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 innity) 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 specied 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 innitesimal 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 satises
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 innity. 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. T...
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 Fall '07
 GROUPTEST
 Energy, Potential Energy, General Relativity, kg

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