Chapter 13. Universal Gravitation

Transcription

1 Chapte 13 Univesal Gavitation CHAPTER OUTLINE 13.1 Newton s Law of Univesal Gavitation 13.2 Measuing the Gavitational Constant 13.3 Fee-Fall Acceleation and the Gavitational Foce 13.4 Keple s Laws and the Motion of Planets 13.5 The Gavitational Field 13.6 Gavitational Potential Enegy 13.7 Enegy Consideations in Planetay and Satellite Motion An undestanding of the law of univesal gavitation has allowed scientists to send spacecaft on impessively accuate jouneys to othe pats of ou sola system. This photo of a volcano on Io, a moon of Jupite, was taken by the Galileo spacecaft, which has been obiting Jupite since The ed mateial has been vented fom below the suface. (Univ. of Aizona/JPL/NASA) 389

2 Befoe 1687, a lage amount of data had been collected on the motions of the Moon and the planets, but a clea undestanding of the foces elated to these motions was not available. In that yea, Isaac Newton povided the key that unlocked the secets of the heavens. He knew, fom his fist law, that a net foce had to be acting on the Moon because without such a foce the Moon would move in a staight-line path athe than in its almost cicula obit. Newton easoned that this foce was the gavitational attaction exeted by the Eath on the Moon. He ealized that the foces involved in the Eath Moon attaction and in the Sun planet attaction wee not something special to those systems, but athe wee paticula cases of a geneal and univesal attaction between objects. In othe wods, Newton saw that the same foce of attaction that causes the Moon to follow its path aound the Eath also causes an apple to fall fom a tee. As he put it, I deduced that the foces which keep the planets in thei obs must be ecipocally as the squaes of thei distances fom the centes about which they evolve; and theeby compaed the foce equisite to keep the Moon in he ob with the foce of gavity at the suface of the Eath; and found them answe petty nealy. In this chapte we study the law of univesal gavitation. We emphasize a desciption of planetay motion because astonomical data povide an impotant test of this law s validity. We then show that the laws of planetay motion developed by Johannes Keple follow fom the law of univesal gavitation and the concept of consevation of angula momentum. We conclude by deiving a geneal expession fo gavitational potential enegy and examining the enegetics of planetay and satellite motion Newton s Law of Univesal Gavitation You may have head the legend that Newton was stuck on the head by a falling apple while napping unde a tee. This alleged accident supposedly pompted him to imagine that pehaps all objects in the Univese wee attacted to each othe in the same way the apple was attacted to the Eath. Newton analyzed astonomical data on the motion of the Moon aound the Eath. Fom that analysis, he made the bold assetion that the foce law govening the motion of planets was the same as the foce law that attacted a falling apple to the Eath. This was the fist time that eathly and heavenly motions wee unified. We shall look at the mathematical details of Newton s analysis in this section. In 1687 Newton published his wok on the law of gavity in his teatise Mathematical Pinciples of Natual Philosophy. Newton s law of univesal gavitation states that The law of univesal gavitation evey paticle in the Univese attacts evey othe paticle with a foce that is diectly popotional to the poduct of thei masses and invesely popotional to the squae of the distance between them. 390

3 SECTION 13.1 Newton s Law of Univesal Gavitation 391 If the paticles have masses m 1 and m 2 and ae sepaated by a distance, the magnitude of this gavitational foce is F g G m 1m 2 2 (13.1) whee G is a constant, called the univesal gavitational constant, that has been measued expeimentally. Its value in SI units is G Nm 2 /kg 2 (13.2) The fom of the foce law given by Equation 13.1 is often efeed to as an invesesquae law because the magnitude of the foce vaies as the invese squae of the sepaation of the paticles. 1 We shall see othe examples of this type of foce law in subsequent chaptes. We can expess this foce in vecto fom by defining a unit vecto ˆ 12 (Fig. 13.1). Because this unit vecto is diected fom paticle 1 towad paticle 2, the foce exeted by paticle 1 on paticle 2 is F 12 G m 1m 2 2 ˆ12 (13.3) whee the negative sign indicates that paticle 2 is attacted to paticle 1, and hence the foce on paticle 2 must be diected towad paticle 1. By Newton s thid law, the foce exeted by paticle 2 on paticle 1, designated F 21, is equal in magnitude to F 12 and in the opposite diection. That is, these foces fom an action eaction pai, and F 21 F 12. Seveal featues of Equation 13.3 deseve mention. The gavitational foce is a field foce that always exists between two paticles, egadless of the medium that sepaates them. Because the foce vaies as the invese squae of the distance between the paticles, it deceases apidly with inceasing sepaation. Anothe impotant point that we can show fom Equation 13.3 is that the gavitational foce exeted by a finite-size, spheically symmetic mass distibution on a paticle outside the distibution is the same as if the entie mass of the distibution wee concentated at the cente. Fo example, the magnitude of the foce exeted by the Eath on a paticle of mass m nea the Eath s suface is F g G M Em R E 2 (13.4) PITFALL PREVENTION 13.1 Be Clea on g and G The symbol g epesents the magnitude of the fee-fall acceleation nea a planet. At the suface of the Eath, g has the value 9.80 m/s 2. On the othe hand, G is a univesal constant that has the same value eveywhee in the Univese. m 1 F 21 ˆ 12 F 12 m 2 Active Figue 13.1 The gavitational foce between two paticles is attactive. The unit vecto ˆ12 is diected fom paticle 1 towad paticle 2. Note that F 21 F 12. At the Active Figues link at you can change the masses of the paticles and the sepaation distance between the paticles to see the effect on the gavitational foce. whee M E is the Eath s mass and R E its adius. This foce is diected towad the cente of the Eath. In fomulating his law of univesal gavitation, Newton used the following easoning, which suppots the assumption that the gavitational foce is popotional to the invese squae of the sepaation between the two inteacting objects. He compaed the acceleation of the Moon in its obit with the acceleation of an object falling nea the Eath s suface, such as the legenday apple (Fig. 13.2). Assuming that both acceleations had the same cause namely, the gavitational attaction of the Eath Newton used the invese-squae law to eason that the acceleation of the Moon towad the Eath (centipetal acceleation) should be popotional to 1/ M 2, whee M is the distance between the centes of the Eath and the Moon. Futhemoe, the acceleation of the apple towad the Eath should be popotional to 1/R a 2, whee R a is the distance between the centes of the Eath and the apple. Because the apple is located at the suface of the eath, R a R E, the adius of the Eath. Using the values M m and R E m, Newton pedicted that the atio of the Moon s acceleation a M to the apple s acceleation g would be a M g (1/ M) 2 (1/R E ) 2 R E M m m An invese popotionality between two quantities x and y is one in which y k/x, whee k is a constant. A diect popotion between x and y exists when y kx.

4 392 CHAPTER 13 Univesal Gavitation Moon a M v g R E Eath M Figue 13.2 As it evolves aound the Eath, the Moon expeiences a centipetal acceleation a M diected towad the Eath. An object nea the Eath s suface, such as the apple shown hee, expeiences an acceleation g. (Dimensions ae not to scale.) Theefoe, the centipetal acceleation of the Moon is a M ( )(9.80 m/s 2 ) m/s 2 Newton also calculated the centipetal acceleation of the Moon fom a knowledge of its mean distance fom the Eath and the known value of its obital peiod, T days s. In a time inteval T, the Moon tavels a distance 2 M, which equals the cicumfeence of its obit. Theefoe, its obital speed is 2 M /T and its centipetal acceleation is a M v 2 (2 M /T) M M M T ( m) ( s) m/s 2 The nealy pefect ageement between this value and the value Newton obtained using g povides stong evidence of the invese-squae natue of the gavitational foce law. Although these esults must have been vey encouaging to Newton, he was deeply toubled by an assumption he made in the analysis. To evaluate the acceleation of an object at the Eath s suface, Newton teated the Eath as if its mass wee all concentated at its cente. That is, he assumed that the Eath acted as a paticle as fa as its influence on an exteio object was concened. Seveal yeas late, in 1687, on the basis of his pioneeing wok in the development of calculus, Newton poved that this assumption was valid and was a natual consequence of the law of univesal gavitation. We have evidence that the gavitational foce acting on an object is diectly popotional to its mass fom ou obsevations of falling objects, discussed in Chapte 2. All objects, egadless of mass, fall in the absence of ai esistance at the same acceleation g nea the suface of the Eath. Accoding to Newton s second law, this acceleation is given by g F g /m, whee m is the mass of the falling object. If this atio is to be the same fo all falling objects, then F g must be diectly popotional to m, so that the mass cancels in the atio. If we conside the moe geneal situation of a gavitational foce between any two objects with mass, such as two planets, this same agument can be applied to show that the gavitational foce is popotional to one of the masses. We can choose eithe of the masses in the agument, howeve; thus, the gavitational foce must be diectly popotional to both masses, as can be seen in Equation Quick Quiz 13.1 The Moon emains in its obit aound the Eath athe than falling to the Eath because (a) it is outside of the gavitational influence of the Eath (b) it is in balance with the gavitational foces fom the Sun and othe planets (c) the net foce on the Moon is zeo (d) none of these (e) all of these.

