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

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

(Parte 2 de 4)

using the values found in Appendix C. (b) Similarly, for the Jupiter escape energy (divided by that for Earth) we obtain

32. (a) The potential energy at the surface is (according to the graph) –5.0 × 109 J, so (since U is inversely proportional to r – see Eq. 13-21) at an r-value a factor of 5/4 times what it was at the surface then U must be a factor of 4/5 what it was. Thus, at r = 1.25Rs U = – 4.0 × 109 J. Since mechanical energy is assumed to be conserved in this problem, we have K + U = –2.0 × 109 J at this point. Since U = – 4.0 × 109 J here, then

(b) To reach the point where the mechanical energy equals the potential energy (that is, where U = – 2.0 × 109 J) means that U must reduce (from its value at r = 1.25Rs) by a factor of 2 – which means the r value must increase (relative to r = 1.25Rs) by a corresponding factor of 2. Thus, the turning point must be at r = 2.5Rs .

3. The equation immediately preceding Eq. 13-28 shows that K = –U (with U evaluated at the planet’s surface: –5.0 × 109 J) is required to “escape.” Thus, K = 5.0 × 109 J.

CHAPTER 13 34. The gravitational potential energy is

r r which we differentiate with respect to m and set equal to zero (in order to minimize). Thus, we find M − 2m = 0 which leads to the ratio m/M = 1/2 to obtain the least potential energy.

Note that a second derivative of U with respect to m would lead to a positive result regardless of the value of m − which means its graph is everywhere concave upward and thus its extremum is indeed a minimum.

35. (a) The work done by you in moving the sphere of mass mB equals the change in the potential energy of the three-sphere system. The initial potential energy is

A C B CA Bi Gm m Gm mGm mU and the final potential energy is

.A C B CA Bf Gm m Gm mGm mU

The work done is f i B A C B A C B A C

W U U Gm m m d L d L d d

L d d L L dGm m m Gm m m d L d d L d d L d

(b) The work done by the force of gravity is −(Uf − Ui) = −5.0 × 10−13 J.

36. (a) From Eq. 13-28, we see that 0/2EvGMR= in this problem. Using energy conservation, we have mvo2 – GMm/RE = – GMm/r which yields r = 4RE/3. So the multiple of RE is 4/3 or 1.3.

(b) Using the equation in the textbook immediately preceding Eq. 13-28, we see that in this problem we have Ki = GMm/2RE, and the above manipulation (using energy conservation) in this case leads to r = 2RE. So the multiple of RE is 2.0.

(c) Again referring to the equation in the textbook immediately preceding Eq. 13-28, we see that the mechanical energy = 0 for the “escape condition.”

37. (a) We use 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 fired upward. The initial kinetic energy is 212mv. The particle just escapes if its kinetic energy is zero when it is infinitely far from the asteroid. The final potential and kinetic energies are both zero. Conservation of energy yields

−GMm/R + ½mv2 = 0.

We replace GM/R with agR, where ag is the acceleration due to gravity at the surface. Then, the energy equation becomes −agR + ½v2 = 0. We solve for v:

(b) Initially the particle is at the surface; the potential energy is Ui = −GMm/R and the kinetic energy is Ki = ½mv2. Suppose the particle is a distance h above the surface when it momentarily comes to rest. The final potential energy is Uf = −GMm/(R + h) and the final kinetic energy is Kf = 0. Conservation of energy yields

GMm GMmmv

We replace GM with agR2 and cancel m in the energy equation to obtain a R a R v

The solution for h is a R h R a R v

(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

CHAPTER 13 asteroid its potential energy is Uf = −GMm/R. Write 21 2fmv for the final kinetic energy.

Conservation of energy yields

We substitute agR2 for GM and cancel m, obtaining a R a R v

The solution for v is a R v a R

R h

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


where M = 5.0 × 1023 kg, r1 = R = 3.0 × 106 m and m = 10 kg. (a) If K1 = 5.0 × 107 J and r2 = 4.0 × 106 m, then the above equation leads to

(b) In this case, we require K2 = 0 and r2 = 8.0 × 106 m, and solve for K1:

39. (a) The momentum of the two-star system is conserved, and since the stars have the same mass, their speeds and kinetic energies are the same. We use the principle of conservation of energy. The initial potential energy is Ui = −GM2/ri, where M is the mass of either star and ri is their initial center-to-center separation. The initial kinetic energy is zero since the stars are at rest. The final potential energy is Uf = −2GM2/ri since the final separation is ri/2. We write Mv2 for the final kinetic energy of the system. This is the sum of two terms, each of which is ½Mv2. Conservation of energy yields

The solution for v is

GMv r

(b) Now the final separation of the centers is rf = 2R = 2 × 105 m, where R is the radius of either of the stars. The final potential energy is given by Uf = −GM2/rf and the energy equation becomes −GM2/ri = −GM2/rf + Mv2. The solution for v is

f i v GM r r

40. (a) The initial gravitational potential energy is 1 3 2

A B i


(b) We use conservation of energy (with Ki = 0):

11328(6.6710 m/skg)(20 kg)(10 kg)1.710

which yields K = 5.6 × 10−9 J. Note that the value of r is the difference between 0.80 m and 0.20 m.

