81 Newton s Law of Universal Gravitation


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1 81 Newton s Law of Univesal Gavitation One of the most famous stoies of all time is the stoy of Isaac Newton sitting unde an apple tee and being hit on the head by a falling apple. It was this event, so the stoy goes, that led Newton to ealize that the same foce that bought the apple down on his head was also esponsible fo keeping the Moon in its obit aound the ath, and fo keeping all the planets of the sola system, including ou own planet ath, in obit aound the Sun. This foce is the foce of gavity. It is had to ovestate the impact of Newton s wok on gavity. Pio to Newton, it was widely thought that thee was one set of physical laws that explained how things woked on ath (explaining why apples fall down, fo instance), and a completely diffeent set of physical laws that explained the motion of the stas in the heavens. Amed with the insight that events on ath, as well as the behavio of stas, can be explained by a elatively simple equation (see the box below), humankind awoke to the undestanding that ou fates ae not detemined by the whims of gods, but depend, in fact, on the way we inteact with the ath, and in the way the ath inteacts with the Moon and the Sun. This simple, yet poweful idea, that we have some contol ove ou own lives, helped tigge a eal enlightenment in many aeas of ats and sciences. The foce of gavity does not equie the inteacting objects to be in contact with one anothe. The foce of gavity is an attactive foce that is popotional to the poduct of the masses of the inteacting objects, and invesely popotional to the squae of the distance between them. A gavitational inteaction involves the attactive foce that any object with mass exets on any othe object with mass. The geneal equation to detemine the gavitational foce an object of mass M exets on an object of mass m when the distance between thei centesofmass is is: v GmM FG = ˆ (quation 8.1: Newton s Law of Univesal Gavitation) 11 whee G = N m / kg is known as the univesal gavitational constant. The magnitude of the foce is equal to GmM / while the diection is given by ˆ, which means that the foce is attactive, diected back towad the object exeting the foce. v v At the suface of the ath, should we use FG = mg o Newton s Law of Univesal Gavitation instead? Why is g equal to 9.8 N/kg at the suface of the ath, anyway? The two equations must be equivalent to one anothe, at least at the suface of the ath, because they epesent the same gavitational inteaction. If we set the expessions equal to one anothe we get: GmM GM mg = which gives g =. At the suface of the ath M is the mass of the ath, M = kg, and is the 6 adius of the ath, R = m. So, the magnitude of g at the ath s suface is: g 11 4 GM ( N m /kg )( kg) = = = 9.83 N/kg. R 6 ( m) Fo any object at the suface of the ath, when we use Newton s Law of Univesal Gavitation, the factos G, M, and R ae all constants, so, until this point in the book, we have simply been eplacing the constant value of GM / R by g = 9.8 N/kg. 4 Chapte 8 Gavity Page 1
2 XAMPL 8.1 A twodimensional situation Thee balls, of mass m, m, and 3m, ae placed at the cones of a squae measuing L on each side, as shown in Figue 8.5. Assume this set of thee balls is not inteacting with anything else in the univese. What is the magnitude and diection of the net gavitational foce on the ball of mass m? Figue 8.1: Thee balls placed at the cones of a squae. SOLUTION Let s begin by attaching foce vectos to the ball of mass m. In Figue 8.6 each vecto is colocoded based on the object exeting the foce. The length of each vecto is popotional to the magnitude of the foce it epesents. Figue 8.: Attaching foce vectos to the ball of mass m. We can find the two individual foces acting on the ball of mass m using Newton s Law of Univesal Gavitation. Let s define +x to the ight and +y up. v Gm( m) Fom the ball of mass m: F1 = to the ight. L v Gm(3 m) Fom the ball of mass 3m: 31 at 45 o F = below the xaxis. L + L Finding the net foce is a vectoaddition poblem. In the xdiection, we get: In the ydiection, we get: v v v F F F Gm 3Gm o 3 Gm 1x = 1x + 31x =+ + cos 45 = + v v v F F F L L L 3Gm o 3 Gm 1y = 1y + 31y = 0 sin45 = L L.. The Pythagoean theoem gives the magnitude of the net foce on the ball of mass m: Gm Gm 1 = 1x + 1y = = 3.4 F F F F1 y 3 The angle is given by: tan θ = = =. F1 x So, the angle is 19.1 below the xaxis. Figue 8.3: The tiangle epesenting the vecto addition poblem being solved above. Related ndofchapte xecises: 16, ssential Question 8.1: The Sun has a much lage mass than the ath. Which object exets a lage gavitational foce on the othe, the Sun o the ath? L L. Chapte 8 Gavity Page
