kg 2 ) 1.9!10 27 kg = Gm 1

Transcription

1 Section 6.1: Newtonian Gavitation Tutoial 1 Pactice, page Given: kg; m kg; N; G N m /kg Requied: Analysis: G m ; G m G m Solution: G m N m 6.67!10 11 kg ) 1.0!100 kg.!10 9 N ) ) 3.0!10 0 kg ) 3.0!10 10 m Statement: The distance between the two asteoids is m.. Given: m kg; m; G N m /kg Requied: g Jupite Analysis: Stat with the univesal law of gavitation, Gm, then use F ma to substitute fo with the mass of an object on the suface, m, and the acceleation of the object, which will be the magnitude of the gavitational field stength on the suface of Jupite, g Jupite. G m m a G m g Jupite G Solution: g Jupite G N m 6.67!10 11 kg ) 1.9!10 7 kg 7.0!10 7 m g Jupite 6 m/s ) ) Statement: The magnitude of the gavitational field stength on the suface of Jupite is 6 m/s. Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields 6.1-1

2 3. a) Given: m A 40.0 kg; m B 60.0 kg; m C 80.0 kg; AB 0.50 m; BC 0.75 m; G N m /kg Requied:! F net Analysis: G m Solution: Detemine! F AB. G m F AB Gm m A B AB N i m 6.67! kg kg 0.50 m ) ) 60.0 kg ) F AB 6.403!10 7 N two exta digits caied) Detemine! F BC. G m F BC Gm m B C BC N m 6.67!10 11 ) 60.0 kg kg 0.75 m ) ) 80.0 kg ) F BC 5.69!10 7 N two exta digits caied)! F net! F AB +! F BC 6.403!10 7 N [left]+ 5.69!10 7 N [ight] 6.403!10 7 N [left] 5.69!10 7 N [left]! F net 7.1!10 8 N [left] Statement: The net foce acting on B is N [left]. b) Given: m A 40.0 kg; m B 60.0 kg; m C 80.0 kg; AB 0.50 m; BC 0.75 m; G N m /kg Requied:! F net Analysis: G m ; detemine the angle using the invese tan function. Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields 6.1-

3 Solution: Detemine! F AB. G m F AB Gm m A B AB N m 6.67!10 11 ) 40.0 kg kg 0.50 m ) ) 60.0 kg ) F AB 6.403!10 7 N two exta digits caied) Detemine! F BC. G m F BC Gm m B C BC N i m 6.67! kg kg 0.75 m ) ) 80.0 kg ) F BC 5.69!10 7 N two exta digits caied)! F net! F AB +! F BC F net F AB + F BC 6.403!10 7 N) !10 7 N) F net 8.6!10 8 N F! tan 1 BC F AB 5.69 )10 7 N tan )10 7 N! 4 Statement: The net foce acting on B is N [W 4 S]. Tutoial Pactice, page Given: m; m white dwaf kg; G N m /kg Requied: g white dwaf Analysis: g white dwaf Gm white dwaf Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields 6.1-3

4 Solution: g white dwaf Gm white dwaf N m 6.67!10 11 kg ) 1.!10 30 kg 7.0!10 6 m g white dwaf 1.6!10 6 N/kg ) ) Statement: The suface gavitational field stength of the white dwaf is N/kg, which is ove times that of Eath.. Given: Satun Requied: g Analysis: g Gm Solution: g Gm Gm ) Satun 1! Gm 4 Satun g 1 4 g Satun Statement: The suface gavitational field stength would be one quate of the old suface gavitational field stength. Reseach This: Gavitational Field Maps and Unmanned Undewate Vehicles, page 95 A. Sample answes: A gavitational field map descibes the stength of the gavitational field at points acoss Eath. The map is ceated by using satellites to detect fine density diffeences in the cust, which cause inceases o deceases in the gavitational foce. This infomation can be used by a UUV to detect whee it is on the planet based on the gavitational foce. B. Diagams may vay depending on the type of UUV chosen. Students should highlight the key featue of the UUV they choose, such as the populsion system a popelle is most common), the powe souce battey poweed), the navigation system, and the sensos, which will vay with pupose of the UUV, but may include depths sensos, sona, o sensos to measue concentation of compounds in the wate. C. Answes may vay. Students epots should explain the how UUVs use gavitational field maps to compae with measuements collected by the UUV on about the diection, angle, and stength of Eath s magnetic field at its position. Students may also discuss the usefulness of navigation by magnetic fields because UUVs tavel to fa fo emote contol and do not have access to satellites fo GPS navigation. Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields 6.1-4

5 Section 6.1 Questions, page Fo you weight to be one half you weight on the suface, the magnitude of the gavitational acceleation must be one half of g. g Gm E 1 g Gm E 1 g Gm ) E The altitude fom Eath s cente is E, o about 1.41 E. Theefoe, the altitude above Eath s suface is 1.41 E 1 E 0.41 E.. Given: m; kg; m kg; G N m /kg Requied: Analysis: G m Solution: G m N m 6.67!10 11 ) 1.67!10 7 kg kg 5.3!10 11 m 3.6!10 47 N ) 9.11!10 31 kg ) ) Statement: The magnitude of the gavitational attaction between the poton and the electon is N. 3. a) The value fo is squaed in the denominato, so as inceases, the gavitational foce deceases. b) G m Gm 4 1 ) G m ! 16 G m 1 The gavitational foce changes by a facto of Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields 6.1-5