5 SECTION 13.2 Measuing the Gavitational Constant 393 Quick Quiz 13.2 A planet has two moons of equal mass. Moon 1 is in a cicula obit of adius. Moon 2 is in a cicula obit of adius 2. The magnitude of the gavitational foce exeted by the planet on moon 2 is (a) fou times as lage as that on moon 1 (b) twice as lage as that on moon 1 (c) equal to that on moon 1 (d) half as lage as that on moon 1 (e) one fouth as lage as that on moon 1. Example 13.1 Billiads, Anyone? Inteactive Thee kg billiad balls ae placed on a table at the cones of a ight tiangle, as shown in Figue Calculate the gavitational foce on the cue ball (designated m 1 ) esulting fom the othe two balls. Solution Fist we calculate sepaately the individual foces on the cue ball due to the othe two balls, and then we find the vecto sum to obtain the esultant foce. We can see gaphically that this foce should point upwad and towad the ight. We locate ou coodinate axes as shown in Figue 13.3, placing ou oigin at the position of the cue ball. The foce exeted by m 2 on the cue ball is diected upwad and is given by F 21 G m 2m ĵ N ĵ ( Nm 2 /kg 2 ) (0.300 kg)(0.300 kg) (0.400 m) 2 ĵ This esult shows that the gavitational foces between eveyday objects have extemely small magnitudes. The foce exeted by m 3 on the cue ball is diected to the ight: m m m F 31 G m 3m î N î ( Nm 2 /kg 2 ) (0.300 kg)(0.300 kg) (0.300 m) 2 î Theefoe, the net gavitational foce on the cue ball is y x F 21 F θ F 31 m m m 3 F F 21 F 31 (6.67î 3.75ĵ) 1011 N and the magnitude of this foce is F F 2 21 F 2 31 (3.75) 2 (6.67) N N Figue 13.3 (Example 13.1) The esultant gavitational foce acting on the cue ball is the vecto sum F 21 F 31. Fom tan 3.75/ , the diection of the net gavitational foce is 29.3 counteclockwise fom the x axis. At the Inteactive Woked Example link at you can move balls 2 and 3 to see the effect on the net gavitational foce on ball Measuing the Gavitational Constant The univesal gavitational constant G was measued in an impotant expeiment by Heny Cavendish ( ) in The Cavendish appaatus consists of two small sphees, each of mass m, fixed to the ends of a light hoizontal od suspended by a fine fibe o thin metal wie, as illustated in Figue When two lage sphees, each of mass M, ae placed nea the smalle ones, the attactive foce between smalle and lage sphees causes the od to otate and twist the wie suspension to a new equilibium oientation. The angle of otation is measued by the deflection of a light beam eflected fom a mio attached to the vetical suspension. The deflection of the light beam is an effective technique fo amplifying the motion. The expeiment is caefully epeated with diffeent masses at vaious sepaations. In addition to poviding a value M Mio m Light souce Figue 13.4 Cavendish appaatus fo measuing G. The dashed line epesents the oiginal position of the od.

6 394 CHAPTER 13 Univesal Gavitation fo G, the esults show expeimentally that the foce is attactive, popotional to the poduct mm, and invesely popotional to the squae of the distance Fee-Fall Acceleation and the Gavitational Foce In Chapte 5, when defining mg as the weight of an object of mass m, we efeed to g as the magnitude of the fee-fall acceleation. Now we ae in a position to obtain a moe fundamental desciption of g. Because the magnitude of the foce acting on a feely falling object of mass m nea the Eath s suface is given by Equation 13.4, we can equate mg to this foce to obtain (13.5) Now conside an object of mass m located a distance h above the Eath s suface o a distance fom the Eath s cente, whee R E h. The magnitude of the gavitational foce acting on this object is F g G M Em 2 m g G M Em R E 2 g G M E R E 2 G M E m (R E h) 2 The magnitude of the gavitational foce acting on the object at this position is also F g mg, whee g is the value of the fee-fall acceleation at the altitude h. Substituting this expession fo F g into the last equation shows that g is Vaiation of g with altitude g GM E 2 GM E (R E h) 2 (13.6) Thus, it follows that g deceases with inceasing altitude. Because the weight of an object is mg, we see that as :, its weight appoaches zeo. Coutesy NASA Astonauts F. Stoy Musgave and Jeffey A. Hoffman, along with the Hubble Space Telescope and the space shuttle Endeavo, ae all in fee fall while obiting the Eath.

7 SECTION 13.3 Fee-Fall Acceleation and the Gavitational Foce 395 Quick Quiz 13.3 Supeman stands on top of a vey tall mountain and thows a baseball hoizontally with a speed such that the baseball goes into a cicula obit aound the Eath. While the baseball is in obit, the acceleation of the ball (a) depends on how fast the baseball is thown (b) is zeo because the ball does not fall to the gound (c) is slightly less than 9.80 m/s 2 (d) is equal to 9.80 m/s 2. Example 13.2 Vaiation of g with Altitude h The Intenational Space Station opeates at an altitude of 350 km. When final constuction is completed, it will have a weight (measued at the Eath s suface) of N. What is its weight when in obit? Solution We fist find the mass of the space station fom its weight at the suface of the Eath: This mass is fixed it is independent of the location of the space station. Because the station is above the suface of the Eath, howeve, we expect its weight in obit to be less than its weight on the Eath. Using Equation 13.6 with h 350 km, we obtain g m Fg g N 9.80 m/s kg GM E (R E h) 2 ( Nm 2 /kg 2 )( kg) ( m m) m/s 2 Because this value is about 90% of the value of g at the Eath suface, we expect that the weight of the station at an altitude of 350 km is 90% of the value at the Eath s suface. Using the value of g at the location of the station, the station s weight in obit is mg ( kg)(8.83 m/s 2 ) Table 13.1 Fee-Fall Acceleation g at Vaious Altitudes Above the Eath s Suface Altitude h (km) g (m/s 2 ) N Values of g at othe altitudes ae listed in Table Example 13.3 The Density of the Eath Using the known adius of the Eath and the fact that g 9.80 m/s 2 at the Eath s suface, find the aveage density of the Eath. Solution Fom Eq. 1.1, we know that the aveage density is whee M E is the mass of the Eath and V E is its volume. Fom Equation 13.5, we can elate the mass of the Eath to the value of g : g G M E R 2 9: M E gr 2 E E G Substituting this into the definition of density, we obtain E M E V E M E V E (gr E 2 /G) 4 R E g GR E m/s 2 ( Nm 2 /kg 2 )( m) kg/m 3 What If? What if you wee told that a typical density of ganite at the Eath s suface wee kg/m 3 what would you conclude about the density of the mateial in the Eath s inteio? Answe Because this value is about half the density that we calculated as an aveage fo the entie Eath, we conclude that the inne coe of the Eath has a density much highe than the aveage value. It is most amazing that the Cavendish expeiment, which detemines G and can be done on a tabletop, combined with simple fee-fall measuements of g povides infomation about the coe of the Eath!

8 396 CHAPTER 13 Univesal Gavitation 13.4 Keple s Laws and the Motion of Planets Johannes Keple Geman astonome ( ) The Geman astonome Keple is best known fo developing the laws of planetay motion based on the caeful obsevations of Tycho Bahe. (At Resouce) People have obseved the movements of the planets, stas, and othe celestial objects fo thousands of yeas. In ealy histoy, scientists egaded the Eath as the cente of the Univese. This so-called geocentic model was elaboated and fomalized by the Geek astonome Claudius Ptolemy (c. 100 c. 170) in the second centuy A.D. and was accepted fo the next yeas. In 1543 the Polish astonome Nicolaus Copenicus ( ) suggested that the Eath and the othe planets evolved in cicula obits aound the Sun (the heliocentic model ). The Danish astonome Tycho Bahe ( ) wanted to detemine how the heavens wee constucted, and thus he developed a pogam to detemine the positions of both stas and planets. It is inteesting to note that those obsevations of the planets and 777 stas visible to the naked eye wee caied out with only a lage sextant and a compass. (The telescope had not yet been invented.) The Geman astonome Johannes Keple was Bahe s assistant fo a shot while befoe Bahe s death, wheeupon he acquied his mento s astonomical data and spent 16 yeas tying to deduce a mathematical model fo the motion of the planets. Such data ae difficult to sot out because the Eath is also in motion aound the Sun. Afte many laboious calculations, Keple found that Bahe s data on the evolution of Mas aound the Sun povided the answe. Keple s complete analysis of planetay motion is summaized in thee statements known as Keple s laws: Keple s laws 1. All planets move in elliptical obits with the Sun at one focus. 2. The adius vecto dawn fom the Sun to a planet sweeps out equal aeas in equal time intevals. 3. The squae of the obital peiod of any planet is popotional to the cube of the semimajo axis of the elliptical obit. We discuss each of these laws below. Keple s Fist Law 1 2 F 1 y c a F 2 Active Figue 13.5 Plot of an ellipse. The semimajo axis has length a, and the semimino axis has length b. Each focus is located at a distance c fom the cente on each side of the cente. At the Active Figues link at you can move the focal points o ente values fo a, b, c, and e to see the esulting elliptical shape. b x We ae familia with cicula obits of objects aound gavitational foce centes fom ou discussions in this chapte. Keple s fist law indicates that the cicula obit is a vey special case and elliptical obits ae the geneal situation. This was a difficult notion fo scientists of the time to accept, because they felt that pefect cicula obits of the planets eflected the pefection of heaven. Figue 13.5 shows the geomety of an ellipse, which seves as ou model fo the elliptical obit of a planet. An ellipse is mathematically defined by choosing two points F 1 and F 2, each of which is a called a focus, and then dawing a cuve though points fo which the sum of the distances 1 and 2 fom F 1 and F 2, espectively, is a constant. The longest distance though the cente between points on the ellipse (and passing though both foci) is called the majo axis, and this distance is 2a. In Figue 13.5, the majo axis is dawn along the x diection. The distance a is called the semimajo axis. Similaly, the shotest distance though the cente between points on the ellipse is called the mino axis of length 2b, whee the distance b is the semimino axis. Eithe focus of the ellipse is located at a distance c fom the cente of the ellipse, whee a 2 b 2 c 2. In the elliptical obit of a planet aound the Sun, the Sun is at one focus of the ellipse. Thee is nothing at the othe focus. The eccenticity of an ellipse is defined as e c/a and descibes the geneal shape of the ellipse. Fo a cicle, c 0, and the eccenticity is theefoe zeo. The smalle b is than a, the shote the ellipse is along the y diection compaed to its extent in the x diection in Figue As b deceases, c inceases, and the eccenticity e inceases.