3 Answe to ssential Question 8.1: Newton s thid law tells us that the gavitational foce the Sun exets on the ath is equal in magnitude (and opposite in diection) to the gavitational foce the ath exets on the Sun. This follows fom quation 8.1, because, whethe we look at the foce exeted by the Sun o the ath, the factos going into the equation ae the same. 8 The Pinciple of Supeposition XPLORATION 8. Thee objects in a line Thee balls, of mass m, m, and 3m, ae equally spaced along a line. The spacing between the balls is. We can aange the balls in thee diffeent ways, as shown in Figue 8.. In each case the balls ae in an isolated egion of space vey fa fom anything else. Figue 8.: Thee diffeent aangements of thee balls of mass m, m, and 3m placed on a line with a distance between neighboing balls. Step 1 How many foces does each ball expeience in each case? ach ball expeiences two gavitational foces, one fom each of the othe balls. We can neglect any othe inteactions. Step Conside Case 1. Is the foce that the ball of mass m exets on the ball of mass 3m affected by the fact that the ball of mass m lies between the othe two balls? Inteestingly, no. To find the net foce on any object, we simply add the individual foces acting on an object as vectos. This is known as the pinciple of supeposition, and it applies to many diffeent physical situations. In case 1, fo instance, we find the foce the ball of mass m applies to the ball of mass 3m as if the ball of mass m is not pesent. The net foce on the ball of mass 3m is the vecto addition of that foce and the foce on the 3m ball fom the ball of mass m. Step 3 In which case does the ball of mass m expeience the lagestmagnitude net foce? Ague qualitatively. Let s attach aows to the ball of mass m, as in Figue 8.3, to epesent the two foces the ball expeiences in each case. The length of each aow is popotional to the foce. Figue 8.3: Attaching foce vectos to the ball of mass m. The vectos ae colocoded based on the colo of the object exeting the foce. The length of each vecto is dawn in units of Gm /. Chapte 8 Gavity Page 3
4 In case 1, the two foces patly cancel, and, in case, the foces add but give a smalle net foce than that in case 3. Thus, the ball of mass m expeiences the lagestmagnitude net foce in Case 3. Step 4 Calculate the foce expeienced by the ball of mass m in each case. To do this, we will make extensive use of Newton s Univesal Law of Gavitation. Let s define ight to be the positive diection, and use the notation F v 1 fo the foce that the ball of mass m expeiences fom the ball of mass m. In each case: v v v F = F + F, net 1 3 Case 1: Case : Case 3: v v v Gm( m) G( m)(3 m) Gm 6Gm 4Gm F, net = F1 + F3 = + = + =+ v v v Gm( m) G( m)(3 m) Gm 3Gm 7Gm F, net = F1 + F3 =+ + =+ + =+ ( ) v v v Gm( m) G( m)(3 m) Gm 6Gm 13Gm F, net = F1 + F3 = = = ( ) This appoach confims that the ball of mass m expeiences the lagestmagnitude net foce in case 3. Step 5  Rank the thee cases, fom lagest to smallest, based on the magnitude of the net foce exeted on the ball in the middle of the set of thee balls. Let s extend ou pictoial method by attaching foce vectos to each ball in each case, as in Figue 8.4. Figue 8.4: Attaching foce vectos to the balls in each case. The foce vectos ae colocoded accoding to the ball applying the foce. The length of each vecto is dawn in units of Gm /. Again, when consideing the net foce on the middle ball, we need to add the individual foces as vectos. Refeing to Figue 8.4, anking the cases based on the magnitude of the net foce exeted on the middle ball gives Case 1 > Case 3 > Case. Key idea about the pinciple of supeposition: The net foce acting on an object can be found using the pinciple of supeposition, adding all the individual foces togethe as vectos and emembeing that each individual foce is unaffected by the pesence of othe foces. Related nd of Chapte xecises: 15 and 7. ssential Question 8.: In the xploation above, which ball expeiences the lagestmagnitude net foce in (i) Case 1 (ii) Case (iii) Case 3? Chapte 8 Gavity Page 4
