The Universal Law of Gravity
The Universal Gravitation Constant, G
Gravity and Distance: The Inverse-Square Law
Weight and Weightlessness
Ocean Tides
Tides in the Earth and Atmosphere
Tides on the Moon
Gravitational Fields
The Gravitational Field Inside a Planet
Einstein’s Theory of Gravitation
Black Holes
Universal Gravitation
This chapter begins and ends on an astronomical theme, and makes use of the historical approach. It is prerequisite to the following chapter on satellite motion.
This chapter offers a good place to reiterate the idea of a scientific theory, and comment on the all-too common and mistaken idea that because something has the status of scientific theory, it is somehow short of being valid. This view is evident in those who say, “But it’s only a theory.” Bring the essence of the first and last footnotes in the chapter into your discussion (about scientific homework and not being able to see radically new ways of looking at the world). The last chapter on Cargo Cult Science of Feynman’s book, Surely You’re Joking, Mr. Feynman (Norton, 1985), expands nicely on this.
Kepler’s 3rd law follows logically from Newton
where the speed of the planet is 2p per period T. Cancel and collect terms,
This is Kepler’s 3d law, for GM/4p2 is a constant.
The idea of the force field is introduced in this chapter and is a good background for the electric field treated later in Chapter 22. The gravitational field here is applied to regions outside as well as inside the Earth.In the text I say without explanation that the gravitational field increases linearly with radial distance inside a planet of uniform density. Figure 9.23 shows that the field increases linearly from zero at its center to maximum at the surface. This is also without explanation. The text states that “perhaps your instructor will provide the explanation.” Here it is: We know that the gravitational force F between a particle m and a spherical mass M, when m is outside M is simple F = GmM/d2. But when m is inside a uniform density solid sphere of mass M, the force on m is due only to the mass M’ contained within the sphere of radius r < R, represented by the dotted line in the figure. Contributions from the shell > r cancel out (established in the text near Figure 9.22, and again for the analogous case of the electric field in Figure 22.20, later in the book). So, F = GmM’/r2. From the ratio of M’/M, you can show that M’ = Mr3/R3. [That is, M’/M = V’/V = (4/3 πr3)/(4/3 πR3) = r3/R3.] Substitute M’ in Newton
Interestingly enough, the condition for simple harmonic motion is that the restoring force be proportional to displacement, F = kr—hence the simple harmonic motion of one who falls in the tunnel through the Earth. (Hence also the simple harmonic motion of one who slides without friction to-and-fro along any straight line tunnel through any part of the Earth. The displacement is then the component r sin q.) The period of simple harmonic motion, T = 2π√(R3/GM) is the same as that of a satellite in close circular orbit about the Earth. Note that it is independent of the length of the tunnel. See more detail in a text such as Tipler’s Physics For Scientists and Engineers.
You can compare the pull of the Moon that is exerted on you with the pull exerted by more local masses, via the gravitational equation. Consider the ratio of the mass of the Moon to its distance squared:
7.4 ´ 1022 kg / (4 ´105 km)2 = 5 ´ 1012 kg / km2.
This is a sizeable ratio, one that buildings in your vicinity cannot match. (City buildings of greatest mass are typically on the order of 106 or 107 kilograms.) However, if you stand 1 kilometer away from the foot of a mountain of mass 5 x 1012 kilograms (about the mass of Mount Kilimanjaro), then the pull of the mountain and the pull of the Moon are about the same on you. Simply put, with no friction you would tend to gravitate from your spot toward the mountain—but you experience no tendency at all to gravitate from your spot toward the Moon! That’s because the spot you stand on undergoes the same gravitational acceleration toward the Moon as you do. Both you and the whole Earth are accelerating toward the Moon. Whatever the lunar force on you, it has no tendency to pull you off a weighing scale—which is the essence of Exercises 30 and 32.
The inverse-cube nature of tidal forces follows from subtracting the tidal force on the far side of a body from the tidal force on the near side. Consider a kilogram of water on the side of the Earth nearest the Moon that is gravitationally attracted to the Moon with a greater force than a kilogram of water on the side of the Earth farthest from the Moon. The difference in force per kilogram of mass, ∆F/m, which we’ll call the tidal force TF is
TF = Fd+R - Fd-R
where M is the Moon’s mass, (d+R) is the distance to the far side of Earth, (d-R) is the distance to the near side.When d is very much greater than R, the (d2-R2)2 is very nearly equal to d4. Then the inverse-cube nature of tidal force is evident, for
TF ~ 
.
.
