Knight Chapter 6 Circular Motion, Orbits & Gravity PDF

LOOKING AHEAD • Circular Motion An object moving in a circle has an acceleration toward the center. so there must be a net force toward the center as well. Apparent Forces These riders feel Likethey·re being pushed against the wall. But this force isn't a real force. it's an apparent force. Gravit...

LOOKING AHEAD • Circular Motion An object moving in a circle has an acceleration toward the center. so there must be a net force toward the center as well. Apparent Forces These riders feel Likethey·re being pushed against the wall. But this force isn't a real force. it's an apparent force. Gravity and Orbits The space station appears to float in space. but gravity is actually pulling down on it quite forcefully. How much force does it take to swing the girl in a circle?You'll learn how to solve such problems. This apparent force makes the riders "feel heavy.' You'll learn to calculate their apparent weight. You'll learn Newton's law of gravity, and you'll see how the force of gravity keeps the station in orbit. - To learn about motion in a circle, including orbital motion under the influence of a gravitational force. LOOKING BACK 41 i4Uil,IWMI Centripetal Acceleration In Section 3.7. you learned that an object moving in a circle at a constant speed experiences an acceleration directed toward the center of the circle. I In this chapter, you'll learn how to extend Newton's second law, which relates acceleration to the forces that cause it, to this type of acceleration. A softball pitcher is throwing a pitch. At the instant shown, the ball is moving in a circular arc at a steady speed. At this instant, the acceleration is ii ii J A. B. C. D. E. Directed Directed Directed Directed Zero. up. down. left. right. 176 CHAPTER 6 Circular Motion, Orbits, and Gravity 6.1 Uniform Circular Motion Uniform circular motion at the fair. The riders on this carnival ride are going in a circle at a constant speed, a type of motion that we've called uniform circular motion. We saw in ~SECTION 3.7 that uniform circular motion results in an acceleration directed toward the center of the circle. According to Newton's second law, this means that there must be a net force directed toward the center of the circle. For these riders, it is the tension in the cables and the force of gravity that combine to provide this net force. In this chapter, we'IJ look at objects moving in circles or circular arcs. We'IJ consider the details of the acceleration and the forces that provide this acceleration, combining our discussion and description of motion from Chapters 1-3 with our treatment of Newton's laws and dynamics from Chapters 4 and 5. For now, we consider objects that move at a constant speed; circular motion with changing speed wiJI wait until Chapter 7. Velocity and Acceleration Velocity and acceleration for uniform circular motion. FIGURE 6.1 The instantaneous velocity v is tangent to the circle at all points. V ./ in Uniform Circular Motion Although the speed of a particle in uniform circular motion is constant, its velocity is not constant because the direction of the motion is always changing. FIGURE 6.1 reminds you of the details: There is an acceleration at every point in the motion, with the acceleration vector apointing toward the center of the circle. We called this the centripetal acceleration, and we showed that for uniform circular motion the acceleration is given by \12 a=- a (6.1) r I Centripetal acceleration for uniform circular motion ~ ~ .. QUADRATIC Thus an object's centripetal acceleration depends on both its speed and the distance from the center of its circular path. center of the circle at all points. CONCEPTUAL EXAMPLE 6.1 A car rounding a comer A car is turning a tight corner at a constant speed. A top view of the motion is shown in FIGURE 6.2. The velocity vector for the car points to the east at the instant shown. What is the direction of the acceleration? FIGURE 6.2 Top view of a car turning a corner. I I Centerof curve • REASON The curve that the car is following is a segment of a circle, so this is an example of uniform circular motion. For uniform circular motion. the acceleration is direcred toward the center of the circle, which is 10 the south. This acceleration is due to a change in direction. not a change in speed. And tllis matches your experience in a car: If you tum the wheel 10 the right-as the driver of this car is doing-your car then changes its motion toward the right. in the direction of the cemer of the circle. ASSESS nmi]I ► In this example, the car is following a curve that is only a segment of a circle, not a full circle. At the instant shown, though, the motion follows a circular arc. You can have uniform circular motion without completing a full circle. ◄ 6.1 Uniform Circular Motion 177 Period, Frequency, and Speed Although an object needn't complete a full circle to be in uniform circular motion, in many of the cases we' II consider, objects will complete multiple full circles of motion, one after another. Since the motion is uniform, each time around the circle is just a repeat of the one before, so the motion is periodic. The time interval it takes an object to go around a circle one time, completing one revolution (abbreviated rev), is called the period of the motion. Period is represented by the symbol T. Rather than specify the time for one revolution, we can specify circular motion by its frequency, the number of revolutions per second, for which we use the symbol f An object with a period of one-half second completes 2 revolutions each second. Similarly, an object can make IO revolutions in l s if its period is one-tenth of a second. This shows that frequency is the inverse of the period: 1 T f=- (6.2) Although frequency is often expressed as "revolutions per second," revolutions are not true units but merely the counting of events. Thus the SI unit of frequency is simply inverse seconds, or s- 1. Frequency may also be given in revolutions per minute (rpm) or another time interval, but these usually need to be converted to s- 1 before calculations are done. FIGURE 6.3 shows an object moving at a constant speed in a circular path of radius r. We know the time for one revolution-one period T-and we know the distance traveled, so we can write an equation relating the speed to the period and radius: v=-- 2nT speed. ii •·· (6.3) T Given Equation 6.2 relating frequency and period, we can also write this equation as V FIGURE 6.3 Relating frequency and = 27Tfr ,. ...•···· In one period T, lhe object travels around the circumference of the circle, a distance of 21rr. (6.4) We can combine this equation with Equation 6.1 for acceleration to get an expression for the centripetal acceleration in terms of the frequency or the period for the circular motion: \12 a=-=(27rf°)2r= r EXAMPLE 6.2 (27T)2 r (6.5) T A spinning table saw blade SOLVE The time for one revolution is the period. which is given by Equation 6.2: The circular blade of a table saw is 25 cm in diameter and spins at 3600 rpm. How much time is required for one revolution? How fast is one of the teeth at the edge of the blade moving? What is the tooth ·s acceleration? I I T = - = -= 0.017 s J 60 s- 1 The speed of the tooth is given by Equation 6.4: ,, = 27ifr = 2r.