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Unformatted text preview: CHAPTER 14 GRAVITATION 381 CHAPTER 14 Answer to Checkpoint Questions 1. 2. 3. 4. 5. 6. all tie (a) 1, tie of 2 and 4, then 3; (b) line d negative y direction (a) increase; (b) negative (a) 2; (b) 1 (a) path 1 [decreased E (more negative) gives decreased a]; (b) less than (decreased a gives decreased T ) Answer to Questions (a) between, closer to less massive particle; (b) no; (c) no (other than in nity) Gm2 =r2 , upward 3GM 2 =d2 , leftward increases from zero to a maximum, then decreases to zero b, tie of a and c, then d 1, tie of 2 and 4, 3 b, a, c (a) { (d) zero (a) negative; (b) negative; (c) positive; (d) all tie 10. A and C tie, then B 11. (a) all tie; (b) all tie 12. orbits 2 and 3 (orbits must be about Earth's center)
1. 2. 3. 4. 5. 6. 7. 8. 9. 382 CHAPTER 14 GRAVITATION 13. (a) same; (b) greater Solutions to Exercises & Problems The magnitude of the force of one particle on the other is given by F = Gm1 m2 =r2 , where m1 and m2 are the masses, r is their separation, and G is the universal gravitational constant. Solve for r:
11 3 /s2 r = Gm1 m2 = (6:67 10 2m 10kg)(5:2 kg)(2:4 kg) = 19 m : 12 N F :3
r s 1E (a) The gravitational force exterted on the baby (denoted with subscript b) by the obstetrician (subscript o) is given by 2E Fbo = Gm2o mb = (6:67 10 r
bo 11 N m2 = kg2 )(70 kg)(3 kg) (1 m)2 = 1 10 8 N : (b) and (c) The maximum (minimum) forces exterted by Jupiter on the baby occur when it is separated from the Earth by the shortest (longest) distance rmin (rmax ), respectively. Thus 11 2 2 27 max = GmJ mb = (6:67 10 N m = kg )(2 10 kg)(3 kg) = 1 10 6 N FbJ 2 rmin (6 1011 m)2 and
min FbJ = GmJ mb = (6:67 10 r2 max 11 N m2 = kg2 )(2 1027 kg)(3 kg) (9 1011 m)2 = 5 10 7 N : (d) No. The gravitational force exetrted by Jupiter on the baby is greater than that by the obstetrician by a factor of up to (1 10 6 N=1 10 8 N) = 100. The ratio is
3E 2 2 FSun = GMs Mm =rsm = Ms rem = (1:99 1030 kg)(3:82 108 m)2 = 2:16 : 2 2 FEarth GMe Mm =rem Me rsm (5:98 1024 kg)(1:50 1011 m)2 CHAPTER 14 GRAVITATION 383 Use F = Gms mm =r2 , where ms is the mass of the satellite, mm is the mass of the meteor, and r is the distance between their centers. The distance between centers is r = R + d = 15 m + 3 m = 18 m. Here R is the radius of the satellite and d is the distance from its surface to the center of the meteor. Thus 4E F = (6:67 10 11 m3 /s2 kg)(20 kg)(7:0 kg) = 2:9 10 (18 m)2 11 N: Since F / m(M m), we need to maximize the function f (m) = m(M m). This is done by setting df (m)=dm = M 2m = 0; which yields m=M = 1=2:
2 2 At the point where the forces balance GMe m=r1 = GMs m=r2 , where Me is the mass of the Earth, Ms is the mass of the Sun, m is the mass of the space probe, r1 is the distance from the center of the Earth to the probe, and r2 is the distance from the center of the Sun to the probe. Substitute r2 = d r1 , where d is the distance from the center of the Earth to 2 the center of the Sun, to nd Me =r1 = Ms =(d r1 )2 . Solve for r1 . A little algebra yields 2 (Ms Me )r1 + 2Me dr1 Me d2 = 0. This is a quadratic equation and has the solutions 2 2 4d2 r1 = 2Me d 4Me d + M )Me (Ms Me ) : 2(Ms e
p 5P 6E Since r1 must be positive, use the positive sign. Some values are Ms = 1:99 1030 kg, Me = 5:98 1024 kg, and d = 150 109 m, all taken from the appendix of the text. Clearly you may neglect Me in the expression (Ms Me ). The result is r1 = 2:60 108 m. Let the distance be r. Thus
7E GM Fm = (D m m)2 = FE = GM2e m ; r r
em where m is the mass of the spaceship. Solve for r: r = p Dem Mm =Me + 1 3:82 108 m 8 =p 22 kg)=(5:98 1024 kg) + 1 = 3:44 10 m : (7:36 10 384 CHAPTER 14 GRAVITATION Let am and as be the contribution to the acceleration a of the Earth due to the gravitational pulls from the Moon and the Sun, respectively. The dierence in the Earth's acceleration in the two cases is due to the reversal in the direction of am , i.e., a = 2am . Thus
2 a a = 2GMe Mm =rem = 2 Mm a a GM M =r2 M 8P 7 36 1022 = 2 1::99 1030 kg kg s e s es 1:5 1011 m 3:8 108 m s 2 res rem 2 1:1% : The net force is given by
F2 = Gm2 9P 2000 kg (0400 kg 2 j (0:80 m)2 i + (0500 kg 2 i :50 m) :40 m) 5 Nj; = 3:7 10 in the positive y direction. m1 j m3 i + m4 i 2 2 2 r21 r23 r24 = (6:67 10 11 m3 =kg s2 )(350 kg) y m1 r 21 m3 r 13 m2 O r24 m4 x The gravitational forces on m5 from the two 500kg masses obviously cancel each other. The net force on m5 is therefore caused by that from the remaining two masses: 10P F = (6:67 10 11 m3 =kg s2 )(250 kg)(300 kg p ( 2 cm)2 100 kg) = 0:017 N : The force lies along the line joining the two masses (300 kg and 100 kg) and m5 , pointing toward the 300kg mass. By symmetry, we only need to consider the force components along the line joining M and
11P m4 . Let the side length of the triangle be r, then p p 2 Gm4 m sin230 = GMm42 ; (r= 3) (r= 3)
which gives M = m. CHAPTER 14 GRAVITATION 385 Let the forces exterted by m2 and m3 on m1 be F12 and F13 , respectively. Then the magnitude of the total force on m1 is F = jF12 + F13 j p = (F12 )2 + (F13 )2 + 2F12 F13 cos ; where the angle between F12 and F13 is given by the cosine theorem: cos = [(0:25 m)2 (0:20 m)2 (0:15 m)2 ]=[2(0:25 m)(0:15 m)] = 0. So 12P F13
0.20m m3 0.15m F F23 m2 m1 0.25m F = (F12 )2 + (F13 )2
= (6:67 10 p (800 kg)2 + (600 kg)2 = 4:4 10 6 N : (0:20 m)4 (0:15 m)4 The direction of F can be speci ed by calculating the angle in the gure: 2 1 F12 = tan 1 800=(0:20) = 37 : = tan F 600=(0:15)2 13 Since = tan 1 (0:15=0:20) = 37 = , F is perpendicular to, and points toward, the line joining m1 and m2 .
