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

(Parte 4 de 4) 87. (a) The total energy is conserved, so there is no difference between its values at aphelion and perihelion.

(b) Since the change is small, we use differentials:

E SGM MdU dr which yields ∆U ≈ 1.8 × 1032 J. A more direct subtraction of the values of the potential energies leads to the same result.

(c) From the previous two parts, we see that the variation in the kinetic energy ∆K must also equal 1.8 × 1032 J.

(d) With ∆K ≈ dK = mv dv, where v ≈ 2piR/T, we have

which yields a difference of ∆v ≈ 0.9 km/s in Earth’s speed (relative to the Sun) between aphelion and perihelion.

8. Let the distance from Earth to the spaceship be r. Rem = 3.82 × 108 m is the distance from Earth to the moon. Thus,

GM m GM mF F rR r−

where m is the mass of the spaceship. Solving for r, we obtain

m e

Rr M M

89. We integrate Eq. 13-1 with respect to r from 3RE to 4RE and obtain the work equal to –GMEm(1/(4RE) – 1/(3RE)) = GMEm/12RE .

90. If the angular velocity were any greater, loose objects on the surface would not go around with the planet but would travel out into space.

(a) The magnitude of the gravitational force exerted by the planet on an object of mass m at its surface is given by F = GmM / R2, where M is the mass of the planet and R is its radius. According to Newton’s second law this must equal mv2 / R, where v is the speed of the object. Thus, 2

Replacing M with (4pi/3) ρR3 (where ρ is the density of the planet) and v with 2piR/T (where T is the period of revolution), we find pi piρ

We solve for T and obtain

(b) The density is 3.0 × 103 kg/m3. We evaluate the equation for T:

91. (a) It is possible to use 2202vvay=+∆as we did for free-fall problems in Chapter 2 because the acceleration can be considered approximately constant over this interval. However, our approach will not assume constant acceleration; we use energy conservation:

GM r rGMm GMmmv mv v r r r r

which yields v = 1.4 × 106 m/s.

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(b) We estimate the height of the apple to be h = 7 cm = 0.07 m. We may find the answer by evaluating Eq. 13-1 at the surface (radius r in part (a)) and at radius r + h, being careful not to round off, and then taking the difference of the two values, or we may take the differential of that equation — setting dr equal to h. We illustrate the latter procedure:

92. (a) The gravitational force exerted on the baby (denoted with subscript b) by the obstetrician (denoted with subscript o) is given by

o b bo

Gm mF r

(b) The maximum (minimum) forces exerted by Jupiter on the baby occur when it is separated from the Earth by the shortest (longest) distance rmin (rmax), respectively. Thus

J b bJ Gm mF

(c) And we obtain

J b bJ Gm mF

(d) No. The gravitational force exerted 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.

93. The magnitude of the net gravitational force on one of the smaller stars (of mass m) is

GMm Gmm Gm mM r r

This supplies the centripetal force needed for the motion of the star:

Gm m v rM m v r r T

Plugging in for speed v, we arrive at an equation for period T:

pi =

94. (a) We note that height = R − REarth where REarth = 6.37 × 106 m. With M = 5.98 × 1024 kg, R0 = 6.57 × 106 m and R = 7.37 × 106 m, we have

GmM GmMK U K U m K

which yields K = 3.83 × 107 J. (b) Again, we use energy conservation.

Therefore, we find Rf = 7.40 × 106 m. This corresponds to a distance of 1034.9 km ≈ 1.03 × 103 km above the Earth’s surface.

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

where M = 7.0 × 1024 kg, r2 = R = 1.6 × 106 m and r1 = ∞ (which means that U1 = 0). We are told to assume the meteor starts at rest, so v1 = 0. Thus, K1 + U1 = 0 and the above equation is rewritten as

96. The initial distance from each fixed sphere to the ball is r0 = ∞, which implies the initial gravitational potential energy is zero. The distance from each fixed sphere to the ball when it is at x = 0.30 m is r = 0.50 m, by the Pythagorean theorem.

(a) With M = 20 kg and m = 10 kg, energy conservation leads to

0 0 2i i GmMK U K U K r which yields K = 2GmM/r = 5.3 × 10−8 J.

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(b) Since the y-component of each force will cancel, the net force points in the –x direction, with a magnitude 2Fx = 2 (GmM/r2) cos θ, where θ = tan−1 (4/3) = 53°. Thus,

97. The kinetic energy in its circular orbit is mv2 where v = 2pir/T. Using the values stated in the problem and using Eq. 13-41, we directly find E = –1.87 × 109 J.

