Projectile Motion
Projectile Launched Horizontally
Projectiles Launched at an Angle
Hang time Revisited
Fast-Moving Projectiles—Satellites
Circular Satellite Orbits
Elliptical Orbits
World Monitoring by Satellite
Kepler’s Laws of Planetary Motion
Energy Conservation and Satellite Motion
Escape Speed
In editions previous to the Ninth Edition, projectile motion was treated with linear motion. Kinematics began the sequence of mechanics chapters. Here we postpone projectile motion until after Newton
Regarding 45° as the maximum range for projectiles, keep in mind that this is only true when air resistance can be neglected, and when the launching speed is the same at all angles concerned. Tilt a water hose up 45° and sure enough, for short distances where air resistance is nil, it attains maximum range. The same is true for a slowly-bunted baseball. But for a high-speed ball, air drag is a factor and maximum range occurs for angles between 39° and 42°. For very high speeds where the lesser air resistance of high altitudes is a consideration, angles greater than 45° produce maximum range. During World War I, for example, the German cannon “Big Bertha” fired shells 11.5 km high and attained maximum range at 52°. Air resistance is one factor; launching speed is another. When one throws a heavy object, like a shot-put, its launching speed is less for higher angles simply because some of the launching force must be used to overcome its weight. (You can throw a heavy boulder a lot faster horizontally than you can straight up.) Shot-puts are usually launched at angles slightly less than 40°. The fact that they are launched higher than ground level decreases the angle as well.
Interestingly, the maximum height of a projectile following a parabolic path is nicely given by sketching an isosceles triangle with the base equal to the range of the projectile. Let the two side angles be equal to the launch angle q, as shown in the figure. The maximum height h reached by the projectile is equal to one-half H, the altitude of the triangle. This goodie from Jon Lamoreux and Luis Phillipe Tosi, of Culver Academies, Culver, IN. (For their explanation see page 183 in The Physics Teacher, Vol. 43, March 2005.)
The interesting fact that projectiles launched at a particular angle have the small range if launched at the complementary angle is stated without proof in the chapter, and is shown in Figure 10.9. This fact is shown by the range formula, R = (2v sinq cosq)/g. Since the sine of an angle is the cosine of the complement of that angle, replacing the angle with its complement will give the same range. So the range is the same whether aiming at q or at (90° - q). Maximum range occurs at a projection angle of 45°, where sine and cosine are equal.
The spin of the Earth is helpful in launching satellites, which gives advantage to launching cities closest to the equator. The launch site closest to the equator is Kaurou, French Guiana, in South America , 5° 08', used by the European Space Agency. The U.S. launches from Cape Canaveral , 28° 22', and Vandenberg, 34° 38'. Russia Hawaii
If you haven’t shown the 15-minute oldie but goodie NASA film, “Zero g”, be sure to show it now. It is of footage taken aboard Skylab in 1978, narrated by astronaut Owen Garriott. Newton
Solar photon force: To a small extent, sunlight affects satellites, particularly the large disco-ball-like satellite LAGEOS, which wobbles slightly in its orbit because of unequal heating by sunlight. The side in the Sun radiates infrared photons, the energy of which provides a small, but persistent, rocket effect as the photons eject from the surface. So a net force some 100 billion times weaker than gravity pushes on the satellite in a direction away from its hot end. LAGEOS has 426 prism-shaped mirrors. By reflecting laser beams off its mirrored surface, geophysicists can make precise measurements of tiny displacements in the Earth’s surface.
Asteroids: Of particular interest are asteroids that threaten Planet Earth. Asteroid 2004 MN4 is big enough to flatten Texas
There are 3 OHTs for this chapter; Figures 10.4, 10.7-10.8, and one with Figures 10.9-10.10.
The independence of horizontal and vertical motion is developed in Practicing Physics book. Also,
• Independence
Components of Motion • Satellites in Elliptical Orbit
• Tossed Ball • Mechanics Overview
There are problems for this chapter in the student ancillary, Problem Solving in Conceptual Physics.