11 N m2=kg2 )(2:0 kg) s Lable the four spheres 1 through 4. The net force on m4 is
F4 =
X 13P y m1 r 41 m3 r 43 m4 r 42 m2 3 Thus 41 42 43 where r41 = r41 (cos i + sin j); r43 = r43 j, and r42 = r42 ( cos i sin j): = Gm4 m1 r41 + m2 r42 + m3 r43 ; r3 r3 r3 i=1 F4i x F4;x = Gm4 m1 cos m2 cos 2 2 r41 r42 = (6:67 10 11 m3 =kg s2 )(20 kg) (20 kg)(0:50 m) (40 kg)(1:0 m) [(1:0 m)2 + (0:50 m)2 ]3=2 [(1:0 m)2 + (0:50 m)2 ]3=2 = 5:6 10 8 N 386 CHAPTER 14 GRAVITATION and F4;y = Gm4 m1 sin m2 sin m3 2 2 2 r41 r42 r43 = (6:67 10 11 m3 =kg s2 )(20 kg) (40 5 (20 0 [(1:0 m)2 kg)(1::50m) 2 ]3=2 [(1:0 m)2 kg)(0::50m) 2 ]3=2 + (0 m) + (0 m) = 3:2 10 7 N : q p 60 kg (0:50 m)2 Finally, the magnitude of F4 is F4 = F42;x + F42;y = (5:6 10 8 N)2 + ( 3:2 10 7 N)2 = 3:2 10 7 N :
14P Put the origin of the xaxis at the left end of the rod on the right, with the positive xdirection pointing to the left. Consider an in nitesimal segment of the rod on the left, located between x and x + dx. The mass of the segment is given by dm = M (dx=L), and its separation from the left end of the rod on the right is x. According to the result of sample Problem 142 the gravitational force between the segment and the rod on the right is given by dF = G(dm)M=x(L + x). Thus the force of one rod on the other is F=
15P Z dF = Z GM dm = GM 2 Z L 1 dx : x(L + x) L 0 x(L + x) If the lead sphere were not hollow the magnitude of the force it exerts on m would be F1 = GMm=d2 . Part of this force is due to material that is removed. Calculate the force exerted on m by a sphere that just lls the cavity, at the position of the cavity, and subtract it from the force of the solid sphere. the cavity has a radius r = R=2. The material that lls it has the same density (mass to volume ratio) as the solid sphere. That is Mc =r3 = M=R3 , where Mc is the mass that lls the cavity. Thus Mc = (r3 =R3 )M = (R3 =8R2 )M = M=8. Its center is d r = d R=2 from m, so the force it exerts on m is F2 = G(M=8)m=(d R=2)2 . The force of the hollowed sphere on m is 1 1 GMm 1 1 F = F1 F2 = GMm d2 8(d R=2)2 = d2 8(1 R=2d)2 :
16E m gm = G M2 = (6:67 10 R m 22 11 m3 =kg s2 ) (7:36 10 kg) (1:74 106 m)2 = 1:62 m/s2 : CHAPTER 14 GRAVITATION 387 The acceleration due to gravity is given by g = GM=r2 , where M is the mass of the Earth and r is the distance from the Earth's center. Substitute r = R + h, where R is the radius of the Earth and h is the altitude, to obtain g = GM=(R + h)2 . Solve for h. You p should get h = GM=g R. According the appendix of the text R = 6:37 106 m and M = 5:98 1024 kg, so 17E h = (6:67 10 s 11 m3 =kg s2 )(5:98 1024 kg) 4:9 m/s2 6:37 106 m = 2:6 106 m : Let h = 1350 ft = (1350 ft)(1:00 m=3:28 ft) = 412 m and your mass be m. the dierence in your weight is then 18E m)(120 = 2(412 106 mlb) = 1:6 10 2 lb ; 6:37 where the approximation (1 + x)n 1 + nx (jx 1j) was used. So you will be lighter by 0:016 lb: (a)
19E h GmM W = (R + he)2 GmMe = GmMe 1 + R 2 2 Re Re e e 2 2 = W 1 Rh + 1 Rh W 2 1 e e (b) The nal speed v is given by 1 mv2 = mgh, or 2
p p g = GM = (6:67 10 r2 11 N m2 = kg2 )(1:99 1030 kg) (10 103 m)2 = 1:3 1012 m/s2 : v = 2gh = (2 1012 m/s2 )(1:0 m) = 1:6 106 m/s :
20E (a) The acceleration towards the center of the Earth is given by 2 2 R = 2 a1 e e = T e 86400 s e 3g : = 3:4 10 = !2 R 2 6:37 106 m g 9:80 m/s2 388 CHAPTER 14 GRAVITATION (b) The acceleration towards the Sun is 2 2 2 1:50 1011 m g 02 Des = 2 Des = a2 = !e Te0 (1 y)(3:16 107 s/y) 9:80 m/s2 = 6:1 10 4 g : (c) The acceleration toward the center of the galaxy is 2 2 D = 2 a3 s sg = T sg 8 y)(3:16 107 s/y) (2:5 10 s 11 g : = 1:4 10 = !2 D 2 2:2 1011 m g 9:80 m/s2 (a) Let the mass of the object be m. Its weight on the Moon is given by
2 Wm = mgm = mge gm = We gm = We GMm =Rm ge ge ge (6:67 10 11 N m2 = kg2 )(7:36 1022 kg) = 17 N : = (100 N) (1:74 106 m)2 (9:8 m/s2 ) 21E (b) Let the separation between the object and the center of the Earth be x. Then Wm = GmMe = GmMe Re 2 x2 Re x 2 Re 2 ; = We x which gives x = We = 100 N = 2:5 : Re Wm 16:5 N r r Thus the percentage error in the measurement of t is given by The time t it takes for an object to fall through a distance d in a freefall is given by 1 gt2 = h, or g = 2h=t2 . The error in g , which we call g , is caused by the error in t, called 2 t: 2h = 2h 1 2h d 1 t = 4ht = 2g t : g = t2 t2 dt t2 t3 t 22P t = 1 g = 1 (0:1%) = 0:05% : t 2 g 2 CHAPTER 14 GRAVITATION 389 where = M=V = 3M=(4R3 ) is the density of the planet. (b) For = 3:0 g/cm3 ,
s If the rotation of the planet were any faster, there would not be enough force of gravity to provide the centripetal force of rotation for materials on the equator. (a) Consider a piece of material of mass m on the equator of a uniform planet of mass M and radius R. We have mM m!2 R = m 2 2 R ; G R2 T or r 3 T Tmin = G ; 3 23P Tmin = (6:67 10
