Chapte answes Heinemann Physics 4e Section. Woked example: Ty youself.. GRAVITATIONAL ATTRACTION BETWEEN SMALL OBJECTS Two bowling balls ae sitting next to each othe on a shelf so that the centes of the balls ae 60 cm apat. Ball has a mass of 7.0 kg and ball has a mass of 5.5 kg. Calculate the foce of gavitational attaction between them. Recall the fomula fo Newton s law of univesal gavitation. Identify the infomation equied, and convet values into appopiate units when necessay. G mm m 7.0 kg m 5.5 kg 0.60 m G 6.67 0 N m kg Substitute the values into the equation. Solve the equation. 6.67 0 7. 0 9 N 7.0 5.5 0.60 Woked example: Ty youself.. GRAVITATIONAL ATTRACTION BETWEEN LARGE OBJECTS Calculate the foce of gavitational attaction between the Eath and the Moon, given the following data: m Eath 6.0 0 4 kg m Moon 7.3 0 kg Moon Eath 3.8 0 8 m Recall the fomula fo Newton s law of univesal gavitation. Identify the infomation equied. G mm m 6.0 0 4 kg m 7.3 0 kg 3.8 0 8 m G 6.67 0 N m kg Substitute the values into the equation. Solve the equation. 6.67 0.0 0 0 N 6.0 0 7.3 0 8 ( 3.8 0 ) 4 Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8
Heinemann Physics 4e Woked example: Ty youself..3 ACCELERATION CAUSED BY A GRAVITATIONAL FORCE The foce of gavitational attaction between the Sun and the Eath is appoximately 3.6 0 N. Calculate the acceleation of the Eath and the Sun caused by this foce. Compae these acceleations by calculating the atio a a Use the following data: m Eath 6.0 0 4 kg m Sun.0 0 30 kg Eath Sun. Recall the fomula fo Newton s second law of motion. F ma Tanspose the equation to make a the subject. a F m Substitute values into this equation to find the acceleations of the Eath and the Sun. a Eath a Sun 3.6 0 6.0 0 3.6 0.0 0 30 4 6.0 0 3 m s.8 0 8 m s Compae the two acceleations. 3 aeath 6.0 0 8 3.3 0 5 asun.8 0 The acceleation of the Eath is 3.3 0 5 times geate than the acceleation of the Sun. Woked example: Ty youself..4 GRAVITATIONAL FORCE AND WEIGHT Compae the weight of a.0 kg mass on the Eath s suface calculated using the fomulas mg and G mm Use the following dimensions of the Eath whee necessay: g 9.8 m s m Eath 6.0 0 4 kg Eath 6.4 0 6 m. Apply the weight equation. Apply Newton s law of univesal gavitation. mg 9.8 9.8 N G mm 6.67 0 9.77 N 9.8 N 4 6.0 0 6 ( 6.4 0 ) Compae the two values. Both equations give the same esult to two significant figues. Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8
Heinemann Physics 4e Woked example: Ty youself..5 APPARENT WEIGHT Calculate the appaent weight of a 90 kg peson in an elevato which is acceleating downwads at 0.8 m s. Use g 9.8 m s. Calculate the weight of the peson using mg. Calculate the foce equied to acceleate the peson at 0.8 m s. The net foce that causes the acceleation consists of the nomal eaction foce (upwads) and the weight foce (downwads). Since the elevato is acceleating downwads, > F N. Notice that, as the peson is patially falling in the diection of gavitational acceleation, thee is less contact foce and the peson feels lighte than if standing still. mg 90 9.8 88 N F net ma 90 0.8 7 N F net 7 F N 7 88 F N 7 F N 88 7 Appaent weight 80 N. eview The foce of attaction between any two bodies in the univese is diectly popotional to the poduct of thei masses and invesely popotional to the squae of the distance between them. is the distance between the centes of the two objects. 3 G mm 30 3.0 0 6.4 0 6.67 0.8 0 (. 0 ) N 4 m Mas a Mas.8 0 6.4 0 3 a Mas.8 0 a Mas 3 6.4 0.8 0 3 m s 5 a Note: million km 0 6 km 0 9 m G mm 4 3 6.0 0 6.4 0 6.67 0 0 ( 9.3 0 ) 3.0 0 6 N b G mm 30 4.0 0 6.0 0 6.67 0 0 ( 5.3 0 ) 3.4 0 N c % compaison (3.0 0 6 ) (3.4 0 ) 00 0.000088%. The Mas Eath foce was 0.000088% of the Sun Eath foce. 6 The Moon has a smalle mass than the Eath and theefoe expeiences a lage acceleation fom the same gavitational foce. 7 a g G M g 6.67 0 3.5 m s 8 G mm 6.67 0 40 N 3 3.3 0 ( 500 000) 3 6.4 0 65 6 ( 3.4 0 ) Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8
