kg 2 5.97  1024 kg   v3 6200 m s3  1.05  104 s  175 min. 2πv b)  3.71 m s 2 . T 12.28: Substitution into Eq. (12.14) gives T  6.96  10 3 s, or 116 minutes. 12.29: Using Eq. (12.12), v 6.673  10 N  m kg 5.97  10 11 2 2 24 kg   7.46  10 3 m s. 6.38  10 m  7.80  10 m  6 5
9. 12.30: Applying Newton’s second law to the Earth  F  ma; GmE ms v2  mE r2 r rv 2 2r ms  and v  G TEarth ms  r   2 r 2 TE  4 2 r 3 G GTE2 4 2 (1.50  1011 m)3  (6.67  10 11 Nm 2 kg 2 ) [(365.3d )( 8.6410 s )] 2 4 d  2.01  1030 kg 12.31:  F  mac for the baseball. The net force is the gravity force exerted on the baseball by Deimos, so mmD v2 G 2 m RD RD v  GmD RD  (6.67  1011 N  m 2 kg 2 ) (2.0  1015 kg) (6.0  103 m)  4.7 m s A world-class sprinter runs 100 m in 10 s so have v  10 m s; v  4.7 m s for a thrown baseball is very achieveable.
10. 12.32: Apply Newton’s second law to Vulcan. Gms mv v2  F  ma :  mv r2 r 2r v T 2 Gms  2r    r  T  4 2 r 3 T Gms   4 2 2 (5.79  1010 m) 3 3  11 (6.67  10 Nm 2 kg 2 )(1.99  1030 kg)  1d   4.14  106 s   47.9 days  86,400 s  12.33: a) v  Gm r  (6.673  10 11 N  m 2 kg 2 )(0.85  1.99  1030 kg) ((1.50  1011 m)(0.11))  8.27  10 4 m s. b) 2r v  1.25  106 s (about two weeks). 12.34: From either Eq. (12.14) or Eq. (12.19), 4 2 r 3 4 2 (1.08  1011 m)3 mS   GT 2 (6.673  10 11 N  m 2 kg 2 ) ((224.7 d)(8.64  10 4 s d)) 2  1.98  1030 kg. 12.35: a) The result follows directly from Fig. 12.18. b) (1  0.248)(5.92  1012 m)  4.45  1012 m, (1  0.010)(4.50  1012 m)  4.55  1012 m. c) T  248 y.
11. Gm1m2 12.36: a) r  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. 11 (6.673  10 N  m kg )(1.90  10 kg) 2 2 30 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 N  m 2 kg 2 )(1.99  1030 kg) (43  109 m)  56 km s. The speed doesn’t have this value, so the orbit is not circular. b) The escape speed for any object at this radius is 2 (56 km s)  79 km s , so the spacecraft must be in a closed elliptical orbit.
12. 12.38: a) Divide the rod into differential masses dm at position l, measured from the right end of the rod. Then , dm = dl ( M L ), and Gm dm GmM dl dU =   . lx L lx Integrating, GmM L dl GmM  L  U=  L O lx  L ln1  .  x For x >> L, the natural logarithm is ~ L x  , and U  Gm M x. b) The x-component of the gravitational force on the sphere is δU GmM ( L x 2 ) GmM Fx     2 , δx L (1  ( L x)) ( x  Lx) with the minus sign indicating an attractive force. As x >> L, the denominator in the above expression approaches x 2 , and 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.