In the Next-Time Questions book:
• Ball Toss from Tower • Orbital Speed
• Monkey and Banana • Escape Fuel
• Elliptical Orbit • Moon Face
• Escape Velocity • Dart Gun
• Satellite Speed • Bull’s-Eye
• Satellite Mass • Projectile Speeds
The lab experiment Bull’s Eye in the Laboratory Manual is a particularly nice wrap-up of projectile motion.
SUGGESTED LECTURE PRESENTATION
Pose the situation of the horizontally-held gun and the shooter who drops a bullet at the same times he pulls the trigger, and ask which bullet hits the ground first.
DEMONSTRATION: Show the independence of horizontal and vertical motion with a spring-gun apparatus that will shoot a ball horizontally while at the same time dropping another that falls vertically. Follow this up with the popular “monkey and hunter” demonstration.
CHECK QUESTIONS: Point to some target at the far side of your classroom and ask your class to imagine you are going to project a rock to the target via a slingshot. Ask if you should aim at the target, above it, or below it. Easy stuff. Then ask your class to suppose it takes 1 second for the rock to reach the target. If you aim directly at the target, it will fall beneath and miss. How far beneath the target would the rock hit (supposing the floor weren’t in the way)? Have your students check with their neighbors on this one. Then ask how far above should you aim to hit the target. Do a neighbor check on this one. Now you’re ready to discuss Figure 10.8 (nicely developed in Practice Pages 41 and 42).
Air Resistance: Acknowledge the large effect of air drag on fast-moving objects such as bullets and cannonballs. A batted baseball, for example, travels only about 60 percent as far in air as it would in a vacuum. Its curved path is no longer a parabola, as Figure 10.11 indicates.
Hang Time Again: Ask if one could jump higher if on a moving skateboard or in a moving bus. It should be clear that the answer is no to both. But one can usually jump higher from a running jump. It is a mistake to assume that the horizontal motion is responsible for the higher jump and longer hang time. The action of running likely enables a greater force between the foot and floor, which gives a greater vertical lift-off component of velocity. This greater bound against the floor, and not any holiday by gravity on a horizontally moving body, is the explanation. Stress that the vertical component of velocity alone determines vertical height and hang time.
Projectiles: Point out that the relationship of the curved path of Figure 10.13 and the vertical distance fallen, d = 5t2, of Chapter 3. Stress that the projectile is falling beneath the straight line it would otherwise follow. This idea is important for later understanding of satellite motion.
Discuss Figure 10.13 and ask for the pitching speed if the ball traveled 30 m instead of 20 m. (Note the vertical height is 5 m. If you use any height that does not correspond to an integral number of seconds, you’re diverting your focus from physics to algebra. An interesting problem on this, however, is Problem 6 about the maximum speed of a tennis ball clearing the net.) More interesting is considering greater horizontal distances—great enough for the curvature of the Earth to make a difference in arriving at the answer. It’s easy to see that the time the projectile is in the air increases where the Earth curves beneath the trajectory.
Satellite Motion: Sketch “Newton
CHECK QUESTION: Why is it confusing to ask why a satellite doesn’t fall? [All satellites are continuously falling, in the sense that they fall below the straight line they would travel if they weren’t. Why they don’t crash to Earth is a different question.]
Calculating Satellite Speed: An effective skit that can have your class calculating the speed necessary for close Earth orbit is as follows: Call attention to the curvature of the Earth, Figure 10.17. Consider a horizontal laser standing about a meter above the ground with its beam shining over a level desert. The beam is straight but the desert floor curves 5 m over an 8000 m or 8 km tangent, which you sketch on your chalkboard (or overhead projector). Stress this is not to scale.