(60 s- 1)(0.125 m) = -1-7mis Each tooth of the saw blade is undergoing unifom1 circular motion. so the period. frequency. speed. and acceleration of a tooth are all related by Equations 6.2--6.5. We can then use Equation 6.5 to find the acceleration: PREPARE Before we get started. we need to do a couple of unit conversions. The diameter of the blade is 0.25 m. so its radius is 0.125 m. The frequency is given in rpm; we need 10 conven this to s-I: ASSESS STRATEGIZE rev f= 3600--X min I min --= 60 s rev 60-= s 60 s- 1 The speed of the toOLh is extremely high: 47 mis = 100 mph! Still. this seems plausible for a high-speed saw. The acceleration is also remarkable. almost 2000g. 178 CHAPTER EXAMPLE 6.3 6 Circular Motion, Orbits, and Gravity Finding the period of a carnival ride Ln the Quasar carnival ride. passengers travel in a horizontal 5.0-m-radius circle. For safe operation. the maximum sustained acceleration that riders may experience is 20 rn/s2 . approximately twice the free-fall acceleration. What is the period of the ride when it is being operated at the maximum acceleration? How fast are the riders moving when the ride is operated at this period? STRATEGIZEThe passengers are in uniform circular motion, so their period. speed. and acceleration are related by Equations 6.4 and 6.5. PREPARE We will assume that the cars on the ride are in uniform circular motion. The visual overview of FIGURE 6.4 shows a top view of the motion of the ride. acceleration increases. Setting a maximum value for the acceleration mean setting a minimum value for the period-as fast as the ride can safely go. We can rearrange Equation 6.5 lo find the period in terms of the acceleration. Sening the acceleration equal to the maximum value gives the minimum period: T= 2r. f; ~ .Om ---, = 3.1 s 20 111/s- We can then use Equation 6.4 to find the speed at which the riders move: 2r.r 2r.(5.0 m) T 3.1 s v=--= FIGURE 6.4 = 2r. Visual overview for the Quasar carnival ride. JOmis ASSESS One rotation in just over 3 seconds seems reasonable for a pretty zippy carnival ride. (The period for this particular ride is actually 3.7 s, so it runs a bit slower than the maximum safe speed.) But in this case we can do a quantitative check on our work. If we use our calculated velocity to find the acceleration using Equation 6.1. we find Known r=5.0m a= 20 m/s 2 Find T, I' 112 a= - SOLVE Equation 6.5 shows that the acceleration increases with decreasing period: If the riders go around in a shoner time, the STOP TO THINK 6.1 ,. (I0m/s)2 5.0m = ---- = 20m/s- , This is the acceleration given in the problem statement. so we can have confidence in our work. Rank in order, from largest to smallest, the period of the motion of particles A to D. Hurling the heavy hammer Scottish games involve feats of strength. Here, a man is throwing a 30 lb hammer for distance. He starts by swinging the hammer rapidly in a circle. You can see from how he is leaning that he is providing a large force directed toward the center to produce the necessary cenlTipetal acceleration. When the hammer is heading in the right direction, the man lets go. With no force directed toward the center, the hammer will stop going in a circle and fly in the chosen direction across the field. A. D. C. 6.2 Dynamics of Uniform Circular Motion Riders traveling around on a circular carnival ride are accelerating, as we have seen. Consequently, according to Newton's second law, the riders must have a netforce acting on them. We've already determined the acceleration of a particle in uniform circular motion~the centripetal acceleration of Equation 6.1. Newton's second law tel Is us what the net force must be to cause this acceleration: _ Fnet A video to support a section's topic is embedded in the eText. Video Forces in Circular Motion B. 2 = ) ma_ = (111v -,.-. toward center of circle Net force producing the centripetal acceleration of uniform circular motion (6.6) ~6 ~o, OUAORATlC IIMRSE r 6.2 Dynamics of Uniform Circular Motion In other words, a particle of mass m moving at constant speed v around a circle of radius r must always have a net force of magnitude mv 2/r pointing toward the center of the circle, as in FIGURE 6.5. It is this net force that causes the centripetal acceleration of circular motion. Without such a net force, the particle would move off in a straight line tangent to the circle. The force described by Equation 6.6 is not a new kind of force. The net force will be due to one or more of our familiar forces, such as tension, friction, or the normal force. Equation 6.6 simply tells us how the net force needs to act-how strongly and in which direction-to cause the particle to move with speed v in a circle of radius r. In each example of circular motion that we will consider in this chapter, a physical force or a combination of forces directed toward the center produces the necessary acceleration. CONCEPTUAL EXAMPLE 6.4 The net force required to keep the object moving in a circle is always directed toward the center of' the circle .... __ _ The net fore~ causes a centripetal acceleration. shows a visual overview of the situation. The car is accelerating, even though it is moving at a constant speed, because its direction is changing. When the car is at the bottom of the dip. the center of its circular path is directly above it and so its acceleration vector points straight up. The free-body diagram of Figure 6.6 identifies the only two forces acting on the car as the normal force, pointing upward, and its weight. pointing downward. Which is larger: 11 or II'? Because a points upward, by Newton ·s second law there must be a net force on the car that also points upward. In order for this FIGURE 6.6 Visual overview for the car in a dip. Circular segment •f~'\,,' a ,61'=:; "• FIGURE 6.6 CONCEPTUAL EXAMPLE 6.5 Net force for circular motion. Forces on a car, part I Engineers design cun1es on roads to be segments of circles. They also design dips and peaks in roads to be segments of circles with a radius that depends on expected speeds and other factors. A car is moving at a constant speed and goes into a dip in the road. At the very bottom of the dip. is the normal force of the road on the car greater than. less than, or equal to the car's weight? REASON FIGURE 6.5 179 tt· to be the case, the free-body diagram shows that the magni rude of the normal force must be greater than the weight. ASSESS You have probably experienced this situation. As you drive tlu-ough a dip in the road. you feel "heavier" than normal. As discussed in Section 5.3, this is because your apparent weight-the normal force that supports you-is greater than your true weight. Forces on a car, part II A car is turning a comer at a constant speed. following a segment of a circle. What force provides the necessary centripetal acceleration? The car moves along a circular arc at a constant speed-uniform circular motion-for the quarter-circle necessary to complete the turn. We know that the acceleration is directed toward the center of the circle. What force or forces can we identify that provide this acceleration? Imagine you are driving a car on a frictionless road, such as a very icy road. You would not be able to turn a comer. Turning the steering wheel would be of no use. The car would slide straight ahead. in accordance with both Newton·s first law and the experience of anyone who has ever driven on ice! So it must be frictio11 that causes the car to turn. The top view of the tire in FIGURE 6.7 shows the force