24P 11 m3 =kg s2 )(3:0 103 kg/m3 ) = 6:8 103 s = 1:9 h : (a) The dierence in weights is W = G mMe R2 where R is the distance from the center of the Earth to the center of mass of the lower 3 mass: R Re . Thus Me =R3 Me =Re = (4=3); and 4 = 8Gmh : W = 2hGm 3 3 (b) Let W=W = 1:0 10 6 ; then W = 8Gmh=3 ; h = GmMe 1 1 + R 2 R = 2h GmMe ; R3 mMe G (R + h)2 2 GmMe 1 1 + 2h R2 R W mg which gives h = 8G W 2 )(1 0 6) = 8(6:67 3(9:80 m/s kg s:2 )(5:10 103 kg/m3 ) = 3:2 m : 10 11 m3 = 5 3g W 390 CHAPTER 14 GRAVITATION The forces acting on an object being weighed are the downward force of gravity and the upward force of the spring balance. Let Fg be the magnitude of the force of gravity and let W be the force of the spring balance. The reading on the balance gives the value of W . The object is traveling around a circle of radius R and so has a centripetal acceleration. Newton's second law becomes Fg W = mV 2 =R, where V is the speed of the object as measured in an inertial frame and m is the mass of the object. Now V = R! v, where ! is the angular velocity of the Earth as it rotates and v is the speed of the ship relative to the Earth. Notice that the rst term gives the speed of a point xed to the rotating Earth. The plus sign is used if the ship is traveling in the same direction as the portion of the Earth under it (west to east) and the negative sign is used if the ship is traveling in the opposite direction (east to west). Newton's second law is now Fg W = m(R! v)2 =R. When we expand the parentheses we may neglect the term v2 since v is much smaller than R!. Thus Fg W = m(R2 !2 2R!v)=R and W = Fg mR!2 2m!v. When v = 0 the scale reading is W0 = Fg mR!2 , so W = W0 2m!v. Replace m with W0 =g to obtain W = W0 (1 2!v=g). The upper sign ( ) is used if the ship is sailing eastward and the lower sign (+) is used if the ship is sailing westward. Consider a small piece of matter (with mass m) on the surafce of the neutron star of radius R and mass M . Since it rotates together with the rest of the star with an angular velocity of !, the small piece must be subject to a centripetal force Fcen given by Fcen = m!2 R Fg = GmM ; R2 where Fg is The gravitational attraction exerted on m by the rest of the neutron star. Note that the equality holds true when there is no radially outward force acting on m from its
26P 25P surroundings anymore, which corresponds to the critical case when the star is about to disintegrate, at which point 2 R3 2 (20 103 3 M = Mmin = ! G = (267rad/s) 11 N m2 =m)2 = 4:7 1024 kg : 6: 10 kg (a) From Eq. 1412
27P GMh c4 ag = GMh = (1:001R )2 = (1:001)2GMh =c2 )2 = (2:002)2 GM r2 (2GMh h h 8 m/s)4 (3:00 10 = (2:002)2 (6:67 10 11 m3 =kg s2 )M h 43 kg m/s2 = 3:03 10 : M
h CHAPTER 14 GRAVITATION 391 (b) Since ag / 1=Mh , it will decrease as Mh increases. (c) Use the result of part (a) above to obtain 3 03 1043 kg m/s2 ag = (1:55 : 1012 )(1:99 1030 kg) = 9:82 m/s2 : (d) From Eqs. 1417 and 1412 a 2a dag 2 GMh dr = 2 Rg dr = 2GMg dr2 2 Rh h h =c 2 )(1:70 m)(3:00 108 m/s)2 (9 = (6:67 10:82 m/s=kg s2 )(1:55 1012 )(1:99 1030 kg) 11 m3 = 7:30 10 15 m/s2 :
(e) No, since dag is very small.
28E Use Newton's shell theorem. The results are (a) Fa = G(M1 + M2 )m=a2 : (b) Fb = GM1 m=b2 : (c) Fc = 0.
29E Apply the workenergy theorem to the particle in question. It starts from a point at the surface of the Earth with zero initial speed and arrives at the center of the Earth with nal 2 speed vf . The corresponding change in its kinetic energy, K = 1 mvf , is equal to the R 2 positive work done on it by the Earth's gravitational pull: K = F dr. Thus 1 mv2 = Z 0 F dr = Z 0 ( Kr) dr = 1 KR2 ; 2 f R 2 R where R is the radius of the Earth. Solve for vf to obtain vf = R K=m. To simplify this expression, note that the acceleration of gravity g on the surface of the Earth can be expressed as g = GM=R2 = G(4R3 =3)=R2 , from which we get K=m = 4G=3 = g=R. Thus
p g p vf = R K = R R = gR m p = (9:80 m/s2 )(6:37 106 m) = 7:91 103 m/s : r r 392 CHAPTER 14 GRAVITATION The acceleration of gravity g at the bottom of the mine shaft is caused by the spherical part of the Earth with radius R D and mass M 0 centered at the center of the Earth. Thus 0 0 3 R g = (RGM )2 = (RGv )2 = G(4(=3)(D)2 D) D D R = 4GR 1 D = gs 1 D : 3 R R
31P 30E (a) Since r = 1:5 m > R = 1:0 m; the gravitational force on m is F = GmM = (6:67 10 r2 11 m3 =kg s2 )(1:0 104 kg)m (1:5 m)2 = (3:0 10 7 m) N ; where m is in kilograms. (b) Apply Newton's shell theorem:
0 r 3 F = GmM = GmM R = GmMr r2 r2 R3 11 3 2 )(1: 4 = (6:67 10 m =kg :s0 m)30 10 kg)(0:50 m) = (3:3 10 7 m) N : (1 (c) The general expression is where m is in kilograms and r is in meters.
32P F (r) = GmMr = (6:7 10 7 mr) N ; R3 First, consider the case r < R. Refer to Sample Problem 145. The acceleration of gravity as a function of r is given by
0 G 3 a(r) = GM = r2 4r = Cr ; r2 3 where C = 4G=3 is a constant. Note that on the surface of the sphere a(R) = ag = CR, or C = ag =R, so r a(r) = Cr = ag R : Now set a(r) = ag =3 to nd r = R=3. CHAPTER 14 GRAVITATION 393 r a(r) = ag R p Again set a(r) = ag =3 to nd r = 3R. GM=r2 / 1=r2 , so Now consider the case r > R. In this case, let the mass of the sphere be M , then a(r) = 2 : (a) The magnitude of the force on a particle with mass m at the surface of the Earth is given by F = GMm=R2 , where M is the total mass of the Earth and R is the Earth's radius. The acceleration due to gravity is
11 3 2 kg)(5 98 24 F g = m = GM = (6:67 10 (6m /s 106 m):2 10 kg) = 9:83 m/s2 : R2 :37 33P (b) Now g = GM=R2 , where M is the total mass contained in the core and mantle together and R is the outer radius of the mantle (6:345 106 m, according to Fig. 35). The total mass is M = 1:93 1024 kg + 4:01 1024 kg = 5:94 1024 kg. The rst term is the mass of the core and the second is the mass of the mantle. Thus