Now erase the laser and sketch in a super cannon positioned so it points along the laser line. Consider a cannonball fired at say, 2 km/s, and ask how far downrange will it be at the end of one second. A neighbor check should yield an answer of 2 km, which you indicate with an “X”. But it doesn’t really get to the “X”, you say, for it falls beneath the “X” because of gravity. How far? 5 m if the sand weren’t in the way. Ask if 2 km/s is sufficient for orbiting the Earth. Clearly not, for the cannonball strikes the ground. If the cannonball is not to hit the ground, we’d have to dig a trench first, as you show on your sketch, which now looks like this:
Continue by considering a greater muzzle velocity, say 4 km/s, so the cannonball travels 4 km in one second. Ask if this is fast enough to attain an Earth orbit. Student response should indicate that they realize that the cannonball will hit the ground before 1 second is up. Then repeat the previous line of reasoning, again having to dig a trench, and your sketch looks like this:
Continue by considering a greater muzzle velocity—great enough so the cannonball travels 6 km in 1 second. This is 6 km/s. Ask if this is fast enough not to hit the ground (or equivalently, if it is fast enough for Earth orbit). Then repeat the previous line of reasoning, again having to dig a trench. Now your sketch looks like this:
You’re almost there. Continue by considering a muzzle velocity great enough so the cannonball travels 8 km in one second. (Don’t state the velocity is 8 km/s here as you’ll diminish your punch line.) Repeat your previous reasoning and note that this time you don’t have to dig a trench! After a pause, and with a tone of importance, ask the class with what speed must the cannonball have to orbit the Earth. Done properly, you have led your class into a “derivation” of orbital speed about the Earth with no equations or algebra.
Acknowledge that the gravitational force is less on satellites in higher orbits so they do not need to go so fast. This is acknowledged later in the chapter in a footnote. (Since v = √GM/d, a satellite at 4 times the Earth’s radius needs to travel only half as fast, 4 km/s.)
You can wind up your brief treatment of satellite motion and catch its essence via the following skit: Ask your students to pretend they are encountered by a bright youngster, too young to have much knowledge of physics and mathematics, but who nevertheless asks why satellites seem to defy gravity and stay in orbit. You ask what answer could correctly satisfy the curiosity of the kid, then pose the following dialogue between the kid and the students in your class (you’re effectively suggesting how the student might interact with the bright kid). Ask the kid to observe and then describe what you do, as you hold a rock at arm’s length and then simply drop it. The kid replies, “You dropped the rock and it fell to the ground below,“ to which you respond, “Very good—now what happens this time?”, as you move your hand horizontally and again drop the rock. The kid observes and then says, “The rock dropped again, but because your hand was moving it followed a curved path and fell farther away.” You continue, “Very good—now again—”as you throw the rock still farther. The kid replies, “I note that as your hand moves faster, the path follows a wider curve.” You’re elated at this response, and you ask the kid, “How far away will the rock hit the ground if its curved path matches the curved surface of the Earth?” The kid at first appears very puzzled, but then beams, “Oh—I get it! The stone doesn’t hit at all—it’s in Earth orbit.” Then you interrupt your dialogue and ask the class, “Do YOU get it?” Then back to the kid who asks, “But isn’t it really more complicated than that?”, to which the answer is NO. The essential idea of satellite motion IS that simple.
Begin by discussing the falling apple story and Newton Newton
Moving Perpendicular vs Moving Nonperpendicular to Gravity: Discuss the motion of a cannonball fired horizontally from a mountain top. Suppose the cannonball leaves the cannon at a velocity of say 1 km/s. Ask the class whether the speed when it strikes the ground will be 1 km/s, more than 1 km/s, or less than 1 km/s (neglecting air resistance). The answer is that it strikes at more than 1 km/s because gravity speeds it up. (Toss your keys horizontally from a one-story window and catching them would pose no problem. Now toss your keys horizontally from the top of a mountain. Ask if a person below would care to catch them!) Now draw “Newton
Before answering the question, pose the case of rolling a ball along a bowling alley. Does gravity pull on the ball? [Yes.] Does gravity speed up or slow down the ball? [No.] Why? The answer to this question is the answer to the satellite question. [In both cases, the ball crisscrosses gravity—with no component of the gravitational field in the direction of motion. No change in speed, no work, no change in KE, no change in PE. Aha! The cannonball and the bowling ball simply coast.]