Some interesting results occur when calculating the tidal force of the Moon on planet Earth. TF is 2.2 ´ 10-6 N ⁄ kg. In contrast TF of an overhead Moon on a person on Earth is 3 ´ 10-13 N ⁄ kg, a hundred million times weaker because of the tiny differences in pulls across the body. The tidal force of the Earth on the same person is 6 ´ 10-6 N ⁄ kg, more than the Moon’s influence. And as the text reports, the tidal force due to a 1-kg mass held 1 m above your head is about 200 times as much effective as the Moon! Have those who believe the tidal effects of planets influence people make the calculations themselves.A brief treatment of black holes is included in this chapter. The idea that light is influenced by a gravitational field isn’t treated until Chapter 36, so may merit further explanation. You’ll probably want to acknowledge that light bends in a gravitational field as does a thrown baseball. We say light travels in straight lines much for the same reason that some people say that a high-speed bullet doesn’t curve downward in the first part of its trajectory. Over short distances the bullet doesn’t appear to drop only because of the high speed and short time involved. Likewise with light, which we don’t notice because of the high speed and therefore vast distance involved compared to the brief time it is in the strong part of the Earth’s gravitational field. Note the treatment of this idea in Figure 36.6 near the end of the book. Black holes at the center of galaxies are bigger than those found in binary star systems. The biggest reported galactic black hole, as of January 1997, is 3 billion suns.
Dark matter is briefly mentioned in this chapter and is discussed in Chapter 11. The presence of dark matter kills the oscillating universe scenario speculated about in the earlier editions of this text. The universe is accelerating outward, rather than slowing down. A different form of energy, referred to as the dark energy, apparently works against the force of gravity to accelerate the expansion of the universe. The concepts of dark matter and dark energy are at the forefront of physics at this point, and are quite mysterious. Dark matter is out of sight, but not out of mind.
There are problems for this chapter in the Problem Solving in Conceptual Physics book.
There are 5 OHTs for this chapter: Figures 9.5, 9.7, 9.15, 9.20 and one with Figures 9.21, 22.24, and 24.1.
In the Practicing Physics book:
• Inverse-Square Law
• Our Ocean Tides
In the Next-Time Questions book:
• Earth-Moon Cable • Giant Plane
• Moon Tides • Body Tide
• Solar Black Hole • Gravity Force on Shuttle
• Earth Rise
The activity Trial and Error, in the Laboratory Manual is a dandy.
While the information in this chapter is useful as a background for Chapter 36 (general relativity) and Chapter 22 (the inverse-square law, and the analogy between a gravitational and electric field) it may be skipped without complicating the treatment of other material. This is an interesting chapter, for the material is interesting in itself, is interesting historically, and is closely related to areas of space science that are currently in the public eye.
SUGGESTED LECTURE PRESENTATION
Begin by briefly discussing the simple codes and patterns that underlie the complex things around us, whether musical compositions or DNA molecules, and then briefly describe the harmonious motion of the solar system, the Milky Way and other galaxies in the universe—stating that the shapes of the planets, stars, and galaxies, and their motions are all governed by an extremely simple code, or if you will, a pattern. Then write the gravitational equation on the board. Give examples of bodies pulling on each other to convey a clear idea of what the symbols in the equation mean and how they relate. (Acknowledge that many other texts and references use the symbol r instead of the d in this text. The r is used to indicate the radial distance from a body’s CG, and to emphasize the center-to-center rather than surface-to-surface nature for distance, and to prepare for r as a displacement vector. We don’t set our plow that deep, however, and use d for distance.)
Inverse-Square Law: Discuss the inverse-square law and go over Figures 9.5 and 9.7 or their equivalents with candlelight or radioactivity. Using OHTs T-22 and T-23 are helpful here.
Plot to scale an inverse-square curve on the board, showing the steepness of the curve—1/4, 1/9, and 1/16, for twice, three times, and four times the separation distance occur rather “suddenly.” This is shown in Figure 9.7. (You’ll return to this curve when you explain tides.)
CHECK QUESTIONS: A sheet of photographic film is exposed to a point source of light that is a certain distance away. If the sheet were instead exposed to the same light four times as far away, how would the intensity on the film compare? A radioactive detector registers a certain amount of radioactivity when it is a certain distance away from a small piece of uranium. If the detector is four times as far from the uranium, how will the radioactivity reading compare?
CHECK QUESTIONS: How is the gravitational force between a pair of planets altered when one of the planets is twice as massive? When both are twice as massive? When they are twice as far apart? When they are three times as far apart? Ten times as far apart?
CHECK QUESTION: What do you say to a furniture mover who claims that gravity gets more with increased distance from the Earth, as evident to him when he’s carrying heavy loads up flights of stairs?
