Chapte 1 Univesal Gavitation Pactice Poblem Solutions Student Textbook page 580 1. Conceptualize the Poblem - The law of univesal gavitation applies to this poblem. The gavitational foce, F g, between Eath and the Sun Identify the Vaiables M S = 1.99 10 30 kg M E = 5.98 10 4 kg = 1.49 10 11 m Develop a Stategy Apply the law of F g = G m 1m univesal gavitation. Substitute the F g = 6.67 10 numeical values and solve. F g = 3.575 10 N F g 3.58 10 N F g 1.99 10 30 kg5.98 10 4 kg 1.49 10 11 m The gavitational foce between Eath and the Sun is about 3.58 10 N. Gavitational foce causes Eath to go aound the Sun, so it s expected that the foce will be lage, and it is.. Conceptualize the Poblem - The law of univesal gavitation applies to this poblem. The gavitational foce, F g, between Eath and the Moon Identify the Vaiables M M = 7.36 10 kg M E = 5.98 10 4 kg = 3.84 10 8 m Develop a Stategy Apply the law of univesal gavitation. F g = G m 1m F g Chapte 1 Univesal Gavitation MHR 336
7.36 10 Substitute the F g = 6.67 10 kg5.98 10 4 kg 3.84 10 8 m numeical values and solve. F g = 1.9909 10 0 N F g 1.99 10 0 N The gavitational foce between Eath and Moon is about 1.99 10 0 N. Gavitational foce causes the Moon to obit Eath, so it s expected to be lage, though smalle than in the pevious poblem, and it is. 3. Conceptualize the Poblem - The law of univesal gavitation applies to this poblem. The distance,, between two bowling balls that feel a given foce between them Identify the Vaiables m 1 = m = 7.0 kg F g = 1.5 10 4 N Develop a Stategy Apply the law of univesal gavitation. F g = G m 1m Solve fo the distance. = G m 1m F g = Gm1 m F g 6.67 10 7.0 kg kg Substitute the numeical values and solve. = 1.5 10 4 N = 5.113 10 3 m 5.1 10 3 m The distance between the two bowling balls that feel the given foce is about 5.1 10 3 m. The distance in the law of univesal gavitation is measued fom the cente of an object. This calculated distance, about 0.5 cm, is smalle than the adius of a bowling ball, so it would not be possible to place them at this distance. The units wok out to be metes, as equied. 4. Conceptualize the Poblem - The law of univesal gavitation applies to this poblem. The gavitational foce, F g, between the electon and poton in a hydogen atom Identify the Vaiables m p = 1.67 10 7 kg m e = 1.67 10 7 kg = 5.30 10 11 m F g Chapte 1 Univesal Gavitation MHR 337
Develop a Stategy Apply the law of F g = G m 1m univesal gavitation. Substitute the F g = 6.67 10 numeical values and solve. F g = 3.615 10 47 N F g 3.61 10 47 N The gavitational foce is about 3.61 10 47 N. 1.67 10 7 kg9.11 10 31 kg 5.30 10 11 m The masses ae vey small, so the gavitational foce is also expected to be small, and it is. 5. Conceptualize the Poblem - The foce that povides the weight of the peson is the same as the foce in the law of univesal gavitation. - The law of univesal gavitation applies to this poblem. The mass of Venus, M Identify the Vaiables F g = 57 N M m = 68 kg = 6.31 10 6 m Develop a Stategy Apply the law of univesal gavitation. F g = G Mm Substitute the numeical values and solve. M = F g The mass of Venus is about 5.0 10 4 kg. Gm M = 57 N6.31 10 6 m 6.67 10 68 kg M = 5.013 10 4 kg M 5.0 10 4 kg The calculated value of the mass of Venus is about 84% of Eath s mass, which seems easonable knowing that the two planets have simila masses. 6. Conceptualize the Poblem - The law of univesal gavitation applies to this poblem. The distance,, between the two objects Identify the Vaiables F g = 1.8 10 8 N m 1 = 8.0 kg m = 1.5 kg Chapte 1 Univesal Gavitation MHR 338
Develop a Stategy Apply the law of univesal gavitation. F g = G m 1m Substitute the numeical values = G m 1m F and solve. g = Gm1 m = F g 6.67 10 = 0.501 m 0.5 m 8.0 kg1.5 kg 1.8 10 8 N The sepaation between the centes of the two objects is about 0.5 m. The distance between the two objects seems easonable fo a laboatoy expeiment. 