Circular Orbits: Erase the mountain from your sketch of the world and draw a huge elevated bowling alley that completely circles the world (Figure 10.21). You’re extending Figure 10.20. Show how a bowling ball on such an alley would gain no speed because of gravity. But now cut part of the alley away, so the ball rolls off the edge and crashes to the ground below. Does it gain speed after falling in the gap? [Yes, because a component of its motion is in the direction of the Earth’s gravitational field.] Show how if the ball moves faster it will fall farther before crashing to the ground. Ask what speed would allow it to clear the gap (like a motorcyclist who drives off a ramp and clears a gap to meet a ramp on the other side). [8 km/s, of course.] Can the gap be bigger at this speed? Sketch a gap that nearly circles the world when you ask this question. Then ask, what happens with no alley? And your class sees at 8 km/s no supporting alley is needed. The ball orbits the Earth.
CHECK QUESTION: We say that satellites are falling around the Earth. But communication satellites remain at one place overhead. Isn’t this contradictory? [Communication satellites fall in a wider circle than closer satellites. Their periods are 24 hours, which coincides with the period of the spinning Earth. So from Earth they appear to be motionless.]
CHECK QUESTION: Why is it advantageous to launch rockets close to the equator? [The tangential speed at the equator is 1000 miles an hour, which can be subtracted from the speed needed to put a satellite in orbit. The closer the launch site to the equator, the closer to the 1000 mph free ride.]
Elliptical Orbits: Back to Newton
Continue your sketch and show a closed path—an ellipse. As you draw the elliptical path, show with a sweeping motion of your arm how the satellite slows in receding from the Earth, moving slowest at its farthermost point, then how it speeds in falling towards the Earth, whipping around the Earth and repeating the cycle over and over again. Move to a fresh part of the chalkboard and redraw with the mountain at the bottom, so your sketch is more like Figure 10.25. (It is more comfortable seeing your chalk moving slowest when farthest coincides with the direction “up” in the classroom. I quip that Australians have no trouble seeing it the first way.)
Sketch in larger ellipses for still greater cannon speeds, with the limit being 11.2 km/s, beyond which the path does not close—escape speed.
State that Newton
Kepler’s Laws: Briefly discuss Kepler’s laws. Sketch an elliptical path of a planet about the Sun as in Figure 10.29. Show how the equal areas law means that the planet travels slowest when farthest from the Sun, and fastest when closest. State that Kepler had no idea why this was so. Walk to the side of your room and toss a piece of chalk upward at a slight angle so the class can see the parabolic path it traces. Ask where the chalk is moving slowest? Fastest? Why is it moving slowest at the top? [Because it has been traveling against gravity all the way up!] Why is it moving fastest when it is thrown and when it is caught? [It’s moving fastest when it is caught because it has been traveling in the direction of gravity all the way down!] Speculate how amazed Kepler would have been if the same questions were asked him, and related to the speeds of the planets around the Sun—slowest where they have been traveling against the gravity of the Sun, and fastest where they have been falling back toward the Sun. Kepler would have been amazed to see the physics of a body tossed upward is essentially the physics of satellite motion! Kepler lacked this simple model to guide his thinking. What simple models of tomorrow do we lack today, that finds us presently blind to the common sense of tomorrow?
Work-Energy Relationship for Satellites: You already have sketches on the board of circular and elliptical orbits. Draw sample satellites and then sketch in force vectors. Ask the class to do likewise, and then draw component vectors parallel and perpendicular to instantaneous directions of motion. Then show how the changes in speed are consistent with the work-energy relationship.
Draw a large ellipse on the board with a planet in various positions and ask your class for a comparison of the relative magnitudes of KE and PE along the orbit. You can do this with different size symbols for KE and PE. Stress that the two add up to be the same.
Escape Speed: Distinguish between ballistic speed and sustained speed, and that the value 11.2 km/s refers to ballistic speed. (One could go to the Moon at 1 km/s, given a means of sustaining that speed and enough time to make the trip!) Compare the escape speeds from different bodies via Table 10-1.
Maximum Falling Speed: The idea of maximum falling speed, footnoted in the chapter, is sufficiently interesting for elaboration. Pretend you throw your car keys from ground level to your friend at the top of a building. Throw them too fast and they continue by your friend; throw them too slow and they never reach her. But if you throw them just right, say 11 m/s, they just barely reach her so she has only to grab them at their point of zero speed. Question: It took a speed of 11 m/s to get the keys up to her—if she simply drops them, how fast will they fall into your hands? Aha! If it takes a speed of 11.2 km/s to throw them to her if she is somewhat beyond Pluto, and she similarly drops them, how fast will they fall into your hands? Now your students understand maximum falling speed.