Weight and Weightlessness: Note that we define weight as a support force. Even in a gravity-free region inside a rotating toroid, you’d experience weight. So weight needn’t always be related to gravity. Discuss weightlessness and relate it to the queasy feeling your students experience when in a car that goes too fast over the top of a hill. State that this feeling is what an astronaut is confronted with all the time in orbit! Ask how many of your class would still welcome the opportunity to take a field trip to Cape Canaveral and take a ride aboard the shuttle. What an exciting prospect!
Discuss the fate of Skylab when its orbit decayed and it “scraped its feet” in the atmosphere in 1978, and Mir in 2001. I show the now oldie but goodie NASA film “Zero g” every semester, for it not only is fascinating in its shots of astronaut acrobatics in the orbiting lab, but recaps Newton’s laws as they apply to interesting situations. The film shows the good sense of humor of the astronauts. A must!
Discuss the differences in a baseball game on the Moon, and your favorite gravity-related topics.
Tides: Begin your treatment of tides by asking the class to consider the consequences of someone pulling your coat. If they pulled only on the sleeve, for example, it would tear. But if every part of your coat were pulled equally, it and you would accelerate—but it wouldn’t tear. It tears when one part is pulled harder than another—or it tears because of a difference in forces acting on the coat. In a similar way, the spherical Earth is “torn” into an elliptical shape by differences in gravitational forces by the Moon and Sun.
CHECK QUESTION: Why do the tides not occur at the same time each day? [As the Earth takes 24 hours to rotate, the Moon advances in its orbit one hour ahead of the Earth. If the Moon didn’t move in its orbit, the high-tide bulge would be at the same time each day as the Earth spins beneath the water.]
Misconceptions About the Moon: This is an appropriate place for you to dispel two popular misconceptions about the Moon. One is that since one side of the Moon’s face is “frozen” to the Earth it doesn’t spin like a top about its polar axis; and two, that the crescent shape commonly seen is not the Earth’s shadow. To convince your class that the Moon spins about its polar axis, simulate the situation by holding your eraser at arms length in front of your face. Tell your class that the eraser represents the Moon and your head represents the Earth. Rotate slowly keeping one face of the eraser in your view. Call attention to the class that from your frame of reference, the eraser doesn’t spin as it rotates about you—as evidenced by your observation of only one face, with the backside hidden. But your students occupy the frame of reference of the stars. (Each of them is a star.) From their point of view they can see all sides of the eraser as it rotates because it spins about its own axis as often as it rotates about your head. Show them how the eraser, if not slowly spinning and rotationally frozen with one face always facing the same stars, would show all of its sides to you as it circles around you. See one face, then wait 14 days later and the backside is in your view. The Moon has a spin rate that is the same as its rotational rate (explained in Figure 9.20).
Misconception 2: Draw a half moon on the board. The shadow is along the diameter and is perfectly straight. If that were the shadow of the Earth, then the Earth would have to be flat, or be a big block shape! Discuss playing “flashlight tag” with a suspended basketball in a dark room that is illuminated by a flashlight in various locations. Ask your class if they could estimate the location of the flashlight by only looking at the illumination of the ball. Likewise with the Moon illuminated by the Sun!Sketch the picture on the right on the board and ask what is wrong with it.
[Answer: The Moon is in a daytime position as evidenced by the upper part of the Moon being illuminated. This means the Sun must be above. Dispel notions that the crescent shape of the Moon is a partial eclipse by considering a half moon and the shape of the Earth to cast such a shadow.]
Back to Tides: Explain tides via the accelerating ball of Jell-O as in the text. Equal pulls result in an undistorted ball as it accelerates, but unequal pulls cause a stretching. This stretching is evident in the Earth’s oceans, where the side nearest the Moon is appreciably closer to the Moon than the side farthest away. If you don’t have OHT-24, carefully draw Figure 9.15 on the board, which explains why closeness is so important for tides. The figure shows that the magnitude of DF rather than F itself is responsible for tidal effects. Hence the greater attraction of the distant Sun produces only a small difference in pulls on the Earth, and compared to the Moon makes a small contribution to the tides on Earth.
Explain why the highest high tides occur when the Earth, Moon, and Sun are aligned—at the time of a new and a full Moon.
Discuss tides in the molten Earth and in the atmosphere.
Amplify Figure 9.17 with a comparison of DFs for both the Sun and the Moon as below.
Clearly DF is smaller for the larger but farther Sun.
