Physics 140 HOMEWORK Chapter 13A Q1. In Fig. 13-20, a central

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Physics 140 HOMEWORK Chapter 13A Q1. In Fig. 13-20, a central particle of mass M is surrounded by a square array of other particles, separated by either distance d or distance d/2 along the perimeter of the square. What are the magnitude and direction of the net gravitational force on the central particle due to the other particles? ——— Wow, it looks tedious. But notice that almost every mass on the perimeter is paired with one of equal value at the exact opposite point on the perimeter. We know it’s exact because of the spacing. Therefore, the net force exerted by the pair is zero, since they cancel. Only the 3M is unpaired. Its distance is d, so F = (G(3M )(M )/d2 = 3GM2 /d2 , to the left. Q2. Figure 13-21 shows three arrangements of the same identical particles, with three of them placed on a circle of radius 0.20 m and the fourth one placed at the center of the circle. (a) Rank the arrangements according to the magnitude of the net gravitational force on the central particle due to the other three particles, greatest first. (b) Rank them according to the gravitational potential energy of the four-particle system, least negative first. ——— The three vector forces to sum are all of the same magnitude, but their directions are most nearly parallel in case c. So: (a) c > b > a. The contributions to the potential are each equal, so: (b) a = b = c. Q4. In Fig. 13-23, two particles, of masses m and 2m, are fixed in place on an axis. (a) Where on the axis can a third particle of mass 3m be placed (other than at infinity) so that the net gravitational force on it from the first two particles is zero: to the left of the first two particles, to their right, between them but closer to the more massive particle, or between them but closer to the less massive particle? (b) Does the answer change if the third particle has, instead, a mass of 16m? (c) Is there a point off the axis (other than infinity) at which the net force on the third particle would be zero? ——— (a) between them but closer to the less massive particle. If the third particle is placed to the left of both, then both of the first two exert forces to the right, so canceling is impossible. Similarly for placing it to the right of both. If placed exactly between them, the forces would be in opposite directions, but the force exerted by 2m would be stronger. So: move it closer to the lower-mass particle by the correct amount, and the lower value of r2 in the formula balances the lower mass in the numerator. (b) No. The answer is the same for any value of the third mass. (c) No. If placed above, both the particles will exert forces with downward components, so cancellation is impossible. P4. 4 The Sun and Earth each exert a gravitational force on the Moon. What is the ratio FSun /FEarth of these two forces? (The average Sun – Moon distance is equal to the Sun – Earth distance.) ——— Note that both F ’s refer to the force on the moon. ´ ³ r ´2 ³ M GMSun MMoon /rE2 FSun M Sun = (3.328 × 105 )(6.486 × 10−6 ) = 2.16. = = 2 FEarth GMEarth MMoon /rM MEarth rE The rE is the radius of Earth’s orbit about the Sun, and rM is the radius of the moon’s orbit about Earth. The result begs the question: If the Sun exerts a greater force on the Moon than the Earth does, why doesn’t it just snatch our Moon away? Think about it. Answer in next solution set.