on one of the car's tires as it turns a corner. It must be a static friction force. not kinetic, because the tires are not skidding: The points where the tires touch the road are not moving relative to the surface. lf you skid. your car won·t tum the corner-it will continue in a straight line! FIGURE 6.7 Top views of a car turning a corner. Top view of car REASON \ The path is a segment of the circle. Top view of tire ~ The static friction force points toward the center. This result agrees with your experience. You know that reduced friction makes it harder to turn corners, and you know as weU that skidding on a curve is bad news. So it makes sense that static friction is the force at work. ASSESS 180 c HAP TE R 6 Circular Motion, Orbits, and Gravity II PROBLEM-SOLVING APPROACH 6.1 Video Car Driving Over a Rise Video Ball Leaves Circular Track Circular dynamics problems Circular motion involves an acceleration and thus a net force. We can therefore use techniques very similar to those we've already seen for other Newton ·s second-law problems. STRATEGtZE An object undergoing uniform circular motion has an acceleration directed toward the center of the circle. This acceleration. according to Newton's second Jaw. is caused by a net force directed toward the center. The origin of this force is one or more of the forces we're familiar with, such as tension, weight, or friction. PREPARE Begin your visual overview with a pictorial representation in which you sketch the motion, define symbols, define axes, and identify what the problem is trying to find. There are two common situations: ■ ■ If the motion is in a horizontal plane, like a tabletop, draw the free-body diagram with the circle viewed edge-on. the x-axis pointing toward the center of the circle, and the y-axis perpendicular to the plane of the circle. If the motion is in a vertical plane, like a FeITis wheel, draw the free-body diagram with the circle viewed face-on, the x-axis pointing toward the center of the circle, and the y-axis tangent to the cixcle. Newton's second law for uniform circular motion, Fnet = (mv 2/r. toward center of circle), is a vector equation. Some forces act in the plane of the circle, some act perpendicular to the circle, and some may have components in both directions. In the coordinate system just described, with the x-axis pointing toward the center of the circle, Newton's second law is SOLVE and That is, the net force toward the center of the circle has magnitude mv 2/r while the net force perpendicular to the circle is zero. The components of the forces are found directly from the free-body diagram. Depending on the problem, either: ■ ■ Use the net force to determine the speed v, then use circular kinematics to find frequencies or other details of the motion. Use circular kinematics to determine the speed v, then solve for unknown forces. Make sure your net force points toward the center of the circle. Check that your result has the correct units, is reasonable, and answers the question. ASSESS Exercise 9 EXAMPLE 6.6 Analyzing a runner on a circular track Kinesiologists-scientists who study human motion-can measure the forces that act on runners as they round a curve. The forces on a runner·s feet are a vertical normal force that, on average, serves to counteract gravity, and a static friction force parallel to the ground and pointing toward the center of the runner's circular path. The average value of this inward-pointing force is measured to be 600 N for a 75 kg runner rounding a curve that has a 2.5 m radius. What is this runner's speed? STRATEGIZE We saw in Conceptual Example 6.5 that the force that causes the car's centripetal acceleration is the friction force between the road and the tires. The same situation holds for a runner moving in a circle: It is the static friction force between (ii the track and the runner's shoes that causes the runner to move in a circle. Following the steps of Problem-Solving Approach 6.1. we will use Newton ·s second law to relate this force to the centripetal acceleration, from which we can find the runner·s speed. PREPARE FIGURE 6.8 on the next page shows a visual overview of the problem. The mai11reason for the pictorial representation on the left is to illustrate the relevant geometry and to define the symbols that will be used. A circular dynamics problem usually does not have starting and ending points like a projectile problem, so subscripts such as xi or Yr are usually not needed. Here we need to define the runner's speed r and the radius r of the circle. mi 6.2 Dynamics of Uniform Circular Motion 181 FIGURE 6.8 A visual overview of the runner moving in a circular path. The free-body d(~gram ~ows an edge-on ,iew. T\.e pictorial representation shows a top view. ••·..:, ······=t known r1 m=7S' k-!J r = :i..,,S'm ~runner - Sfa.,fi..c fri..cfi..on fs = 600 r. --it----♦--)( N / r This is the plane of the motion. Ru.nner1.s.S'feed.v The free-body diagram shows the forces i11 the problem. Because the motion is in a horizontal plane, Problem-Solving Approach 6.1 tells us to draw the free-body diagram looking at the edge of the circle. with the x-axis pointing toward the center of the circle and the y-axis perpendicular to the plane of the circle. Three forces are acting on the runner: the weight force iv, the normal force of the ground ii. and the friction force.~. Only the friction force has an x-component, so Newton's second law is , " L.,F, -.fs- - -111\1-r We know the mass. the radius of the circle. and the friction force. so we can solve for 11: v= SOLVE There is no net force in the y-clirection, perpendicular to the circle, so ii· and ii must be equal and opposite. There is a net force in the x-direction, toward the center of the circle. as there must be to cause the centripetal acceleration of circular motion. ... Fnet (600 N)(2.5 m) [J;;-= _ \j--;;; 7:, kg _ = 4.:, mis The speed is about 10 mph, a reasonable value for a runner going around a sharp curve. ASSESS Finding the maximum speed for a car to turn a corner EXAMPLE 6.7 What is the maximum speed with which a 1500 kg car can make a turn around a curve of radius 20 m on a level (unbanked) road without sliding? (This radius turn is about what you might expect at a major intersection in a city.) As Conceptual Example 6.5 showed, the force that causes the car's centripetal acceleration is the static friction between the car's tires and the road. If the car's speed is too high, the friction force needed to cause this centripetal force wi!J exceed its maximum possible value and the car will slide. STRATEGIZE We start with the visual overview in FIGURE 6.9. The car moves along a circular arc at a constant speed-uniform circular motion--<luring the tum. The static friction force causes the centripetal acceleration. so this force must be directed toward the center of the circular arc. We've drawn the free-body diagram, shown from behind the car, with the friction force pointing in PREPARE y Known = 1500 kg r=20 m µ., = 1.0 Find r \'m:u ts ~ {. ~- SOLVE The only force in the x-direction, toward the center of the circle, is static friction. Newton's second law along the x-axis is x s ~ F=f=- 1mr' ,. This is the same equation as in the previous example, because the force toward the center of the circle in that example is also static friction. Newton ·s second law in they-direction is °"F k,/ J = 11 - w = ma)' = 0 so that 11 = ir = mg. The car will slide when the static friction force reaches its maximum value. Recall from Equation 5.7 in Chapter 5 that this maximum force is given by FIGURE 6.9 Visual overview of a car turning a corner. 