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# Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch13

(Parte **3** de 4)

L, y = 0; and x = L/2, 3/2yL=. The x coordinate of the center of mass is xc = (L +

L/2)/3 = L/2 and the y coordinate is ()3/2/3/23cyLL==. The distance from a star to the center of mass is

Once the substitution for R is made Newton’s second law becomes ()2/cos303/GMLMvL°=. This can be simplified somewhat by recognizing that cos303/2°=, and we divide the equation by M. Then, GM/L2 = v2/L and /vGML= .

60. Although altitudes are given, it is the orbital radii which enter the equations. Thus, rA = (6370 + 6370) km = 12740 km, and rB = (19110 + 6370) km = 25480 km

(a) The ratio of potential energies is

U GmM r r U GmM r r

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(b) Using Eq. 13-38, the ratio of kinetic energies is

K GmM r r

(c) From Eq. 13-40, it is clear that the satellite with the largest value of r has the smallest value of |E| (since r is in the denominator). And since the values of E are negative, then the smallest value of |E| corresponds to the largest energy E. Thus, satellite B has the largest energy.

(d) The difference is

GmME E E

Being careful to convert the r values to meters, we obtain ∆E = 1.1 × 108 J. The mass M of Earth is found in Appendix C.

61. (a) We use the law of periods: T2 = (4pi2/GM)r3, where M is the mass of the Sun (1.9 × 1030 kg) and r is the radius of the orbit. The radius of the orbit is twice the radius of

Earth’s orbit: r = 2re = 2(150 × 109 m) = 300 × 109 m. Thus,

GM pi pi−

Dividing by (365 d/y) (24 h/d) (60 min/h) (60 s/min), we obtain T = 2.8 y.

(b) The kinetic energy of any asteroid or planet in a circular orbit of radius r is given by K = GMm/2r, where m is the mass of the asteroid or planet. We note 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 Earth is K/Ke = (m/me) (re/r). We substitute m = 2.0 × 10−4me and r = 2re to obtain K/Ke = 1.0 × 10−4.

62. (a) Circular motion requires that the force in Newton’s second law provide the necessary centripetal acceleration:

2 GmM vm

Since the left-hand side of this equation is the force given as 80 N, then we can solve for the combination mv2 by multiplying both sides by r = 2.0 × 107 m. Thus, mv2 = (2.0 × 107 m) (80 N) = 1.6 × 109 J. Therefore,

(b) Since the gravitational force is inversely proportional to the square of the radius, then

Thus, F = (80 N) (2/3)2 = 36 N.

63. The energy required to raise a satellite of mass m to an altitude h (at rest) is given by

E U GM m and the energy required to put it in circular orbit once it is there is

GM mE mv

Consequently, the energy difference is

E E E GM m

(a) Solving the above equation, the height h0 at which 0E∆= is given by

130 | 3.1910 m. |

(b) For greater height 0hh>, 0E∆> implying 12EE>. Thus, the energy of lifting is greater.

64. (a) From Eq. 13-40, we see that the energy of each satellite is −GMEm/2r. The total energy of the two satellites is twice that result:

A B GM mE E E

(b) We note that the speed of the wreckage will be zero (immediately after the collision), so it has no kinetic energy at that moment. Replacing m with 2m in the potential energy expression, we therefore find the total energy of the wreckage at that instant is

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EGM mE r

(c) An object with zero speed at that distance from Earth will simply fall towards the Earth, its trajectory being toward the center of the planet.

65. (a) From Kepler’s law of periods, we see that T is proportional to r3/2. (b) Eq. 13-38 shows that K is inversely proportional to r.

(c) and (d) From the previous part, knowing that K is proportional to v2, we find that v is proportional to 1/r. Thus, by Eq. 13-31, the angular momentum (which depends on the product rv) is proportional to r/r = r.

6. (a) The pellets will have the same speed v but opposite direction of motion, so the relative speed between the pellets and satellite is 2v. Replacing v with 2v in Eq. 13-38 is equivalent to multiplying it by a factor of 4. Thus,

EGM mK r

67. (a) The force acting on the satellite has magnitude GMm/r2, where M is the mass of 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 is given by v =

/GMr. The radius of the orbit is the sum of Earth’s radius and the altitude of the satellite: r = (6.37 × 106 + 640 × 103) m = 7.01 × 106 m. Thus,

(b) The period is T = 2pir/v = 2pi(7.01 × 106 m)/(7.54 × 103 m/s) = 5.84 × 103 s ≈ 97 min.

(c) If E0 is the initial energy then the energy after n orbits is E = E0 − nC, where C = 1.4 × 105 J/orbit. For a circular orbit the energy and orbit radius are related by E = −GMm/2r, so the radius after n orbits is given by r = −GMm/2E. The initial energy is

the energy after 1500 orbits is

The altitude is h = r − R = (6.78 × 106 m − 6.37 × 106 m) = 4.1 × 105 m. Here R is the radius of Earth. This torque is internal to the satellite-Earth system, so the angular momentum of that system is conserved.