5 Answe to ssential Question 8.: We could detemine the net foce on each object quantitatively, but Figue 8.4 shows that the object expeiencing the lagestmagnitude net foce is the object of mass 3m in cases 1 and, and the object of mass m in case 3. In geneal, in the case of thee objects of diffeent mass aanged in a line the object expeiencing the lagest net foce will be one of the objects at the end of the line, the one with the lage mass. The object in the middle will not have the lagest net foce because the two foces it expeiences ae in opposite diections. 83 Gavitational Field Let s discuss the concept of a gavitational field, which is epesented by g v. So fa, we have efeed to g v as the acceleation due to gavity, but a moe appopiate name is the stength of the local gavitational field. A field is something that has a magnitude and diection at all points in space. One way to define the gavitational field at a paticula point is in tems of the gavitational foce that an object of mass m would expeience if it wee placed at that point: v v F g = G. (quation 8.: Gavitational field) m The units fo gavitational field ae N/kg, o m/s. A special case is the gavitational field outside an object of mass M, such as the ath, that is poduced by that object: v GM g = ˆ, (quation 8.3: Gavitational field fom a point mass) whee is the distance fom the cente of the object to the point. The magnitude of the field is GM /, while the diection is given by ˆ, which means that the field is diected back towad the object poducing the field. One way to think about a gavitational field is the following: it is a measue of how an object, o a set of objects, with mass influences the space aound it. Visualizing the gavitational field It can be useful to daw a pictue that epesents the gavitational field nea an object, o a set of objects, so we can see at a glance what the field in the egion is like. In geneal thee ae two ways to do this, by using eithe field lines o field vectos. The fieldline epesentation is shown in Figue 8.7. If Figue 8.7 (a) epesents the field at the suface of the ath, Figue 8.7 (b) could epesent the field at the suface of anothe planet whee g is twice as lage as it is at the suface of the ath. In both these cases we have a unifom field, because the field lines ae equally spaced and paallel. In Figue 8.7 (c) we have zoomed out fa fom a planet to get a wide pespective on how the planet affects the space aound it, while in Figue 8.7 (d) we have done the same thing fo a diffeent planet with half the mass, but the same adius, as the planet in (c). Chapte 8 Gavity Page 5
6 Figue 8.7: Fieldline diagams fo vaious situations. Diagams a and b epesent unifom gavitational fields, with the field in b two times lage than that in a. Diagams c and d epesent nonunifom fields, such as the fields nea a planet. The field at the suface of the planet in c is two times lage than that at the suface of the planet in d. Question: How is the diection of the gavitational field at a paticula point shown on a fieldline diagam? What indicates the elative stength of the gavitational field at a paticula point on the fieldline diagam? Answe: ach field line has a diection maked on it with an aow that shows the diection of the gavitational field at all points along the field line. The elative stength of the gavitational field is indicated by the density of the field lines (i.e., by how close the lines ae). The moe lines thee ae in a given aea the lage the field. A second method of epesenting a field is to use field vectos. A field vecto diagam has the nice featue of einfocing the idea that evey point in space has a gavitational field associated with it, because a gid made up of equally spaced dots is supeimposed on the pictue and a vecto is attached to each of these gid points. All the vectos ae the same length. The situations epesented by the fieldline pattens in Figue 8.7 ae now edawn in Figue 8.8 using the fieldvecto epesentation. Figue 8.8: Fieldvecto diagams fo vaious situations. In figues a and b the field is unifom and diected down. The field vectos ae dake in figue b, eflecting the fact that the field has a lage magnitude in figue b than in figue a. Figues c and d epesent nonunifom fields, such as those found nea a planet. Again, the fact that each field vecto in figue c is dake than its countepat in figue d tells us that the field at any point in figue c has a lage magnitude than the field at an equivalent point in figue d. Related nd of Chapte xecises: 18, 36. ssential Question 8.3: How is the diection of the gavitational field at a paticula point shown on a fieldvecto diagam? What indicates the elative stength of the gavitational field at a paticula point on the fieldvecto diagam? Chapte 8 Gavity Page 6