11 3 2 kg)(5: 24 g = (6:67 10 (6:m /s106 m)94 10 kg) = 9:84 m/s2 : 2 345 (c) A point 25 km below the surface is at the mantlecrust interface and is on the surface of a sphere with a radius of R = 6:345 106 m. Since the mass is now assumed to be uniformly distributed the mass within this sphere can be found by multiplying the mass 3 per unit volume by the volume of the sphere: M = (R3 =Re )Me , where Me is the total mass of the Earth and Re is the radius of the Earth. Thus 6:345 106 m 3 (5:98 1024 kg) = 5:91 1024 kg : M = 6:37 106 m The acceleration due to gravity is GM = (6:67 10 11 m3 /s2 kg)(5:91 1024 kg) = 9:79 m/s2 : g = R2 (6:345 106 m)2
34E (a) The gravitational potential energy is U = GMm = (6:67 10 r 11 m3 =kg s2 )(5:2 kg)(2:4 kg) 19 m = 4:4 10 11 J : 394 CHAPTER 14 GRAVITATION (b) Since U = GMm 3r GMm = 2 ( 4:4 10 r 3 11 J) = 2:9 10 11 J ; 11 J: the work done by the gravitational force is W = U = 2:9 10 (c) The work done by you is W 0 = U = 2:9 1011 J: (a)
35E Ua = Gm1 m3 Gm1 m4 Gm3 m4 r13 r14 r34 = (6:67 10 11 m3 =kg s2 ) p(400 kg)(2000 kg) 2 + p (400 kg)(500 kg) 2 (0:5 m)2 + (0:8 m) (0:5 m)2 + (0:4 m)
+ p(2000 kg)(500 kg) 2 (0:8 m)2 + (0:4 m) = 1:4 10 4 J : (b) The additional terms in the gravitational potential energy as a result of the introduction of m1 is m1 m2 + m1 m3 + m1 m4 < 0 : Ub = G r r r
12 13 14 (c) The change in gravitational potential energy due to the removal of m1 is Uc = G mr3 m4 Ua = G mr1 m3 + mr1 m4 > 0 : 34 13 14 Therefore the work you have to do is Wc = Uc > 0: (d) Wd = Ub < 0. Since U / F / mM , the condition for U = Umin is the same as that for F = Fmax , i.e., m=M = 1=2. (a)
37E 36E m = Mm =Vm = (0:11) 1:3 104 = 0:74 : e Me =Ve 6:9 103 CHAPTER 14 GRAVITATION 395 (b) Use Eq. 1412: Mm am = GMm = 4R3 =3 4GRm g 2 Rm m 3 = 43 Ge Re m Rm = ae m Rm g e e Re Re 3 6 9 10 = (9:8 m/s2 )(0:736) 1::3 104 3:8 m/s2 :
(c) Use Eq. 1426:
r vm = = ve Mm Me 5:1 km/s : s 2GMm = R
m r 2GMe s Re Mm Me Re Rm
s Re = (11:2 km/s) (0:11) 1:3 104 Rm 6:9 103 The escape speed is
11 3 2 11 )(1:99 30 v = 2GM = 2(6:67 10 (80;m =kg s )(1:4 10 m/ly) 10 kg) R 000 ly)(9:51 1015 = 2:2 105 m/s = 2:2 102 km/s :
r s 38E (a) The amount of energy E needed for a mass m to secape from the surface of a shperical planet of mass M and radius R is given by E = GMm=R. Thus 39E Em = GMm m=Rm = Mm Rr Ee GMe m=Re Me Rm 6 (7 36 1022 = (5::98 1024 kg)(6:37 106 m) = 0:0451 : kg)(1:74 10 m)
(b) Ej = Mj Re = (1:90 1027 kg)(6:37 106 m) = 28:5 : Ee Me Rj (5:98 1024 kg)(7:15 107 m) 396 CHAPTER 14 GRAVITATION The orbiting speed of the Earth satis es
2 me vorb = GMe Ms ; 2 res res 40E which gives vorb = p 2vorb :
41E p GMs =res : Compare this with vesc = 2GMs =res to obtain vesc = p The potential energy of a particle of comet dust of mass m in the gravitational elds of the Earth and the Moon is U = GmMe GmMm : R r Consider the work W done against gravity to remove a mass m from the surface of the star to in nity. Let W = U = GmMs =Rs = mc2 ; we have Rs = GMs =c2 . (a) The work done by you in moving the mass M2 equals the change in the potential energy of the threemass system. the initial potential energy is
1 Ui = Gmd m2 42E 43P Gm1 m3 L Gm1 m3 L
1 Gm2 m3 L d Gm2 m3 : d and the nal potential energy is Uf = Gm1 m2 L d
the work is W = Uf Ui = Gm2 m1 d L d + m3 L d d "
= (6:67 10
11 m3 /s2 kg)(0:10 kg) 1 1 1 = 5:0 10 11 J : 1 1 (0:80 kg) 0:040 m 0:080 m # 1 1 + (0:20 kg) 0:080 m 0:040 m
11 J. (b) The work done by the force of gravity is (Uf Ui ) = 5:0 10 CHAPTER 14 GRAVITATION 397 (a) Apply the principle of conservation of energy. Initially the rocket is at the Earth's surface and the potential energy is Ui = GMm=Re = mgRe , where M is the mass of the Earth, m is the mass of the rocket, and Re is the radius of the Earth. The substitution g = GM=Re was made. The initial kinetic energy is 1 mv2 = 2mgRe , where the substitution 2 p v = 2 gRe was made. If the rocket can escape then conservation of energy must lead to a positive kinetic energy no matter how far from the Earth it gets. Take the nal potential energy to be 0 and let Kf be the nal kinetic energy. Then Ui + Ki = Uf + Kf leads to Kf = Ui + Ki = mgRe + 2mgRe = mgRe . The result is positive and the rocket has enough kinetic energy to escape the gravitational pull of the Earth. p 2 2 (b) Write 1 mvf for the nal kinetic energy. Then 1 mvf = mgRe and vf = 2gRe . 2 2 44P (a) Apply the principle of conservation of energy. Initially the particle is at the surface of the asteroid and has potential energy Ui = GMm=R, where M is the mass of the asteroid, R is its radius, and m is the mass of the particle being red upward. Take the initial kinetic energy to be 1 mv2 . The particle just escapes if its kinetic energy is 0 when it is in nitely 2 far from the asteroid. The nal potential and kinetic energies are both 0. Conservation of energy yields GMm=R + 1 mv2 = 0. Replace GM=R with gR, where g is the acceleration 2 due to gravity at the surface. Then the energy equation becomes gR + 1 v2 = 0. Solve 2 p p for v: v = 2gR = 2(3:0 m/s2 )(500 103 m) = 1:7 103 m/s. (b) Initially the particle is at the surface; the potential energy is Ui = GMm=R and the kinetic energy is Ki = 1 mv2 . Suppose the particle is a distance h above the surface when 2 it momentarily comes to rest. The nal potential energy is Uf = GMm=(R + h) and the nal kinetic energy is Kf = 0. Conservation of energy yields GMm=R + 1 mv2 = 2 GMm=(R + h). Replace GM with gR2 and cancel m in the energy equation to obtain gR + 1 v2 = gR2 =(R + h). The solution for h is 2 2gR2 h = 2gR v2 R 2 )(500 3 m)2 ) 2(3: = 2(3:0 m/s20 m/s 103 m)10 (1000 m/s)2 )(500 = 2:5 105 m : 45P (500 103 m) (c) Initially the particle is a distance h above the surface and is at rest. Its potential energy is Ui = GMm=(R + h) and its initial kinetic energy is Ki = 0. Just before it hits the 2 asteroid its potential energy is Uf = GMm=R. Write 1 mvf for the nal kinetic energy. 