CHECK QUESTIONS: This reviews several chapters of mechanics; draw an elliptical orbit about a planet as shown on the board. Pose the following questions (from the Practice Book): At which position does the satellite have the maximum
(1) gravitational force on it?
(2) speed?
(3) momentum?
(4) kinetic energy?
(5) gravitational potential energy?
(6) total energy?
(7) acceleration?
(8) angular momentum?
Don’t be surprised to find many of your students miss (7), acceleration, even though they answer the first about force correctly. If they use either equation for acceleration as their “guide,” the answer is at hand; that is, from a - F/m, the acceleration is seen to be maximum where the force is maximum—at A. Or from a = (change in v)/t, acceleration is seen to be greatest where most of the change occurs—where the satellite whips around A. This Check Question summarizes important ideas in four chapters. Go over the answers carefully.
Solutions to Chapter 10 Exercises
1. The divers are in near-free-fall, and as Figure 4.11 back in Chapter 4 shows, falling speed is independent of mass (or weight).
2. In accord with the principle of horizontal and vertical projectile motion, the time to hit the floor is independent of the ball’s speed.
3. Yes, it will hit with a higher speed in the same time because the horizontal (not the vertical) component of motion is greater.
4. No, because while in the air the train changes its motion.
5. The crate will not hit the Porsche, but will crash a distance beyond it determined by the height and speed of the plane.
6. The path of the falling object will be a parabola as seen by an observer off to the side on the ground. You, however, will see the object fall straight down along a vertical path beneath you. You’ll be directly above the point of impact. In the case of air resistance, where the airplane maintains constant velocity via its engines while air drag decreases the horizontal component of velocity for the falling object, impact will be somewhere behind the airplane.
7. (a) The paths are parabolas. (b) The paths would be straight lines.
8. There are no forces horizontally (neglecting air drag) so there is no horizontal acceleration, hence the horizontal component of velocity doesn’t change. Gravitation acts vertically, which is why the vertical component of velocity changes.
9. Minimum speed occurs at the top, which is the same as the horizontal component of velocity anywhere along the path.
10. For very slow-moving bullets, the dropping distance is comparable to the horizontal range, and the resulting parabola is easily noticed (the curved path of a bullet tossed sideways by hand, for example). For high speed bullets, the same drop occurs in the same time, but the horizontal distance traveled is so large that the trajectory is “stretched out” and hardly seems to curve at all. But it does curve. All bullets will drop equal distances in equal times, whatever their speed. (It is interesting to note that air resistance plays only a small role, since the air resistance acting downward is practically the same for a slow-moving or fast-moving bullet.)
11. Kicking the ball at angles greater than 45° sacrifices some distance to gain extra time. A kick greater than 45° doesn’t go as far, but stays in the air longer, giving players on the kicker’s team a chance to run down field and be close to the player on the other team who catches the ball.
12. Both balls have the same range (see Figure 10.9). The ball with the initial projection angle of 30°, however, is in the air for a shorter time and hits the ground first.
13. The bullet falls beneath the projected line of the barrel. To compensate for the bullet’s fall, the barrel is elevated. How much elevation depends on the velocity and distance to the target. Correspondingly, the gunsight is raised so the line of sight from the gunsight to the end of the barrel extends to the target. If a scope is used, it is tilted downward to accomplish the same line of sight.
14. The monkey is hit as the dart and monkey meet in midair. For a fast-moving dart, their meeting place is closer to the monkey’s starting point than for a slower-moving dart. The dart and monkey fall equal vertical distances—the monkey below the tree, and the dart below the line of sight—because they both fall with equal accelerations for equal times.
15. Any vertically projected object has zero speed at the top of its trajectory. But if it is fired at an angle, only its vertical component of velocity is zero and the velocity of the projectile at the top is equal to its horizontal component of velocity. This would be 100 m/s when the 141-m/s projectile is fired at 45°.