The text treats tides in terms of forces rather than fields. In terms of the latter, tidal forces are related to differences in gravitational field strengths across a body, and occur only for bodies in a nonuniform gravitational field. The gravitational fields of the Earth, Moon, and Sun, for example, are inverse-square fields—stronger near them than farther away. The Moon obviously experiences tidal forces because the near part to us is in a stronger part of the Earth’s gravitational field than the far part. But even an astronaut in an orbiting space shuttle strictly speaking experiences tidal forces because parts of her body are closer to the Earth than other parts. This tidal force, the difference between the forces on near and far parts of her body, follow an inverse-cube law (as derived back on pages 83 and 84). The micro differences produce microtides. Farther away in deep space, the differences are less. Put another way, the Earth’s gravitational field is more uniform farther away. The “deepness” of a deep-space location can in fact be defined in terms of the amount of microtides experienced by a body there. Or equivalently, by the uniformity of any gravitational field there. There are no microtides in a body located in a strictly uniform gravitational field.
If there are microtides of an astronaut in orbit, would such microtides are even greater on the Earth’s surface? The answer is yes. This brings up Exercise 42 that concerns biological tides. Interestingly enough, microtides in human bodies are popularly attributed to not the Earth, but the Moon. This is because popular knowledge cites that the Moon raises the ocean an average of 1 meter each 12 hours. Point out that the reason the tides are “stretched” by 1 meter is because part of that water is an Earth diameter closer to the Moon than the other part. In terms of fields, the near part of the Earth is in an appreciably stronger part of the Moon’s gravitational field than the far part. To the extent that part of our bodies are closer to the Moon than other parts, there would be lunar microtides—but enormously smaller than the microtides produced by not only the Earth, but massive objects in one’s vicinity.
Is there a way to distinguish between a gravity-free region and orbital free-fall inside the orbiting space shuttle? It turns out there is. Consider a pair of objects placed side by side. If the shuttle were floating in a gravity-free region, the two objects would remain as placed over time. Since the shuttle orbits the Earth, however, each object is in its own orbit about the Earth’s center, in its own orbital plane. All orbital planes pass through the center of the Earth and intersect, which means that depending on the proximity of the objects, they may collide by the time the shuttle makes a quarter orbit—a little more than 20 minutes! If the pair of objects are placed one in front of the other, with respect to their direction of motion, there will be no such effect since they follow the same orbital path in the same orbital plane. If the objects are one above the other, one farther from Earth, they will migrate in seemingly strange ways relative to each other because they are in distinct orbits with different PEs. Gravity makes itself present to astronauts by the strange secondary effects that are not directly related to weight (more on this in Weightlessness and Microgravity, by David Chandler, The Physics Teacher, back in May 1991). This relates also to the “Bob Biker” Practice Pages 49 and 50 for Chapter 8.
CHECK QUESTION: Consider the tiny tidal forces that DO act on our bodies, as a result of parts of our bodies experiencing slightly different gravitational forces. What planetary body is most responsible for microtides in our bodies? [The Earth, by far. When we are standing, there is a greater difference in Earth gravity on our feet compared to our heads than the corresponding differences in gravity due to farther away planetary bodies.]
Simulated Gravity in Space Habitats: The tallness of people in outer space compared to the radius of their rotating space habitats is very important. A gravitational gradient is appreciable in a relatively small structure. If the rim speed is such that the feet are at Earth-normal one g, and the head is at the hub, then the gravitational gradient is a full 1-g. If the head is halfway to the hub, then the gradient is 1/2-g, and so forth. Simulated gravity is directly proportional to the radius. To achieve a comfortable 1/100-g gradient, the radius of the structure must be 100 times that of one’s height. Hence the designs of large structures that rotate to produce Earth-normal gravity.
Saturn’s rings are the result of the strong gravitational gradient about Saturn. Such rings are the debris left over from bodies torn apart by Saturn’s tidal forces. Such would occur for our Moon if it comes closer to the Earth. Too close, and the Earth’s tidal forces would tear the Moon into a billion pieces, likely forming a ring similar to that of Saturn.
Tidal forces reach an extreme in the case of a black hole. The unfortunate fate of an astronaut falling into a black hole is not encountering the singularity, but the tidal forces encountered far before getting that close. Approaching feet first, for example, his closer feet would be pulled with a greater force than his midsection, which in turn would be pulled with a greater force than his head. The tidal forces would stretch him and he would be killed before these forces literally pulled him apart.
[This is a good place to break your lecture until next time.]
Gravitational Fields: Introduce the idea of force/mass for a body, and the gravitational force field. Relate the gravitational field to the more visible magnetic field as seen via iron filings. Since the field strength of the gravitational field is simply the ratio of force per mass, it behaves as force—it follows an inverse-square relationship with distance. Sketch Figure 9.15 on the board, but only the part of the graph that represents the field from the surface of the planet outward—the inverse-square part. This you have practically done before via Figure 9.6.