(d) The speed is

GMv r

(e) The period is 6 pi pi ×

(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. We evaluate this expression for the first orbit. For a complete orbit s = 2pir = 2pi(7.01 × 106 m) = 4.40 × 107 m, and ∆E = −1.4 × 105 J. Thus,

(g) The resistive force exerts a torque on the satellite, so its angular momentum is not conserved.

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(h) The satellite-Earth system is essentially isolated, so its momentum is very nearly conserved.

(a) Using Kepler’s law given in Eq. 13-34, we find the period of the ships to be

GM pi pi−

(b) The speed of the ships is pi pi ×

(c) The new kinetic energy is

(d) Immediately after the burst, the potential energy is the same as it was before the burst. Therefore,

GMmU r

(f) For elliptical orbit, the total energy can be written as (see Eq. 13-42) /2EGMma=−, where a is the semi-major axis. Thus,

GMma E

(g) To find the period, we use Eq. 13-34 but replace r with a. The result is

GM pi pi−

(h) The orbital period T for Picard’s elliptical orbit is shorter than Igor’s by

0 5540s 5370s 170sT T T∆ = − = − = . Thus, Picard will arrive back at point P ahead of Igor by 170 s – 90 s = 80 s.

69. We define the “effective gravity” in his environment as geff = 220/60 = 3.67 m/s2. Thus, using equations from Chapter 2 (and selecting downwards as the positive direction), we find the “fall-time” to be

70. We estimate the planet to have radius r = 10 m. To estimate the mass m of the planet, we require its density equal that of Earth (and use the fact that the volume of a sphere is

m M rm M r R R which yields (with ME ≈ 6 × 1024 kg and RE ≈ 6.4 × 106 m) m = 2.3 × 107 kg. (a) With the above assumptions, the acceleration due to gravity is

Gma r

(b) Eq. 13-28 gives the escape speed:

71. Using energy conservation (and Eq. 13-21) we have

K1 – = K2 – .

Plugging in two pairs of values (for (K1 ,r1) and (K2 ,r2)) from the graph and using the value of G and M (for earth) given in the book, we find

72. (a) The gravitational acceleration ag is defined in Eq. 13-1. The problem is concerned with the difference between ag evaluated at r = 50Rh and ag evaluated at r =

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50Rh + h (where h is the estimate of your height). Assuming h is much smaller than 50Rh then we can approximate h as the dr which is present when we consider the differential of

Eq. 13-1:

|dag| = dr ≈ h = h .

If we approximate |dag| = 10 m/s2 and h ≈ 1.5 m, we can solve this for M. Giving our results in terms of the Sun’s mass means dividing our result for M by 2 × 1030 kg. Thus, admitting some tolerance into our estimate of h we find the “critical” black hole mass should in the range of 105 to 125 solar masses.

(b) Interestingly, this turns out to be lower limit (which will surprise many students) since the above expression shows |dag| is inversely proportional to M. It should perhaps be emphasized that a distance of 50Rh from a small black hole is much smaller than a distance of 50Rh from a large black hole.

73. The magnitudes of the individual forces (acting on mC, exerted by mA and mB respectively) are

Gm m Gm mF F r r where rAC = 0.20 m and rBC = 0.15 m. With rAB = 0.25 m, the angle AFr makes with the x axis can be obtained as

r r r r r

Similarly, the angle BFr makes with the x axis can be obtained as

r r r r r

The net force acting on mC then becomes

74. The key point here is that angular momentum is conserved: Ipωp = Iaωa which leads to 2(/)paparrωω=, but rp = 2a – ra where a is determined by Eq. 13-34 (particularly, see the paragraph after that equation in the textbook). Therefore, ωp = = 9.24 × 10−5 rad/s . 75. (a) Using Kepler’s law of periods, we obtain

(b) The speed is constant (before she fires the thrusters), so vo = 2pir/T = 1.23 × 104 m/s.

(c) A two percent reduction in the previous value gives v = 0.98vo = 1.20 × 104 m/s. (d) The kinetic energy is K = ½mv2 = 2.17 × 1011 J.

(e) The potential energy is U = −GmM/r = −4.53 × 1011 J.

(f) Adding these two results gives E = K + U = −2.35 × 1011 J. (g) Using Eq. 13-42, we find the semi-major axis to be

(h) Using Kepler’s law of periods for elliptical orbits (using a instead of r) we find the new period is

This is smaller than our result for part (a) by T − T = 1.2 × 103 s. (i) Elliptical orbit has a smaller period.