Heinemann Physics 4e 9 On Eath, weight is the gavitational foce acting on an object nea the Eath s suface wheeas appaent weight is the contact foce between the object and the Eath s suface. In many situations, these two foces ae equal in magnitude but ae in opposite diections. This is because appaent weight is a eaction foce to the weight of an object esting on the gound. Howeve, in an elevato acceleating upwads, the appaent weight of an object would be geate than its weight since an additional foce would be equied to cause the object to acceleate upwads. 0 a mg 50 9.8 490 N When acceleating upwads at. m s, the net foce is F net ma 50. 60 N, and F N >. F net F N 60 N F N 60 + 490 550 N. The peson s appaent weight is 550 N. b When the peson is moving at a constant speed, thei appaent weight is equal to thei weight. F N mg 50 9.8 490 N Section. Woked example: Ty youself.. INTERPRETING GRAVITATIONAL FIELD DIAGRAMS The diagam below shows the gavitational field of a planet. A B C (a) Use aows to indicate the diection of the gavitational foce acting at points A, B and C. The diection of the field aows indicates the diection of the gavitational foce, which is inwads towads the cente of the planet. A B C Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8
Heinemann Physics 4e (b) Indicate the elative stength of the gavitational field at each point. The close the field lines, the stonge the foce. A weakest field B stongest field mediumstength field C Woked example: Ty youself.. CALCULATING GRAVITATIONAL FIELD STRENGTH A student uses a sping balance to measue the weight of a piece of wood as.5 N. If the piece of wood is thought to have a mass of 60 g, calculate the gavitational field stength indicated by this expeiment. Recall the equation fo gavitational field stength. Substitute in the appopiate values. g m g.5 0.6 Solve the equation. g 9.6 N kg Woked example: Ty youself..3 CALCULATING GRAVITATIONAL FIELD STRENGTH AT DIFFERENT ALTITUDES Commecial ailines typically fly at an altitude of 000 m. Calculate the gavitational field stength of the Eath at this height using the following data: Eath 6.38 0 6 m m Eath 5.97 0 4 kg Recall the fomula fo gavitational field stength. g G M Add the altitude of the plane to the adius of the Eath. 6.38 0 6 + 000 m 6.39 0 6 m Substitute the values into the fomula. g G M 6.67 0 9.75 N kg 4 5.97 0 ( 6.39 0 ) 6 Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8
Heinemann Physics 4e Woked example: Ty youself..4 GRAVITATIONAL FIELD STRENGTH ON ANOTHER PLANET OR MOON Calculate the stength of the gavitational field on the suface of Mas. m Mas 6.4 0 3 kg Mas 3390 km Give you answe coect to two significant figues. Recall the fomula fo gavitational field stength. g G M Convet Mas adius to m. 3390 km 3.39 0 6 m Substitute values into the fomula. g G M 6.67 0 3.7 N kg 3 6.4 0 ( 3.39 0 ) 6. eview N kg g m.4 9.3 N kg 0.5 3 The distance has been inceased thee times fom 400 km to 00 km so, in tems of the invese squae law, and the oiginal distance, : F ( 3 ) ( 9 ) of the oiginal 9 4 a g G M 6.67 0 5.67 N kg 5.97 0 ( ) 4 ( 6380 + 000 0 ) 3 b g G M 6.67 0.48 N kg 5.97 0 4 ( ) ( 6380 + 0 000 0 ) 3 c g G M 6.67 0 0.56 N kg 5.97 0 4 ( ) ( 6380 + 0 00 0 ) 3 d g G M 6.67 0 0. N kg 5.97 0 4 ( ) ( 6380 + 35 786 0 ) 3 Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8
Heinemann Physics 4e 5 g G M 6.67 0 0 900 0.0008 N kg o 8 0 4 N kg 3 6 g G M 3 3.30 0 Mecuy: g 6.67 0 (.44 0) 6 5.69 0 Satun: g 6.67 0 7 (6.03 0) 7.90 0 Jupite: g 6.67 0 (7.5 0) 7 g G M 6.67 0 3.0 0 30 (0 0) 3 7 6 3.7 N kg 0.4 N kg 4.8 N kg 0 N kg 8 g poles G M 8 6.67 0 M 3 0 4 kg M 5 000 000 g equato G M 4 6.67 3 0 6 000 000 5.6 N kg 8 5.6.4. The gavitational field stength at the poles is.4 times that at the equato. (Altenatively, the invese squae law could also be used to find this elationship.) 9 Let x be the distance fom the cente of the Eath whee the Eath s gavity equals the Moon s gavity. Then: 4 6.67 0 6 0 g Eath x 6.67 0 7.3 0 g Moon 8 ( 3.8 0 x) Equating these two expessions gives: 6 0 4 7.3 0 x 8 ( 3.8 0 x) 8. x 3.8 0 8 x ( ) Taking squae oots of both sides gives: 9.07 x 3.8 0 8 x ( ) Inveting both sides gives: x 9.07 3.8 08 x x 3.45 0 9 9.07x 0.07x 3.45 0 9 x 3.4 0 8 m 0 g is popotional to, so if g becomes 0 times means a distance of 0 Eath adii. 00 th of its value, must become 0 times its value so that becomes 00. Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8