7 Answe to ssential Question 8.3: The diection of the gavitational field at a paticula point is epesented by the diection of the field vecto at that point (o the ones nea it if the point does not coespond exactly to the location of a field vecto). The elative stength of the field is indicated by the dakness of the aow. The lage the field s magnitude, the dake the aow. 84 Gavitational Potential negy The expession we have been using fo gavitational potential enegy up to this point, UG = mgh, applies when the gavitational field is unifom. In geneal, the equation fo gavitational potential enegy is: GmM UG =. (quation 8.4: Gavitational potential enegy, in geneal) This gives the enegy associated with the gavitational inteaction between two objects, of mass m and M, sepaated by a distance. The minus sign tells us the objects attact one anothe. Conside the diffeences between the mgh equation fo gavitational potential enegy and the moe geneal fom. Fist, when using quation 8.4 we ae no longe fee to define the potential enegy to be zeo at some convenient point. Instead, the gavitational potential enegy is zeo when the two objects ae infinitely fa apat. Second, when using quation 8.4 we find that the gavitational potential enegy is always negative, which is cetainly not what we found with mgh. That should not woy us, howeve, because what is citical is how potential enegy changes as objects move with espect to one anothe. If you dop you pen and it falls to the floo, fo instance, both foms of the gavitational potential enegy equation give consistent esults fo the change in the pen s gavitational potential enegy. quation 8.4 also einfoces the idea that, when two objects ae inteacting via gavity, neithe object has its own gavitational potential enegy. Instead, gavitational potential enegy is associated with the inteaction between the objects. XPLORATION 8.4 Calculate the total potential enegy in a system Thee balls, of mass m, m, and 3m, ae placed in a line, as shown in Figue What is the total gavitational potential enegy of this system? Figue 8.10: Thee equally spaced balls placed in a line. To detemine the total potential enegy of the system, conside the numbe of inteacting pais. In this case thee ae thee ways to pai up the objects, so thee ae thee tems to add togethe to find the total potential enegy. Because enegy is a scala, we do not have to woy about diection. Using a subscipt of 1 fo the ball of mass m, fo the ball of mass m, and 3 fo the ball of mass 3m, we get: Gm(3 m) G( m)(3 m) Gm( m) 10Gm UTotal = U13 + U3 + U1 = =. Key ideas fo gavitational potential enegy: Potential enegy is a scala. The total gavitational potential enegy of a system of objects can be found by adding up the enegy associated with each inteacting pai of objects. Related ndofchapte xecises: 5, 9, 40. Chapte 8 Gavity Page 7
8 XAMPL 8.4 Applying consevation ideas A ball of mass 1.0 kg and a ball of mass 3.0 kg ae initially sepaated by 4.0 m in a egion of space in which they inteact only with one anothe. When the balls ae eleased fom est, they acceleate towad one anothe. When they ae sepaated by.0 m, how fast is each ball going? SOLUTION Figue 8.11: The initial situation shows the balls at est. The foce of gavity causes them to acceleate towad one anothe. Figue 8.11 shows the balls at the beginning and when they ae sepaated by.0 m. Analyzing foces, we find that the foce on each ball inceases as the distance between the balls deceases. This makes it difficult to apply a foce analysis. negy consevation is a simple appoach. Ou enegy equation is: U + K + W = U + K. i i nc f f In this case, thee ae no nonconsevative foces acting, and in the initial state the kinetic enegy is zeo because both objects ae at est. This gives Ui = U f + K f. The final kinetic enegy epesents the kinetic enegy of the system, the sum of the kinetic enegies of the two objects. Let s solve this geneally, using a mass of m and a final speed of v1 fo the 1.0 kg ball, and a mass of 3m and a final speed of v fo the 3.0 kg ball. The enegy equation becomes: Gm(3 m) Gm(3 m) 1 1 = + mv1 + (3 m ) v. 4.0 m.0 m 3Gm 3Gm 1 3 Canceling factos of m gives: = + v1 + v. 4.0 m.0 m 3 Multiplying though by, and combining tems, gives: Gm + = v v..0 m Because thee is no net extenal foce, the system s momentum is conseved. Thee is no initial momentum. Fo the net momentum to emain zeo, the two momenta must always be equalandopposite. Defining ight to be positive, momentum consevation gives: 0=+ mv1 3mv, which we can simplify to v1 = 3v. Substituting this into the expession we obtained fom applying enegy consevation: 3Gm + = (3 v) + 3v = 1v.0 m Gm Gm This gives v =, and v1 = 3v = m 8.0 m 6 6 Using m = 1.0 kg, we get v =.9 10 m / s and v 1 = m / s. Related ndofchapte xecises: Poblems ssential Question 8.4: Retun to the pevious xample. If you epeat the expeiment with balls of mass.0 kg and 6.0 kg instead, would the final speeds change? If so, how? Chapte 8 Gavity Page 8
