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 1 de 4)

Chapter 13

1. The magnitude of the force of one particle on the other is given by F = Gm1m2/r2, where m1 and m2 are the masses, r is their separation, and G is the universal gravitational constant. We solve for r:

Gmmr F − −

2. We use subscripts s, e, and m for the Sun, Earth and Moon, respectively. Plugging in the numerical values (say, from Appendix C) we find sm s m sm s em

em e m em e sm

F Gm m r m r F Gm m r m r

3. The gravitational force between the two parts is

Gm M m GF mM m r r which we differentiate with respect to m and set equal to zero:

This leads to the result m/M= 1/2.

4. The gravitational force between you and the moon at its initial position (directly opposite of Earth from you) is

GM mF

where mM is the mass of the moon, MER is the distance between the moon and the Earth, and ER is the radius of the Earth. At its final position (directly above you), the gravitational force between you and the moon is

GM mF

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(a) The ratio of the moon’s gravitational pulls at the two different positions is m ME E ME E

m ME E ME E

GM m R RF R R F GM m R R R R

Therefore, the increase is 0.06898, or approximately, 6.9%. (b) The change of the gravitational pull may be approximated as m m m m m E E

GM m GM m GM m GM m GM mRR RF F R R R R R R R R R

On the other hand, your weight, as measured on a scale on Earth is

2E g E

E GM mF mg R

Since the moon pulls you “up,” the percentage decrease of weight is m E

g E ME

5. We require the magnitude of force (given by Eq. 13-1) exerted by particle C on A be equal to that exerted by B on A. Thus,

We substitute in mB = 3mA and mB = 3mA, and (after canceling “mA”) solve for r. We find r = 5d. Thus, particle C is placed on the x axis, to left of particle A (so it is at a negative value of x), at x = –5.00d.

6. Using F = GmM/r2, we find that the topmost mass pulls upward on the one at the origin with 1.9 × 10−8 N, and the rightmost mass pulls rightward on the one at the origin with

1.0 × 10−8 N. Thus, the (x, y) components of the net force, which can be converted to polar components (here we use magnitude-angle notation), are

(a) The magnitude of the force is 2.13 × 10−8 N. (b) The direction of the force relative to the +x axis is 60.6°.

7. At the point where the forces balance 2 12//esGMmrGMmr=, where Me is the mass of

Earth, Ms is the mass of the Sun, m is the mass of the space probe, r1 is the distance from the center of Earth to the probe, and r2 is the distance from the center of the Sun to the probe. We substitute r2 = d − r1, where d is the distance from the center of Earth to the center of the Sun, to find22 1 1

= .e sM M r d r−

Taking the positive square root of both sides, we solve for r1. A little algebra yields d M r

Values for Me, Ms, and d can be found in Appendix C.

8. The gravitational forces on m5 from the two 5.00g masses m1 and m4 cancel each other. Contributions to the net force on m5 come from the remaining two masses:

The force is directed along the diagonal between m2 and m3, towards m2. In unit-vector notation, we have

net net

9. The gravitational force from Earth on you (with mass m) is

E GM mF mgR where 2/9.8m/s.EEgGMR== If r is the distance between you and a tiny black hole of mass 11110kgbM=× that has the same gravitational pull on you as the Earth, then

2 .bg GM mF mgr

Combining the two equations, we obtain

CHAPTER 13

9.8 m/s b bEE

GM m GMGM mmg r R r g

10. (a) We are told the value of the force when particle C is removed (that is, as its position x goes to infinity), which is a situation in which any force caused by C vanishes (because Eq. 13-1 has r2 in the denominator). Thus, this situation only involves the force exerted by A on B:

Since mB = 1.0 kg, then this yields mA = 0.25 kg.

(b) We note (from the graph) that the net force on B is zero when x = 0.40 m. Thus, at that point, the force exerted by C must have the same magnitude (but opposite direction) as the force exerted by A (which is the one discussed in part (a)). Therefore

= 4.17 × 10−10 N⇒ mC = 1.0 kg.

1. (a) The distance between any of the spheres at the corners and the sphere at the center is where l is the length of one side of the equilateral triangle. The net (downward) contribution caused by the two bottom-most spheres (each of mass m) to the total force on m4 has magnitude

This must equal the magnitude of the pull from M, so

Gm m Gm m =

l l which readily yields m = M.

(b) Since m4 cancels in that last step, then the amount of mass in the center sphere is not relevant to the problem. The net force is still zero.

12. All the forces are being evaluated at the origin (since particle A is there), and all forces (except the net force) are along the location-vectors which point to particles B and C. We note that the angle for the location-vector pointing to particle B is 180º – 30.0º = 150º (measured ccw from the +x axis). The component along, say, the x axis of one of the force-vectors is simply Fx/r in this situation (where F is the magnitude of ). Since the force itself (see Eq. 13-1) is inversely proportional to r2 then the aforementioned x component would have the form GmMx/r3; similarly for the other components. With mA = 0.0060 kg, mB = 0.0120 kg, and mC = 0.0080 kg, we therefore have and Fnet y = + = (2.7 × 10−14 N)sin(−163.8º) where rB = dAB = 0.50 m, and (xB, yB) = (rBcos(150º), rBsin(150º)) (with SI units understood). A fairly quick way to solve for rC is to consider the vector difference between the net force and the force exerted by A, and then employ the Pythagorean theorem. This yields rC = 0.40 m.