2 Conservation of energy yields GMm=(R + h) = GMm=R + 1 mv2 . Replace GM with 2 398 CHAPTER 14 GRAVITATION gR2 and cancel m to obtain gR2 =(R + h) = gR + 1 v2 . The solution for v is 2
2 2 v = 2gR RgR h + r 2(3:0 2 )(500 3 m)2 = 2(3:0 m/s2 )(500 103 m) 500 m/sm + 100010 103 m 103 3 m/s : = 1:4 10
46P
r (a) Use conservation of energy: Ki + Ui = Kf + Uf : The kinetic energy is given by 1 Kf =Ki + Ui Uf = Ki GMm r i =5:0 107 J (6:67 10 =2:2 107 J : (b) The initial kinetic energy needed is rf 1 11 m3 =kg s2 )(5:0 1023 kg)(10 kg) 1 3:0 106 m 1 4:0 106 m K 0 = U = GmM 1 R rmax 11 3 2 23 6 6 kg)(5 = (6:67 10 m =kg s )(10 106:0 10 kg)(8:0 10 m 3:0 10 m) (3:0 m)(8:0 106 m) = 6:9 107 J : 1 (a) Consider either of the two stars. The gravitational force exerted on it from its companion star (located a distance of 2r away from it) must be equal to the centripetal force needed for it to move around the center of mass of the system: 47P M2 Fg = G (2r)2 = Fcen = M!2 r ;
which gives
11 m3 30 s2 ! = GM = (6:67 10 4(1:0 = kg 11 )(3:30 10 kg) = 2:2 10 7 rad/s : 4r3 10 m)
r s CHAPTER 14 GRAVITATION 399 (b) The initial gravitational potential energy of the meteorite (with mass m) as it passes through the center of the doublestar system is given by Ui = 2GMm=r. If it has just barely enough speed (vmin ) to leave the system, it will have no kinetic energy left upon reaching in nity. Thus the nal energy of the meteorite is zero. Since mechanical energy is conserved in this case, its initial value, Ei = Ki + Ui , should also be zero: 2 Ki + Ui = 1 mvmin 2GMm=r = 0. Solve for vmin : 2 vmin = 4GM = 4r GM = 4!r r 4r3 = 4(2:2 10 7 rad/s)(1:0 1011 m) = 9:0 104 m/s :
48P r r (a) The momentum of the twostar system is conserved, and since the stars have the same mass, their speeds and kinetic energies are the same. Use the principle of conservation of energy. The initial potential energy is Ui = GM 2 =ri , where M is the mass of either star and ri is their initial centertocenter separation. The initial kinetic energy is 0 since the stars are at rest. The nal potential energy is Uf = 2GM 2 =ri since the nal separation is ri =2. Write Mv2 for the nal kinetic energy of the system. This is the sum of two terms, each of which is 1 Mv2 . Conservation of energy yields GM 2 =ri = 2GM 2 =ri + Mv2 . 2 The solution for v is
11 m3 2 30 v = GM = (6:67 10 1010/s kg)(10 kg) = 8:2 104 m/s : ri m (b) Now the nal separation of the centers is rf = 2R = 2 105 m, where R is the radius of either of the stars. The nal potential energy is given by Uf = GM 2 =rf and the energy equation becomes GM 2 =ri = GM 2 =rf + Mv2 . The solution for v is
r s v = GM r1
s s f ri
11 m3 /s2 kg)(1030 kg) 1 = (6:67 10 = 1:8 107 m/s :
49P 1 2 105 m 1010 kg 1 Use conservation of energy for the projectile of mass m and initial speed vi rising to a maximum height of h: 1 mv2 = U = GmMe GmMe : K = 2 i Re + h Re 400 CHAPTER 14 GRAVITATION Solve for h: v2 2 h = 2GM i Rev2 R
e (104 m/s)2 (6:37 106 m)2 = 2(6:67 10 11 m3 =kg s2 )(5:98 1024 kg) (104 m/s)2 (6:37 106 m) = 2:6 107 m = 2:6 104 km : i e (a) 50P F=GMm/r2 [( r3 b3)/(a3b3 )] The gravitational force as a function of r is illustrated in the gure shown. Note that from Newton's shell theorem F (r a) = GMm r2 and where Also, F (0 r b) = 0: is the mass of the portion of the shell that is enclosed in a spherical area of radius r centered at the origin. Thus GmM r3 b3 : F (b ra) = r2 a3 b3 F=GMm/r 2 F=0 b a r F=GM'm/r2 F (b ra) = GM 0 m=r2 ;
4 3 3 V 0 M = 3 (r b ) M = r3 b3 M 0= M V 4 (a3 b3 ) a3 b3 3 CHAPTER 14 GRAVITATION 401 (b)
U(r) b a r U=GMm/r U= const. (a) 51P
1 Ui = Gmx m2 = (6:67 10 11 N m2 = kg2 )(20 kg)(10 kg) (b) Since 0:8 m = 1:67 10 8 J : we have x Uf = Gm10m2 = Ui x0 x 0:80 m 8 J) 8 = (1:67)(10 0:80 m 0:20 m = 2:23 10 J ; Kf = U = Ui Uk = (1:67)(10 8 J) + (2:23)(10 8 J) = 5:6 10 9 J : (a) The gravitational attraction is dU (x) = d Z p Gm dM = d pGMm = GMm ; F (x) = dx dx ring R2 + x2 dx R2 + x2 (R2 + x2 )3=2 where the minus sign in F (x) indicates that the force is attractive. (b) Use conservation of energy equation 1 mv2 = U (x) U (0) to nd speed v: 2 52P* v= r 1 = 2GM R s 2 [U (x) U (0)] = m s m 2 GMm R pGMm 2 R2 + x p 21 2 : R +x 402 CHAPTER 14 GRAVITATION Since Ui = 0 and Uf = GmM=d, 1 MV 2 + 1 mv2 = U = GmM ; 2 2 d where V and v are the speeds of the two objects in question. Conservation of momentum requires that P = MV = mv which, when substituted into the above equation, yields 2 2 2 P = d(GM+m ) ; M m from which we calculate the relative speed:
s 53P* P p Vr = v + V = m + M = 2G(Md + m) :