111 this direction. Because the motion is in a horizontal plane, we've again chosen an x-axis toward the center of the circle and a y-axis perpendicular to the plane of motion. X f, max= JJ.,,1!= µ,,mg The maximum speed occurs when the static friction force reaches its maximum value, or when ? M1Vmax- fsm~x=--1'- Top view of car Rear view of car Conlinued 182 CHAPTER 6 Circular Motion, Orbits, and Gravity ASSESS 14 m/s = 30 mph, which seems like a reasonable upper limit for the speed at which a car can go around a curve without sliding. There are two other things to note about the solution: Using t11eknown value of lsmax· we find ■ The car·s mass canceled out. The maximum speed does 1101 depend on the mass of the vehicle. though this may seem surprising. ■ The final expression for vmaxdoes depend on the coefficient of fric1ion and the radius of the rum. Bolh of these faciors make sense. You know. from experience, thai the speed al which you can take a rum decreases if J.Lsis less (!he road is wet or icy) or if r is smaller (!he rum is tigh1er). Rearranging. we gel ? "max - = µ,gr For rubber tires oo pavemem. we find from Table 5.2 tha! µ 5 = 1.0. We then have "ma.,= v;;:;.= V( 1.0)(9.8 m/s 2 )(20 m) = 14 m/s FIGURE6.10 Modifications to increase the maximum speed around a curve. (a) Wings on a race car. (b) A banked turn on a racetrack. EXAMPLE 6.8 Because Vrnax depends on µ,s and because /1,s depends on road conditions, the maximum safe speed through turns can vary dramatically. A car that easily handles a curve in dry weather can suddenly slide out of control when the pavement is wet. Icy conditions are even worse. If you lower the value of the coefficient of friction in Example 6.7 from LO (dry pavement) to 0.1 (icy pavement), the maximum speed for the turn goes down to 4.4 mis-about lO mph! Race cars turn corners at much higher speeds than normal passenger vehicles. One design modification of the cars to allow this is the addition of wings, as on the car in FIGURE6.10a. The wings provide an additional force pushing the car down onto the pavement by deflecting air upward. This extra downward force increases the normal force, thus increasing the maximum static friction force and making faster turns possible. There are also design modifications of the track that allow race cars to take corners at high speeds. If the track is banked by raising the outside edge of curved sections, the normal force can provide some of the force necessary to produce the centripetal acceleration, as we will see in the next example. The curves on racetracks may be quite sharply banked (FIGURE 6.10b). Curves on ordinary highways are often banked as well, though at more modest angles suiting the lower speeds. Finding a car's speed on a banked tum A curve on a racerrackof radius 70 rn is banked at a I 5° angle. At what speed can a car 1akemis curve wit1loutassistance from friction? STRATEGIZE With no friction acting, it is the horizontal component of the nonual force that causes me centripetal acceleration. PREPAREAfler drawing the pictorial representation in FIGURE6.11, we use 1he force identification diagram to find 1hat, FIGURE 6.11 Visual overview for the car on a banked turn. Weight ii· ~ The normal force is perpendicular to the surface. )' . t Road II given that there is no friction acting. the only two forces are the normal force and 1he car's weight. We can then construct the free-body diagram. making sure thal we draw the normal force perpendicular to the road· s surface. Even though the car is tilted, ic is still moving in a hori::.0111al circle. Thus, following Problem-Solving Approach 6. l. we choose the x-axis to be horizontal and pointing 1oward the center of the circle. SOLVEWithout friction. 11_. = 11sin 0 is !he only component of force toward 1he center of the circle. It is !his inward component of 1he normal force on the car char causes it to turn 1he corner. Newton·s second law is Normal ii LF,. = ncos0- w= O 6.2 where 0 is the angle at which the road is banked. and we·ve assumed that the car is traveling at the correct speed v. From the _,·-equation. w mg n=--=-cos0 cos0 Substituting d1is into the x-equation and olving for v give Dynamics of Uniform Circular Motion ( ) r = 14 m/s This is =30 mph. a reasonable speed. Only at this exact speed can the turn be negotiated without reliance on friction forces. ASSESS The friction force provides the necessary centripetal acceleration not only for cars turning corners, but also for bicycles, horses, and, as we saw in Example 6.6, humans. The cyclists in FIGURE s.12 are going through a tight turn; you can tell by how they lean. The road exerts both a vertical normal force and a horizontal friction force on their tires. A vector sum of these forces points at an angle. The cyclists lean to the side so that the sum of the road forces points along the line of their bikes and their bodies; this keeps them in balance. The horses in the photo that opened the chapter are leaning into the turn for a similar reason. Maximum Walking Speed mv-' mg -sin0=mgtan0=-cos0 v = Vrgtan0 183 FIGURE 6.12 Road forces on a cyclist leaning into a turn. 4iJ Humans and other two-legged animals have two basic gaits: walking and running. At slow speeds, you walk. When you need to go faster, you run. Why don't you just walk faster? There is an upper limjt to the speed of walking, and this limit is set by the physics of circular motion. Think about the motion of your body as you take a walking stride. You put one foot forward, then push off with your rear foot. Your body pivots over your front foot, and you bring your rear foot forward to take the next stride. As you can see in FIGURE 6.13a, the path that your body takes during this stride is the arc of a circle. In a walking gait, your body is in circular motion as you pivot on your forward foot. A force toward the center of the circle is required for this circular motion, as shown in Figure 6.13. FIGURE 6.13b shows the forces acting on the woman's body during the midpoint of the stride: her weight, directed down, and the normal force of the ground, directed up. Newton's second law for the x-axis is FIGURE 6.13 Analysis of a walking stride. (a) Walking stride During each stride, her hip undergoes circular motion. ? mv- "' L., F.., = w - n = -- r Because of her circular motion, the net force must point toward the center of the circle, or, in this case, down. In order for the net force to point down, the normal force must be less than her weight. Your body t:J.iesto "lift off" as it pivots over your foot, decreasing the normal force exetted on you by the ground. The normal force becomes smaller as you walk faster, but n. cannot be less than zero. Thus the maximum possible walking speed Vmax occurs when n. = 0. Setting n = 0 in Newton's second law gives lnVmax 2 w=mg=--,-.