16. Hang time depends only on the vertical component of your lift-off velocity. If you can increase this vertical component from a running position rather than from a dead stop, perhaps by bounding harder against the ground, then hang time is also increased. In any case, hang time depends only on the vertical component of your lift-off velocity.
17. The hang time will be the same, in accord with the answer to the preceding exercise. Hang time is related to the vertical height attained in a jump, not on horizontal distance moved across a level floor.
18. The Moon’s tangential velocity is what keeps the Moon coasting around the Earth rather than crashing into it. If its tangential velocity were reduced to zero, then it would fall straight into the Earth!
19. Yes, the shuttle is accelerating, as evidenced by its continual change of direction. It accelerates due to the gravitational force between it and the Earth. The acceleration is toward the Earth’s center.
20. From Kepler’s third law, T2 ~ R3, the period is greater when the distance is greater. So the periods of planets farther from the Sun are longer than our year.
21. Neither the speed of a falling object (without air resistance) nor the speed of a satellite in orbit depends on its mass. In both cases, a greater mass (greater inertia) is balanced by a correspondingly greater gravitational force, so the acceleration remains the same (a = F/m, Newton
22. Speed does not depend on the mass of the satellite (just as free-fall speed doesn’t).
23. Gravitation supplies the centripetal force on satellites.
24. Mars or any body in Earth’s orbit would take the same time to orbit. A satellite, like a freely-falling object, does not depend on mass.
25. The initial vertical climb lets the rocket get through the denser, retarding part of the atmosphere most quickly, and is also the best direction at low initial speed, when a large part of the rocket’s thrust is needed just to support the rocket’s weight. But eventually the rocket must acquire enough tangential speed to remain in orbit without thrust, so it must tilt until finally its path is horizontal.
26. Gravity changes the speed of a cannonball when the cannonball moves in the direction of Earth gravity. At low speeds, the cannonball curves downward and gains speed because there is a component of the force of gravity along its direction of motion. Fired fast enough, however, the curvature matches the curvature of the Earth so the cannonball moves at right angles to the force of gravity. With no component of force along its direction of motion, its speed remains constant.
27. The Moon has no atmosphere (because escape velocity at the Moon’s surface is less than the speeds of any atmospheric gases). A satellite 5 km above the Earth’s surface is still in considerable atmosphere, as well as in range of some mountain peaks. Atmospheric drag is the factor that most determines orbiting altitude.
28. Consider “Newton
29. Consider “Newton
30. Rockets for launching satellites into orbit are fired easterly to take advantage of the spin of the Earth. Any point on the equator of the Earth moves at nearly 0.5 km/s with respect to the center of the Earth or the Earth’s polar axis. This extra speed does not have to be provided by the rocket engines. At higher latitudes, this “extra free ride” is less.
31. Upon slowing it spirals in toward the Earth and in so doing has a component of gravitational force in its direction of motion which causes it to gain speed. Or put another way, in circular orbit the perpendicular component of force does no work on the satellite and it maintains constant speed. But when it slows and spirals toward Earth there is a component of gravitational force that does work to increase the KE of the satellite.
32. Hawaii
33. A satellite travels faster when closest to the body it orbits. Therefore Earth travels faster about the Sun in December than in June.
34. At midnight you face away from the Sun, and therefore cannot see the planets closest to the Sun—Mercury and Venus (which lie inside the Earth’s orbit).
35. When descending, a satellite meets the atmosphere at almost orbital speed. When ascending, its speed through the air is considerably less and it attains orbital speed well above air drag.
36. Yes, a satellite needn’t be above the surface of the orbiting body. It could orbit at any distance from the Earth’s center of mass. Its orbital speed would be less because the effective mass of the Earth would be that of the mass below the tunnel radius. So interestingly, a satellite in circular orbit has its greatest speed near the surfa
38. In circular orbit there is no component of force along the direction of the satellite’s motion so no work is done. In elliptical orbit, there is always a component of force along the direction of the satellite’s motion (except at the apogee and perigee) so work is done on the satellite.