6 4. a) Given: 5 kg; d m; m kg; E m; G N m /kg Requied: Analysis: G m Gm E d + E ) Solution: Gm E d + E ) N m 6.67!10 11 ) 5 kg kg 8.6!10 6 m !10 6 m 399 N ) 5.98!10 4 kg ) ) Statement: The gavitational foce is 399 N towad Eath s cente. b) Given: 5 kg; d m; m E kg; E m; G N m /kg Requied: g Analysis: g Gm Solution: g Gm Gm E d + E ) N m 6.67!10 11 kg ) 5.98!10 4 kg 8.6!10 6 m !10 6 m ) ) g 1.77 m/s Statement: The esulting acceleation is 1.77 m/s towad Eath s cente. 5. Given: g Titan 1.3 N/kg; m kg; G N m /kg Requied: Analysis: g Gm Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields 6.1-6

7 Solution: g Titan Gm Gm g Titan ) N m 6.67!10 11 kg ) 1.3!103 kg 1.3 N/ kg ).6!10 6 m Statement: The adius of Titan is m. 6. Given: g E 9.8 N/kg; g 3.0 N/kg Requied: Analysis: Use the equation g Gm to detemine the change in given the change in the value of g. Solution: g 3.0 N/kg g E 9.8 N/kg g g E g g E g 16! 49 Gm E Gm Gm E! 7 4 E Statement: The acceleation due to gavity is 3.0 N/kg at 7 4 E fom Eath s cente, o 0.75 E above Eath s suface, 7. Given: m Sun kg; m; G N m /kg Requied: g Analysis: g Gm Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields 6.1-7

8 Solution: g Gm N i m 6.67!10 11 kg.0!10 30 kg 1.5!10 11 m ) ) g 5.9!10 3 N/kg Statement: The gavitational field stength of the Sun at a distance of m fom its cente is N/kg. 8. Let be the lage mass, and let x be the distance fom to the location of zeo net foce. Set the two gavitational field stengths equal to each othe, and develop a quadatic equation. Solve fo x. G Gm x! x) G! x) G m x! x) m x! x + x m x 1! m x! x + 0 Use the quadatic fomula: x!b ± b! 4ac a!!) ±!)! 4 1! m 1! m ± 4! m 1! m ) Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields 6.1-8

9 ± 4 m 1! m ± m 1! m 1± m m x 1 1! m Since the geate value will not be between the two masses but will be the othe side of m fom : 1! m m x 1 1! m 1! m m x 1 1! m 1+ m 1+ m 1! m m x 1 1! m 1+ m x 1+ m The location of zeo foce is 1+ m fom the lage object,. Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields 6.1-9

10 9. a) Given: 537 kg; m E kg; m; G N m /kg Requied: g Analysis: g Gm Solution: g Gm Gm E N m 6.67!10 11 kg ) 5.98!10 4 kg.5!10 7 m ) ) g 0.64 m/s Statement: The esulting acceleation is 0.64 m/s towad Eath s cente. b) Given: 537 kg; m E kg; m; G N m /kg Requied: Analysis: G m Solution: G m Gm E N m 6.67!10 11 ) 537 kg kg.5!10 7 m 340 N ) 5.98!10 4 kg ) ) Statement: The gavitational foce is 340 N towad Eath s cente. 10. Given: m; m Mecuy kg; G N m /kg Requied: g Mecuy Analysis: g Mecuy Gm Mecuy Solution: g Mecuy Gm Mecuy N m 6.67!10 11 kg ) 3.8!10 3 kg.44!10 6 m g Mecuy 3.67 N/kg ) ) Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields

11 Statement: The suface gavitational field stength on Mecuy is 3.67 N/kg. The value povided in Table is 3.7 N/kg, which is the same the value that I calculated to two significant digits. 11. a) Given: g 5.3 N/kg; m E kg; E m; G N m /kg Requied: Analysis: g Gm Solution: g Gm Gm g ) N m 6.67!10 11 kg ) 5.98!104 kg 5.3 N/ kg ) 8.675!10 6 m two exta digits caied) Calculate the altitude above Eath s suface: m m m Statement: The altitude of the satellite is m. b) Given: 60 kg; m E kg; satellite m; G N m /kg Requied: Analysis: G m Solution: G m N m 6.67!10 11 ) 60 kg kg 8.675!10 6 m 3.3!10 3 N ) 5.98!10 4 kg ) ) Statement: The gavitational foce on the satellite is N towad Eath s cente. 1. The motion of the Moon depends on Eath s mass and G though the univesal law of gavitation. Using data on the mass and obital adius of the Moon and G, we can detemine Eath s mass using the univesal law of gavitation, G m acceleation, F c mv, since the two foces ae equal., and the equation fo centipetal Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields

12 F c G m Gm Eath m Moon mv m v Moon Gm Eath v m Eath v G 13. Fom question 8, the location of zeo foce is 1+ m cente to cente distance between the Moon and Eath. Since m E kg and m Moon kg: 0 1+ m 7.36! ! The mass should be 0.9 fom the cente of the Eath, o 10 fom the lage object,. is the fom the cente of the Moon. Copyight 01 Nelson Education Ltd. Chapte 6: Gavitational Fields 6.1-1