CHAPTER 8 SOLUTIONS TO PROBLEMS AND QUESTIONS

Citation: Jacqueline D. Spears and Dean Zollman, Instructor's Guide for The Fascination of Physics, (The Benjamin /Cummings Publishing Company, Inc., Menlo Park, CA 1985). Permission granted by the publisher.

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A. Review of Chapter Material

A1. The gravitational interaction occurs between any two bodies with mass. It is directly proportional to the masses of the objects and inversely proportional to the square of the distance between them. The electrical interaction occurs between any two bodies with electrical charge. It is directly proportional to the magnitude of the charges on the objects and inversely proportional to the square of the distance between them. The force is one of attraction if the two objects have opposite charge and one of repulsion if they have the same type of charge. The two nuclear interactions describe forces that occur over the very short distances within the nucleus of the atom. The strong nuclear interaction is responsible for holding the nucleus together. The weak nuclear interaction is needed to explain why certain nuclei fall apart.

A2. The force which pulls the apple to the ground is the same force which pulls the moon toward the earth and, thus, keeps it in its orbit. Gravitational interactions explain both phenomena.

A3. The gravitational interaction is directly proportional to the masses of the objects and inversely proportional to the square of the distance between them.

A4. The quantity which is not yet known is the total mass of the universe. If the value of this mass is great enough, then the force of attraction among the masses will be large enough to reverse the expansion that is presently taking place. If the value of this mass is not large enough, the gravitational force will be insufficient to halt the present expansion.

A5. The electrical interaction is directly proportional to the magnitude of the electrical charges on the objects and inversely proportional to the square of the distance between them. The force is one of attraction if the two objects have opposite charge and one of repulsion if they have the same type of charge.

A6. The gravitational and electrical interactions are similar in that they each depend on the inverse of the square of the distance between the objects involved and that they are each directly proportional to a property of the object (mass in the gravitational case and electric charge in the electrical case). They differ in that the gravitational force is always attractive whereas the electrical force can be either attractive or repulsive. They also differ in strength. The electrical interaction is a much stronger interaction.

A7. Rutherford relied on the fact that charged objects will change their motion due to the force applied by another charged object. He directed positively charged alpha particles at a target of thin gold foil, then inferred the structure of the atom from the changes in the motion of these alpha particles.

A8. All protons in a nucleus have positive charge. If the only force acting in the nucleus were due to the electrical interaction, these protons would be repelled from one another; and the nucleus would fall apart. Because the nucleus does not fall apart, we conclude that another interaction must be present.

A9. The weak nuclear interaction explains the disintegration of the neutron and certain other nuclei.

A10. The process by which objects exchange particles leads to an interaction which we can interpret in terms of forces exerted at a distance. The range of the interaction can be related to the mass of the exchange particle.

B. Using the Chapter Material

B1. The force of attraction is greater on the planet which is closer to the sun. Because the masses are identical, the ratio of the forces is the ratio of the inverse squares of their distances.

Ratio of A to B = (8000)2/(4000)2 = 4

The force exerted on planet A is four times the force exerted on planet B.

B2. Because both satellites are located at the same distance from the earth, the only variable that differs is the mass of the satellites. Satellite C has a larger mass and experiences a greater gravitational force. The ratio of the forces exerted on C to D is simply the ratio of their masses, 2 to 1. The gravitational force exerted on satellite C is twice the gravitational force exerted on satellite D.

B3. Force = G (Mass of Skylab) x (Mass of earth) (distance between earth and Sky lab) 2

The distance between the center of the earth and Sky lab is 6380 km + 432 km = 6812 km = 6,812,000 m.

Force = (6.67 x 10-11 N m2/kg2) x 90606 kg x 5.98 x 1024 kg = 7.79 x 105 N (6,812,000 m)2

B4. In traveling from the earth to the moon, the Apollo spacecraft felt gravitational forces from both planets. Neither force became zero. At one point, however, the net force acting on the satellite momentarily became zero. The gravitational force exerted by the earth equaled the gravitational force exerted by the moon.

B5. Force = G (Mass of proton) x (Mass of proton) (distance between protons) '

Force = (6.67 x 10-11 N m2/kg2) x (1.672 x 10-27 kg)(1.672 x 10-27 kg) (10-15 m)2

Force = 1.86 x 10-34 N. This is an attractive force.

Electrical Force = k (charge 1) x (charge 2) distance

Electrical Force = (9 x 109 N m2/C2) (1.6 x 10-19 C) x (1.6 x 10-19 C) (10-15 m)2

Electrical Force = 230 N. This is a repulsive force.

The electrical force is much larger in magnitude than the gravitational force. If these two forces were the only forces that acted in the nucleus, the net force acting on the protons would be repulsive and the nucleus would fall apart.

B6. Because the distances are equal for the two sets of objects, the set with the larger electrical charges will have the greater force. The +24 C charges will have a greater repulsion than the +12 C charges. (b) Because the charges are the same for each set of objects, the ones which are closer together will have the greater force. The objects separated by 6 m will experience a greater force than the ones separated by 20 m.

B7. Electrical Force = k (charge 1) x (charge 2) (distance)2

Electrical Force = (9 x 109 N m2/C2) (-1.6 x 10-19 C) x (1.6 x 10-19 C) (5.3 x 10-11 m)2

Electrical Force = 8.2 x 18-8 N, toward the nucleus. The nucleus exerts attractive force 8.2 x 10-8 N on the electron. The electron exerts an attractive force of 8.2 x 10-8 N on the nucleus.

B8. Let's consider a closed system consisting of the comb and hair. Initially, the net electrical charge of the system is zero. Once the comb rubs the hair, it acquires a net electrical charge of -3 x 10-6 C. In order to conserve electrical charge, the hair must now have a net electrical charge of +3 x 10-6 C. While both the comb and the hair have a net electrical charge, the net electrical charge of the system is still zero.