Heinemann Physics 4e Section.3 Woked example: Ty youself.3. WORK DONE FOR A CHANGE IN GRAVITATIONAL POTENTIAL ENERGY Calculate the wok done (in MJ) to lift a weathe satellite of 3. tonnes fom the Eath s suface to the limit of the atmosphee, which ends at the Kaman line (exactly 00 km up fom the suface of the Eath). Assume g 9.8 N kg. Convet the values into the appopiate units. m 3. tonnes 300 kg h 00 km 00 0 3 m Substitute the values into E g mgδh. Remembe to give you answe in MJ to two significant figues. The wok done is equal to the change in gavitational potential enegy. E g mgδh 300 9.8 00 0 3 3.36 0 9 J 3. 0 3 MJ W E 3. 0 3 MJ Woked example: Ty youself.3. SPEED OF A FALLING OBJECT Calculate how fast a 450 g hamme would be going as it hit the gound if it was dopped fom a height of.4 m on Eath, whee g 9.8 N kg. Calculate the gavitational potential enegy of the hamme on Eath. Assume that when the hamme hits the suface of the Eath, all of its gavitational potential enegy has been conveted into kinetic enegy. Use the definition of kinetic enegy to calculate the speed of the hamme as it hits the gound. E g mgδh 0.45 9.8.4 6. J E k E g 6. J E k mv 6. 0.45 v 6. 0.45 v v 5. m s Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8
Heinemann Physics 4e Woked example: Ty youself.3.3 CHANGE IN GRAVITATIONAL POTENTIAL ENERGY USING A FORCE DISTANCE GRAPH A 500 kg lump of space junk is plummeting towads the Moon. The Moon has a adius of.7 0 6 m. Using the foce distance gaph, detemine the decease in gavitational potential enegy of the junk as it falls to the Moon s suface. m 500 kg v 50 m s.7 0 6 m.7 0 6 m Gavitational foce on space junk (N) 900 600 300.5.0.5 3.0.7.7 Distance fom cente of Moon ( 0 6 m) Count the numbe of shaded squaes. (Only count squaes that ae at least 50% shaded.) Numbe of shaded squaes 5 Calculate the aea (enegy value) of each squae. E squae 0. 0 6 00 0 7 J To calculate the enegy change, multiply the numbe of shaded squaes by the enegy value of each squae. ΔE g 5 ( 0 7 ) 5. 0 8 J Woked example: Ty youself.3.4 CHANGE IN GRAVITATIONAL POTENTIAL ENERGY USING A GRAVITATIONAL FIELD STRENGTH DISTANCE GRAPH A 3000 kg Soyuz ocket moves fom an obital height of 300 km above the Eath s suface to dock with the Intenational Space Station at a height of 500 km. Use the gaph of the gavitational field stength of the Eath below to detemine the appoximate change in gavitational potential enegy of the ocket. Gavitational field stength (N kg ).0 0.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0.0.0 0.0 0 50 00 50 00 50 300 350 400 450 500 550 600 Altitude (km) Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8
Heinemann Physics 4e Count the numbe of shaded squaes. Only count squaes that ae at least 50% shaded. Numbe of shaded squaes 36 Calculate the enegy value of each squae. E squae 50 0 3 m N kg 5 0 4 J kg To calculate the enegy change, multiply the numbe of shaded squaes by the enegy value of each squae and the mass of the ocket. E g 36 5 0 4 3000 5.4 0 9 J.3 eview C. A stable obit suggests that the object is in a unifom gavitational field, hence its gavitational potential enegy does not change. Its speed will also emain the same in a stable obit. g inceases fom point A to point D. 3 The meteo is unde the influence of the Eath s gavitational field which will cause it to acceleate at an inceasing ate as it appoaches the Eath. 4 A, B and C ae all coect. The total enegy of the system does not change. 5 W E g 3000000 9.8 67000.0 0 J 6 E g mg h 0.4 6. 