38. In circular orbit there is no component of force along the direction of the satellite’s motion so no work is done. In elliptical orbit, there is always a component of force along the direction of the satellite’s motion (except at the apogee and perigee) so work is done on the satellite.
39. When the velocity of a satellite is everywhere perpendicular to the force of gravity, the orbital path is a circle (see Figure 10.20).
40. The period of any satellite at the same distance from Earth as the Moon would be the same as the Moon’s, 28 days.
41. No way, for the Earth’s center is a focus of the elliptical path (including the special case of a circle), so an Earth satellite orbits the center of the Earth. The plane of a satellite coasting in orbit always intersects the Earth’s center.
42. No, for an orbit in the plane of the Arctic Circle does not intersect the Earth’s center. All Earth satellites orbit in a plane that intersects the center of the Earth. A satellite may pass over the Arctic Circle , but cannot remain above it indefinitely, as a satellite can over the equator.
43. The plane of a satellite coasting in orbit intersects the Earth’s center. If its orbit were tilted relative to the equator, it would be sometimes over the Northern Hemisphere, sometimes over the Southern Hemisphere. To stay over a fixed point off the equator, it would have to be following a circle whose center is not at the center of the Earth.
44. Singapore Singapore , so a satellite can be located directly above Singapore San Francisco
45. Period is greater for satellites farthest from Earth.
46. If a wrench or anything else is “dropped” from an orbiting space vehicle, it has the same tangential speed as the vehicle and remains in orbit. If a wrench is dropped from a high-flying jumbo jet, it too has the tangential speed of the jet. But this speed is insufficient for the wrench to fall around and around the Earth. Instead it soon falls into the Earth.
47. It could be dropped by firing it straight backward at the same speed of the satellite. Then its speed relative to Earth would be zero, and it would fall straight downward.
48. When a capsule is projected rearward at 7 km/s with respect to the spaceship, which is itself moving forward at 7 km/s with respect to the Earth, the speed of the capsule with respect to the Earth will be zero. It will have no tangential speed for orbit. What will happen? It will simply drop vertically to Earth and crash.
49. If the speed of the probe relative to the satellite is the same as the speed of the satellite relative to the Moon, then, like the projected capsule that fell to Earth in the previous question, it will drop vertically to the Moon. If fired at twice the speed, it and the satellite would have the same speed relative to the Moon, but in the opposite direction, and might collide with the satellite after half an orbit.
50. This is similar to Exercises 39 and 40. The tangential velocity of the Earth about the Sun is 30 km/s. If a rocket carrying the radioactive wastes were fired at 30 km/s from the Earth in the direction opposite to the Earth’s orbital motion about the Sun, the wastes would have no tangential velocity with respect to the Sun. They would simply fall into the Sun.
51. Communication satellites only appear motionless because their orbital period coincides with the daily rotation of the Earth.
52. The half brought to rest will fall vertically to Earth. The other half, in accord with the conservation of linear momentum will have twice the initial velocity, overshoot the circular orbit, and enter an elliptical orbit whose apogee (highest point) is farther from the Earth’s center.
53. The design is a good one. Rotation would provide a centripetal force on the occupants. Watch for this design in future space faring.
54. The principle advantage is bypassing expensive rockets, which are often single-time lift vehicles. The same aircraft can be used to repeatedly launch space vehicles.
55. The escape speeds from various planets refer to “ballistic speeds”—to the speeds attained after the application of an applied force at low altitude. If the force is sustained, then a space vehicle could escape the Earth at any speed, so long as the force is applied sufficiently long.
56. Maximum falling speed by virtue only of the Earth’s gravity is 11.2 km/s (see the footnote in the chapter).
57. This is similar to the previous exercise. In this case, Pluto’s maximum speed of impact on the Sun, by virtue of only the Sun’s gravity, would be the same as the escape speed from the surface of the Sun, which according to Table 10.1 in the text is 620 km/s.
58. Acceleration is maximum where gravitational force is maximum, and that’s when Earth is closest to the Sun, at the perigee. At the apogee, force and acceleration are minimum.