- Thus The circular motion requires a force directed toward the center of the circle. The radius of the circular motion is the length of the leg from the foot to the hip. (b) Forces in the stride V max = ~ I (6.7) The maximum possible walking speed is limjted by ,; the length of the leg, and g, the free-fall acceleration. This formula is a good approximation of the maximum walking speed for humans and other animals. Giraffes, with their very long legs, can walk at high speeds. Animals such as mice with very short legs have such a low maximum walking speed that they rarely use this gait. For humans, the length of the leg is approximately 0.7 111, giving V111ax = 2.6 mis = 6 mph. You can walk this fast, though it becomes energetically unfavorable to walk at speeds above 4 mph. Most people make a transition to a running gait at about this speed. ii ---&.---y The .r-axis points down, toward the center of the circle. ,v •, •••• ••••.•• ;.I X Side view (same as photo) 184 c HAP TE R 6 Circular Motion, Orbits, and Gravity A block on a string spins in a horizontal circle on a frictionless table. Rank in order, from largest to smallest, the tensions TA to TE acting on the blocks A to E. STOP TO THINK 6.2 A. B. D. C. G r= 10cm f= 50 rpm r = 20 cm f= 50 rpm r= /= 20 cm 100 rpm r= 40 cm /= LOOrpm r= /= 40 cm 200 rpm 6.3 Apparent Forces in Circular Motion FIGURE 6.14 shows a carnival tide that spins the riders around inside a large cylinder. The people are "stuck" to the inside wall of the cylinder! As you probably know from experience, the riders feel that tbey are being pushed outward, into the wall. But our analysis has found that an object in circular motion must have an inward force to create the centripetal acceleration. How can we explain this apparent difference? FIGURE 6.14 Inside the Gravitron, a rotating circular room. Centrifugal Force? Bird's-eye view of a passenger in a car turning a corner. FIGURE 6.15 Without the door, you'd keep moving .... straight ahead. ••••• ...• The door provides the center-directed force that makes you ..... move in a circle. ••• 1 •' Cente>of curve II Video What the Physics? Shake It Off If you are a passenger in a car that turns a comer quickly, you may feel "thrown" by some mysterious force against the door. But is there really such a force? FIGURE 6.15 shows a bird's-eye view of you riding in a car as it makes a left turn. Just before the tum, you were moving in a straight line so, according to Newton's first law, as the car begins to turn, you want to continue moving along that same line. However, the door moves into your path and so runs into you! You feel the force of the door because it is this force, pushing inward toward the center of the curve, that is causing you to tum the corner. But you were not "thrown" into the door; the door ran into you. A "force" that seems to push an object to the outside of a circle is called a centrifugal force. Despite having a name, there really is no such force. What you feel is your body trying to move ahead in a straight line (which would take you away from the center of the circle) as outside forces act to turn you in a circle. The only real forces, those that appem· on free-body diagrams, are the ones pushing inward toward the center. A centrifugal force will never appear on a free-body diagram and never be included in Newton's laws. With this in mind, let's revisit the rotating carnival ride. A person watching from above would see the riders in the cylinder moving in a circle with the walls providing the inward force that causes their centripetal acceleration. The riders feel as if they're being pushed outward because their natural tendency to move in a straight line is being resisted by the wa!J of the cylinder, which keeps getting in the way. But feelings aren't forces. The only actual force is the contact force of the cylinder wall pushing im;,ard. Apparent Weight in Circular Motion Imagine swinging a bucket of water over your head. If you swing the bucket fast enough, the water stays in. But you'll get a shower if you swing too siowly. Why does the water stay in the bucket? Or think about a roller coaster that does a loopthe-loop. How does the car stay on the track when it's upside down? You might have said that there was a centrifugal force holding the water in the bucket and the car on the track, but we have seen that there really isn't a centrifugal force. Analyzing these questions will tell us a lot about forces in general and circular motion in particular. 6.3 Apparent shows a roller coaster car going around a vertical loop-the-loop of radius r. If you've ever ridden a roller coaster, you know that your sensation of weight changes as you go over the crests and through the dips. To understand why, let's look at the forces on passengers going through the loop. To simplify our analysis, we will assume that the speed of the car stays constant as it moves through the loop. FIGURE s.1sb shows a passenger's free-body diagram at the top and the bottom of the loop. Let's start by exarnining the forces on the passenger at the bottom of the loop. The only forces acting on her are her weight and the normal force nof the seat pushing up on her. But recall from •SECTION 5.3 that you don't feel the weight force. The force you feel, your apparent weight, is the magnitude of the contact force that supports you. Here the seat is supporting the passenger with the normal force ii, so her apparent weight is Wapp = n. Based on our understanding of circular motion, we can say: FIGURE 6.16a Forces in Circular Motion 185 roller coaster car going around a loop-the-loop. FIGURE 6.16 A w ■ ■ ■ She's moving in a circle, so there must be a net force directed toward the center of the circle-currently directly above her head-to provide the centripetal acceleration. The net force points upward, so it must be the case that 11 > w. Her apparent weight is Wapp= 11, so her apparent weight is greater than her true weight ( Wapp > w). Thus she "feels heavy" at the bottom of the ciscle. (b) Altop: II fnct This situation is the same as for the car driving through a dip in Conceptual Example 6.4. To analyze the situation quantitatively, we'll apply the steps of ProblemSolving Approach 6.1. As always, we choose the x-axis to point toward the center of the circle or, in this case, vertically upward. Then Newton's second law is ~ F" = n"+ ff➔ l Toward centerof circle ! 1\/ 1 ~ ..... The r-ax1s always points towardthe x ~ ..... center of the circle. At bottom: mv-? W.r= n. - w = -,.- ii t Toward I centerof From this equation, the passenger's apparent weight is 11' Wapp = /1 = mv 2 W + -,.- (6.8) Her apparent weight at the bottom is greater than her true weight w, which agrees with your experience when you go through a dip or a valley. Now let's look at the roller coaster car as it crosses the top of the loop. Things are a little trickier here. As Figure 6.16b shows, whereas the normal force of the seat pushes up when the passenger is at the bottom of the circle, it pushes down. when she is at the top and the seat is above her. It's worth thinking carefully about this diagram to make sure you understand what it is showing. The passenger is still moving in a circle, so there must be a net force downward, toward the center of the circle, to provide her centripetal acceleration. As always, we define the x-axis to be toward the center of the circle, so here the x-axis points vertically downward. Newton's second law gives circle II Video Figure6.16 Note that W.ris now positive because the x-axis is directed downward. We can solve for the passenger's apparent weight: mv 2 Wapp = /1. = -/'- - W (6.9) If vis sufficiently large, her apparent weight can exceed the true weight, just as it did at the bottom of the track. But