P5. Miniature black holes. Left over from the big-bang beginning of the universe, tiny black holes might still wander through the universe. If one with a mass of 1 × 1011 kg (and a radius of only 1 × 10−16 m reached Earth, at what distance from your head would its gravitational pull on you match that of Earth’s? ——— The forces are equal means that GMBH Mhead /d2 = Mhead g. Solve for d: p p d = GMBH /g = (6.67 × 10−11 N m2 /kg2 )(1011 kg)/(9.8 m/s2 ) = 0.825 m. P10. Two dimensions. In Fig. 13-34, three point particles are fixed in place in an xy plane. Particle A has mass mA , particle B has mass 2.00 mA , and particle C has mass 3.00 mA . A fourth particle D, with mass 4.00 mA , is to be placed near the other three particles. In terms of distance d, at what (a) x-coordinate and (b) y-coordinate should particle D be placed so that the net gravitational force on particle A from particles B, C, and D is zero? ——— This is similar to the force table experiment, done in one of the early labs. We will get the forces on A due to B and C, get their vector sum, then determine the angle and distance to produce the opposite force, and finally convert that position to x- and y-coordinates. All forces that we consider act on A, so the subscript will indicate to which particle it is due. FB = G(2MA )(MA )/d2 ˆ = 2 GMA2 /d2 ˆ. FC = G(3MA )(MA )/(1.5d)2 (−ˆı) = −1.333 GMA2 /d2 ˆı. Combining these, the force to be canceled has a magnitude of 2.404 GMA2 /d2 at an angle of 123.7◦ . To accomplish this, particle D must be placed at an angle of −56.3◦ at a distance h which satisfies 2 2 2 G(4M pA )(MA )/h = 2.404 GMA /d , or h = 2.404/4d = 0.775 d. Use trig to get xy-coordinates. (a) xD = 0.775 d cos(−56.3◦ )d = +0.430 d. (b) yD = 0.775 d sin(−56.3◦ )d = −0.645 d. P13. Figure 13-37 shows a spherical hollow inside a lead sphere of radius R = 4.00 cm; the surface of the hollow passes through the center of the sphere and “touches” the right side of the sphere. The mass of the sphere before hollowing was M = 2.95 kg. With what gravitational force does the hollowed-out lead sphere attract a small sphere of mass m = 0.431 kg that lies at a distance d = 9.00 cm from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow? ——— This is a job for negative mass. We take the force the solid sphere would exert, then subtract the force of the material which makes up the hollow. In effect, we treat the hollow as negative mass. F = GM m/d2 − G(M/8)m/(d − R/2)2 = 1.047 × 10−8 N − 2.16 × 10−9 N = 8.31 × 10−9 N The factor of 0.125 is because the hollow has 1/8 the volume of the full sphere. The center of the hollow is at a distance of d − R/2 = 0.07 m from the smaller mass. P28. Assume a planet is a uniform sphere of radius R that (somehow) has a narrow radial tunnel through its center (Fig. 13-7). Also assume we can position an apple anywhere along the tunnel or outside the sphere. Let FR be the magnitude of the gravitational force on the apple when it is located at the planet’s surface. How far from the surface is there a point where the magnitude is (1/2)FR if we move the apple (a) away from the planet and (b) into the tunnel? ——— (a) To √ halve the force, we need to double the r2 in the denominator. ′ r = 2 R, so the distance to move is 0.414 R. (b) Inside the sphere we apply Newton’s shell theorem: Material in the spherical shell outside the location of the apple exerts zero force; and material inside exerts a force as though it were concentrated at the center. Therefore, we only need to calculate the force from mass of the sphere at the radius of the apple. Use ρ for the density of the planet. F = G[(4/3)πr3 ρ]mapple /r2 = (4/3)πρGmapple r.

Hence, the force decreases linearly to zero at the center of the planet. To decrease to (1/2)FR , the apple needs to be moved to r = R/2, or a distance of 0.5 R. The formula above applies anywhere inside the sphere, or on its surface. It shows that the gravitational acceleration on the surface of a planet is proportional to its radius and to its density. P31. The mean diameters of Mars and Earth are 6.9 × 103 km and 1.3 × 104 km, respectively. The mass of Mars is 0.11 times Earth’s mass. (a) What is the ratio of the mean density (mass per unit volume) of Mars to that of Earth? (b) What is the value of the gravitational acceleration on Mars? (c) What is the escape speed on Mars? ——— (a) Using V for volume, where V = (4/3)πr3 : ρM /ρE = [MM /VM ]/[ME /VE ] = (MM /ME )(RE /RM )3 = (0.11)(13.0/6.9)3 = 0.736. (b) For an object of mass m at the surface of a planet, the gravitational force is: 2 2 mg = GmMplanet /Rplanet ⇒ g = GMplanet /Rplanet . 2 gM /gE = (MM /ME ) (RE /RM ) = (0.11)(3.55) = 0.391 ⇒ gM = 0.391 · 9.8 m/s2 = 3.83 m/s2 . p p (c) vescape = 2GM/r = 2 · (6.67 × 10−11 )(6.42 × 1023 )/(3.386 × 106 ) = 5030 m/s. I found the mass and radius of Mars on a NASA fact sheet.

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Physics 140 HOMEWORK Chapter 13A Q1. In Fig. 13-20, a central

Physics 140 HOMEWORK Chapter 13A Q1. In Fig. 13-20, a central particle of mass M is surrounded by a square array of other particles, separated by eith...

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