It’s interesting to consider the field strength inside a planet, and to consider a tunnel bored clear through the Earth. It’s easy to convince your students that the gravitational force on a body located at the exact center of the tunnel would be zero—a chalkboard sketch showing a few symmetrical force vectors will do this. Hence the gravitational field at the Earth’s center is zero. Then consider the magnitude of force the body would experience between the center of the Earth and the surface. A few more carefully drawn vectors will show that the forces don’t cancel to zero. The gravitational field is between zero and the value at the surface. You’d like to easily show that it’s half for an Earth of uniform density, to establish the linear part of the graph of Figure 9.23. Careful judgment should be exercised at this point. For most classes I would think the geometrical explanation would constitute “information overload” and it would be best to simply say “It can be shown by geometry that halfway to the center the field is half that at the surface…”and get on with your lecture. For highly motivated students it may be best to develop the geometrical explanation (given on page 90 in this manual). Then the pedagogical question is raised; how many students profit from your display of the derivation and how many will not?
Class time might better be spent on speculating further about the hole drilled through the Earth. Show with the motion of your hand how if somebody fell in such a tunnel they would undergo simple harmonic motion—and that this motion keeps perfect pace with a satellite in close circular orbit about the Earth. The time for orbit, nearly 90 minutes, is the time to make a to-and-fro trip in the tunnel. Consider going further and explain how ideally the period of oscillation of a body traveling in such a tunnel under the influence of only gravity would be the same for any straight tunnel—whether from New York to Australia , or from New York to Hawaii or China
Gravitational Field Inside a Hollow Planet: Consider the case of a body at the center of a completely hollow planet. Again, the field at the center is zero. Then show that the field everywhere inside is zero—by careful explanation of the following sketch. [Consider sample point P, twice as far from side A than side B. A solid cone defines area A and area B. Careful thought shows A has 4 times the area of B, and therefore has 4 times as much mass as B. That would mean 4 times as much gravitational pull, but being twice as far has only 1/4 as much pull. 1/4 of 4 gives the same gravitational pull as the pull toward B. So the forces cancel out (as they of course do in the center). The forces cancel everywhere inside the shell provided it is of uniform composition. If you stress this material (which will likely be on the heavy-duty side for many students) the following Check Question will measure the worth of your lecture effort.]
CHECK QUESTION: Sketch a graph similar to that in Figure 9.23 to represent the gravitational field inside and outside a hollow sphere. (The graphical answers should look like the following: A thin shelled planet is on the left, a thick shelled one is on the right.)
Speculate about the living conditions of a civilization inside a hollow planet. Expanding on Exercise 54 that considers a hollow planet, consider what happens to the g field inside when a massive spaceship lands on the outside surface of the hollow planet. The situation is interesting!
Black Holes: Begin by considering an indestructible person standing on a star, as in Figure 9.26. Write the gravitational equation next to your sketch of the person on the star, and show how only the radius changes in the equation as the star shrinks, and how the force therefore increases. Stress that the force on the person who is able to remain at distance R as the star shrinks experiences no change in force—the field there is constant as the star shrinks, even to a black hole. It is near the shrinking surface where the huge fields exist.
CHECK QUESTION: Consider a satellite companion to a star that collapses to become a black hole. How will the orbit of the companion satellite be affected by the star’s transformation to a black hole? [Answer is not at all. No terms in the gravitation equation change. What does happen, though, is that matter streams from the visible star to the black hole companion, emitting x-rays as it accelerates toward the black hole, providing evidence of its existence.]
Solutions to Chapter 9 Exercises
1. Nothing to be concerned about on this consumer label. It simply states the universal law of gravitation, which applies to all products. It looks like the manufacturer knows some physics and has a sense of humor.
2. The reason that a heavy body doesn’t fall faster than a light body is because the greater gravitational force on the heavier body (its weight), acts on a correspondingly greater mass (inertia). The ratio of gravitational force to mass is the same for every body—hence all bodies in free fall accelerate equally. And it’s true not just near the Earth, but anywhere. (This is illustrated in Figures 4.11 and 4.12.)
3. In accord with the law of inertia, the Moon would move in a straight-line path instead of circling both the Sun and Earth.
4. The force of gravity is the same on each because the masses are the same, as Newton
5. The force of gravity is the same on each because the masses are the same, as Newton
6. The force decreases as the square of increasing distance, or force increases with the square of decreasing distance.
7. Newton Newton
8. The force of gravity on moon rocks at the Moon’s surface is considerably stronger than the force of gravity of the distant Earth. Rocks dropped on the Moon fall onto the Moon surface. (The force of the Moon’s gravity is about 1/6 of the weight the rock would have on Earth; the force of the Earth’s gravity at that distance is only about 1/3600 of the rock’s Earth-weight.)