let's look at what happens if the car goes slower. Notice from Equation 6.9 that, as v decreases, there comes a point when mv 2 Ir= w and n becomes zero. At that point, the seat is not pushing against the passenger at all! Instead, she is able to complete the circle because her weight force alone provides sufficient centripetal acceleration. When "down" is up (iJYou can tell, even with your eyes closed, what direction is down. Organs in your innerear contain small crystalsof calcium carbonate,called 010/iths. These crystals are supportedby a sensitive membrane.Yourbrain interprets"down" as the opposite of the directionof the normal force of the membraneon the otoLiths. At the top of a loop in a roller coaster,this normal force is directed down, so your inner ear tells you that "down" is up. Youare upside down, but it doesn't feel that way. 186 c HAP TE R 6 Circular Motion, Orbits, and Gravity The speed for which n = 0 is called the critical speed vc. Because for n to be zero we must have mv}lr = w, the critical speed is V C (6.10) What happens if the speed is slower than the critical speed? In this case, Equation 6.9 gives a negative value for n. if v < Ve. But that is physically impossible. The seat can push against the passenger ( n > 0), but it can't pull on her, so the slowest possible speed is the speed for which n = 0 at the top. Thus, the critical speed is the slowest speed at which the car can complete the circle. If v < Ve, the passenger cannot turn the full loop but, instead, will fall from the car as a projectile! (This is why you're always strapped into a roller coaster.) Water stays in a bucket swung over your head for the same reason. The bottom of the bucket pushes against the water to provide the inward force that causes circular motion. If you swing the bucket too slowly, the force of the bucket on the water drops to zero. At that point, the water leaves the bucket and becomes a projectile following a parabolic trajectory onto your head! A fast-spinning world Saturn, a gas giant planet composed largelyof fluid matter, is quite a bit larger thanthe earth. It also rotates much more quickly,completingone rotation in just under 11 hours.The rapid rotation decreases the apparentweight al the equator enough to distort the fluid surface. The planet has a noticeable ovalshape,as the red circle shows. The diameterat the equator is 11% greater than the diameterat the poles. EXAMPLE 6.9 = \jr,:; -;;;= \j{rmi -;;; = \Ii; How slow can you go? A motorcyclist in the Globe of Death, pictured here. rides in a 2.2-m-radius venical loop. To keep control of the bike. the rider wants the normal force on his tires at the top of the loop to equal or exceed his and the bike's combined weight. What is the minimum speed at wbich the rider can take the loop? At the top of the loop both the normal force and the weight point downward. Together they provide the net force that determines the cenaipetal acceleration. FIGURE 6.17 Kn.wn V' "'-'F ..£..i .t" =w+n=-- 2 11111 r Centrifuges Find ~ii )(. The minimum acceptable speed occurs when 11 = w: thus 2w=2mg=--- 111\lmin 2 r Solving for the speed, we find PREPARE We wilJ consider the forces at the top point of the loop. Because the x-axis points downward, Newton's second law is n;;,.w- w F,,,1- ii l'min FIGURE 6.17. SOLVE r;t.1.m !I STRATEGIZE The visual overview for Ibis problem is shown in In accordance with Problem-Solving Approach 6.1, we've chosen the x-axis to point toward the cemer of the circle. Riding in a vertical loop around the Globe of Death. = ~= V2(9.8 m/s 2 )(2.2 m) = 6.6 mis The minimum speed is = 15 mph, wh.ich isn't all that fast: the bikes can easily reach this speed. But normally several bikes are in the globe at one Lime.The big challenge is to keep all of the riders in the cage moving at this speed in synchrony. The period for the circular motion at this speed is T = Zr.riv = 2 s, leaving little room for error! ASSESS 4iJ The centrifuge, an important biological application of circular motion, is used to separate the components of a liquid that have different densities. Typically these are different types of cells, or the components of cells, suspended in water. You probably know that small particles suspended in water will eventually settle to the bottom. However, the downward motion due to gravity for extremely small objects such as cells is so slow that it could take days or even months for the cells to settle out. It's not practical to wait for biological samples to separate due to gravity alone. 6.4 Circular Orbits and Weightlessness The separation would go faster if the force of gravity could be increased. Although we can't change gravity, we can increase the apparent weight of objects in the sample by spinning them very fast, and that is what the centrifuge shown in FIGURE s.1sdoes. The centrifuge produces centripetal accelerations that are thousands of times greater than free-fall acceleration. As the centrifuge effectively increases gravity to thousands of times its normal value, the cells or cell components settle out and separate by density in a matter of minutes or hours. EXAMPLE 6.10 187 FIGURE 6.18 The operation of a centrifuge. The high centripetal acceleration requires a large normal force. which leads,-:0 a large apparent weight. Analyzing the ultracentrifuge An 18-cm-diameter ultracentrifuge produces an extraordinarily large centripetal acceleration of 250,000g, where g is the free-falJ acceleration due to gravity. What is its frequency in rpm? What is the apparent weight of a sample with a mass of 0.0030 kg? STRATEGIZEWe will use Equation 6.5 for the relationship between the centripetal acceleration and the frequency and radius. PREPARE The acceleration in SJ units is a= 250,000(9.80 m/s 2) The radius is half the diameter. or r = 9.0 cm = 2.45 X 106 m/s 2 = 0.090 m. SOLVE We can rearrange Equation 6.5 to find the frequency given the centripetal acceleration: f = _l_ ~ = _!__fa.45 X 2,, V~ 2,, V• I06 m/s2 = 830 rev/s o.o9o m Converting to rpm. we find rev 60 s 830 - X --. s I mm = 50,000 rpm The acceleration is so high that every force is negligible except for the force that provides the centripetal acceleration. The net force is simply equal to the inward force, which is also the sampte·s apparent weight: = F0,. 11·app =ma= (3.0 X 10- 3 kg)(2.45 X 106 m/s 2 ) = 7.4 X 103 N 4iJ The 3 gram sample has an effective weight of about 1700 pounds! ASSESS Because the acceleration is 250.000g. the apparent weight is 250.000 times the acrual weight. This makes sense. as does the fact that we calculated a very high frequency. which is necessary to give the large acceleration. 141,iJl-ii!iiJj§#I A car is rolling over the top of a hill at constant speed v. At this instant, A. 11 > w. B. n< w. C. n= w. D. We can't tell about n without knowing v. ,i \I, 6.4 Circular Orbits and Weightlessness The International Space Station orbits the earth in a circular path at a speed of over 17,000 mi Jes per hour. What forces act on it? Why does it move in a circle? Before we start considering the physics of orbital motion, let's return, for a moment, to projectile motion. Projectile motion occurs when the only force on an object is gravity. Our analysis of projectiles made an implicit assumption that the earth is flat and that the freefall acceleration, due to gravity, is everywhere straight down. This is an acceptable approximation for projectiles of limited range, such as baseballs or cannon balls, but there comes a point where we can no longer ignore tbe curvature of the e,uth. Human centrifuge If you spin your arm rapidly in a vertical circle, the motion will produce an effect like that in a centrifuge. The motion will assist outbound blood flow in your arteries and retard inbound blood flow in your veins. There will be a buildup of fluid in your hand that you will be able to see (and feel!) quite easily. 