9 Answe to ssential Question 8.4: If we double each mass, the analysis above still woks. Plugging m =.0 kg into ou speed equations shows that the speeds incease by a facto of. 85 xample Poblems XAMPL 8.5A Whee is the field zeo? Locations whee the net gavitational field is zeo ae special, because an object placed whee the field is zeo expeiences no net gavitational foce. Let s place a ball of mass m at the oigin, and place a second ball of mass 9m on the xaxis at x = +4a. Find all the locations nea the balls whee the net gavitational field associated with these balls is zeo. SOLUTION A diagam of the situation is shown in Figue 8.9. Let s now appoach the poblem conceptually. At evey point nea the balls thee ae two gavitational fields, one fom each ball. The net field is zeo only whee the two fields ae equalandopposite. These fields ae in exactly opposite diections only at locations on the xaxis between the balls. If we get too close to the fist ball it dominates, and if we get too close to the second ball it dominates; thee is just one location between the balls whee the fields exactly balance. Figue 8.9: The two balls in xample 8.5A. An equivalent appoach is to use foces. Imagine having a thid ball (we geneally call this a test mass) and placing it nea the othe two balls. The thid ball expeiences two foces, one fom each of the oiginal balls, and these foces have to exactly balance. This happens at one location between the oiginal two balls. Whethe we think about fields o foces, the appoach is equivalent. The special place whee the net field is zeo is close to the ball with the smalle mass. To make up fo a facto of 9, epesenting the atio of the two masses, we need to have a facto of 3 (which gets squaed to 9) in the distances. In othe wods, we need to be thee times futhe fom the ball with a mass of 9m than we ae fom the ball of mass m fo the fields to be of equal magnitude. This occus at x = +a. We can also get this answe using a quantitative appoach. Using the subscipt 1 fo the ball of mass m, and fo the ball of mass 9m, we can expess the net field as: g v = g v + g v = net 1 0. Define ight to be positive. If the point we e looking fo is between the balls a distance x fom the ball of mass m, it is (4a x) fom the ball of mass 9m. Using the definition of g v gives: Gm G(9 m) + = 0. x (4 a x) Canceling factos of G and m, and eaanging gives: 1 = 9. x (4 a x) Cossmultiplying leads to: (4 a x) = 9 x. We could use the quadatic equation to solve fo x, but let s instead take the squae oot of both sides of the equation. When we take a squae oot the esult can be eithe plus o minus: 4a x=± 3 x. Chapte 8 Gavity Page 9
10 Using the positive sign, we get 4a= + 4x, so x = + a. This is the coect solution, lying between the balls and close to the ball with the smalle mass. Because it is thee times fathe fom the ball of mass 9m than the ball of mass m, and because the distance is squaed in the equation fo field, this exactly balances the facto of 9 in the masses. Using a minus sign gives a second solution, x = a. This location is thee times fathe (6a) fom the ball of mass 9m than fom the ball of mass m (a). Thus at x = a the two fields have the same magnitude, but they point in the same diection so they add athe than canceling. Related ndofchapte xecises: 13, 14, 0. XAMPL 8.5B scape fom ath When you thow a ball up into the ai, it comes back down. How fast would you have to launch a ball so that it neve came back down, but instead it escaped fom the ath? The minimum speed equied to do this is known as the escape speed. SOLUTION A diagam is shown in Figue 8.1. Let s assume the ball stats at the suface of the ath and that we can neglect ai esistance (this would be fine if we wee escaping fom the Moon, but it is a poo assumption if we e escaping fom ath  let s not woy about that, howeve). We ll also assume the ath is the only object in the Univese. So, this is an inteesting calculation but the esult will only be a ough appoximation of eality. Let s apply the enegy consevation equation: U + K + W = U + K. i i nc f f Figue 8.1: negy ba gaphs ae shown in addition to the pictues showing the initial and final situations. We e neglecting any wok done by nonconsevative foces, so W nc = 0. The final gavitational potential enegy is negligible, because the distance between the ball and ath is vey lage (we can assume it to be infinite). What about the final kinetic enegy? Because we e looking fo the minimum initial speed let s use the minimum possible speed of the ball when it is vey fa fom ath, which we can assume to be zeo. This leads to an equation in which eveything on the ighthand side is zeo: Ui + Ki = 0. GmM 1 + mvescape = 0. R The mass of the ball does not matte, because it cancels out. This gives: 11 4 GM ( N m / kg )( kg) vescape = = = 11. km / s. 6 R m This is athe fast, and explains why objects we thow up in the ai come down again! Related ndofchapte xecises: 41, 4. ssential Question 8.5: Let s say we wee on a diffeent planet that had the same mass as ath but twice ath s adius. How would the escape speed compae to that on ath? Chapte 8 Gavity Page 10