B9. Neutrons have no electrical charge. Thus, the net force between two neutrons is due entirely to the strong nuclear force. Protons have electrical charge. In the nucleus, protons experience an attractive force due to the strong nuclear interaction and a repulsive force due to their electrical charge. The net force that acts between two protons will be the vector sum of these two forces, which is less than the strong nuclear force by itself.

B10. The range of the nuclear force is extremely small. In our every day endeavors we do not notice interactions at such a short distance.

B11. As discussed in the chapter the greater the mass of the exchange particle is, the smaller is the range of the force. If the pion mass increased, then the range of the nuclear force would decrease.

C. Extensions to New Situations

C1. (a) The closed system includes the book and the earth. (b) The gravitational interaction is the fundamental interaction involved. (c) The book attracts the earth, so there is a force acting on the earth. This force must cause a change in the momentum of the earth. (d) The mass of the earth is extremely large, so the change in velocity needed to conserve momentum is extremely small.

C2. (a) Newton's third law assures us that if the earth exerts a gravitational force on the moon, the moon exerts a gravitational force on the earth. (b) The strength of the gravitational force is inversely proportional to the separation between the two objects. Since point A lies closest to the moon, the force exerted by the moon will be the greatest. (c)See Figure

(d) As that part of the earth rotates, it moves closer to the moon. The stronger gravitational force causes the ocean to rise toward the moon. (e) The tides occur because of a gravitational attraction between the water on the earth and the moon. When the water is closer W the moon, it is attracted more and rises. As it moves away from the moon, the water falls. (f) All parts of the earth are attracted to the moon by the gravitational interaction. The water at point C is attracted the least because it is located at the greatest distance from the moon. The parts of the earth between points B and D are attracted more than the water at C. Thus, the material is attracted away from the water at C leaving it behind as a slight rise.

C3. (a) The greatest high tide will occur when the moon is in position (iii). At that position the gravitational forces exerted by both the sun and the moon are pulling the water in the same direction. (b) At point (i) the high tide will be the least because the sun and the moon are pulling the water in the opposite directions. (c) The magnitude of the high tide is determined by the location of the sun relative to the moon. When the sun and the moon are on the same side of the earth, they pull in the same direction and cause the greatest net force on the water.

C4. Each planet's orbit will be determined by the gravitational interaction between it and other massive bodies. The sun exerts the largest single gravitational force on the planets. However, other planets also exert gravitational forces on any given planet. If the forces due to all known objects are added as vectors, we should be able to predict the motion of each planet. A motion which differs from this prediction would indicate that another, yet undiscovered, object is also applying a force on the planet. That undiscovered object is likely to be another planet.

C5. (a) The electrons on the center wire are attracted toward the positively charged cylinder because of the electrical interaction. Some of the electrons will not strike the cylinder but will pass through the hole. The result is a beam of electrons that move toward the screen. When they hit the screen, the electrons will cause light to be emitted and create a bright spot. (b) As the electrons pass between the two plates, they are attracted by the positively charged plate and repelled by the negatively charged one. The net force is up, so the electrons move upward. However, they still maintain their horizontal motion (Newton's first law). Thus, the motion is:See Figure

(c) This change will cause a gradual change in the force applied to the electrons. As the charge on the plates decreases, the electrons will be deflected less from their straight line motion. At some point the charge will be zero and the electrons will pass straight through. Then, the bottom plate will gradually build up a positive charge and the electrons will be deflected downward. The net effect will be a sweep of the electrons from the top of the screen to the bottom of the screen in 1/60 of a second.

C6. (a) The electrons in the paper will be attracted toward the positive charges on the comb. They will move in the direction of the comb. (b)See Figure

The comb has a net positive charge on it, so it has more positive charges than negative ones. The electrons in the paper are pulled toward these positive charges while the positively charged nuclei are repelled slightly. The positive and negative charges in the paper are slightly separated, as shown. (c) The negative charges are attracted toward the comb while the positive charges are repelled from the comb. (d) The magnitude of the force on the negative charges will be greater than the magnitude of the force on the positive charges because the negative charges are closer to the comb than the positive charges. The piece of paper will be attracted to the comb. (e) If the comb has a net negative charge, the electrons in the paper will be repelled by the comb. Then, the magnitude of the force on the positive charges will be greater than the magnitude of the force on the negative charges. However, the net effect is still that the piece of paper will feel a force of attraction. (f) Objects with no net electrical charge are composed of electrical charges that can be separated slightly. When a charged object is brought near an uncharged object, the charges with the sign opposite to that on the charged object always end up closer to the charged object. Because the electrical interaction depends inversely on the square of the distances between the charges, the net force is always one of attraction.

C7. (a) Two forces act between protons. One is due to the strong nuclear interaction while the other is due to the electrical interaction. No electrical interaction exists between two neutrons or between a neutron and a proton. The only force which acts is the strong nuclear interaction. (b) The net force between two protons is the vector sum of the strong nuclear attraction and the electrical repulsion. The net force for a proton and a neutron is just the strong nuclear force. Because the strength of the strong nuclear force is the same for two protons as it is for a proton and a neutron, the difference in the net forces is due to the electrical interaction for the two proton case. Thus, the net attractive force between a proton and a neutron is greater than the net attractive force between two protons. (c) If several protons were very close together, it might be possible for the sum of the repulsive forces of all protons on all others to be great enough to overcome the attractive force of the strong nuclear interaction. (d) The neutrons can help keep the nucleus together by providing a contribution to the attractive force without providing any additional repulsive force. They can also get "in the way" and keep the protons further apart, thus diminishing the magnitude of the electrical repulsion.