9 SECTION 13.4 Keple s Laws and the Motion of Planets 397 Sun PITFALL PREVENTION 13.2 Whee is the Sun? The Sun is located at one focus of the elliptical obit of a planet. It is not located at the cente of the ellipse. Cente Obit of Pluto (a) Obit of Comet Halley (b) Sun Cente Figue 13.6 (a) The shape of the obit of Pluto, which has the highest eccenticity (e 0.25) among the planets in the sola system. The Sun is located at the lage yellow dot, which is a focus of the ellipse. Thee is nothing physical located at the cente (the small dot) o the othe focus (the blue dot). (b) The shape of the obit of Comet Halley. Thus, highe values of eccenticity coespond to longe and thinne ellipses. The ange of values of the eccenticity fo an ellipse is 0 e 1. Eccenticities fo planetay obits vay widely in the sola system. The eccenticity of the Eath s obit is 0.017, which makes it nealy cicula. On the othe hand, the eccenticity of Pluto s obit is 0.25, the highest of all the nine planets. Figue 13.6a shows an ellipse with the eccenticity of that of Pluto s obit. Notice that even this highesteccenticity obit is difficult to distinguish fom a cicle. This is why Keple s fist law is an admiable accomplishment. The eccenticity of the obit of Comet Halley is 0.97, descibing an obit whose majo axis is much longe than its mino axis, as shown in Figue 13.6b. As a esult, Comet Halley spends much of its 76-yea peiod fa fom the Sun and invisible fom the Eath. It is only visible to the naked eye duing a small pat of its obit when it is nea the Sun. Now imagine a planet in an elliptical obit such as that shown in Figue 13.5, with the Sun at focus F 2. When the planet is at the fa left in the diagam, the distance between the planet and the Sun is a c. This point is called the aphelion, whee the planet is the fathest away fom the Sun that it can be in the obit. (Fo an object in obit aound the Eath, this point is called the apogee). Convesely, when the planet is at the ight end of the ellipse, the point is called the peihelion (fo an Eath obit, the peigee), and the distance between the planet and the Sun is a c. Keple s fist law is a diect esult of the invese squae natue of the gavitational foce. We have discussed cicula and elliptical obits. These ae the allowed shapes of obits fo objects that ae bound to the gavitational foce cente. These objects include planets, asteoids, and comets that move epeatedly aound the Sun, as well as moons obiting a planet. Thee could also be unbound objects, such as a meteooid fom deep space that might pass by the Sun once and then neve etun. The gavitational foce between the Sun and these objects also vaies as the invese squae of the sepaation distance, and the allowed paths fo these objects include paabolas (e 1) and hypebolas (e 1). Keple s Second Law Keple s second law can be shown to be a consequence of angula momentum consevation as follows. Conside a planet of mass M P moving about the Sun in an elliptical obit (Fig. 13.7a). Let us conside the planet as a system. We will model the Sun to be

10 398 CHAPTER 13 Univesal Gavitation Sun M S Sun F g (a) (b) da M P v d = vdt Active Figue 13.7 (a) The gavitational foce acting on a planet is diected towad the Sun. (b) As a planet obits the Sun, the aea swept out by the adius vecto in a time inteval dt is equal to half the aea of the paallelogam fomed by the vectos and d v dt. At the Active Figues link at you can assign a value of the eccenticity and see the esulting motion of the planet aound the Sun. so much moe massive than the planet that the Sun does not move. The gavitational foce acting on the planet is a cental foce, always along the adius vecto, diected towad the Sun (Fig. 13.7a). The toque on the planet due to this cental foce is clealy zeo, because F is paallel to. That is F F()ˆ 0 Recall that the extenal net toque on a system equals the time ate of change of angula momentum of the system; that is, dl/dt. Theefoe, because 0, the angula momentum L of the planet is a constant of the motion: L p M P v constant We can elate this esult to the following geometic consideation. In a time inteval dt, the adius vecto in Figue 13.7b sweeps out the aea da, which equals half the aea d of the paallelogam fomed by the vectos and d. Because the displacement of the planet in the time inteval dt is given by d v dt, we have da 1 2 d 1 v dt L 2 da dt L 2M P constant 2M P dt (13.7) whee L and M P ae both constants. Thus, we conclude that the adius vecto fom the Sun to any planet sweeps out equal aeas in equal times. It is impotant to ecognize that this esult is a consequence of the fact that the gavitational foce is a cental foce, which in tun implies that angula momentum of the planet is constant. Theefoe, the law applies to any situation that involves a cental foce, whethe invese-squae o not. v M S M P Keple s Thid Law It is infomative to show that Keple s thid law can be pedicted fom the invesesquae law fo cicula obits. 2 Conside a planet of mass M P that is assumed to be moving about the Sun (mass M S ) in a cicula obit, as in Figue Because the gavitational foce povides the centipetal acceleation of the planet as it moves in a cicle, we use Newton s second law fo a paticle in unifom cicula motion, GM S M P 2 M P v 2 The obital speed of the planet is 2/T, whee T is the peiod; theefoe, the peceding expession becomes Figue 13.8 A planet of mass M P moving in a cicula obit aound the Sun. The obits of all planets except Mecuy and Pluto ae nealy cicula. whee K S is a constant given by GM S 2 (2/T ) 2 T GM S 3 K S 3 K S s 2 /m 3 GM S 2 The obits of all planets except Mecuy and Pluto ae vey close to being cicula; hence, we do not intoduce much eo with this assumption. Fo example, the atio of the semimino axis to the semimajo axis fo the Eath s obit is b/a

11 SECTION 13.4 Keple s Laws and the Motion of Planets 399 Table 13.2 Useful Planetay Data Peiod of Mean Distance Body Mass (kg) Mean Radius (m) Revolution (s) fom Sun (m) T 2 3 (s2 /m 3 ) Mecuy Venus Eath Mas Jupite Satun Uanus Neptune Pluto Moon Sun This equation is also valid fo elliptical obits if we eplace with the length a of the semimajo axis (Fig. 13.5): T GM S a 3 K S a 3 (13.8) Keple s thid law Equation 13.8 is Keple s thid law. Because the semimajo axis of a cicula obit is its adius, Equation 13.8 is valid fo both cicula and elliptical obits. Note that the constant of popotionality K S is independent of the mass of the planet. Equation 13.8 is theefoe valid fo any planet. 3 If we wee to conside the obit of a satellite such as the Moon about the Eath, then the constant would have a diffeent value, with the Sun s mass eplaced by the Eath s mass, that is, K E 4 2 /GM E. Table 13.2 is a collection of useful planetay data. The last column veifies that the atio T 2 / 3 is constant. The small vaiations in the values in this column ae due to uncetainties in the data measued fo the peiods and semimajo axes of the planets. Recent astonomical wok has evealed the existence of a lage numbe of sola system objects beyond the obit of Neptune. In geneal, these lie in the Kuipe belt, a egion that extends fom about 30 AU (the obital adius of Neptune) to 50 AU. (An AU is an astonomical unit the adius of the Eath s obit.) Cuent estimates identify at least objects in this egion with diametes lage than 100 km. The fist KBO (Kuipe Belt Object) was discoveed in Since then, many moe have been detected and some have been given names, such as Vauna (diamete about km, discoveed in 2000), Ixion (diamete about km, discoveed in 2001), and Quaoa (diamete about 800 km, discoveed in 2002). A subset of about KBOs ae called Plutinos because, like Pluto, they exhibit a esonance phenomenon, obiting the Sun two times in the same time inteval as Neptune evolves thee times. Some astonomes even claim that Pluto should not be consideed a planet but should be identified as a KBO. The contempoay application of Keple s laws and such exotic poposals as planetay angula momentum exchange and migating planets 4 suggest the excitement of this active aea of cuent eseach. Quick Quiz 13.4 Pluto, the fathest planet fom the Sun, has an obital peiod that is (a) geate than a yea (b) less than a yea (c) equal to a yea. 3 Equation 13.8 is indeed a popotion because the atio of the two quantities T 2 and a 3 is a constant. The vaiables in a popotion ae not equied to be limited to the fist powe only. 4 Malhota, R., Migating Planets, Scientific Ameican, Septembe 1999, volume 281, numbe 3.

12 400 CHAPTER 13 Univesal Gavitation Quick Quiz 13.5 An asteoid is in a highly eccentic elliptical obit aound the Sun. The peiod of the asteoid s obit is 90 days. Which of the following statements is tue about the possibility of a collision between this asteoid and the Eath? (a) Thee is no possible dange of a collision. (b) Thee is a possibility of a collision. (c) Thee is not enough infomation to detemine whethe thee is dange of a collision. Quick Quiz 13.6 A satellite moves in an elliptical obit about the Eath such that, at peigee and apogee positions, its distances fom the Eath s cente ae espectively D and 4D. The elationship between the speeds at these two positions is (a) v p v a (b) v p 4v a (c) v a 4v p (d) v p 2v a (e) v a 2v p. Example 13.4 The Mass of the Sun Calculate the mass of the Sun using the fact that the peiod of the Eath s obit aound the Sun is s and its distance fom the Sun is m. Solution Using Equation 13.8, we find that M S GT kg 4 2 ( m) 3 ( Nm 2 /kg 2 )( s) 2 In Example 13.3, an undestanding of gavitational foces enabled us to find out something about the density of the Eath s coe, and now we have used this undestanding to detemine the mass of the Sun! What If? Suppose you wee asked fo the mass of Mas. How could you detemine this value? Answe Keple s thid law is valid fo any system of objects in obit aound an object with a lage mass. Mas has two moons, Phobos and Deimos. If we ewite Equation 13.8 fo these moons of Mas, we have T GM M a 3 whee M M is the mass of Mas. Solving fo this mass, M M 4 2 G a 3 Phobos has an obital peiod of 0.32 days and an almost cicula obit of adius km. The obit of Deimos is even moe cicula, with a adius of km and an obital peiod of 1.26 days. Let us calculate the mass of Mas using each of these sets of data: Phobos: Deimos: T 2 ( kgs 2 /m 3 ) a 3 M M ( kgs 2 /m 3 ) ( m) 3 (0.32 d) 2 1 d s kg M M ( kgs 2 /m 3 ) ( m) 3 (1.26 d) 2 1 d s kg These two calculations ae within 1% of each othe and both ae within 0.5% of the value of the mass of Mas given in Table T Nm 2 /kg 2 a 3 T 2 Example 13.5 A Geosynchonous Satellite Inteactive Conside a satellite of mass m moving in a cicula obit aound the Eath at a constant speed v and at an altitude h above the Eath s suface, as illustated in Figue (A) Detemine the speed of the satellite in tems of G, h, R E (the adius of the Eath), and M E (the mass of the Eath). Solution Conceptualize by imagining the satellite moving aound the Eath in a cicula obit unde the influence of the gavitational foce. The satellite must have a centipetal acceleation. Thus, we categoize this poblem as one involving Newton s second law, the law of univesal gavitation, and cicula motion. To analyze the poblem, note that the only extenal foce acting on the satellite is the gavitational foce, which acts towad the cente of the Eath and keeps the satellite in its cicula obit. Theefoe, the net foce on the satellite is the gavitational foce F F g G M Em 2 Fom Newton s second law and the fact that the acceleation of the satellite is centipetal, we obtain G M Em 2 m v 2