7. Conceptualize the Poblem - The law of univesal gavitation applies to this poblem. - This poblem can be solved using atios. The gavitational foce on Uanus, F U, compaed to that on Eath, F E Identify the Vaiables U = 4.3 E G = 6.673 10 F U M U = 14.7M E Develop a Stategy 1 altenate method Apply the law of univesal gavitation. Divide one by the othe. Substitute the numeical values and solve. F U = G M Um U F E = G M Em E F U FE F U FE = GM Um U = M U M E E U F U = 14.7M E FE M E F U FE = 14.7 4.3 F U FE = 0.795 F U 0.80F E E GM E m E 4.3 E F U = G M Um U F U = G 14.7M Em 4.3 E F U = 14.7 4.3 F U = 0.795F E F U 0.80F E G M Em The gavitational foce on the suface of Uanus is about 0.80 times the gavitational foce on the suface of Eath. Both the mass and adius of Uanus ae lage than that of Eath. Howeve, because the law of gavitation depends on the squae of the adius, the gavitational foce on the suface of Uanus is smalle than that of Eath. E Chapte 1 Univesal Gavitation MHR 339
8. Conceptualize the Poblem - The object is to be placed whee it would feel equal gavitational foces fom Eath and the Moon so that the net foce on the object is zeo. - The law of univesal gavitation applies to this poblem. - Make a sketch of the poblem and indicate the distances. Let the Eath-Moon distance be d, the distance fom the Eath to the point be x and the distance fom the Moon to the point be d-x. The distance, x, fom the cente of Eath, whee a point between Eath and the Moon would feel equal and opposite foces Identify the Vaiables M M = 0.013 M E G = 6.673 10 x Develop a Stategy Apply Newton s second law to a test mass placed at the point. F net = F E F M = ma = 0 F E = F M G M Em x = G M Mm d x Let d = 1.0. Then the distance, x, will be quoted as a faction of the Eath- Moon distance. F M F E d x = M M M E x d x M M x M = 0 E d dx + x M M x M = 0 E x 1 M M dx + d = 0 M E A quadatic equation is obtained which 0.9877x x + 1 = 0 can be solved using the quadatic fomula: x = b ± b 4ac a Choose the solution that is less than 1.0 because x should be less than d. x = ± 40.98771 0.9877 x = 0.900 o 1.14 x 0.900 A point 0.9 times the Eath-Moon distance will feel equal and opposite gavitational foces fom Eath and the Moon. The Eath is much moe massive than the Moon, so it s expected that the point whee the gavitational foces cancel will be close to the Moon than to Eath. Pactice Poblem Solutions Student Textbook page 586 9. Conceptualize the Poblem - Keple s thid law, combined with Newton s law of univesal gavitation, yields an equation that elates the peiod and obital adius of a satellite to the mass of the body aound which the satellite is obiting. The mass of Jupite, M J Chapte 1 Univesal Gavitation MHR 340
Identify the Vaiables T = 1.769 days G = 6.673 10 = 4.16 10 8 m Develop a Stategy T Wite Keple s thid 3 = 4π GM J law, using the M J = 4π 3 constant deived fom GT Newton s law of univesal gavitation. Solve fo the mass. Substitute the M J = 4π 4.16 10 8 m 6.673 3 numeical values 10 1.769 days 4 h 1 day 3600 s 1h and solve. Convet days M J = 1.898 10 7 kg to seconds. The mass of Jupite is about 1.898 10 7 kg. Fom efeence tables, the mass of Jupite is about 317.8 times the mass of Eath. This is in excellent ageement 1.898 10 7 kg 317.8 = 5.97 10 4 kg mass of Eath with the above calculation. 10. Conceptualize the Poblem - Keple s thid law, combined with Newton s law of univesal gavitation, yields an equation that elates the peiod and obital adius of a satellite to the mass of the body aound which the satellite is obiting. The mass of Pluto, M P Identify the Vaiables T = 6.387 days G = 6.673 10 = 1.9640 10 7 m Develop a Stategy T Wite Keple s thid 3 = 4π GM law, using the P M P = 4π constant deived fom 3 GT Newton s law of univesal gavitation. Solve fo the mass. Substitute the M P = 4π 1.9640 10 7 m 6.673 3 numeical values 10 6.387 days 4 h 1 day 3600 s 1h and solve. Convet days M P = 1.47 10 kg to seconds. The mass of Pluto is about 1.47 10 kg. M J M P Chapte 1 Univesal Gavitation MHR 341