7000 700 J E k mv 700 0.4 v v 700 0.4 9 m s 7 a 00 km above the Eath s suface is a distance of 6.4 0 6 m + 00000 m 6.5 0 6 m. Accoding to the gaph, F is between 9 N and 9. N at this height. b Accoding to the gaph, 5 N occus at appoximately 9.0 0 6 m fom the cente of the Eath. So, the height above the Eath s suface 9.0 0 6 6.4 0 6.6 0 6 m o 600 km. 8 a Convet km s to m s then apply the ule: E k mv 4000 8 0 6 J b E k E g E g aea unde the gaph aea 9 squaes 0.5 0 6.9 0 7 J c new E k stating E k + E k 8 0 6 +.9 0 7.7 0 7 J d new speed.7 0 7 7348 m s o 7.3 km s 9 600 km above the Eath s suface 6.4 0 6 + 600 000 7.0 0 6 m o 7000 km Aea unde the gaph between 7000 km and 8000 km is appoximately 7 squaes. As the satellite comes to a stop, the change in kinetic enegy ove the distance is the same as the E k at its launch. E k aea unde the gaph mass of the satellite 7 squaes 0.5 0 6 40.7 0 9 J Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8
Heinemann Physics 4e 0 600 km above the Eath s suface 6.4 0 6 + 600 000 7.0 0 6 m o 7000 km 600 km above the Eath s suface 6.4 0 6 + 600 000 9.0 0 6 m o 9000 km. The aea unde the gaph between 7000 km and 9000 km is appoximately 6 squaes. E g aea unde the gaph mass of the satellite 6 squaes 0.5 0 6 0 000.6 0 J CHAPTER REVIEW G mm 4 6.0 0 75 6.67 0 6 ( 6.4 0 ) 730 N G mm.79 0 0 6.67 0 3 F ma Sun 6.67 0.05 0 5.69 0 378 000 000 m 3.78 0 8 m 6.79 0 0.05 0 5.69 0 6 a Sun F m 4. 0.0 0 3 30. 0 7 m s 4 a The foce exeted on Jupite by the Sun is equal to the foce exeted on the Sun by Jupite. b The acceleation of Jupite caused by the Sun is geate than the acceleation of the Sun caused by Jupite. 5 g G M 6.67 0 3.7 m s 3 6.4 0 ( 3400 000) 6 a mg 50 9.8 490 N When acceleating downwads at 0.6 m s, the net foce is F net ma 50 0.6 30 N and > F N. F N 30 490 F N 30 F N 490 30 460 N b When the peson is moving at a constant speed, thei appaent weight is equal to thei weight: F N 490 N 7 a F G mm 7 (6.67 0.9 0 000) 7 (7.5 0).48 0 4 N b The magnitude of the gavitational foce that the comet exets on Jupite is equal to the magnitude of the gavitational foce that Jupite exets on the comet.48 0 4 N. net c a F m 4.48 0 000 4.8 m s net d a F m 4.48 0 7.90 0.3 0 3 m s Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8
Heinemann Physics 4e 8 D. At a height of two Eath adii above the Eath s suface, a peson is a distance of thee Eath adii fom the cente of the Eath. 900 900 Then F 00 N 3 9 9 a D. F net F N ma F N 80 30 + 80 9.8 384 N o 300 N b B. Fom pat (a), the appaent weight is geate than the weight of the astonaut. c C. Tue weight is unchanged duing lift-off as g is constant. d A. Duing obit, the astonaut is in fee fall. e D. ma 80 8. 656 N o 660 N 0 When epesenting a gavitational field with a field diagam, the diection of the aowhead indicates the diection of the gavitational foce and the space between the aows indicates the magnitude of the field. In gavitational fields, the field lines always point towads the souces of the field. 600 g 9.76 N kg m 6.5 a g G M 6.67 0 5.97 0 4 (6378 000) 9.79 N kg b g G M 4 6.67 0 5.97 0 (6357 000) 9.85 N kg % 9.85 00 00.6% 9.79 3 a g G M 6 6.67 0.0 0 7 (.48 0). N kg b C. It will acceleate at a ate given by the gavitational field stength, g. 4 G M ( 0.8R) M m 0.64 0.04 M m G m ( 0.R) 0.64 0.04 6 5 a Incease in E k aea unde the gaph between 3 0 6 m and.5 0 6 m 6 squaes 0 0.5 0 6 3 0 7 J b E k(initial) mv 0 000 0 7 J E k(new) 0 7 + 3 0 7 4 0 7 J 7 Ek 4 0 c v 000 m s o km s m 0 d Fom the gaph, F 70 N mg g 70 3.5 N kg 0 6 300 km 300 000 m o 3 0 5 m Fom the gaph, g 9 N kg at this altitude. 7 D. The units on the gaph ae N m kg, which ae the same as J kg 8 C. As the satellite falls, its gavitational potential enegy deceases. The units on the gaph ae J kg, so theefoe C is coect. 9 Incease in E k aea unde the gaph mass of the satellite 35 squaes 0 5 000 3.5 0 9 J 0 No. Ai esistance will play a majo pat as the satellite e-entes the Eath s atmosphee. Copyight Peason Austalia 06 (a division of Peason Austalia Goup Pty Ltd) ISBN 978 4886 7 8