9. If gravity between the Moon and its rocks vanished, the rocks, like the Moon, would continue in their orbital path around the Earth. The assumption ignores the law of inertia.
10. Astronauts are weightless because they lack a support force, but they are well in the grips of Earth gravity, which accounts for them circling the Earth rather than going off in a straight line in outer space.
11. Nearer the Moon.
12. The forces between the apple and Earth are the same in magnitude. Only the corresponding accelerations of each are different.
13. In accord with Newton
14. The Earth and Moon equally pull on each other in a single interaction. In accord with Newton
15. Although the forces are equal, the accelerations are not. The much more massive Earth has much less acceleration than the Moon. Actually Earth and Moon do rotate around a common point, but it’s not midway between them (which would require both Earth and Moon to have the same mass). The point around which Earth and Moon rotate (called the barycenter) is within the Earth about 4600 km from the Earth’s center.
16. Less, because an object there is farther from Earth’s center.
17. Letting the equation for gravitation guide your thinking, twice the diameter is twice the radius, which corresponds to 1/4 the astronaut’s weight at the planet’s surface.
18. Letting the equation for gravitation guide your thinking, twice the mass means twice the force, and twice the distance means one-quarter the force. Combined, the astronaut weighs half as much.
19. Your weight would decrease if the Earth expanded with no change in its mass and would increase if the Earth contracted with no change in its mass. Your mass and the Earth’s mass don’t change, but the distance between you and the Earth’s center does change. Force is proportional to the inverse square of this distance.
20. For the planet half as far from the Sun, light would be four times as intense. For the planet ten times as far, light would be 1/100th as intense.
21. By the geometry of Figure 9.4, tripling the distance from the small source spreads the light over 9 times the area, or 9 m2. Five times the distance spreads the light over 25 times the area or 25 m2, and for 10 times as far, 100 m2.
22. The gravitational force on a body, its weight, depends not only on mass but distance. On Jupiter, this is the distance between the body being weighed and Jupiter’s center—the radius of Jupiter. If the radius of Jupiter were the same as that of the Earth, then a body would weigh 300 times as much because Jupiter is 300 times more massive than Earth. But Jupiter is also much bigger than the Earth, so the greater distance between its center and the CG of the body reduces the gravitational force. The radius is great enough to make the weight of a body only 3 times its Earth weight. How much greater is the radius of Jupiter? That will be Problem 2.
23. The high-flying jet plane is not in free fall. It moves at approximately constant velocity so a passenger experiences no net force. The upward support force of the seat matches the downward pull of gravity, providing the sensation of weight. The orbiting space vehicle, on the other hand, is in a state of free fall. No support force is offered by a seat, for it falls at the same rate as the passenger. With no support force, the force of gravity on the passenger is not sensed as weight.
24. A person is weightless when the only force acting is gravity, and there is no support force. Hence the person in free fall is weightless. But more than gravity acts on the person falling at terminal velocity. In addition to gravity, the falling person is “supported” by air drag.
25. In a car that drives off a cliff you “float” because the car no longer offers a support force. Both you and the car are in the same state of free fall. But gravity is still acting on you, as evidenced by your acceleration toward the ground. So, by definition, you would be weightless (until air resistance becomes important).
26. Gravitational force is indeed acting on a person who falls off a cliff, and on a person in a space shuttle. Both are falling under the influence of gravity.
27. The two forces are the normal force and mg, which are equal when the elevator doesn’t accelerate, and unequal when the elevator accelerates.
28. The pencil has the same state of motion that you have. The force of gravity on the pencil causes it to accelerate downward alongside of you. Although the pencil hovers relative to you, it and you are falling relative to the Earth.
29. The jumper is weightless due to the absence of a support force.
30. If Earth gained mass you’d gain weight. Since Earth is in free fall around the Sun, the Sun contributes nothing to your weight. Earth gravitation presses you to Earth; solar gravitation doesn’t press you to Earth.
31. You disagree, for the force of gravity on orbiting astronauts is almost as strong as at Earth’s surface. They feel weightless because of the absence of a support force.