54E r Since T / r3=2 (Kepler's law of periods), we nd the period of revolution for Mars: rms 3=2 = (1:00 y)(1:52)3=2 = 1:87 y : Tm = Te r es Here the subscript e denotes the Earth. The period T and orbit radius r are related by the law of periods: T 2 = (42 =GM )r3 , where M is the mass of Mars. The period is 7 h 39 min, which is 2:754 104 s. Solve for M: (42 )(9:4 106 m)3 2 3 = 6:5 1023 kg : M = 4GTr2 = 11 m3 /s2 kg)(2:754 104 s)2 (6:67 10
56E 55E From T 3 = (4=GMe )r3 we solve for Me : 2 3 Me = 4GTr2 = (6:67 10 = 5:93 1024 kg : 42 (3:82 108 m)3 11 m3 =kg s2 )[(27:3 da)(86400 s/da)]2 CHAPTER 14 GRAVITATION 403 Let N be the number of stars in the galaxy, M be the mass of the Sun, and R be the radius of the galaxy. The total mass in the galaxy is NM and the magnitude of the gravitational force acting on the Sun is F = GNM 2 =R2 . The force points toward the galactic center. The magnitude of the Sun's acceleration is a = v2 =R, where v is its speed. If T is the period of the Sun's motion around the galactic center then v = 2R=T and a = 42 R=T 2 . Newton's second law yields GNM 2 =R2 = 42 MR=T 2 . The solution for N is 2 3 N = 4 R : the period is 2:5 108 y, which is 7:88 1015 s, so (42 )(2:2 1020 m)3 = 5:1 1010 : N= (6:67 10 11 m3 /s2 kg)(7:88 1015 s)2 (2:0 1030 kg) Since T / r3=2 , the period of the satellite is rs 3=2 = T 1 3=2 = 0:35T ; Ts = Tm r m 2 m m 57E GT 2 M 58E which is about 0.35 lunar months. (a) If R is the radius of the orbit then the magnitude of the gravitational force acting on the satellite is given by GMm=R2 , where M is the mass of the Earth and m is the mass of the satellite. The magnitude of the acceleration of the satellite is given by v2 =R, where v is its speed. Newton's second law yields GMm=R2 = mv2 =R. Since the radius of the Earth is 6:37 106 m the orbit radius is R = 6:37 106 m + 160 103 m = 6:53 106 m. The solution for v is r r GM = (6:67 10 11 m3 =kg s2 )(5:98 1024 kg) = 7:82 103 m/s : v= R 6:53 106 m (b) Since the circumference of the circular orbit is 2R, the period is T = 2R=v = (2)(6:53 106 m)=(7:82 103 m/s) = 5:25 103 s. This is 87:5 min. From the gure, we estimate the average distance from the asteroid to the Sun to be about 2.5 times the EarthSun distance. Thus from Kelper's law of periods, the period of the asteroid is approximately
60E 59E Ta = Te (2:5)3=2 = (1:0 y)(2:5)3=2 4:0 y : 404 CHAPTER 14 GRAVITATION (a) The greatest distance between the satellite and the Earth's center (the apogee distance) is Ra = 6:37 106 m + 360 103 m = 6:73 106 m. The least distance (perigee distance) is Rp = 6:37 106 m + 180 103 m = 6:55 106 m. Here 6:37 106 m is the radius of the Earth. Look at Fig. 1414 to see that the semimajor axis is a = (Ra + Rp )=2 = (6:73 106 m + 6:55 106 m)=2 = 6:64 106 m. (b) The apogee and perigee distances are related to the eccentricity e by Ra = a(1 + e) and Rp = a(1 e). Thus 6 6 73 106 R e = Ra + Rp = 6::73 106 m + 6::55 106 m = 0:0136 : m 6 55 10 m a Rp The separation between the two focuses is obtained from Fig. 1414 to be d = 2ea = 2(0:0167)(1:50 1011 m) = 5:01 109 m 9 = 5::01 108 m ( solar radii) = 7:20 solar radii : 6 69 10 m Apply Keppler's third law (Eq. 1431) to the EarthSun system to solve for G: G = (42 =T 2 M )r3 , where T = 1 year, r = 1 AU, and M = 1Ms . Thus G = 42 (AU)3 = y2 Ms 39:5 (AU)3 = y2 Ms : ~ ~ ~ ~ ~ (b) Let T = T y, r = r AU and M = MMs , where T , r and M are dimensionless quantities, ~ then Keppler's third law can be written in the form 2 2 3 ~3 2 = T 2 y2 = 4 r3 = 4 ( AU) r : T ~ ~ GM GMs M Using the result for G in (a), this becomes 3 ~ ~ T 2 = r~ : M The period T of a geosynchronous orbit is equal to 86400 s (one day). From the law of periods T 2 = (42 =GMe )r3 , we may solve for r: T 2 GMe 1=3 r = 42 2 = (86400 s) (6:67 10 61E 62E 63E 64E 11 m3 =kg s2 )(5:98 1024 kg) 1=3 42 = 4:23 107 m = 4:23 104 km : CH AP TER Thu
65E 14 s its GR =r Use Re = (a) F the res 4:2 10 10 rom E ult of 10 4 k m (b) s, whi q. 14 Sampl ch y 33 a e Pr 6:37 ields = ( oble 10 3 a = GM T 2 m 14 km = In te 1:9 =4 2 8. 3:58 rms 10 13 ) 1=3 of R Ra = (1 m. . In p , th + e) our is is 66E case Ra = a = (1 + T (a) U 0:11 (2:1 1994 se E )(1:9 10 13 q. 14 10 13 574 )(R 31 t = 14 p =59 m) = o sol 20 01 ve fo 2:1 M = 4 2 0 10) rM 10 13 r3 = 3: (b) : m: GT 2 = 5Rp : (6:67 10 4 2 11 m 3= (100 km 3 kg s 2 ) )[(27 = M 67P h)(3 61 V 600 s /h)] 2 = 14; 1 0 16 kg 61 00 km 3 =4 0 16k 10 3 g: kg/m 3 :
horizontal line Chicago altit AV ITA TIO ude N is h R r' satellite 47.5o From dir the r theoection a gure rem nd t show south ilted n w up, m e see akin that Sinc g an the er= angl satel 4:23 e w lite a 10 7 ith t nten cos(9 m (f he h na sh 0 o+ rom orizo ould ) = r 02 64E) and r 2 + ntal leve be poin l. Th ted 2Rr 0 R 2 en fr in th : om t e sou he c th osine 406 CHAPTER 14 GRAVITATION r0 = R2 + r2 2Rr cos ;
we nd p = cos 1 r02 r2 + R2 2Rr0 90 = 35:4 : (a) 68P loga 9.6 9.2 8.8 8.4 8.0 5.0 5.2 5.4 5.6 5.8 6.0 6.2 logT (b) The slope of the line is 2=3, as expected from Kepler's third law. (c) The intercept is obtained directly from the plot above to be b = 0:47. Since T 2 = (42 =GM )a3 ; we have log a = 1:5 log T (1=3) log(42 =GM ): the intercept is thus b = (1=3) log(42 =GM ), which gives 42 103b = (42 )[103(0:47) ] = 2:0 1027 kg : M= G 5:0 10 25 Note that we have converted G from 6:67 10 11 m3 =kgs2 to 5:0 10 unit of length is 108 m and that of time is days.