13 SECTION 13.5 The Gavitational Field 401 Solving fo v and emembeing that the distance fom the cente of the Eath to the satellite is R E h, we obtain (1) v GM E GM E R E h h (B) If the satellite is to be geosynchonous (that is, appeaing to emain ove a fixed position on the Eath), how fast is it moving though space? Solution In ode to appea to emain ove a fixed position on the Eath, the peiod of the satellite must be 24 h and the satellite must be in obit diectly ove the equato. Fom Keple s thid law (Equation 13.8) with a and M S : M E, we find the adius of the obit: Substituting numeical values and noting that the peiod is T 24 h s, we find 3 ( N m 2 /kg 2 )( kg)( s) m To find the speed of the satellite, we use Equation (1): v GM E m/s T GM E 3 3 GM E T ( Nm 2 /kg 2 )( kg) m To finalize this poblem, it is inteesting to note that the value of calculated hee tanslates to a height of the satel- R E m Figue 13.9 (Example 13.5) A satellite of mass m moving aound the Eath in a cicula obit of adius with constant speed v. The only foce acting on the satellite is the gavitational foce F g. (Not dawn to scale.) lite above the suface of the Eath of almost km. Thus, geosynchonous satellites have the advantage of allowing an eathbound antenna to be aimed in a fixed diection, but thee is a disadvantage in that the signals between Eath and the satellite must tavel a long distance. It is difficult to use geosynchonous satellites fo optical obsevation of the Eath s suface because of thei high altitude. What If? What if the satellite motion in pat (A) wee taking place at height h above the suface of anothe planet moe massive than the Eath but of the same adius? Would the satellite be moving at a highe o a lowe speed than it does aound the Eath? Answe If the planet pulls downwad on the satellite with moe gavitational foce due to its lage mass, the satellite would have to move with a highe speed to avoid moving towad the suface. This is consistent with the pedictions of Equation (1), which shows that because the speed v is popotional to the squae oot of the mass of the planet, as the mass inceases, the speed also inceases. F g v You can adjust the altitude of the satellite at the Inteactive Woked Example link at The Gavitational Field When Newton published his theoy of univesal gavitation, it was consideed a success because it satisfactoily explained the motion of the planets. Since 1687 the same theoy has been used to account fo the motions of comets, the deflection of a Cavendish balance, the obits of binay stas, and the otation of galaxies. Nevetheless, both Newton s contempoaies and his successos found it difficult to accept the concept of a foce that acts at a distance, as mentioned in Section 5.1. They asked how it was possible fo two objects to inteact when they wee not in contact with each othe. Newton himself could not answe that question. An appoach to descibing inteactions between objects that ae not in contact came well afte Newton s death, and it enables us to look at the gavitational inteaction in a diffeent way, using the concept of a gavitational field that exists at evey point in space. When a paticle of mass m is placed at a point whee the gavitational

14 402 CHAPTER 13 Univesal Gavitation (a) Figue (a) The gavitational field vectos in the vicinity of a unifom spheical mass such as the Eath vay in both diection and magnitude. The vectos point in the diection of the acceleation a paticle would expeience if it wee placed in the field. The magnitude of the field vecto at any location is the magnitude of the fee-fall acceleation at that location. (b) The gavitational field vectos in a small egion nea the Eath s suface ae unifom in both diection and magnitude. (b) field is g, the paticle expeiences a foce F g mg. In othe wods, the field exets a foce on the paticle. The gavitational field g is defined as Gavitational field g Fg m (13.9) That is, the gavitational field at a point in space equals the gavitational foce expeienced by a test paticle placed at that point divided by the mass of the test paticle. Notice that the pesence of the test paticle is not necessay fo the field to exist the Eath ceates the gavitational field. We call the object ceating the field the souce paticle. (Although the Eath is clealy not a paticle, it is possible to show that we can appoximate the Eath as a paticle fo the pupose of finding the gavitational field that it ceates.) We can detect the pesence of the field and measue its stength by placing a test paticle in the field and noting the foce exeted on it. Although the gavitational foce is inheently an inteaction between two objects, the concept of a gavitational field allows us to facto out the mass of one of the objects. In essence, we ae descibing the effect that any object (in this case, the Eath) has on the empty space aound itself in tems of the foce that would be pesent if a second object wee somewhee in that space. 5 As an example of how the field concept woks, conside an object of mass m nea the Eath s suface. Because the gavitational foce acting on the object has a magnitude GM E m/ 2 (see Eq. 13.4), the field g at a distance fom the cente of the Eath is g Fg m GM E 2 ˆ (13.10) whee ˆ is a unit vecto pointing adially outwad fom the Eath and the negative sign indicates that the field points towad the cente of the Eath, as illustated in Figue 13.10a. Note that the field vectos at diffeent points suounding the Eath vay in both diection and magnitude. In a small egion nea the Eath s suface, the downwad field g is appoximately constant and unifom, as indicated in Figue 13.10b. Equation is valid at all points outside the Eath s suface, assuming that the Eath is spheical. At the Eath s suface, whee R E, g has a magnitude of 9.80 N/kg. (The unit N/kg is the same as m/s 2.) 5 We shall etun to this idea of mass affecting the space aound it when we discuss Einstein s theoy of gavitation in Chapte 39.

15 SECTION 13.6 Gavitational Potential Enegy Gavitational Potential Enegy In Chapte 8 we intoduced the concept of gavitational potential enegy, which is the enegy associated with the configuation of a system of objects inteacting via the gavitational foce. We emphasized that the gavitational potential-enegy function mgy fo a paticle Eath system is valid only when the paticle is nea the Eath s suface, whee the gavitational foce is constant. Because the gavitational foce between two paticles vaies as 1/ 2, we expect that a moe geneal potential-enegy function one that is valid without the estiction of having to be nea the Eath s suface will be significantly diffeent fom U mgy. Befoe we calculate this geneal fom fo the gavitational potential enegy function, let us fist veify that the gavitational foce is consevative. (Recall fom Section 8.3 that a foce is consevative if the wok it does on an object moving between any two points is independent of the path taken by the object.) To do this, we fist note that the gavitational foce is a cental foce. By definition, a cental foce is any foce that is diected along a adial line to a fixed cente and has a magnitude that depends only on the adial coodinate. Hence, a cental foce can be epesented by F()ˆ whee ˆ is a unit vecto diected fom the oigin towad the paticle, as shown in Figue Conside a cental foce acting on a paticle moving along the geneal path to in Figue The path fom to can be appoximated by a seies of steps accoding to the following pocedue. In Figue 13.11, we daw seveal thin wedges, which ae shown as dashed lines. The oute bounday of ou set of wedges is a path consisting of shot adial line segments and acs (gay in the figue). We select the length of the adial dimension of each wedge such that the shot ac at the wedge s wide end intesects the actual path of the paticle. Then we can appoximate the actual path with a seies of zigzag movements that altenate between moving along an ac and moving along a adial line. By definition, a cental foce is always diected along one of the adial segments; theefoe, the wok done by F along any adial segment is dw F d F() d By definition, the wok done by a foce that is pependicula to the displacement is zeo. Hence, the wok done in moving along any ac is zeo because F is pependicula to the displacement along these segments. Theefoe, the total wok done by F is the sum of the contibutions along the adial segments: W f i F() d whee the subscipts i and f efe to the initial and final positions. Because the integand is a function only of the adial position, this integal depends only on the initial and final values of. Thus, the wok done is the same ove any path fom to. Because the wok done is independent of the path and depends only on the end points, we conclude that any cental foce is consevative. We ae now assued that a potential enegy function can be obtained once the fom of the cental foce is specified. Recall fom Equation 8.15 that the change in the gavitational potential enegy of a system associated with a given displacement of a membe of the system is defined as the negative of the wok done by the gavitational foce on that membe duing the displacement: U U f U i f F() d (13.11) We can use this esult to evaluate the gavitational potential enegy function. Conside a paticle of mass m moving between two points and above the Eath s suface (Fig ). The paticle is subject to the gavitational foce given by Equation We can expess this foce as i F() GM Em 2 O ˆ i ˆ F f Radial segment Figue A paticle moves fom to while acted on by a cental foce F, which is diected adially. The path is boken into a seies of adial segments and acs. Because the wok done along the acs is zeo, the wok done is independent of the path and depends only on f and i. Wok done by a cental foce R E M E i F g F g Ac Figue As a paticle of mass m moves fom to above the Eath s suface, the gavitational potential enegy changes accoding to Equation f m