188 c HAP TE A 6 Circular Motion, Orbits, and Gravity Orbital Motion FIGURE 6.19 Projectiles being launched at increasing speeds from height hon a smooth, airless planet. A Projectile motion B The ground is curving away from the projectile. C D This projectile "falls" all the way around the planet because the curvature of its trajectory matches the planet's. FIGURE 6.19 shows a perfectly smooth, spherical, airless planet with a vertical tower of height h. A projectile is launched from this tower with initial speed v; parallel to the ground. If 11;is very small, as in trajectory A, the "flat-earth approximation" is valid and the problem is identical to Example 3.9 in which a dog ran off the end of a dock. The projectile simply falls to the ground along a parabolic trajectory. As the initial speed v; is increased, it seems from the projectile's perspective that the ground is curving out from beneath it. It is still falling the entire time, always getting closer to the ground, but the distance that the projectile travels before finally reaching the ground-that is, its range-increases because the projectile must "catch up" with the ground that is curving away from it. Trajectories B and C are like this. If the launch speed v; is sufficiently large, there comes a point at which the curve of the trajectory and the curve of the earth are parallel. In this case, the projectile "falls" but it never gets any closer to the ground! This is the situation for trajectory D. The projectile returns to the point from which it was launched, at the same speed at which it was launched, making a closed trajectory. Such a closed trajectory around a planet or star is called an orbit. The most important point of this qualitative analysis is that, in the absence of air resistance, an orbiting projectile is in free fall. This is, admittedly, a strange idea, but one worth careful thought. An orbiting projectile is really no different from a thrown baseball or a dog jumping off a dock. The only force acting on it is gravity, but its tangential velocity is so great that the curvature of its trajectory matches the curvature of the earth. When this happens, the projectile "falls" under the influence of gravity but never gets any closer to the surface, which curves away beneath it. When we first studied free fall in Chapter 2, we said that free-fall acceleration is always directed vertically downward. As we see in FIGURE 6.20. "downward" really means "toward the center of the earth." For a projectile in orbit, the direction of the force of gravity changes, always pointing toward the center of the earth. FIGURE 6.20 The force of gravity is really directed toward the center of the earth. (b) (a) (c) Projectile Parabolic trajectory For a projectile near the surface of the earth, the force of gravity is directed vertically downward. Zooming out, we see that the force of gravity is really directed toward the center of the earth. For a projectile in orbit, as the projectile moves around the earth. the direction or the force of gravity changes as well. As you have learned, a force of constant magnitude that always points toward the center of a circle causes the centripetal acceleration of uniform circular motion. Because the only force acting on the orbiting projectile in Figure 6.20 is gravity, and we're assuming the projectile is very near the surface of the earth, we can write (6.11) 6.4 An object moving in a circle of radius r at speed acceleration if Vorbit Circular Orbits and Weightlessness 189 will have this centripetal (6.12) That is, if an object moves parallel to the surface with the speed (6.13) then the free-fall acceleration provides exactly the centripetal acceleration needed for a circular orbit of radius r. An object with any other speed will not follow a circular orbit. The earth's radius is r =Re= 6.37 X 106 m. The orbital speed of a projectile just skimming the surface of a smooth, airless earth is llorbit =½Re,= V(9.80 m/s 2 )(6.37 We can use Vorbit 6 X 10 m) = 7900 m/s = 18,000 mph to calculate the period of the satellite's orbit: 27Tr T= --= Vorbil 27T fi- g (6.14) For this earth-skimming orbit, T = 5065 s = 84.4 mjn_ Of course, actual satellites must orbit at some height above the surface to be above mountains and trees-and most of the atmosphere, so there is little drag. Most people envision the International Space Station orbiting far above the earth's surface, but the average height is just over 250 miles, giving a value of r that is only 6% greater than the earth's radius-not too far from skimming the surface. At this sljghtly larger value of r, Equation 6.14 gives T= 87 min. In fact, the International Space Station orbits with a period of about 93 min, going around the earth more than 15 times each day. Weightlessness in Orbit When we discussed weigh1fessness in ◄ SECTION s.3, we saw that it occurs during free fall. We asked whether astronauts and their spacecraft are in free fall. We can now give an affirmative answer: They are, indeed, in free fall. They are falling continuously around the earth, under the influence of only the gravitational force, but never getting any closer to the ground because the earth's surface curves beneath them. Weightlessness in space is no different from the weightlessness in a free-falling elevator. Weightlessness does not occur from an absence of weight or an absence of gravity. Instead, the astronaut, the spacecraft, and everything in it are "weightless" (i.e., their apparenl weight is zero) because they are all falling together. We know that the free-fall acceleration doesn't depend on mass, so the astronaut and the station follow exactly the same orbit. Zero apparent weight in space. The Orbit of the Moon The moon moves in an orbit around the earth that is approximately circular. The force that holds the moon in its orbit, that provides the necessary centripetal acceleration, is the gravitational attraction of the earth. The moon, like all satellites, is simply "falling" around the earth. But if we use the distance to the moon, r = 3.84 X I 08 m, in Equation 6.14 to predict the period of the moon's orbit, we get a period of approximately l l hours. This is clearly wrong; you know that the period of the moon's orbit is about one month. What went wrong? In using Equation 6.14, we assumed that the free-fall acceleration g is the same at the distance of the moon as it is on or near the earth's surface. But if gravity is the force of the earth pulling on an object, it seems plausible that the size of that force, and thus the size of g, should diminish with increasing distance from the earth. And, indeed, the force of gravity does decrease with distance, in a manner we'll explore in the next section. @liJ Rotating space stations The weightlessness astronauts experience in orbit has serious physiological consequences. Astronauts who spend time in weightless environments lose bone and muscle mass and suffer other adverse effects. One solution is to introduce "artificial gravity." On a space station, the easiest way to do this would be to make the station rotate, producing an apparent weight. The designers of this space station model for the movie 2001: A Space Odyssey made it rotate for just that reason. 