32. First of all, it would be incorrect to say that the gravitational force of the distant Sun on you is too small to be measured. It’s small, but not immeasurably small. If, for example, the Earth’s axis were supported such that the Earth could continue turning but not otherwise move, an 85-kg person would see a gain of 1/2 newton on a bathroom scale at midnight and a loss of 1/2 newton at noon. The key idea is support. There is no “Sun support” because the Earth and all objects on the Earth—you, your bathroom scale, and everything else—are continually falling around the Sun. Just as you wouldn’t be pulled against the seat of your car if it drives off a cliff, and just as a pencil is not pressed against the floor of an elevator in free fall, we are not pressed against or pulled from the Earth by our gravitational interaction with the Sun. That interaction keeps us and the Earth circling the Sun, but does not press us to the Earth’s surface. Our interaction with the Earth does that.
33. The gravitational force varies with distance. At noon you are closer to the Sun. At midnight you are an extra Earth diameter farther away. Therefore the gravitational force of the Sun on you is greater at noon.
34. As stated in the preceding answer, our “Earth weight” is due to the gravitational interaction between our mass and that of the Earth. The Earth and its inhabitants are freely falling around the Sun, the rate of which does not affect our local weights. (If a car drives off a cliff, the Earth’s gravity, however strong, plays no role in pressing the occupant against the car while both are falling. Similarly, as the Earth and its inhabitants fall around the Sun, the Sun plays no role in pressing us to the Earth.)
35. The gravitational pull of the Sun on the Earth is greater than the gravitational pull of the Moon. The tides, however, are caused by the differences in gravitational forces by the Moon on opposite sides of the Earth. The difference in gravitational forces by the Moon on opposite sides of the Earth is greater than the corresponding difference in forces by the stronger pulling but much more distant Sun.
36. Just as differences in tugs on your shirt will distort the shirt, differences in tugs on the oceans distort the ocean and produce tides.
37. No. Tides are caused by differences in gravitational pulls. If there are no differences in pulls, there are no tides.
38. Ocean tides are not exactly 12 hours apart because while the Earth spins, the Moon moves in its orbit and appears at its same position overhead every 25 hours, instead of every 24 hours. So the two-high-tide cycle occurs at about 25-hour intervals, making high tides about 12.5 hours apart.
39. Lowest tides occur along with highest tides—spring tides. So the spring tide cycle consists of higher-than-average high tides followed by lower-than-average low tides (best for digging clams!).
40. Whenever the ocean tide is unusually high, it will be followed by an unusually low tide. This makes sense, for when one part of the world is having an extra high tide, another part must be donating water and experiencing an extra low tide. Or as the hint in the exercise suggests, if you are in a bathtub and slosh the water so it is extra deep in front of you, that’s when it is extra shallow in back of you—“conservation of water”!
41. Because of its relatively small size, different parts of the Mediterranean Sea are essentially equidistant from the Moon (or from the Sun). As a result, one part is not pulled with any appreciably different force than any other part. This results in extremely tiny tides. The same argument applies, with even more force, to smaller bodies of water, such as lakes, ponds, and puddles. In a glass of water under a full Moon you’ll detect no tides because no part of the water surface is closer to the Moon than any other part of the surface. Tides are caused by appreciable differences in pulls.
42. Tides are produced by differences in forces, which relate to differences in distance from the attracting body. One’s head is appreciably closer than one’s feet to the overhead melon. The greater proportional difference for the melon out-tides the more massive but more distant Moon. One’s head is not appreciably closer to the Moon than one’s feet.
43. Yes, the Earth’s tides would be due only to the Sun. They’d occur twice per day (every 12 hours instead of every 12.5 hours) due to the Earth’s daily rotation.
44. Tides would be greater if the Earth’s diameter were greater because the difference in pulls would be greater. Tides on Earth would be no different if the Moon’s diameter were larger. The gravitational influence of the Moon is as if all the Moon’s mass were at its CG. Tidal bulges on the solid surface of the Moon, however, would be greater if the Moon’s diameter were larger—but not on the Earth.
45. From the nearest body, the Earth.
46. Tides occur in the Earth’s crust and the Earth’s atmosphere for the same reason they occur in the Earth’s oceans. Both are large enough so there are appreciable differences in distances to the Moon and Sun, with corresponding gravitational differences as well.
47. In accord with the inverse-square law, twice as far from the Earth’s center diminishes the value of g to 1/4 its value at the surface or 2.45 m/s2.
48. For a uniform-density planet, g inside at half the Earth’s radius would be 4.9 m/s2. This can be understood via the spherical shell idea discussed on page 166. Halfway to the center of the Earth, the mass of the Earth in the outer shell can be neglected—the gravitational contribution of all parts of the shell cancels to zero. Only the mass of the Earth “beneath” contributes to g, the mass in the sphere of radius r/2. This sphere of half radius has only 1/8 the volume and only 1/8 the mass of the whole Earth (volume varies as r3). This effectively smaller mass alone would find the acceleration due to gravity 1/8 that of g at the surface. But consider the closer distance to the Earth’s center as well. This twice-as-close distance alone would make g four times as great (inverse-square law). Combining both factors, 1/8 of 4 = 1/2, so the acceleration due to gravity at r/2 is g/2.