16 404 CHAPTER 13 Univesal Gavitation whee the negative sign indicates that the foce is attactive. Substituting this expession fo F() into Equation 13.11, we can compute the change in the gavitational potential enegy function: Gavitational potential enegy of the Eath paticle system fo R E M E Eath U f U i GM E m f U f U i GM E m 1 f 1 i i (13.12) As always, the choice of a efeence configuation fo the potential enegy is completely abitay. It is customay to choose the efeence configuation fo zeo potential enegy to be the same as that fo which the foce is zeo. Taking U i 0 at i, we obtain the impotant esult U() GM Em d 2 GM E m 1 f (13.13) This expession applies to the Eath paticle system whee the paticle is sepaated fom the cente of the Eath by a distance, povided that R E. The esult is not valid fo paticles inside the Eath, whee R E. Because of ou choice of U i, the function U is always negative (Fig ). Although Equation was deived fo the paticle Eath system, it can be applied to any two paticles. That is, the gavitational potential enegy associated with any pai of paticles of masses m 1 and m 2 sepaated by a distance is i U U Gm 1m 2 (13.14) O GM E m R E Figue Gaph of the gavitational potential enegy U vesus fo an object above the Eath s suface. The potential enegy goes to zeo as appoaches infinity R E Figue Thee inteacting paticles This expession shows that the gavitational potential enegy fo any pai of paticles vaies as 1/, wheeas the foce between them vaies as 1/ 2. Futhemoe, the potential enegy is negative because the foce is attactive and we have taken the potential enegy as zeo when the paticle sepaation is infinite. Because the foce between the paticles is attactive, we know that an extenal agent must do positive wok to incease the sepaation between them. The wok done by the extenal agent poduces an incease in the potential enegy as the two paticles ae sepaated. That is, U becomes less negative as inceases. When two paticles ae at est and sepaated by a distance, an extenal agent has to supply an enegy at least equal to Gm 1 m 2 / in ode to sepaate the paticles to an infinite distance. It is theefoe convenient to think of the absolute value of the potential enegy as the binding enegy of the system. If the extenal agent supplies an enegy geate than the binding enegy, the excess enegy of the system will be in the fom of kinetic enegy of the paticles when the paticles ae at an infinite sepaation. We can extend this concept to thee o moe paticles. In this case, the total potential enegy of the system is the sum ove all pais of paticles. 6 Each pai contibutes a tem of the fom given by Equation Fo example, if the system contains thee paticles, as in Figue 13.14, we find that U total U 12 U 13 U 23 G m 1m 2 m 1m 3 m 2m (13.15) The absolute value of U total epesents the wok needed to sepaate the paticles by an infinite distance. 6 The fact that potential enegy tems can be added fo all pais of paticles stems fom the expeimental fact that gavitational foces obey the supeposition pinciple. 23

17 SECTION 13.7 Enegy Consideations in Planetay and Satellite Motion 405 Example 13.6 The Change in Potential Enegy A paticle of mass m is displaced though a small vetical distance y nea the Eath s suface. Show that in this situation the geneal expession fo the change in gavitational potential enegy given by Equation educes to the familia elationship U mg y. Solution We can expess Equation in the fom U GM E m 1 1 f i GM Em f i i f If both the initial and final positions of the paticle ae close to the Eath s suface, then f i y and i f R 2 E. (Recall that is measued fom the cente of the Eath.) Theefoe, the change in potential enegy becomes U GM Em R E 2 y mg y whee we have used the fact that g GM E /R 2 E (Eq. 13.5). Keep in mind that the efeence configuation is abitay because it is the change in potential enegy that is meaningful. What If? Suppose you ae pefoming uppe-atmosphee studies and ae asked by you supeviso to find the height in the Eath s atmosphee at which the suface equation U mg y gives a 1.0% eo in the change in the potential enegy. What is this height? Answe Because the suface equation assumes a constant value fo g, it will give a U value that is lage than the value given by the geneal equation, Equation Thus, a 1.0% eo would be descibed by the atio Substituting the expessions fo each of these changes U, we have whee i R E and f R E y. Substituting fo g fom Equation 13.5, we find (GM E /R E 2 )R E (R E y) GM E Thus, U suface U geneal mg y GM E m(y/ i f ) gi f GM E R E y R E y 0.010R E 0.010( m) 1 y R E m 63.7 km 13.7 Enegy Consideations in Planetay and Satellite Motion Conside an object of mass m moving with a speed v in the vicinity of a massive object of mass M, whee M m. The system might be a planet moving aound the Sun, a satellite in obit aound the Eath, o a comet making a one-time flyby of the Sun. If we assume that the object of mass M is at est in an inetial efeence fame, then the total mechanical enegy E of the two-object system when the objects ae sepaated by a distance is the sum of the kinetic enegy of the object of mass m and the potential enegy of the system, given by Equation 13.14: 7 E K U E 1 2 mv 2 GMm (13.16) 7 You might ecognize that we have ignoed the kinetic enegy of the lage body. To see that this simplification is easonable, conside an object of mass m falling towad the Eath. Because the cente of mass of the object Eath system is effectively stationay, it follows fom consevation of momentum that mv M E v E. Thus, the Eath acquies a kinetic enegy equal to 1 2 M Ev E m 2 M E v 2 m M E K whee K is the kinetic enegy of the object. Because M E m, this esult shows that the kinetic enegy of the Eath is negligible.

18 406 CHAPTER 13 Univesal Gavitation Figue An object of mass m moving in a cicula obit about a much lage object of mass M. v M m This equation shows that E may be positive, negative, o zeo, depending on the value of v. Howeve, fo a bound system, 8 such as the Eath Sun system, E is necessaily less than zeo because we have chosen the convention that U : 0 as :. We can easily establish that E 0 fo the system consisting of an object of mass m moving in a cicula obit about an object of mass M m (Fig ). Newton s second law applied to the object of mass m gives GMm 2 ma mv 2 Multiplying both sides by and dividing by 2 gives 1 2 mv 2 GMm 2 Substituting this into Equation 13.16, we obtain E GMm 2 GMm (13.17) Total enegy fo cicula obits E GMm 2 (cicula obits) (13.18) This esult clealy shows that the total mechanical enegy is negative in the case of cicula obits. Note that the kinetic enegy is positive and equal to half the absolute value of the potential enegy. The absolute value of E is also equal to the binding enegy of the system, because this amount of enegy must be povided to the system to move the two objects infinitely fa apat. The total mechanical enegy is also negative in the case of elliptical obits. The expession fo E fo elliptical obits is the same as Equation with eplaced by the semimajo axis length a: Total enegy fo elliptical obits E GMm 2a (elliptical obits) (13.19) Futhemoe, the total enegy is constant if we assume that the system is isolated. Theefoe, as the object of mass m moves fom to in Figue 13.12, the total enegy emains constant and Equation gives E 1 2 mv i 2 GMm i 1 2 mv f 2 GMm f (13.20) Combining this statement of enegy consevation with ou ealie discussion of consevation of angula momentum, we see that both the total enegy and the total angula momentum of a gavitationally bound, two-object system ae constants of the motion. Quick Quiz 13.7 A comet moves in an elliptical obit aound the Sun. Which point in its obit (peihelion o aphelion) epesents the highest value of (a) the speed of the comet (b) the potential enegy of the comet Sun system (c) the kinetic enegy of the comet (d) the total enegy of the comet Sun system? 8 Of the thee examples povided at the beginning of this section, the planet moving aound the Sun and a satellite in obit aound the Eath ae bound systems the Eath will always stay nea the Sun, and the satellite will always stay nea the Eath. The one-time comet flyby epesents an unbound system the comet inteacts once with the Sun but is not bound to it. Thus, in theoy the comet can move infinitely fa away fom the Sun.

19 SECTION 13.7 Enegy Consideations in Planetay and Satellite Motion 407 Example 13.7 Changing the Obit of a Satellite The space shuttle eleases a 470-kg communications satellite while in an obit 280 km above the suface of the Eath. A ocket engine on the satellite boosts it into a geosynchonous obit, which is an obit in which the satellite stays diectly ove a single location on the Eath. How much enegy does the engine have to povide? Solution We fist detemine the initial adius (not the altitude above the Eath s suface) of the satellite s obit when it is still in the shuttle s cago bay. This is simply R E 280 km m i In Example 13.5, we found that the adius of the obit of a geosynchonous satellite is f m. Applying Equation 13.18, we obtain, fo the total initial and final enegies, E i GM Em 2 i The enegy equied fom the engine to boost the satellite is E E f E i GM Em 2 1 f 1 i E f GM Em 2 f ( Nm 2 /kg 2 )( kg)(470 kg) m m 6 This is the enegy equivalent of 89 gal of gasoline. NASA enginees must account fo the changing mass of the spacecaft as it ejects buned fuel, something we have not done hee. Would you expect the calculation that includes the effect of this changing mass to yield a geate o lesse amount of enegy equied fom the engine? If we wish to detemine how the enegy is distibuted afte the engine is fied, we find fom Equation that the change in kinetic enegy is K (GM E m/2)(1/ f 1/ i ) J (a decease), and the coesponding change in potential enegy is U GM E m(1/ f 1/ i ) J (an incease). Thus, the change in obital enegy of the system is E K U J, as we aleady calculated. The fiing of the engine esults in a tansfomation of chemical potential enegy in the fuel to obital enegy of the system. Because an incease in gavitational potential enegy is accompanied by a decease in kinetic enegy, we conclude that the speed of an obiting satellite deceases as its altitude inceases J Escape Speed Suppose an object of mass m is pojected vetically upwad fom the Eath s suface with an initial speed v i, as illustated in Figue We can use enegy consideations to find the minimum value of the initial speed needed to allow the object to move infinitely fa away fom the Eath. Equation gives the total enegy of the system at any point. At the suface of the Eath, v v i and i R E. When the object eaches its maximum altitude, v v f 0 and f max. Because the total enegy of the system is constant, substituting these conditions into Equation gives v f = 0 h 1 2 mv i 2 GM Em R E GM Em max v i max Solving fo v i 2 gives v i 2 2GM E 1 R E 1 max (13.21) R E m Theefoe, if the initial speed is known, this expession can be used to calculate the maximum altitude h because we know that M E h max R E We ae now in a position to calculate escape speed, which is the minimum speed the object must have at the Eath s suface in ode to appoach an infinite sepaation distance fom the Eath. Taveling at this minimum speed, the object continues to Figue An object of mass m pojected upwad fom the Eath s suface with an initial speed v i eaches a maximum altitude h.