(a) By solving the above equations, the x coordinate of particle C is xC = −0.20 m.

(b) Similarly, the y coordinate of particle C is yC = −0.35 m.

13. If the lead sphere were not hollowed 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. We calculate the force exerted on m by a sphere that just fills 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 fills 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 fills the cavity. The common factor 4pi/3 has been canceled. Thus, r R M M M

The center of the cavity is d − r = d − R/2 from m, so the force it exerts on m is

G M m F

The force of the hollowed sphere on m is

GMmF F F GMm d d R R d

CHAPTER 13

=and = – .

14. Using Eq. 13-1, we find

Since the vector sum of all three forces must be zero, we find the third force (using magnitude-angle notation) is

= (2.404 ∠ –56.3º) .

This tells us immediately the direction of the vector (pointing from the origin to particle

2.404 =

D), but to find its magnitude we must solve (with mD = 4mA) the following equation:

This yields r = 1.29d. In magnitude-angle notation, then, = (1.29 ∠ –56.3º) , with SI

= (, ) .

units understood. The “exact” answer without regard to significant figure considerations is (a) In (x, y) notation, the x coordinate is x =0.716d.

(b) Similarly, the y coordinate is y = −1.07d.

dimensions, the Pythagorean theorem becomes r =The component along, say, the x

15. All the forces are being evaluated at the origin (since particle A is there), and all forces are along the location-vectors which point to particles B, C and D. In three axis of one of the force-vectors is simply Fx/r in this situation (where F is the magnitude of ). Since the force itself (see Eq. 13-1) is inversely proportional to r2 then the aforementioned x component would have the form GmMx/r3; similarly for the other components. For example, the z component of the force exerted on particle A by particle B is

In this way, each component can be written as some multiple of GmA2/d2. For the z component of the force exerted on particle A by particle C, that multiple is –9/196. For the x components of the forces exerted on particle A by particles B and C, those multiples are 4/27 and –3/196, respectively. And for the y components of the forces exerted on particle A by particles B and C, those multiples are 2/27 and 3/98, respectively. To find the distance r to particle D one method is to solve (using the fact that the vector add to zero)

(where mD = 4mA) for r. This gives r = 4.357d. The individual values of x, y and z (locating the particle D) can then be found by considering each component of the

GmAmD/r2 force separately.

(a) The x component of rrwould be

GmA mD x/r3 = –(4/27 –3/196)GmA2/d2, which yields x = –1.88d.

(b) Similarly, y = −3.90d,

(c) and z = 0.489d. In this way we are able to deduce that (x, y, z) = (1.88d, 3.90d, 0.49d).

16. Since the rod is an extended object, we cannot apply Equation 13-1 directly to find the force. Instead, we consider a small differential element of the rod, of mass dm of thickness dr at a distance r from 1m. The gravitational force between dm and 1mis

where we have substituted (/)dmMLdr= since mass is uniformly distributed. The direction of dFr is to the right (see figure). The total force can be found by integrating over the entire length of the rod:

GmM GmM GmMdrF dF

L r L L d d d L d

Substituting the values given in the problem statement, we obtain

GmMF d L d

17. The acceleration due to gravity is given by ag = GM/r2, where M is the mass of Earth and r is the distance from Earth’s center. We substitute r = R + h, where R is the radius of Earth and h is the altitude, to obtain ag = GM /(R + h)2. We solve for h and obtain

/ghGMaR=−. According to Appendix C, R = 6.37 × 106 m and M = 5.98 × 1024 kg, so

18. We follow the method shown in Sample Problem 13-3. Thus,

CHAPTER 13

GM GMa da dr

which implies that the change in weight is

( )top bottom .gW W m da− ≈ But since Wbottom = GmME/R2 (where R is Earth’s mean radius), we have

GmM drmda dr W R R for the weight change (the minus sign indicating that it is a decrease in W). We are not including any effects due to the Earth’s rotation (as treated in Eq. 13-13).

19. (a) The gravitational acceleration at the surface of the Moon is gmoon = 1.67 m/s2 (see Appendix C). The ratio of weights (for a given mass) is the ratio of g-values, so

Wmoon = (100 N)(1.67/9.8) = 17 N.

(b) For the force on that object caused by Earth’s gravity to equal 17 N, then the free-fall acceleration at its location must be ag = 1.67 m/s2. Thus,

so the object would need to be a distance of r/RE = 2.4 “radii” from Earth’s center.