69P
25 units, where the Let the force holding the Moon be F = crn , where c is a constant, r is the EarthMoon distance, and n is a number to be determined. Thus F 2 = crn = Fc = Mm 42 Tm r; CHAPTER 14 GRAVITATION 407 2 2 or Tm = (42 Mm =c)r1 n / r1 n . Comparing this with Kepler's third law: Tm / r3 , we 2: get 3 = 1 n, or n = 2. Thus F / r The centripetal force is porvided by the gravitational force:
2 2 Gm2 Fc = ms v = 4Tms r = Fg = (2r)s ; 2 2 r 70P which yields T 2 = (162 r3 =GMs ) = 1 Te2 ; i.e., T = Te = 2 0:71 y : 2 Consider the rotation of one of the stars. Again, the centripetal force Fc is provided by the gravitational force Fg . The only thing new here is that Fg now includes two terms, one from the star at the center of the rotation, and the other from the companion star which is also in rotation. Thus
2 2 Gm2 Fc = mv = 4T mr = GMm + (2r)2 = GMe m ; 2 r r2 r2 p 71P where Me = M + m=4 is the eective mass, which means that if the star in consideration were to rotate about a single star of mass Me it would have the same period as in the current case. Thus from Keppler's third law
3 r T = 2 GM = 2 G(M r m=4) : + e
r s (a) The escape speed is
11 3 2 30 v = 2GMs = 2(6:67 10 1:m =kg s11)(1:99 10 kg) R 50 10 m 4 m/s = 42:1 km/s : = 4:21 10
r r 72P (b) The Earth's orbital speed is equal to v= 2 (see 40E). Thus the additional speed needed is 1 1 v = 1 p (42:1 km/s) = 12:3 km/s : v = 1 p 2 2 (c) Let the initial speed required for the object of mass m be v0 . When the object is far away 2 2 from the Earth, its speed (relative to the Earth) reduces to v1 , where K = 1 mv0 1 mv1 = 2 2 p 408 CHAPTER 14 GRAVITATION U = GmMe =Re : But v1 should be large just enough so that v1 = v = 12:3 km/s: Thus the speed required is v0 = (v)2 + 2GMe R
r r 11 3 kg s2 24 = (1:23 104 m/s)2 + 2(6:67 10 6m = 106)(5:98 10 kg) :37 m 4 m/s = 16:6 km/s : = 1:66 10 e Each star is attracted toward each of the other two by a force of magnitude GM 2 =L2 , along the line that joins the stars. The net force on each star has magnitude (GM 2 =L2 ) cos 30 and is directed toward the center of the triangle. This is a centripetal force and keeps the stars on the same circular orbit if their speeds are appropriate. If R is the radius of the orbit Newton's second law yields (GM 2 =L2 ) cos 30 = Mv2 =R. The stars rotate about their center of mass, at the intersection of the perpendicular bisectors y of the triangle sides, and the radius of the orbit is the distance from a star to the center of mass of the threestar system. Put a coordinate system as shown in the diagram to the right, with its origin at the leftmost p The altitude of star. L L an equilateral triangle is ( 3=2)L, so the stars are located atp = 0, y = 0; x = L, y = 0; and x x = L=2, y = 3L=2. The x coordinate of the + and center of mass is xc = (L p L=2)=3 = L=2 p the y coordinate is yc = ( 3L=2)=3 = L=2 3. x L The distance from a star to the center of mass p p 2 is p = x2 + yc = (L2 =4) + (L2 =12) = R c L= 3. 2 2 Once p the substitution for R is made Newton's second law becomes (2GM =Lp) cos 30 = 3Mv2 =L. This can be simpli ed somewhat by recognizing that cos 30 = 3=2. Also p divide the equation by M . Then GM=L2 = v2 =L and v = GM=L. Use T = 42 =(GMe r3=2 : For the intended geosynchronous orbit T = T0 = 86400 s and r = r0 = 4:23 104 km (see 64E), and for the actual orbit
s 73P 74P* p T= 42 (r + r)3=2 = T 1 + r 3=2 T 1 + 3r : 0 0 GM 0 r 2r e 0 0 CHAPTER 14 GRAVITATION 409 Thus the dierence in periods is T = T T0 = T0 (3r=2r0 ): This means that the satellite will move slower than the one in the geosynchronous orbit. Its projected point on Earth will therefore move from east to west, with a speed of v = 2Re T 2Re 2Re T = 3Re r T0 T02 r0 T0 0 6 m)(1:0 103 m) = 3(6:37 10 7 m)(86400 s) = 1:6 10 2 m/s = 1:6 cm/s : (4:23 10 (a) Use the law of periods: T 2 = (42 =GM )r3 , where M is the mass of the Sun (1:99 1030 kg) and r is the radius of the orbit. The radius of the orbit is twice the radius of the Earth's orbit: r = 2re = 2 150 109 m = 300 109 m. Thus 75E T= r 42 r3 s GM = 42 (300 109 m)3 = 8:96 107 s ; 11 m3 /s2 kg)(1:99 1030 kg) (6:67 10 which is 2:8 y. (b) The kinetic energy of any satellite in a circular orbit of radius r is given by K = GMm=r, where m is the mass of the satellite . Notice that it is proportional to m and inversely proportional to r. The ratio of the kinetic energy of the asteroid to the kinetic energy of the Earth is K=Ke = (m=m2 )(re =r). Substitute m = 2:0 10 4 me and r = 2re to obtain K=Ke = 1:0 10 4 . (a) The energy before collision is EA + EB = 2EA = GMe m=r: (b) Immediately after collision v = 0. Thus E = U = GMe (2m)=r: (c) The wreckage will fall radially towards the Earth. (a)
77P 76E (b) Since K = U=2 (see Eqs. 1420 and 1440), UB = GmMe =rB = rA = 4000 + 4000 = 1 : UA GmMe =rA rB 4000 + 12000 2 KB = UB =2 = 1 : KA UA =2 2 410 CHAPTER 14 GRAVITATION (c) EA = KA + KB = U2A = GmMe 2rA 11 N m2 = kg2 )(5:98 1024 kg)(14:6 kg) 10 = (6:67 2(4; 000 mi + 4; 000 mi)(1609 m/mi) = 2:3 108 J ; EB = U2B = UB EA = EA = 1:1 108 J : UA 2
Thus EB > EA , and EB EA = EA =2 = 1:1 108 J: The total energy is given by E = GMm=2a, where M is the mass of the central attracting body (the Sun, for example), m is the mass of the object (a planet, for example), and a is the semimajor axis of the orbit. If the object is a distance r from the central body the potential energy is U = GMm=r. Write 1 mv2 for the kinetic energy. Then E = K + U 2 becomes GMm=2a = 1 mv2 GMm=r. Solve for v2 . The result is 2
78P 2 1 v = GM r a :
2 (a) For r = Rp ,
2 79P 2 1 vp = GMs R a p 2 1 11 3 2 30 = (6:67 10 m =kg s )(1:99 10 kg) 8:9 1010 m 2:7 1012 m = 5:4 104 m/s : (b) For r = Ra ,
2 2 1 va = GMs R a a 2 1 11 3 2 30 = (6:67 10 m =kg s )(1:99 10 kg) 5:3 1012 m 2:7 1012 m = 9:6 102 m/s : CHAPTER 14 GRAVITATION 411 (c) Since L = mvr = mva Ra = mvp Rp = const., we have va =vp = Rp =Ra . The energy required to put a satellite of mass m to an orbit of altitude h is given by