20 408 CHAPTER 13 Univesal Gavitation PITFALL PREVENTION 13.3 You Can t Really Escape Although Equation povides the escape speed fom the Eath, complete escape fom the Eath s gavitational influence is impossible because the gavitational foce is of infinite ange. No matte how fa away you ae, you will always feel some gavitational foce due to the Eath. move fathe and fathe away fom the Eath as its speed asymptotically appoaches zeo. Letting max : in Equation and taking v i v esc, we obtain v esc 2GM E R E (13.22) Note that this expession fo v esc is independent of the mass of the object. In othe wods, a spacecaft has the same escape speed as a molecule. Futhemoe, the esult is independent of the diection of the velocity and ignoes ai esistance. If the object is given an initial speed equal to v esc, the total enegy of the system is equal to zeo. This can be seen by noting that when :, the object s kinetic enegy and the potential enegy of the system ae both zeo. If v i is geate than v esc, the total enegy of the system is geate than zeo and the object has some esidual kinetic enegy as :. Example 13.8 Escape Speed of a Rocket Calculate the escape speed fom the Eath fo a kg spacecaft, and detemine the kinetic enegy it must have at the Eath s suface in ode to move infinitely fa away fom the Eath. Solution Using Equation gives v esc 2GM E R E 2( Nm 2 /kg 2 )( kg) m This is equivalent to about gal of gasoline. What If? What if we wish to launch a kg spacecaft at the escape speed? How much enegy does this equie? Answe In Equation 13.22, the mass of the object moving with the escape speed does not appea. Thus, the escape speed fo the kg spacecaft is the same as that fo the kg spacecaft. The only change in the kinetic enegy is due to the mass, so the kg spacecaft will equie one fifth of the enegy of the kg spacecaft: m/s K 1 5 ( J) J This coesponds to about mi/h. The kinetic enegy of the spacecaft is K 1 2 mv 2 esc 1 2 ( kg)( m/s) J Table 13.3 Escape Speeds fom the Sufaces of the Planets, Moon, and Sun Planet v esc (km/s) Mecuy 4.3 Venus 10.3 Eath 11.2 Mas 5.0 Jupite 60 Satun 36 Uanus 22 Neptune 24 Pluto 1.1 Moon 2.3 Sun 618 Equations and can be applied to objects pojected fom any planet. That is, in geneal, the escape speed fom the suface of any planet of mass M and adius R is v esc 2GM R (13.23) Escape speeds fo the planets, the Moon, and the Sun ae povided in Table Note that the values vay fom 1.1 km/s fo Pluto to about 618 km/s fo the Sun. These esults, togethe with some ideas fom the kinetic theoy of gases (see Chapte 21), explain why some planets have atmosphees and othes do not. As we shall see late, at a given tempeatue the aveage kinetic enegy of a gas molecule depends only on the mass of the molecule. Lighte molecules, such as hydogen and helium, have a highe aveage speed than heavie molecules at the same tempeatue. When the aveage speed of the lighte molecules is not much less than the escape speed of a planet, a significant faction of them have a chance to escape. This mechanism also explains why the Eath does not etain hydogen molecules and helium atoms in its atmosphee but does etain heavie molecules, such as oxygen and nitogen. On the othe hand, the vey lage escape speed fo Jupite enables that planet to etain hydogen, the pimay constituent of its atmosphee.

21 SECTION 13.7 Enegy Consideations in Planetay and Satellite Motion 409 Black Holes In Example 11.7 we biefly descibed a ae event called a supenova the catastophic explosion of a vey massive sta. The mateial that emains in the cental coe of such an object continues to collapse, and the coe s ultimate fate depends on its mass. If the coe has a mass less than 1.4 times the mass of ou Sun, it gadually cools down and ends its life as a white dwaf sta. Howeve, if the coe s mass is geate than this, it may collapse futhe due to gavitational foces. What emains is a neuton sta, discussed in Example 11.7, in which the mass of a sta is compessed to a adius of about 10 km. (On Eath, a teaspoon of this mateial would weigh about 5 billion tons!) An even moe unusual sta death may occu when the coe has a mass geate than about thee sola masses. The collapse may continue until the sta becomes a vey small object in space, commonly efeed to as a black hole. In effect, black holes ae emains of stas that have collapsed unde thei own gavitational foce. If an object such as a spacecaft comes close to a black hole, it expeiences an extemely stong gavitational foce and is tapped foeve. The escape speed fo a black hole is vey high, due to the concentation of the mass of the sta into a sphee of vey small adius (see Eq ). If the escape speed exceeds the speed of light c, adiation fom the object (such as visible light) cannot escape, and the object appeas to be black; hence the oigin of the teminology black hole. The citical adius R S at which the escape speed is c is called the Schwazschild adius (Fig ). The imaginay suface of a sphee of this adius suounding the black hole is called the event hoizon. This is the limit of how close you can appoach the black hole and hope to escape. Although light fom a black hole cannot escape, light fom events taking place nea the black hole should be visible. Fo example, it is possible fo a binay sta system to consist of one nomal sta and one black hole. Mateial suounding the odinay sta can be pulled into the black hole, foming an accetion disk aound the black hole, as suggested in Figue Fiction among paticles in the accetion disk esults in tansfomation of mechanical enegy into intenal enegy. As a esult, the obital height of the mateial above the event hoizon deceases and the tempeatue ises. This hightempeatue mateial emits a lage amount of adiation, extending well into the x-ay egion of the electomagnetic spectum. These x-ays ae chaacteistic of a black hole. Seveal possible candidates fo black holes have been identified by obsevation of these x-ays. R S Black hole Event hoizon Figue A black hole. The distance R S equals the Schwazschild adius. Any event occuing within the bounday of adius R S, called the event hoizon, is invisible to an outside obseve. Figue A binay sta system consisting of an odinay sta on the left and a black hole on the ight. Matte pulled fom the odinay sta foms an accetion disk aound the black hole, in which matte is aised to vey high tempeatues, esulting in the emission of x-ays.

22 410 CHAPTER 13 Univesal Gavitation H. Fod et al. & NASA Figue Hubble Space Telescope images of the galaxy M87. The inset shows the cente of the galaxy. The wide view shows a jet of mateial moving away fom the cente of the galaxy towad the uppe ight of the figue at about one tenth of the speed of light. Such jets ae believed to be evidence of a supemassive black hole at the galaxy cente. Thee is also evidence that supemassive black holes exist at the centes of galaxies, with masses vey much lage than the Sun. (Thee is stong evidence of a supemassive black hole of mass 2 3 million sola masses at the cente of ou galaxy.) Theoetical models fo these bizae objects pedict that jets of mateial should be evident along the otation axis of the black hole. Figue shows a Hubble Space Telescope photogaph of galaxy M87. The jet of mateial coming fom this galaxy is believed to be evidence fo a supemassive black hole at the cente of the galaxy. SUMMARY Take a pactice test fo this chapte by clicking on the Pactice Test link at Newton s law of univesal gavitation states that the gavitational foce of attaction between any two paticles of masses m 1 and m 2 sepaated by a distance has the magnitude F g G m 1m 2 2 (13.1) whee G N m 2 /kg 2 is the univesal gavitational constant. This equation enables us to calculate the foce of attaction between masses unde a wide vaiety of cicumstances. An object at a distance h above the Eath s suface expeiences a gavitational foce of magnitude mg, whee g is the fee-fall acceleation at that elevation: g GM E 2 GM E (R E h) 2 (13.6) In this expession, M E is the mass of the Eath and R E is its adius. Thus, the weight of an object deceases as the object moves away fom the Eath s suface.

23 Questions 411 Keple s laws of planetay motion state that 1. All planets move in elliptical obits with the Sun at one focus. 2. The adius vecto dawn fom the Sun to a planet sweeps out equal aeas in equal time intevals. 3. The squae of the obital peiod of any planet is popotional to the cube of the semimajo axis of the elliptical obit. Keple s thid law can be expessed as T GM S a 3 (13.8) whee M S is the mass of the Sun and a is the semimajo axis. Fo a cicula obit, a can be eplaced in Equation 13.8 by the adius. Most planets have nealy cicula obits aound the Sun. The gavitational field at a point in space is defined as the gavitational foce expeienced by any test paticle located at that point divided by the mass of the test paticle: g Fg m (13.9) The gavitational foce is consevative, and theefoe a potential enegy function can be defined fo a system of two objects inteacting gavitationally. The gavitational potential enegy associated with two paticles sepaated by a distance is U Gm 1m 2 (13.14) whee U is taken to be zeo as :. The total potential enegy fo a system of paticles is the sum of enegies fo all pais of paticles, with each pai epesented by a tem of the fom given by Equation If an isolated system consists of an object of mass m moving with a speed v in the vicinity of a massive object of mass M, the total enegy E of the system is the sum of the kinetic and potential enegies: E 1 2 mv 2 GMm (13.16) The total enegy is a constant of the motion. If the object moves in an elliptical obit of semimajo axis a aound the massive object and if M m, the total enegy of the system is E GMm 2a (13.19) Fo a cicula obit, this same equation applies with a. The total enegy is negative fo any bound system. The escape speed fo an object pojected fom the suface of a planet of mass M and adius R is v esc 2GM R (13.23) QUESTIONS 1. If the gavitational foce on an object is diectly popotional to its mass, why don t objects with lage masses fall with geate acceleation than small ones? 2. The gavitational foce exeted by the Sun on you is downwad into the Eath at night, and upwad into the sky duing the day. If you had a sensitive enough bathoom scale,