20. The free-body diagram of the force acting on the plumb line is shown on the right. The mass of the sphere is

The force between the “spherical” mountain and the plumb line is 2/FGMmr=. Suppose at equilibrium the line makes an angle θ with the vertical and the net force acting on the line is zero. Therefore,

0 sin sin

0 cosx y

GMmF T F T r

F T mg

The two equations can be combined to give 2tanFGM mg gr θ==. The distance the lower end moves toward the sphere is

GMx l l gr θ

21. (a) The gravitational acceleration is 2==7.6 m/s.g GMa

(b) Note that the total mass is 5M. Thus, ( )

h h g h h

GM GM ca MR GGM c which yields ag = (3.02 × 1043 kg·m/s2) /Mh. (b) Since Mh is in the denominator of the above result, ag decreases as Mh increases.

(c) With Mh = (1.5 × 1012) (1.9 × 1030 kg), we obtain ag = 9.82 m/s2.

(d) This part refers specifically to the very large black hole treated in the previous part. With that mass for M in Eq. 13–16, and r = 2.002GM/c2, we obtain g GM cda dr dr

GMGMc − − where dr → 1.70 m as in Sample Problem 13-3. This yields (in absolute value) an acceleration difference of 7.30 × 10−15 m/s2.

(e) The miniscule result of the previous part implies that, in this case, any effects due to the differences of gravitational forces on the body are negligible.

23. From Eq. 13-14, we see the extreme case is when “g” becomes zero, and plugging in Eq. 13-15 leads to

CHAPTER 13

Thus, with R = 20000 m and ω = 2pi rad/s, we find M = 4.7 × 1024 kg ≈ 5 × 1024 kg.

24. (a) What contributes to the GmM/r2 force on m is the (spherically distributed) mass M contained within r (where r is measured from the center of M). At point A we see that M1 + M2 is at a smaller radius than r = a and thus contributes to the force:

G M M m F

(b) In the case r = b, only M1 is contained within that radius, so the force on m becomes GM1m/b2.

(c) If the particle is at C, then no other mass is at smaller radius and the gravitational force on it is zero.

25. (a) The magnitude of the force on a particle with mass m at the surface of Earth is given by F = GMm/R2, where M is the total mass of Earth and R is Earth’s radius. The acceleration due to gravity is g F GMa

m R

(b) Now ag = 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. 13-43). The total mass is M = (1.93 × 1024 kg + 4.01 × 1024 kg ) = 5.94 × 1024 kg.

The first term is the mass of the core and the second is the mass of the mantle. Thus,

(c) A point 25 km below the surface is at the mantle-crust 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 per unit volume by the volume of the sphere: 3(/),eeMRRM= where Me is the total mass of Earth and Re is the radius of Earth. Thus,

The acceleration due to gravity is g GMa

26. (a) Using Eq. 13-1, we set GmM/r2 equal to GmM/R2, and we find r = RThus, the

distance from the surface is ( – 1)R = 0.414R.

(b) Setting the density ρ equal to M/V where V = piR3, we use Eq. 13-19:

Gmr Gmr M GMmr GMmF r R

R R R pi ρ pi pi

27. Using the fact that the volume of a sphere is 4piR3/3, we find the density of the sphere:

1.0 mMRρ pi pi

When the particle of mass m (upon which the sphere, or parts of it, are exerting a gravitational force) is at radius r (measured from the center of the sphere), then whatever mass M is at a radius less than r must contribute to the magnitude of that force (GMm/r2).

(a) At r = 1.5 m, all of Mtotal is at a smaller radius and thus all contributes to the force:

GmMF mr

(b) At r = 0.50 m, the portion of the sphere at radius smaller than that is

Thus, the force on m has magnitude GMm/r2 = m (3.3 × 10−7 N/kg). (c) Pursuing the calculation of part (b) algebraically, we find

Gm r

F mr r

CHAPTER 13

28. The difference between free-fall acceleration g and the gravitational acceleration ga at the equator of the star is (see Equation 13.14):

2ga g Rω− = where

is the angular speed of the star. The gravitational acceleration at the equator is

Therefore, the percentage difference is a g R

29. (a) The density of a uniform sphere is given by ρ = 3M/4piR3, where M is its mass and R is its radius. The ratio of the density of Mars to the density of Earth is

(b) The value of ag at the surface of a planet is given by ag = GM/R2, so the value for Mars is

g gE E M

(c) If v is the escape speed, then, for a particle of mass m

For Mars, the escape speed is

30. (a) The gravitational potential energy is

GMmU r

(b) Since the change in potential energy is

GMm GMmU r r the work done by the gravitational force is W = − ∆U = −2.9 × 10−1 J.

(c) The work done by you is W = ∆U = 2.9 × 10−1 J.

31. The amount of (kinetic) energy needed to escape is the same as the (absolute value of the) gravitational potential energy at its original position. Thus, an object of mass m on a planet of mass M and radius R needs K = GmM/R in order to (barely) escape. (a) Setting up the ratio, we find

(Parte 1 de 4)

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