49. Your weight would be less in the mine shaft. One way to explain this is to consider the mass of the Earth above you which pulls upward on you. This effect reduces your weight, just as your weight is reduced if someone pulls upward on you while you’re weighing yourself. Or more accurately, we see that you are effectively within a spherical shell in which the gravitational field contribution is zero; and that you are being pulled only by the spherical portion below you. You are lighter the deeper you go, and if the mine shaft were to theoretically continue to the Earth’s center, your weight moves closer to zero.
50. The increase in weight indicates that the Earth is more compressed—more compact—more dense—toward the center. The weight that normally would be lost when in the deepest mine shafts from the upward force of the surrounding “shell” is more than compensated by the added weight gained due to the closeness to the more dense center of the Earth. (Referring to our analysis of Exercise 38, if the mine shaft were deep enough, reaching halfway to the center of the Earth, you would, in fact, weigh less at the bottom of the shaft than on the surface, but more than half your surface weight.)
51. More fuel is required for a rocket that leaves the Earth to go to the Moon than the other way around. This is because a rocket must move against the greater gravitational field of the Earth most of the way. (If launched from the Moon to the Earth, then it would be traveling with the Earth’s field most of the way.)
52. On a shrinking star, all the mass of the star pulls in a noncancelling direction (beneath your feet)—you get closer to the overall mass concentration and the force increases. If you tunnel into a star, however, there is a cancellation of gravitational pulls; the matter above you pulls counter to the matter below you, resulting in a decrease in the net gravitational force.
53. F ~ m1 m2/d2, where m2 is the mass of the Sun (which doesn’t change when forming a black hole), m1 is the mass of the orbiting Earth, and d is the distance between the center of mass of the Earth and the Sun. None of these terms change, so the force F that holds the Earth in orbit does not change. (There may in fact be black holes in the galaxy around which stars or planets orbit.)
54. Letting the gravitational force equation be a guide to thinking, we see that gravitational force and hence one’s weight does not change if the mass and radius of the Earth do not change. (Although one’s weight would be zero inside a hollow uniform shell, on the outside one’s weight would be no different than if the same-mass Earth were solid.)
55. The misunderstanding here is not distinguishing between a theory and a hypothesis or conjecture. A theory, such as the theory of universal gravitation, is a synthesis of a large body of information that encompasses well-tested and verified hypothesis about nature. Any doubts about the theory have to do with its applications to yet untested situations, not with the theory itself. One of the features of scientific theories is that they undergo refinement with new knowledge. (Einstein’s general theory of relativity has taught us that in fact there are limits to the validity of Newton
56. There is no contradiction because gravitational forces, like all forces, can cancel one another when they are equal and opposite. And this is the case for the gravitational force of a uniform hollow shell acting on any object inside. The shell, however, does not shield the influences of other massive bodies beyond or within the shell.
57. You weigh a tiny bit less in the lower part of a massive building because the mass of the building above pulls upward on you.
58. Tidal forces occur when there is a difference in gravitational field strength across a body. Nearing the singularity of a black hole feet first, the feet of the unfortunate astronaut would be pulled with so much more force than his midsection that separation would likely occur.
59. There is no gravitational field change at the spaceship’s location because there’s no changes in the terms of the gravitational equation. The mass of the black hole is the same before and after collapse.
60. Open-ended.
Chapter 9 Problem Solutions
1. From F = GmM/d2, 1/5 of d squared is 1/25th d2, which means the force is 25 times greater.
2. In accord with the inverse-square law, four times as far from the Earth’s center diminishes the value of g to g/42, or g/16, or 0.6 m/s2.
3.
4. It is g = GM/r2 = (6.67´ 10-11) (3.0 ´1030)/(8.0 ´103)2 = 3.1 ´ 1012 m/s2, 300 billion times g on Earth.
5. g = = = 9.24 N/kg or 9.24 m/s2; 9.24/9.8 = 0.94 or 94%.
6. (a) Moon:
TF = = ≈ 3 ´ 10-13 N/kg.
(b) Melon: TF = = ≈ 3 ´10-11 N/kg.
(c) We see that the tidal force due to the melon is about 100 times greater! Since tidal forces due to planets are far less than from the Moon, there is no substance to the claim that planetary tidal forces have an influence on people.