59. The satellite experiences the greatest gravitational force at A, where it is closest to the Earth; and the greatest speed and the greatest velocity at A, and by the same token the greatest momentum and greatest kinetic energy at A, and the greatest gravitational potential energy at the farthest point C. It would have the same total energy (KE + PE) at all parts of its orbit, likewise the same angular momentum because it’s conserved. It would have the greatest acceleration at A, where F/m is greatest.
60. In accord with the work-energy relationship, Fd = ∆KE, for a constant thrust F, the maximum change in KE will occur when d is maximum. The rocket will travel the greatest distance d during the brief firing time when it is traveling fastest—at the perigee.
Chapter 10 Problem Solutions
1. One second after being thrown, its horizontal component of velocity is 10 m/s, and its vertical component is also 10 m/s. By the Pythagorean theorem, V = √(102 + 102) = 14.1 m/s. (It is moving at a 45° angle.)
2. (a) From y = 5t2 = 5(30)2 = 4,500 m, or 4.5 km high (4.4 km if we use g = 9.8 m/s2).
(b) In 30 seconds the falling engine travels horizontally 8400 m (d = vt = 280 m/s ´ 30 s = 8400 m).
(c) The engine is directly below the airplane. (In a more practical case, air resistance is overcome for the plane by its engines, but not for the falling engine, so the engine’s speed is reduced by air drag and it covers less than 8400 horizontal meters, landing behind the plane.)
3. 100 m/s. At the top of its trajectory, the vertical component of velocity is zero, leaving only the horizontal component. The horizontal component at the top or anywhere along the path is the same as the initial horizontal component, 100 m/s (the side of a square where the diagonal is 141).
4. The distance wanted is horizontal velocity ´ time. We find the time from the vertical distance the ball falls to the top of the can. This distance is 1.5 m – 0.2 m = 1.3 m. The time is found using g = 10 m/s2 and d = 1.3 m = (1/2)gt2. Solving for t we get 0.51 s. Horizontal travel is then d = vt = (4.0 m/s)(0.51 s) = 2.04 m. (For g = 9.8 m/s2, d = 2.06 m. The width of the can should make either setting successful. If the height of the can is not subtracted from the 1.5 m vertical distance between floor and tabletop, the calculated d will equal 2.2 m, the can will be too far away, and the ball will miss!)
5. John and Tracy
d = 5t2, where rearrangement gives t = √ = √= 4 s.
So to travel 20 m horizontally in this time means John and Tracy should jump horizontally with a velocity of 20 m/4 s = 5 m/s. But this would put them at the edge of the pool, so they should jump a little faster. If we knew the length of the pool, we could calculate how much faster without hitting the far end of the pool. (John and Tracy would be better advised to take the elevator.)
6. The maximum horizontal speed of the ball clearing the net is 26.6 m/s (any faster will put it outside the court). Horizontal speed is v = d/t where d is the 12.0-m horizontal distance and time t, the time of flight of the ball, is found by considering the same time for a vertical 1.0-m drop; t = √2d/g = √2(1.0)/10 = 0.45 s. So v = d/t = 12.0 m/0.45 s = 26.6 m/s.
7. Hang time depends only on the vertical component of initial velocity and the corresponding vertical distance attained. From d = 5t2 a vertical 1.25 m drop corresponds to 0.5 s (t = √2d/g = √2(1.25)/10 = 0.5 s). Double this (time up and time down) for a hang time of 1 s. Hang time is the same whatever the horizontal distance traveled.
8. One way is: v = distance/time where distance is the circumference of the Earth’s orbit and time is 1 year. Then
v = = = =
3 ´ 104 m/s = 30 km/s.
Another way is using the equation shown in the footnote on page 195:
v = √= = 3 ´ 104 m/s.
9. v = √= √= 1026 m/s.
10. In accord with the work-energy theorem (Chapter 7) W = ∆KE the work done equals energy gained. The KE gain is 8 - 5 billion joules = 3 billion joules. The potential energy decreases by the same amount that the kinetic energy increases, 3 billion joules.
ce of the Earth, and decreases with both decreasing and increasing distances.
37. The component along the direction of motion does work on the satellite to change its speed. The component perpendicular to the direction of motion changes its direction of motion.