11 Answe to ssential Question 8.5: Since v escape GM doubling the adius educes the escape speed by a facto of. =, keeping the mass the same while R 86 Obits Imagine that we have an object of mass m in a cicula obit aound an object of mass M. An example could be a satellite obiting the ath. What is the total enegy associated with this object in its cicula obit? The total enegy is the sum of the potential enegy plus the kinetic enegy: GmM 1 = U + K = + mv. This is a lovely equation, but it doesn t tell us much. Let s conside foces to see if we can shed moe light on what s going on. Fo the object of mass m to expeience unifom cicula motion about the lage mass it must expeience a net foce diected towad the cente of the cicle (i.e., towad the object of mass M). This is the gavitational foce exeted by the object of mass M. Applying Newton s Second Law gives: v v mv Σ F = ma =, diected towad the cente. GmM mv GmM =, which tells us that mv =. Substituting this esult into the enegy expession gives: GmM GmM GmM = + =. This esult is geneally tue fo the case of a lighte object taveling in a cicula obit aound a moe massive object. We can make a few obsevations about this. Fist, the magnitude of the total enegy equals the kinetic enegy; the kinetic enegy has half the magnitude of the gavitational potential enegy; and the total enegy is half of the gavitational potential enegy. All this is tue when the obit is cicula. Second, the total enegy is negative, which is tue fo a bound system (a system in which the components emain togethe). Systems in which the total enegy is positive tend to fly apat. What happens when an object has a velocity othe than that necessay to tavel in a cicula obit? One way to think of this is to stat the obiting object off at the same place, with a velocity diected pependicula to the line connecting the two objects, and simply vay the speed. If the speed necessay to maintain a cicula obit is denoted by v, let s conside what happens if the speed is 0% less than v ; 0% lage than v ; the special case of v ; and 1.5v. The obits followed by the object in these cases ae shown in Figue Unless the object s initial speed is too small, causing it to eventually collide with the moe massive object, an initial speed that is less than v will poduce an elliptical obit whee the initial point tuns out to be the futhest the object eve gets fom the moe massive object. The initial point is special because at that point the object s velocity is pependicula to the gavitational foce the object expeiences. Chapte 8 Gavity Page 11
12 If the initial speed is lage than v the esult depends on how much lage it is. When the initial speed is v that is the escape speed, and is thus a special case. The shape of the obit is paabolic, and this path maks the bounday between the elliptical paths in which the object emains in obit and the highespeed hypebolic paths in which the object escapes fom the gavitational pull of the massive object. Figue 8.13: The obits esulting fom stating at a paticula spot, the ightmost point on each obit, with initial velocities diected the same way (up in the figue) but with diffeent initial speeds. The dak blue obit epesents the almostcicula obit of the ath, whee the distances on each axis ae in units of metes and the Sun is not shown but is located at the intesection of the axes. If the ath s speed wee suddenly educed by 0% the ath would instead follow the light puple obit, coming athe close to the Sun. If instead the ath s speed wee inceased by 0% the esulting elliptical obit would take us quite a long way fom the Sun befoe coming back again. Inceasing the ath s speed to times its cuent speed (an incease of a little moe than 40%) the ath would be moving at the escape speed and we would follow the light blue paabolic obit to infinity (and beyond). Any initial speed lage than this would esult in a hypebolic obit to infinity, such as that shown in the dak puple. Note that the speeds given in the key to the ight of the gaph epesent initial speeds, the speed the ath would have at the ightmost point in the obit to follow the coesponding path. Related ndofchapte xecises: 47, 59, and 60. ssential Question 8.6: Is linea momentum conseved fo any of these obits? If so, which? Chapte 8 Gavity Page 1
Revision Guide for Chapter 11
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