(b) Although a close answer may be gotten by using the constant acceleration equations of Chapter 2, we show the more general approach (using energy conservation):

CHAPTER 13 o oK U K U+ = + where Ko = 0, K = ½mv2 and U given by Eq. 13-21. Thus, with ro = 10001 m, we find

7. We note that rA (the distance from the origin to sphere A, which is the same as the separation between A and B) is 0.5, rC = 0.8, and rD = 0.4 (with SI units understood). The force kFr that the kth sphere exerts on mB has magnitude 2/kBkGmmr and is directed from the origin towards mk so that it is conveniently written as

k k k k

Gm m x y Gm mF x y

Consequently, the vector addition (where k equals A,B and D) to obtain the net force on mB becomes

k k k k m x m yF F Gm r r

78. (a) We note that rC (the distance from the origin to sphere C, which is the same as the separation between C and B) is 0.8, rD = 0.4, and the separation between spheres C and D is rCD = 1.2 (with SI units understood). The total potential energy is therefore

r r r using the mass-values given in the previous problem.

(b) Since any gravitational potential energy term (of the sort considered in this chapter) is necessarily negative (−GmM/r2 where all variables are positive) then having another mass to include in the computation can only lower the result (that is, make the result more negative).

(c) The observation in the previous part implies that the work I do in removing sphere A (to obtain the case considered in part (a)) must lead to an increase in the system energy; thus, I do positive work.

(d) To put sphere A back in, I do negative work, since I am causing the system energy to become more negative.

79. We use F = Gmsm/r2, where ms is the mass of the satellite, m 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,

80. (a) Since the volume of a sphere is 4piR3/3, the density is total total

R Rρ pi pi

When we test for gravitational acceleration (caused by the sphere, or by parts of it) at radius r (measured from the center of the sphere), the mass M which is at radius less than r is what contributes to the reading (GM/r2). Since M = ρ(4pir3/3) for r ≤ R then we can write this result as

3 total 3

total 2 3

R GM r r R when we are considering points on or inside the sphere. Thus, the value ag referred to in the problem is the case where r = R:

total2=g GMa , and we solve for the case where the acceleration equals ag/3:

total total

GM GM r Rr R R = ⇒ =

(b) Now we treat the case of an external test point. For points with r > R the acceleration is GMtotal/r2, so the requirement that it equal ag/3 leads to total total

GM GM r R

81. Energy conservation for this situation may be expressed as follows:

CHAPTER 13 where M = 5.98 × 1024 kg, r1 = R = 6.37 × 106 m and v1 = 10000 m/s. Setting v2 = 0 to find the maximum of its trajectory, we solve the above equation (noting that m cancels in the process) and obtain r2 = 3.2 × 107 m. This implies that its altitude is r2 − R = 2.5 × 107 m.

82. (a) Because it is moving in a circular orbit, F/m must equal the centripetal acceleration:

50 kg v r

But v = 2pir/T, where T = 21600 s, so we are led to

which yields r = 1.9 × 107 m.

(b) From the above calculation, we infer v2 = (1.6 m/s2)r which leads to v2 = 3.0 × 107 m2/s2. Thus, K = ½mv2 = 7.6 × 108 J.

(c) As discussed in § 13-4, F/m also tells us the gravitational acceleration:

We therefore find M = 8.6 × 1024 kg.

83. (a) We write the centripetal acceleration (which is the same for each, since they have identical mass) as rω2 where ω is the unknown angular speed. Thus,

G M M GM Mr r ω= =

(b) To barely escape means to have total energy equal to zero (see discussion prior to Eq. 13-28). If m is the mass of the meteoroid, then

84. See Appendix C. We note that, since v = 2pir/T, the centripetal acceleration may be written as a = 4pi2r/T2. To express the result in terms of g, we divide by 9.8 m/s2.

(a) The acceleration associated with Earth’s spin (T = 24 h = 86400 s) is

(b) The acceleration associated with Earth’s motion around the Sun (T = 1 y = 3.156 ×

(c) The acceleration associated with the Solar System’s motion around the galactic center (T = 2.5 × 108 y = 7.9 × 1015 s) is

85. We use m1 for the 20 kg of the sphere at (x1, y1) = (0.5, 1.0) (SI units understood), m2 for the 40 kg of the sphere at (x2, y2) = (−1.0, −1.0), and m3 for the 60 kg of the sphere at

(x3, y3) = (0, −0.5). The mass of the 20 kg object at the origin is simply denoted m. We note that 121.25,2rr==, and r3 = 0.5 (again, with SI units understood). The force nFr that the nth sphere exerts on m has magnitude 2/nnGmmr and is directed from the origin towards mn, so that it is conveniently written as

n n n n

Gm m x y Gm mF x y

Consequently, the vector addition to obtain the net force on m becomes

n n n n m x m yF F Gm r r in SI units. Therefore, we find the net force magnitude is 7 net 3.2 10 NF −= ×r .

86. We apply the work-energy theorem to the object 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 final speed vf. The corresponding increase in its kinetic energy, ½mvf2, is equal to the work done on it by Earth’s gravity: ()FdrKrdr=−∫∫ (using the notation of that Sample Problem referred to in the problem statement). Thus,

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where R is the radius of Earth. Solving for the final speed, we obtain vf = R /Km. We note that the acceleration of gravity ag = g = 9.8 m/s2 on the surface of Earth is given by where ρ is Earth’s average density. This permits us to write K/m = 4piGρ/3 = g/R. Consequently,

(Parte **3** de 4)