80P 1 E1 = U = GMe m R e 1 Re + h ; and the energy required to put it in circular orbit once it is there is 1 2 e E2 = 2 mvorb = 2(GM+m ) : R h
e the dierence in these two energy is E = E1 (a) Since 1 1 E2 = GMe m R e 3 2(Re + h) : we have E1 = E2 . (c) Since 3 1 = 4000 mi Re 2(Re + h) the answer is no (E1 < E2 ): (b) Since 3 1 1 = 4000 mi R 2(R + h)
e e 3 2(4000 mi + 1000 mi) < 0 3 2(4000 mi + 2000 mi) = 0 3 2(4000 mi + 3000 mi) > 0 3 1 = 4000 mi Re 2(Re + h) the answer is yes (E1 > E2 ): 1
81P (a) From Eq. 1431 we may nd the speed p of an object of mass m orbiting in an orbit v of radius r about the Earth: v = 2r=T = GM=r. This expression is valid for both the satellite and the pellet, as v is independent ofp mass m. The relative speed of the pellet the approaching the satellite is thus vr = 2v = 2 GM=r. Therefore the kinetic energy of the pellet relative to the satellite is given by 1 2 Kp = 2 mp vr
11 N 2 2 )(5: 1 1024 = 2 (4:0 kg)(4) (6:67 :10 106m = kg:00 98 m kg) = 4:6 108 J : 6 37 m+5 105 412 CHAPTER 14 GRAVITATION (b) The kinetic energy of the bullet is 1 2 1 Kb = 2 mb vb = 2 (4:0 kg)(950 m/s)2 = 1:8 106 J ; i.e., Kp = (4:6 108 ) J = 2:6 102 : k (1:8 106 ) J
b (a) T / r3=2 (from Kepler's law of periods). 1 2 (b) K = 2 mvorb = GMre m / r 1 : 2 1=2 r / (r 1 )1=2 r = r 1=2 : (c) L = mvorb r / K (d) vorb / K 1=2 / r 1=2 : (a) The force acting on the satellite has magnitude GMm=r2 , where M is the mass of the Earth, m is the mass of the satellite, and r is the radius of the orbit. The force points toward the center of the orbit. Since the acceleration of the satellite is v2 =r, where v is its speed, Newton's second law yields GMm=r2 = mv2 =r and the speed p is given by v = GM=r. The radius of the orbit is the sum of the Earth's radius and p altitude of the satellite: r = 6:37 106 + 640 103 = 7:01 106 m. Thus the v = (6:67 10 11 )(5:98 1024 )=(7:01 106 ) = 7:54 103 m/s. (b) The period is T = 2=v = (2)(7:01 106 )=(7:54 103 ) = 5:84 103 s. This is 97:4 min. (c) If the total energy is known E = GMm=2r can be used to calculate the radius of the orbit. If E0 is the initial energy then the energy after n orbits is E = E0 nC , where C = 1:4 105 J/orbit, and the radius after n orbits is r = GMn=2E . The initial energy is
11 :98 1024 E0 = ( 6:67 10 :)(5 106 ) )(220) = 6:26 109 J ; (2)(7 01 82P 83P the energy after 1500 orbits is E = E0 nC = 6:26 109 1500 1:4 105 = 6:47 109 J :
and the orbit radius after 1500 orbits is 11 98 14 r = (6:67 10 6:)(5: 10910 )(220) = 6:78 106 m : 26 the altitude is h = r R = 6:78 106 6:37 106 = 4:1 105 m. Here R is the radius of the Earth. CHAPTER 14 GRAVITATION 413 r GM = (6:67 10 11 )(5:98 1024 ) = 7:67 103 m/s: v= r 6:786 (e) The period is T = 2r=v = (2)(6:78 106 )=(7:67 103 ) = 5:6 103 s. This is 93 min. (f) Let F be the magnitude of the average force and s be the distance traveled by the satellite. Then the work done by the force is W = Fs. This is the change in energy: Fs = E . Thus F = E=s. Evaluate this expression for the rst orbit. For a complete orbit s = 2r = 2(7:01 106 ) = 4:40 107 m and E = 1:4 105 J. Thus F = 1:4 105 =4:40 107 = 3:3 10 3 N. r (d) The speed is (g) The resistive force exerts a torque on the satellite, so its angular momentum is not conserved. The total angular momentum of the system consisting of the satellite and the source of the resistive force is conserved. (a) There should be no change in the total energy, since it is conserved. (b)
84P 1 1 GM M r e s U = GMe Ms r r r2 1 2 11 m3 =kg s2 )(5:98 1024 kg)(1:99 1030 kg)(5:0 109 m) = (6:67 10 (1:50 1011 m)2 = 1:8 1032 J : (c) Since E = K + U = 0; we have jK j = jU j = 1:8 1032 J: (d) Use J = mvr =const., which gives J=m = vr + rv = 0: Thus 109 jvj vr = (29:7 km/s)(5:011 m m) = 0:99 km/s : r 1:50 10
(a) Use Eq. 1431:
85P s r3 = 2 (6:80 106 m)3 3 T = 2 GM (6:67 10 11 m3 =kg s2 )(5:98 1024 kg) = 5:58 10 s : r
:80 10 r v = 2T = 2(658 103 sm) = 7:66 103 s : 5:
6 (b) (c) Now v becomes v0 = (1 1:00%)(7:66 103 s) = 7:58 103 s. 414 CHAPTER 14 GRAVITATION (d) (e) 1 1 K = 2 mv02 = 2 (2000 kg)(7:58 103 s)2 = 5:75 1010 J :
11 U = GMm = (6:67 10 r
(f) (g) From Eq. 1444 m3 =kg s2 )(5:98 1024 kg)(2000 kg) = 11:7 1010 J : 6:80 106 m E = K + U = 5:75 1010 J 11:7 1010 J = 5:98 1010 J :
11 a = GMm = (6:67 10 2E m3 =kg s2 )(5:98 1024 kg)(2000 kg) = 6:67 106 m : 2( 5:98 1010 J) (h) The new period of motion is given by s a3 = 2 (6:67 106 m)3 3 T 0 = 2 GM (6:67 10 11 m3 =kg s2 )(5:98 1024 kg) = 5:42 10 s ; r so the period of motion of the new orbital is smaller than that of the original one by T = 5:58 103 s 5:42 103 s = 1:6 102 s. The apparent acceleration of gravity as observed by the person in the elevator is ga = 220 N=60 kg = 3:67 m/s2 . Thus from s = 1 ga t2 we nd the timeof
ight to be 2 1 m) t = 2s = 32(2:m/s2 = 1:1 s : ga :67
87P 86E r s In the reference frame xed on the spaceship the acceleration of the electron is given by a = (2:5 m/s2 ) i. The timeof
ight for the electron moving across the ship is t = w=v0 , where w = 3:0 cm. (a) The xcomponent of the displacement of the electron is given by 1 :0 1 x = 2 at2 = 2 ( 2:5 m/s2 ) 0340 cm : m/s 2 = 7:0 mm ; while the ycomponent is simply y = w = 3:0 cm. Thus the displacement vector is r = x i + y j = ( 7:0 mm) i + (3:0 cm) j. CHAPTER 14 GRAVITATION 415 (c) Just before hitting the wall, the xcomponent of the velocity of the electron is given by 3:0 cm vx = at = ( 2:5 m/s ) 0:40 m/s = 0:19 m/s ;
2 while the ycomponent remains vy = v0 = 0:40 m/s. Thus v = vx i + vy j = ( 0:19 m/s) i + (0:40 m/s) j. (a) speed increased from 3:51 104 m/s to 3:78 104 m/s (v = 2:7 103 m/s), energy increased from 3:69 1012 J to 3:09 1012 J (E = 6:0 1011 J), semimajor axis: 1:29 1011 m, semiminor axis: 1:27 1011 m; (b) speed increased from 2:73 104 m/s to 2:98 104 m/s (v = 2:5 103 m/s), energy increased from 3:09 1012 J to 2:66 1012 J (E = 4:3 1011 J) (a) 1:98 1030 kg; (b) 1:96 1030 kg
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.
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