Fom efeence tables, the mass of Pluto is about 0.006 times the mass of Eath. This gives a value of 1.47 10 kg = 5.66 10 4 kg fo the mass of Eath, 0.006 within about 5.0% of the accepted value. 11. Conceptualize the Poblem - Keple s thid law, combined with Newton s law of univesal gavitation, yields an equation that elates the peiod and obital adius of a satellite to the mass of the body aound which the satellite is obiting. The altitude, h, of the satellite above Eath s suface Identify the Vaiables T = 90.0 m M E = 5.98 10 4 kg h E = 6.38 10 6 m Develop a Stategy Wite Keple s thid law, using the constant T 3 = 4π GM deived fom Newton s law of univesal E gavitation. Solve fo the obital adius. 3 = GM PT 4π Substitute the numeical values and solve. Convet minutes to seconds. = 6.67 10 5.98 10 4 kg 4π = 6.654 10 6 m The obital adius is about 6.654 10 6 m. Find the altitude above Eath s suface by subtacting Eath s adius fom the obital adius. 90.0 min 60 s 1 min 1 3 h = E h = 6.654 10 6 m 6.38 10 6 m h =.74 10 5 m The satellite obits at.74 10 5 m above Eath s suface, o, about 74 km. Knowing that Eath s mesosphee extends up to 80 km, and the uppe atmosphee has a much lowe density, the esult is easonable. Check the units: 1 N m kgs 3 = kg m s m kgs 1 3 = m 3 1 3 = m 1. Conceptualize the Poblem - The gavitational foce between Eath and Moon povides the centipetal foce to keep the Moon in its obit. - Assume the Moon is obiting Eath in a cicle. The speed of the Moon, v, as it obits Eath Chapte 1 Univesal Gavitation MHR 34
Identify the Vaiables T = 90.0 m v M E = 5.98 10 4 kg M = 3.84 10 5 km = 3.84 10 8 m Develop a Stategy Apply the law of univesal F = G M Em M = m M a = m v M gavitation with Newton s G M E = v second law, noting that the acceleation is centipetal GM E acceleation. 6.67 10 5.98 10 4 kg kg Substitute the numeical values 3.84 10 8 m and solve. Convet minutes to seconds. 1.019 10 3 m s v 1.0 10 3 m s The obital speed of the Moon is about 1.0 10 3 m s Compaing the esult with Eath s obital speed, v E = GMS Eobit = 6.67 10 1.49 10 11 m 1.99 10 30 kg o 1.0 km s = 9.8 km s.. Eath feels a significantly geate gavitational foce fom the Sun than the Moon does fom Eath see Pactice Poblems 1 and, so it is expected that Eath s obital speed will be geate than the Moon s. 1 = m Check the units: N m kg 1 m 1 = kg m s m kg 1 m s 1 = m s 13. Conceptualize the Poblem - Keple s thid law, combined with Newton s law of univesal gavitation, yields an equation that elates the peiod and obital adius of a satellite to the mass of the body aound which the satellite is obiting. - The gavitational foce between the Moon and the Apollo luna module povides the centipetal foce to keep the module in its obit. - Assume the Apollo luna module is obiting the Moon in a cicle. a The obital peiod, T, of the luna module b The obital speed, v, of the luna module Identify the Vaiables M M = 7.36 10 kg T h = 60.0 km v M = 7738 km Chapte 1 Univesal Gavitation MHR 343
Develop a Stategy Note that the obital adius is the altitude plus the Moon s adius. Wite Keple s thid law, using the constant deived fom Newton s law of univesal gavitation. Solve fo the obital peiod. T 3 = 4π GM M T = 4π 3 GM M T = 4π 3 GM M Substitute the numeical T = 4π 60.0 km 1000 m 1000 m + 7738 km 1km 1km values and solve. 6.67 10 7.36 10 kg T = 6.175 10 4 s T = 6.175 10 4 s 1h 3600 s Multiply by the numbe of seconds in one hou T = 17. h to obtain the esult in hous. a The obital peiod of the luna module is about 6.18 10 4 s o about 17. h. Apply the law of univesal F = G M Mm module = m module a = m v module gavitation with Newton s G M M = v second law, noting that the acceleation is centipetal GM M acceleation. Substitute the numeical 6.67 10 7.36 10 kg values and solve. kg 60.0 10 3 m + 7738 10 3 m 7.93 10 m s b The module has an obital speed of about 7.93 10 m s. 3 1 Check the units fo the peiod: m 3 N m = kg The units fo speed wee checked in the pevious question. m 3 kg m s m kg 1 = s 1 = s 14. Conceptualize the Poblem - The gavitational foce between the cente of the galaxy and the sta povides the centipetal foce to keep the sta in its obit. - Assume the sta is obiting the galaxy in a cicle. The mass of the Andomeda galaxy, M Identify the Vaiables.0 10 km s M =.0 10 5 m s = 5 10 9 AU Chapte 1 Univesal Gavitation MHR 344