98. (a) From Ch. 2, we have 2202vvax=+∆, where a may be interpreted as an average acceleration in cases where the acceleration is not uniform. With v0 = 0, v = 10 m/s and ∆x = 220 m, we find a = 2.75 × 105 m/s2. Therefore,

(b) The acceleration is certainly deadly enough to kill the passengers.

(d) Energy conservation gives the craft’s speed v (in the absence of friction and other dissipative effects) at altitude h = 700 km after being launched from R = 6.37 × 106 m

(the surface of Earth) with speed v0 = 7000 m/s. That altitude corresponds to a distance from Earth’s center of r = R + h = 7.07 × 106 m.

GMm GMmmv mv

With M = 5.98 × 1024 kg (the mass of Earth) we find v = 6.05 × 103 m/s. But to orbit at that radius requires (by Eq. 13-37)

The difference between these is v − v = 1.46 × 103 m/s 31.510 m/s≈×, which presumably is accounted for by the action of the rocket engine.

9. (a) All points on the ring are the same distance (r = ) from the particle, so the gravitational potential energy is simply U = –GMm/, from Eq. 13-21. The corresponding force (by symmetry) is expected to be along the x axis, so we take a (negative) derivative of U (with respect to x) to obtain it (see Eq. 8-20). The result for the magnitude of the force is GMmx(x2 + R2)−3/2.

(b) Using our expression for U, then the magnitude of the loss in potential energy as the particle falls to the center is GMm(1/R −1/). This must “turn into” kinetic energy ( mv2 ), so we solve for the speed and obtain

100. Consider that we are examining the forces on the mass in the lower left-hand corner of the square. Note that the mass in the upper right-hand corner is 20 = 28 cm = 0.28 m away. Now, the nearest masses each pull with a force of GmM / r2 = 3.8 × 10−9 N, one upward and the other rightward. The net force caused by these two forces is (3.8 × 10−9,

3.8 × 10−9) → (5.3 × 10−9 ∠ 45°), where the rectangular components are shown first -- and then the polar components (magnitude-angle notation). Now, the mass in the upper right-hand corner also pulls at 45°, so its force-magnitude (1.9 × 10−9) will simply add to the magnitude just calculated. Thus, the final result is 7.2 × 10−9 N.

101. (a) Their initial potential energy is −Gm2/Ri and they started from rest, so energy conservation leads to 2 2 2

total total . 0.5i i i

(b) They have equal mass, and this is being viewed in the center-of-mass frame, so their speeds are identical and their kinetic energies are the same. Thus, total1 .

2 2 i GmK K R

(c) With K = ½ mv2, we solve the above equation and find v = /iGmR.

(d) Their relative speed is 2v = 2 /iGmR. This is the (instantaneous) rate at which the gap between them is closing.

(e) The premise of this part is that we assume we are not moving (that is, that body A acquires no kinetic energy in the process). Thus, Ktotal = KB and the logic of part (a) leads to KB = Gm2/Ri.

(f) And 21 2BBmvK= yields vB = 2/iGmR.

(g) The answer to part (f) is incorrect, due to having ignored the accelerated motion of “our” frame (that of body A). Our computations were therefore carried out in a

CHAPTER 13 noninertial frame of reference, for which the energy equations of Chapter 8 are not directly applicable.

102. Gravitational acceleration is defined in Eq. 13-1 (which we are treating as a positive quantity). The problem, then, is asking for the magnitude difference of ag net when the contributions from the Moon and the Sun are in the same direction (ag net = agSun + agMoon) as opposed to when they are in opposite directions (ag net = agSun – agMoon). The difference (in absolute value) is clearly 2agMoon. In specifically wanting the percentage change, the problem is requesting us to divide this difference by the average of the two ag net values being considered (that average is easily seen to be equal to agSun), and finally multiply by 100% in order to quote the result in the right format. Thus,

= 2= 2 = 0.011 = 1.1%.

103. (a) Kepler’s law of periods is

With M = 6.0 × 1030 kg and T = 300(86400) = 2.6 × 107 s, we obtain r = 1.9 × 1011 m. (b) That its orbit is circular suggests that its speed is constant, so

104. Using Eq. 13-21, the potential energy of the dust particle is U = –GmME/R – GmMm/r = –Gm(ME/R + Mm/r) .

(Parte 4 de 4)