24 412 CHAPTER 13 Univesal Gavitation would you expect to weigh moe at night than duing the day? Note also that you ae fathe away fom the Sun at night than duing the day. Would you expect to weigh less? 3. Use Keple s second law to convince youself that the Eath must move faste in its obit duing Decembe, when it is closest to the Sun, than duing June, when it is fathest fom the Sun. 4. The gavitational foce that the Sun exets on the Moon is about twice as geat as the gavitational foce that the Eath exets on the Moon. Why doesn t the Sun pull the Moon away fom the Eath duing a total eclipse of the Sun? 5. A satellite in obit is not tuly taveling though a vacuum. It is moving though vey, vey thin ai. Does the esulting ai fiction cause the satellite to slow down? 6. How would you explain the fact that Jupite and Satun have peiods much geate than one yea? 7. If a system consists of five paticles, how many tems appea in the expession fo the total potential enegy? How many tems appea if the system consists of N paticles? 8. Does the escape speed of a ocket depend on its mass? Explain. 9. Compae the enegies equied to each the Moon fo a kg spacecaft and a kg satellite. 10. Explain why it takes moe fuel fo a spacecaft to tavel fom the Eath to the Moon than fo the etun tip. Estimate the diffeence. 11. A paticula set of diections foms the celestial equato. If you live at 40 noth latitude, these diections lie in an ac acoss you southen sky, including hoizontally east, hoizontally west, and south at 50 above the hoizontal. In ode to enjoy satellite TV, you need to install a dish with an unobstucted view to a paticula point on the celestial equato. Why is this equiement so specific? 12. Why don t we put a geosynchonous weathe satellite in obit aound the 45th paallel? Wouldn t this be moe useful in the United States than one in obit aound the equato? 13. Is the absolute value of the potential enegy associated with the Eath Moon system geate than, less than, o equal to the kinetic enegy of the Moon elative to the Eath? 14. Explain why no wok is done on a planet as it moves in a cicula obit aound the Sun, even though a gavitational foce is acting on the planet. What is the net wok done on a planet duing each evolution as it moves aound the Sun in an elliptical obit? 15. Explain why the foce exeted on a paticle by a unifom sphee must be diected towad the cente of the sphee. Would this be the case if the mass distibution of the sphee wee not spheically symmetic? 16. At what position in its elliptical obit is the speed of a planet a maximum? At what position is the speed a minimum? 17. If you ae given the mass and adius of planet X, how would you calculate the fee-fall acceleation on the suface of this planet? 18. If a hole could be dug to the cente of the Eath, would the foce on an object of mass m still obey Equation 13.1 thee? What do you think the foce on m would be at the cente of the Eath? 19. In his 1798 expeiment, Cavendish was said to have weighed the Eath. Explain this statement. 20. The Voyage spacecaft was acceleated towad escape speed fom the Sun by Jupite s gavitational foce exeted on the spacecaft. How is this possible? 21. How would you find the mass of the Moon? 22. The Apollo 13 spacecaft developed touble in the oxygen system about halfway to the Moon. Why did the mission continue on aound the Moon, and then etun home, athe than immediately tun back to Eath? PROBLEMS 1, 2, 3 = staightfowad, intemediate, challenging = full solution available in the Student Solutions Manual and Study Guide = coached solution with hints available at = compute useful in solving poblem = paied numeical and symbolic poblems Section 13.1 Newton s Law of Univesal Gavitation Poblem 17 in Chapte 1 can also be assigned with this section. 1. Detemine the ode of magnitude of the gavitational foce that you exet on anothe peson 2 m away. In you solution state the quantities you measue o estimate and thei values. 2. Two ocean lines, each with a mass of metic tons, ae moving on paallel couses, 100 m apat. What is the magnitude of the acceleation of one of the lines towad the othe due to thei mutual gavitational attaction? Teat the ships as paticles. 3. A 200-kg object and a 500-kg object ae sepaated by m. (a) Find the net gavitational foce exeted by these objects on a 50.0-kg object placed midway between them. (b) At what position (othe than an infinitely emote one) can the 50.0-kg object be placed so as to expeience a net foce of zeo? 4. Two objects attact each othe with a gavitational foce of magnitude N when sepaated by 20.0 cm. If the total mass of the two objects is 5.00 kg, what is the mass of each?

25 Poblems Thee unifom sphees of mass 2.00 kg, 4.00 kg, and 6.00 kg ae placed at the cones of a ight tiangle as in Figue P13.5. Calculate the esultant gavitational foce on the 4.00-kg object, assuming the sphees ae isolated fom the est of the Univese. ( 4.00, 0) m 6.00 kg (0, 3.00) m F 64 O Figue P13.5 y 2.00 kg F 24 x 4.00 kg Section 13.4 Keple s Laws and the Motion of Planets 12. The cente-to-cente distance between Eath and Moon is km. The Moon completes an obit in 27.3 days. (a) Detemine the Moon s obital speed. (b) If gavity wee switched off, the Moon would move along a staight line tangent to its obit, as descibed by Newton s fist law. In its actual obit in 1.00 s, how fa does the Moon fall below the tangent line and towad the Eath? 13. Plaskett s binay system consists of two stas that evolve in a cicula obit about a cente of mass midway between them. This means that the masses of the two stas ae equal (Fig. P13.13). Assume the obital speed of each sta is 220 km/s and the obital peiod of each is 14.4 days. Find the mass M of each sta. (Fo compaison, the mass of ou Sun is kg.) 220 km/s M 6. Duing a sola eclipse, the Moon, Eath, and Sun all lie on the same line, with the Moon between the Eath and the Sun. (a) What foce is exeted by the Sun on the Moon? (b) What foce is exeted by the Eath on the Moon? (c) What foce is exeted by the Sun on the Eath? CM Section Measuing the Gavitational Constant In intoductoy physics laboatoies, a typical Cavendish balance fo measuing the gavitational constant G uses lead sphees with masses of 1.50 kg and 15.0 g whose centes ae sepaated by about 4.50 cm. Calculate the gavitational foce between these sphees, teating each as a paticle located at the cente of the sphee. 8. A student poposes to measue the gavitational constant G by suspending two spheical objects fom the ceiling of a tall cathedal and measuing the deflection of the cables fom the vetical. Daw a fee-body diagam of one of the objects. If two kg objects ae suspended at the lowe ends of cables m long and the cables ae attached to the ceiling m apat, what is the sepaation of the objects? Section Fee-Fall Acceleation and the Gavitational Foce When a falling meteooid is at a distance above the Eath s suface of 3.00 times the Eath s adius, what is its acceleation due to the Eath s gavitation? The fee-fall acceleation on the suface of the Moon is about one sixth of that on the suface of the Eath. If the adius of the Moon is about 0.250R E, find the atio of thei aveage densities, Moon / Eath. 11. On the way to the Moon the Apollo astonauts eached a point whee the Moon s gavitational pull became stonge than the Eath s. (a) Detemine the distance of this point fom the cente of the Eath. (b) What is the acceleation due to the Eath s gavitation at this point? 14. A paticle of mass m moves along a staight line with constant speed in the x diection, a distance b fom the x axis (Fig. P13.14). Show that Keple s second law is satisfied by showing that the two shaded tiangles in the figue have the same aea when t 4 t 3 t 2 t m v 0 b O M y 220 km/s Figue P13.13 t 1 t 2 t 3 t 4 Figue P13.14 Io, a moon of Jupite, has an obital peiod of 1.77 days and an obital adius of km. Fom these data, detemine the mass of Jupite. 16. The Exploe VIII satellite, placed into obit Novembe 3, 1960, to investigate the ionosphee, had the following obit paametes: peigee, 459 km; apogee, km (both distances above the Eath s suface); peiod, min. Find the atio v p /v a of the speed at peigee to that at apogee. x

26 414 CHAPTER 13 Univesal Gavitation 17. Comet Halley (Figue P13.17) appoaches the Sun to within AU, and its obital peiod is 75.6 yeas. (AU is the symbol fo astonomical unit, whee 1AU m is the mean Eath Sun distance.) How fa fom the Sun will Halley s comet tavel befoe it stats its etun jouney? 21. Suppose the Sun s gavity wee switched off. The planets would leave thei nealy cicula obits and fly away in staight lines, as descibed by Newton s fist law. Would Mecuy eve be fathe fom the Sun than Pluto? If so, find how long it would take fo Mecuy to achieve this passage. If not, give a convincing agument that Pluto is always fathe fom the Sun. 22. As themonuclea fusion poceeds in its coe, the Sun loses mass at a ate of kg/s. Duing the y peiod of ecoded histoy, by how much has the length of the yea changed due to the loss of mass fom the Sun? Suggestions: Assume the Eath s obit is cicula. No extenal toque acts on the Eath Sun system, so its angula momentum is conseved. If x is small compaed to 1, then (1 x) n is nealy equal to 1 nx AU Sun x Section 13.5 The Gavitational Field 23. Thee objects of equal mass ae located at thee cones of a squae of edge length as in Figue P Find the gavitational field at the fouth cone due to these objects. 2a y Figue P13.17 m m 18. Two planets X and Y tavel counteclockwise in cicula obits about a sta as in Figue P The adii of thei obits ae in the atio 3:1. At some time, they ae aligned as in Figue P13.18a, making a staight line with the sta. Duing the next five yeas, the angula displacement of planet X is 90.0, as in Figue P13.18b. Whee is planet Y at this time? O Figue P13.23 m x (a) Y X Figue P13.18 Y X (b) 24. A spacecaft in the shape of a long cylinde has a length of 100 m, and its mass with occupants is kg. It has stayed too close to a black hole having a mass 100 times that of the Sun (Fig. P13.24). The nose of the spacecaft points towad the black hole, and the distance between the nose and the cente of the black hole is 10.0 km. (a) Detemine the total foce on the spacecaft. (b) What is the diffeence in the gavitational fields acting on the occupants in the nose of the ship and on those in the ea of the ship, fathest fom the black hole? This diffeence in acceleations gows apidly as the ship appoaches the black hole. It puts the body of the ship unde exteme tension and eventually teas it apat. 19. A synchonous satellite, which always emains above the same point on a planet s equato, is put in obit aound Jupite to study the famous ed spot. Jupite otates about its axis once evey 9.84 h. Use the data of Table 13.2 to find the altitude of the satellite. 20. Neuton stas ae extemely dense objects that ae fomed fom the emnants of supenova explosions. Many otate vey apidly. Suppose that the mass of a cetain spheical neuton sta is twice the mass of the Sun and its adius is 10.0 km. Detemine the geatest possible angula speed it can have so that the matte at the suface of the sta on its equato is just held in obit by the gavitational foce. 100 m Figue P km Black hole 25. Compute the magnitude and diection of the gavitational field at a point P on the pependicula bisecto of the line joining two objects of equal mass sepaated by a distance 2a as shown in Figue P13.25.