41. Let m = 0.020 kg and d = 0.600 m (the original edge-length, in terms of which the final edge-length is d/3). The total initial gravitational potential energy (using Eq. 13-21 and some elementary trigonometry) is

Ui = – – .

Since U is inversely proportional to r then reducing the size by 1/3 means increasing the magnitude of the potential energy by a factor of 3, so

Uf = 3Ui ⇒ ∆U = 2Ui = 2(4 + ) = – 4.82 × 10–13 J . 42. (a) Applying Eq. 13-21 and the Pythagorean theorem leads to

CHAPTER 13 where M is the mass of particle B (also that of particle C) and m is the mass of particle A. The value given in the problem statement (for infinitely large y, for which the second term above vanishes) determines M, since D is given. Thus M = 0.50 kg.

(b) We estimate (from the graph) the y = 0 value to be Uo = – 3.5 × 10−10 J. Using this, our expression above determines m. We obtain m = 1.5 kg.

43. The period T and orbit radius r are related by the law of periods: T2 = (4pi2/GM)r3, where M is the mass of Mars. The period is 7 h 39 min, which is 2.754 × 104 s. We solve for M:

GT pi pi−

4. From Eq. 13-37, we obtain v = /GMr for the speed of an object in circular orbit (of radius r) around a planet of mass M. In this case, M = 5.98 × 1024 kg and r = (700 + 6370)m = 7070 km = 7.07 × 106 m.

The speed is found to be v = 7.51 × 103 m/s. After multiplying by 3600 s/h and dividing by 1000 m/km this becomes v = 2.7 × 104 km/h.

(a) For a head-on collision, the relative speed of the two objects must be 2v = 5.4 × 104 km/h.

(b) A perpendicular collision is possible if one satellite is, say, orbiting above the equator and the other is following a longitudinal line. In this case, the relative speed is given by the Pythagorean theorem: 2νν+ = 3.8 × 104 km/h.

45. 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 N M and the magnitude of the gravitational force acting on the Sun is F = GNM2/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 = 2piR/T and a = 4pi2R/T2.

Newton’s second law yields GNM2/R2 = 4pi2MR/T2. The solution for N is pi =

The period is 2.5 × 108 y, which is 7.8 × 1015 s, so

46. Kepler’s law of periods, expressed as a ratio, is

where we have substituted the mean-distance (from Sun) ratio for the semi-major axis ratio. This yields TM = 1.87 y. The value in Appendix C (1.8 y) is quite close, and the small apparent discrepancy is not significant, since a more precise value for the semi-major axis ratio is aM/aE = 1.523 which does lead to TM = 1.8 y using Kepler’s law. A question can be raised regarding the use of a ratio of mean distances for the ratio of semi-major axes, but this requires a more lengthy discussion of what is meant by a ”mean distance” than is appropriate here.

47. (a) The greatest distance between the satellite and Earth’s center (the apogee distance) and the least distance (perigee distance) are, respectively,

Ra = (6.37 × 106 m + 360 × 103 m) = 6.73 × 106 m Rp = (6.37 × 106 m + 180 × 103 m) = 6.5 × 106 m.

Here 6.37 × 106 m is the radius of Earth. From Fig. 13-13, we see that the semi-major axis is

(b) The apogee and perigee distances are related to the eccentricity e by Ra = a(1 + e) and

Rp = a(1 − e). Add to obtain Ra + Rp = 2a and a = (Ra + Rp)/2. Subtract to obtain Ra − Rp = 2ae. Thus, a p a p

a p

R R R R e

48. Kepler’s law of periods, expressed as a ratio, is

21lunar month s s s m m r T T which yields Ts = 0.35 lunar month for the period of the satellite.


49. (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 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 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

GMv r

(b) Since the circumference of the circular orbit is 2pir, the period is

This is equivalent to 87.5 min.