190 CHAPTER 6 Circular Motion, Orbits, and Gravity STOP TO THINK 6.4 A satellite is in a low earth orbit. Which of the following changes would increase the orbital period? A. Increasing the mass of the satellite B. Increasing the height of the satellite about the surface C. Increasing the value of g 6.5 Newton's Law of Gravity Our current understanding of the force of gravity begins with Isaac Newton. The popular image of Newton coming to a key realization about gravity after an apple fell on his head is at least close to the truth: Newton himself said that the "notion of gravitation" came to him as he "sat in a contemplative mood" and "was occasioned by the fall of an apple." The important notion that came to Newton is this: Gravity is a universal force that affects all objects in !he universe. The force that causes the fall of an apple is the same force that keeps the moon in orbit. This is something widely accepted now, but at the time this was a revolutionary idea, and there were some important details for Newton to work out-in particular, the way that the force varies with distance. Gravity Obeys an Inverse-Square Law Newton proposed that eve,y object in the universe attracts every other object with a force that has the following properties: 1. The force is inversely proprntional to the square of the distance between the objects. 2. The force is directly prop011ionalto the product of the masses of the two objects. FIGURE 6.21 The gravitational on masses 111.1 and m2 . r is the distance between the .. centers. ·····:,,r forces FIGURE s.21 shows two spherical objects with masses m I and rn.2 separated by distance r. Each object exerts an attractive force on the other, a force that we call the gravitational force. These two forces form an action/reaction pair, so FI0112 is equal in magnitude and opposite in direction to F20111. The magnitude of the forces is given by Newton's law of gravity. :JZ)"' \ F1on 2 <-···········•· Tl1e forces are an I action/reaction pair. on JI/ Newton's law of gravity If two objects with masses m 1 and m 2 are a distance r apart, the objects exert attractive forces on each other of magnitude (6.15) F ~r The forces are directed along the line joining the two objects. The constant G is called the gravitational constant. In SI units, INVERSESQUARE G = 6.67 X 10- 11 N • m2 /kg 2 lmIEI ► Strictly speaking, Newton's law of gravity applies to particles with masses 1111 and 111 2. However, it can be shown that the law also applies to the force between two spherical objects if r is the distance between their centers. ◄ As the distance r between two objects increases, the gravitational force between them decreases. Because the distance appears squared in the denominator, Newton's law of gravity is what we call an inverse-square law. Doubling the distance between two masses causes the force between them to decrease by a factor of 4. This mathematical form is one we will see again, so it is worth our time to explore it in more detail. 6.5 lL Newton's 191 Law of Gravity Inverse-square relationships Two quantities have an inverse-square relationship if y is inversely proportional to the square of .r. We w1ite the mathematical relationship a y 4,.\ ...•····· When .r is hah·ed. y increases .,,.······ b) a factor of 4. A y= x:! ../_.--Whenxis I. y is A. y=, A .••• •····· A .C lA ::r..i.::i..:..::..J::"·==···=· ==-x J is inversely proponional 10 x~. 0 ½I Here. A is a constant. This relationship is sometime w1irten as y cc 1!x 1 . As the graph shows. inverse-square scaling means. for example: RATIOS SCALING • If you double .r. you decrease y by a factor of 4. ■ If you halve .r. you increase y by a factor of-+. • ff you increase x by a factor of 3. you decrease y by a factor of 9. • lf you decrease .r by a factor of 3. you increase y by a factor of 9. Generally. if x increases by a factor of C.y decreases by a factor of C 2 . If x decreases by a factor of C, y increases by a factor of C1 . When .r is doubled. y is reduced by a factor of 4 (2 squared) . 2 3 4 For any two values of.r-say. Y1=, A and XI x 1and .r1-we y 1 =, have A Xi Dividing the y 1-equation by the )'requation. we find .\' I A!x;2 Y1 A!xl ? A Xf ? -'"t A ? Xi ? x.- That is. the ratio of y-values is the inverse of the ratio of the squares of the corresponding values of .r. As .r becomes large. y becomes very small; as x becomes small. _rbecomes very large. LIMITS Exercises 19, 20 CONCEPTUAL EXAMPLE6. 11 The gravitational force between two spheres TI1egravitational force between two giant lead spheres is 0.010 N when the centers of the spheres are 20 m apart. What is the distance between their centers when the gravitational force bet ween them is 0.160 ? the force is related to the inverse square of the distance. The force i11creasesby a factor of (0.160 N)/(0.010 N) = 16. so the distance must decrease by a factor of Vl6 = 4. The distance is thus (20 m)/4 = 5.0 m. We can solve this problem without knowing the masses of the two spheres. The key is to consider the ratios of forces and distances. Gravity is an inverse-square relationship: ASSESS REASON EXAMPLE 6.12 Finding the gravitational force between two people You are seated in your physics class next 10 another s1Uden1 0.60 m away. Estimate the magni1Ude of the gravitational force between you. Assume that you each have a mass of 65 kg. STRATEGIZE This type of ratio reasoning is a very good way to get a quick handle on the solution to a problem. F() ou) on (olher<tudcnt) = Gill ~·ou111othersludent ,--, (6.67 X 10- 11N • m2/kg 2 )(65 kg)(65 kg) We will estimate the force using Newton·s law of (0.60 m)2 gravity. PREPARE We model each of you as a sphere. This is not a particularly good model. but it will do for making an estimate. We then take 0.60 m as the distance between your centers. SOLVE The gravitational force is given by Equation 6.15: = 7.8 X I0- 7 ASSESS The force is quite small. roughly the weight of one hair on your head. This seems reasonable: you don·t nom1ally sense this attractive force! There is a gravitational force between all objects in the universe, but the gravitational force between two ordinary-sized objects is extremely small. Only when one (or both) of the masses is exceptionally large does the force of gravity become important. The downward force of the earth on you-your weight-is large because the earth has an enormous mass. And the attraction is mutual: By Newton's third law, you exert an upward force on the earth that is equal to your weight. However, the large mass of the earth makes the effect of this force on the earth negligible. jg 192 c HAP TE R 6 EXAMPLE 6.13 Circular Motion, Orbits, and Gravity Finding the gravitational force of the earth on a person What is lhe magnitude of the gravitational force of the earth on a 60 kg person? The earth has mass 5.98 X 102-1 kg and radius 6.37 X 106 m. GMem =~ Fcarth on peISOn e (6.67 X 10- 11 N • m2 /kg 2 )(5.98 STRATEGIZE We will find the force between the person and the (6.37 earth by using Newton's law of gravity. X 6 X I0 m) 102-1 kg)(60 kg) 2 =590N PREPARE We again model the person as a sphere. The dist