27 Poblems 415 Section 13.6 Assume U 0 at. a M M Figue P13.25 Gavitational Potential Enegy 26. A satellite of the Eath has a mass of 100 kg and is at an altitude of m. (a) What is the potential enegy of the satelliteeath system? (b) What is the magnitude of the gavitational foce exeted by the Eath on the satellite? (c) What If? What foce does the satellite exet on the Eath? 27. How much enegy is equied to move a kg object fom the Eath s suface to an altitude twice the Eath s adius? 28. At the Eath s suface a pojectile is launched staight up at a speed of 10.0 km/s. To what height will it ise? Ignoe ai esistance and the otation of the Eath. 29. Afte ou Sun exhausts its nuclea fuel, its ultimate fate may be to collapse to a white dwaf state, in which it has appoximately the same mass as it has now, but a adius equal to the adius of the Eath. Calculate (a) the aveage density of the P white dwaf, (b) the fee-fall acceleation, and (c) the gavitational potential enegy of a 1.00-kg object at its suface. 30. How much wok is done by the Moon s gavitational field as a kg meteo comes in fom oute space and impacts on the Moon s suface? 31. A system consists of thee paticles, each of mass 5.00 g, located at the cones of an equilateal tiangle with sides of 30.0 cm. (a) Calculate the potential enegy of the system. (b) If the paticles ae eleased simultaneously, whee will they collide? 32. An object is eleased fom est at an altitude h above the suface of the Eath. (a) Show that its speed at a distance fom the Eath s cente, whee R E R E h, is given by (b) Assume the elease altitude is 500 km. Pefom the integal to find the time of fall as the object moves fom the elease point to the Eath s suface. The negative sign appeas because the object is moving opposite to the adial diection, so its speed is v d/dt. Pefom the integal numeically. Section 13.7 v 2GM E 1 1 R E h t f dt f i Enegy Consideations in Planetay and Satellite Motion 33. A space pobe is fied as a pojectile fom the Eath s suface with an initial speed of m/s. What will its speed be when it is vey fa fom the Eath? Ignoe fiction and the otation of the Eath. i d v By pemission of John Hat and Ceatos Syndicate, Inc. Figue P13.35

28 416 CHAPTER 13 Univesal Gavitation 34. (a) What is the minimum speed, elative to the Sun, necessay fo a spacecaft to escape the sola system if it stats at the Eath s obit? (b) Voyage 1 achieved a maximum speed of km/h on its way to photogaph Jupite. Beyond what distance fom the Sun is this speed sufficient to escape the sola system? 35. A teetop satellite (Fig. P13.35) moves in a cicula obit just above the suface of a planet, assumed to offe no ai esistance. Show that its obital speed v and the escape speed fom the planet ae elated by the expession v esc 2 v. 36. A 500-kg satellite is in a cicula obit at an altitude of 500 km above the Eath s suface. Because of ai fiction, the satellite eventually falls to the Eath s suface, whee it hits the gound with a speed of 2.00 km/s. How much enegy was tansfomed into intenal enegy by means of fiction? 37. A satellite of mass 200 kg is placed in Eath obit at a height of 200 km above the suface. (a) With a cicula obit, how long does the satellite take to complete one obit? (b) What is the satellite s speed? (c) What is the minimum enegy input necessay to place this satellite in obit? Ignoe ai esistance but include the effect of the planet s daily otation. 38. A satellite of mass m, oiginally on the suface of the Eath, is placed into Eath obit at an altitude h. (a) With a cicula obit, how long does the satellite take to complete one obit? (b) What is the satellite s speed? (c) What is the minimum enegy input necessay to place this satellite in obit? Ignoe ai esistance but include the effect of the planet s daily otation. At what location on the Eath s suface and in what diection should the satellite be launched to minimize the equied enegy investment? Repesent the mass and adius of the Eath as M E and R E. 39. A kg satellite obits the Eath at a constant altitude of 100 km. How much enegy must be added to the system to move the satellite into a cicula obit with altitude 200 km? 40. The planet Uanus has a mass about 14 times the Eath s mass, and its adius is equal to about 3.7 Eath adii. (a) By setting up atios with the coesponding Eath values, find the fee-fall acceleation at the cloud tops of Uanus. (b) Ignoing the otation of the planet, find the minimum escape speed fom Uanus. 41. Detemine the escape speed fo a ocket on the fa side of Ganymede, the lagest of Jupite s moons (Figue P13.41). The adius of Ganymede is m, and its mass is Ganymede v kg. The mass of Jupite is kg, and the distance between Jupite and Ganymede is m. Be sue to include the gavitational effect due to Jupite, but you may ignoe the motion of Jupite and Ganymede as they evolve about thei cente of mass. 42. In Robet Heinlein s The Moon is a Hash Mistess, the colonial inhabitants of the Moon theaten to launch ocks down onto the Eath if they ae not given independence (o at least epesentation). Assuming that a ail gun could launch a ock of mass m at twice the luna escape speed, calculate the speed of the ock as it entes the Eath s atmosphee. (By luna escape speed we mean the speed equied to move infinitely fa away fom a stationay Moon alone in the Univese. Poblem 61 in Chapte 30 descibes a ail gun.) 43. An object is fied vetically upwad fom the suface of the Eath (of adius R E ) with an initial speed v i that is compaable to but less than the escape speed v esc. (a) Show that the object attains a maximum height h given by h R E v 2 i v2 esc v 2 i (b) A space vehicle is launched vetically upwad fom the Eath s suface with an initial speed of 8.76 km/s, which is less than the escape speed of 11.2 km/s. What maximum height does it attain? (c) A meteoite falls towad the Eath. It is essentially at est with espect to the Eath when it is at a height of m. With what speed does the meteoite stike the Eath? (d) What If? Assume that a baseball is tossed up with an initial speed that is vey small compaed to the escape speed. Show that the equation fom pat (a) is consistent with Equation Deive an expession fo the wok equied to move an Eath satellite of mass m fom a cicula obit of adius 2R E to one of adius 3R E. 45. A comet of mass kg moves in an elliptical obit aound the Sun. Its distance fom the Sun anges between AU and 50.0 AU. (a) What is the eccenticity of its obit? (b) What is its peiod? (c) At aphelion what is the potential enegy of the comet Sun system? Note: 1AU one astonomical unit the aveage distance fom Sun to Eath m. 46. A satellite moves aound the Eath in a cicula obit of adius. (a) What is the speed v 0 of the satellite? Suddenly, an explosion beaks the satellite into two pieces, with masses m and 4m. Immediately afte the explosion the smalle piece of mass m is stationay with espect to the Eath and falls diectly towad the Eath. (b) What is the speed v i of the lage piece immediately afte the explosion? (c) Because of the incease in its speed, this lage piece now moves in a new elliptical obit. Find its distance away fom the cente of the Eath when it eaches the othe end of the ellipse. Additional Poblems Jupite Figue P The Sola and Heliospheic Obsevatoy (SOHO) spacecaft has a special obit, chosen so that its view of the Sun is neve eclipsed and it is always close enough to the Eath to

29 Poblems 417 tansmit data easily. It moves in a nea-cicle aound the Sun that is smalle than the Eath s cicula obit. Its peiod, howeve, is just equal to 1 y. It is always located between the Eath and the Sun along the line joining them. Both objects exet gavitational foces on the obsevatoy. Show that its distance fom the Eath must be between m and m. In 1772 Joseph Louis Lagange detemined theoetically the special location allowing this obit. The SOHO spacecaft took this position on Febuay 14, Suggestion: Use data that ae pecise to fou digits. The mass of the Eath is kg. 48. Let g M epesent the diffeence in the gavitational fields poduced by the Moon at the points on the Eath s suface neaest to and fathest fom the Moon. Find the faction g M /g, whee g is the Eath s gavitational field. (This diffeence is esponsible fo the occuence of the luna tides on the Eath.) 49. Review poblem. Two identical had sphees, each of mass m and adius, ae eleased fom est in othewise empty space with thei centes sepaated by the distance R. They ae allowed to collide unde the influence of thei gavitational attaction. (a) Show that the magnitude of the impulse eceived by each sphee befoe they make contact is given by [Gm 3 (1/2 1/R)] 1/2. (b) What If? Find the magnitude of the impulse each eceives if they collide elastically. 50. Two sphees having masses M and 2M and adii R and 3R, espectively, ae eleased fom est when the distance between thei centes is 12R. How fast will each sphee be moving when they collide? Assume that the two sphees inteact only with each othe. 51. In Lay Niven s science-fiction novel Ringwold, a igid ing of mateial otates about a sta (Fig. P13.51). The tangential speed of the ing is m/s, and its adius is m. (a) Show that the centipetal acceleation of the inhabitants is 10.2 m/s 2. (b) The inhabitants of this ing wold live on the stalit inne suface of the ing. Each peson expeiences a nomal contact foce n. Acting alone, this nomal foce would poduce an inwad acceleation of 9.90 m/s 2. Additionally, the sta at the cente of the ing exets a gavitational foce on the ing and its inhabitants. The diffeence between the total acceleation and the acceleation povided by the nomal foce is due to the gavitational attaction of the cental sta. Show that the mass of the sta is appoximately kg. 52. (a) Show that the ate of change of the fee-fall acceleation with distance above the Eath s suface is NASA dg d 2GM E 3 This ate of change ove distance is called a gadient. (b) If h is small in compaison to the adius of the Eath, show that the diffeence in fee-fall acceleation between two points sepaated by vetical distance h is (c) Evaluate this diffeence fo h 6.00 m, a typical height fo a two-stoy building. 53. A ing of matte is a familia stuctue in planetay and stella astonomy. Examples include Satun s ings and a ing nebula. Conside a unifom ing of mass kg and adius m. An object of mass kg is placed at a point A on the axis of the ing, m fom the cente of the ing (Figue P13.53). When the object is eleased, the attaction of the ing makes the object move along the axis towad the cente of the ing (point B). (a) Calculate the gavitational potential enegy of the object ing system when the object is at A. (b) Calculate the gavitational potential enegy of the system when the object is at B. (c) Calculate the speed of the object as it passes though B. B R E g 2GM Eh R 3 E Sta A n Figue P13.51 F g Figue P Voyages 1 and 2 suveyed the suface of Jupite s moon Io and photogaphed active volcanoes spewing liquid sulfu to heights of 70 km above the suface of this moon. Find