50. (a) The distance from the center of an ellipse to a focus is ae where a is the semimajor axis and e is the eccentricity. Thus, the separation of the foci (in the case of Earth’s orbit) is

(b) To express this in terms of solar radii (see Appendix C), we set up a ratio:

51. (a) The period of the comet is 1420 years (and one month), which we convert to T = 4.48 × 1010 s. Since the mass of the Sun is 1.9 × 1030 kg, then Kepler’s law of periods gives

(b) Since the distance from the focus (of an ellipse) to its center is ea and the distance from center to the aphelion is a, then the comet is at a distance of when it is farthest from the Sun. To express this in terms of Pluto’s orbital radius (found in Appendix C), we set up a ratio:

52. To “hover” above Earth (ME = 5.98 × 1024 kg) means that it has a period of 24 hours (86400 s). By Kepler’s law of periods,

r r GM

Its altitude is therefore r − RE (where RE = 6.37 × 106 m) which yields 3.58 × 107 m. 53. (a) If we take the logarithm of Kepler’s law of periods, we obtain

where we are ignoring an important subtlety about units (the arguments of logarithms cannot have units, since they are transcendental functions). Although the problem can be continued in this way, we prefer to set it up without units, which requires taking a ratio. If we divide Kepler’s law (applied to the Jupiter-moon system, where M is mass of Jupiter) by the law applied to Earth orbiting the Sun (of mass Mo), we obtain

where TE = 365.25 days is Earth’s orbital period and rE = 1.50 × 1011 m is its mean distance from the Sun. In this case, it is perfectly legitimate to take logarithms and obtain

E E Mr T a T M

(written to make each term positive) which is the way we plot the data (log (rE/a) on the vertical axis and log (TE/T) on the horizontal axis).


(b) When we perform a least-squares fit to the data, we obtain log (rE/a) = 0.6 log (TE/T) + 1.01, which confirms the expectation of slope = 2/3 based on the above equation.

(c) And the 1.01 intercept corresponds to the term 1/3 log (Mo/M) which implies

Plugging in Mo = 1.9 × 1030 kg (see Appendix C), we obtain M = 1.86 × 1027 kg for Jupiter’s mass. This is reasonably consistent with the value 1.90 × 1027 kg found in Appendix C.

54. (a) The period is T = 27(3600) = 97200 s, and we are asked to assume that the orbit is circular (of radius r = 100000 m). Kepler’s law of periods provides us with an approximation to the asteroid’s mass:

(b) Dividing the mass M by the given volume yields an average density equal to ρ = 6.3 × 1016/1.41 × 1013 = 4.4 × 103 kg/m3, which is about 20% less dense than Earth.

5. In our system, we have m1 = m2 = M (the mass of our Sun, 1.9 × 1030 kg). With r =

2r1 in this system (so r1 is one-half the Earth-to-Sun distance r), and v = pir/T for the speed, we have

r TGmm rm T r r GM

With r = 1.5 × 1011 m, we obtain T = 2.2 × 107 s. We can express this in terms of Earth-years, by setting up a ratio:

56. The two stars are in circular orbits, not about each other, but about the two-star system’s center of mass (denoted as O), which lies along the line connecting the centers of the two stars. The gravitational force between the stars provides the centripetal force necessary to keep their orbits circular. Thus, for the visible, Newton’s second law gives

Gmm mvF r r = =

where r is the distance between the centers of the stars. To find the relation between r and 1r, we locate the center of mass relative to 1m. Using Equation 9-1, we obtain

(0)m m r m r m mr r r m m m m m

On the other hand, since the orbital speed of 1m is 12/vrTpi=, then 1/2rvTpi= and the expression for r can be rewritten as m m vTr m pi

Substituting r and 1r into the force equation, we obtain

Gmm mvF m m v T T pi pi = =

or m v T

m m G M pi pi − where 301.9910kgsM=× is the mass of the sun. With 16smM=, we write 2smMα= and solve the following cubic equation for α:


The equation has one real solution: 9.3α=, which implies 2/9smM≈.

57. From Kepler’s law of periods (where T = 2.4(3600) = 8640 s), we find the planet’s mass M: 2

But we also know ag = GM/R2 = 8.0 m/s2 so that we are able to solve for the planet’s radius:

58. (a) We make use of m v T

where m1 = 0.9MSun is the estimated mass of the star. With v = 70 m/s and T = 1500 days (or 1500 × 86400 = 1.3 × 108 s), we find

Since MSun ≈ 2.0 × 1030 kg, we find m2 ≈ 7.0 × 1027 kg. Dividing by the mass of Jupiter (see Appendix C), we obtain m ≈ 3.7mJ.

(b) Since v = 2pir1/T is the speed of the star, we find

2 2 vTr pi pi for the star’s orbital radius. If r is the distance between the star and the planet, then r2 = r − r1 is the orbital radius of the planet, and is given by

Dividing this by 1.5 × 1011 m (Earth’s orbital radius, rE) gives r2 = 2.5rE.

59. Each star is attracted toward each of the other two by a force of magnitude GM2/L2, along the line that joins the stars. The net force on each star has magnitude 2(GM2/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 (GM2/L2) cos 30° = Mv2/R.

The stars rotate about their center of mass (marked by a circled dot on the diagram above) at the intersection of the perpendicular bisectors of the triangle sides, and the radius of the orbit is the distance from a star to the center of mass of the three-star system. We take the coordinate system to be as shown in the diagram, with its origin at the left-most star. The altitude of an equilateral triangle is ()3/2L, so the stars are located at x = 0, y = 0; x =

(Parte 2 de 4)