Develop a Stategy Apply the law of univesal gavitation with Newton s second law, noting that the acceleation is centipetal acceleation. Solve fo the mass of the galaxy. Substitute the numeical values and solve. F = G Mm sta = m sta a = m v sta G M = v M = v G 5 10 9 AU 1.49 1011 m 1AU M = M = 4.4678 10 41 kg M 4 10 41 kg 6.67 10.0 10 5 m s The mass of the galaxy is about 4 10 41 kg. Check the units fo the mass: m m s N m = m m s kg m = kg m s The mass of the galaxy can be compaed to the mass of the Sun: 4 10 41 kg.0 10 30 kg = 10 11. Lage galaxies ae known to have masses of moe than 100 billion times the sun s mass, so this esult is easonable. Pactice Poblem Solutions Student Textbook page 591 15. Conceptualize the Poblem - The peiod is elated to the velocity of the satellite. - The peiod of an obit is the invese of the fequency, f. - The velocity and altitude of the satellite ae detemined by the amount of centipetal foce that is causing the satellite to emain on a cicula path. - Eath s gavity povides the centipetal foce fo satellite motion. The obital speed, v, and altitude, h, of the satellite Identify the Vaiables M E = 5.98 10 4 kg v f = 14.1 obits/day h Chapte 1 Univesal Gavitation MHR 345
Develop a Stategy Detemine the obital peiod in seconds fom the fequency the numbe of obits pe day. Wite Keple s thid law, using the constant deived fom Newton s law of univesal gavitation. Solve fo the obital adius. T = 1 f 4 h 1 day 1 day T = 3600 s 1h 14.1 obits T = 6.176 10 3 s T 3 = 4π GM E 3 = GM ET 4π Substitute the numeical values and solve. 6.67 10 5.98 10 4 kg6.176 10 3 s = 4π 1 3 = 7.39 10 6 m 7.4 10 6 m Detemine the altitude of the obit by subtacting the adius of Eath fom the obital adius. Apply the law of univesal gavitation with Newton s second law, noting that the acceleation is centipetal acceleation. Substitute known values and solve. G M E h = E h = 7.39 10 6 m 6.38 10 6 m h = 8.588 10 5 m h 8.59 10 5 m F = G M Em satellite = m satellite a = m v satellite = v GM E 6.67 10 7.39 10 6 m 5.98 10 4 kg 7.49 10 3 m s v 7.4 10 3 m s The obital speed of the satellite is about 7.4 10 3 m s. Based on the esult of poblem 11, it is expected that a satellite that obits Eath seveal times pe day will have an altitude of seveal hunded km above Eath s suface, so the esult hee 859 km seems easonable. 16. Conceptualize the Poblem - The peiod is elated to the velocity of the satellite. - The velocity and altitude of the satellite ae detemined by the amount of centipetal foce that is causing the satellite to emain on a cicula path. - Eath s gavity povides the centipetal foce fo satellite motion. The obital speed, v, and peiod, T, of the space station Chapte 1 Univesal Gavitation MHR 346
Identify the Vaiables h = 6 km v M = 5.98 10 4 kg T E = 6.38 10 6 m Develop a Stategy Apply the law of F = G M Em Space station = m Space station a = m v Space station univesal gavitation G M E = v with Newton s second law, noting that the GM E acceleation is centipetal acceleation. 6.67 10 5.98 10 4 kg The obital adius is the kg adius of Eath plus 6 10 3 m + 6.38 10 6 m the altitude = E + h. 7.77 10 3 m s The Space Station has an obital speed of about 7.77 10 3 m s. Wite Keple s thid law, using the constant deived fom Newton s law of univesal gavitation. Solve fo the obital peiod. Multiple by the numbe of seconds in an hou to obtain the peiod in hous. T 3 = 4π GM M T = 4π 3 GM M T = 4π 3 GM M T = 4π 3 + 6.38 106 m 6 km 1000 m 1km 6.67 10 5.98 10 4 kg T = 5.34 10 3 s T = 5.34 10 3 s T = 1.48 h 1h 3600 s The obital peiod of the satellite is about 5.34 10 3 s o 1.48 h. The altitude and peiod of the Intenational Space Station agee with the esults of poblem 11, so the esults hee ae easonable. 17. Conceptualize the Poblem - Keple s thid law elates the obital adius and obital peiod of a satellite o planet, to the mass of the object it obits. - The peiod is elated to the velocity of the planet. - The velocity of the planet is detemined by the amount of centipetal foce that is causing the planet to emain on a cicula path. - The Sun s gavity povides the centipetal foce fo planet motion. a The obital peiod, T and speed, v, of the planet Neptune b The numbe of obits, N, that Neptune has completed since its discovey Chapte 1 Univesal Gavitation MHR 347
Identify the Vaiables M = 1.99 10 30 kg v N = 4.50 10 1 m T N Develop a Stategy Wite Keple s thid law, using the constant deived fom Newton s law of univesal gavitation. Solve fo the obital peiod. Multiply by the numbe of seconds in a yea to obtain the numbe of yeas. T 3 = 4π GM S T = 4π 3 GM S T = 4π 3 GM S T = 4π 4.50 10 1 m 6.67 3 10 1.99 10 30 kg T = 5.1 10 9 s T = 165 yeas 1yea 3.156 10 7 s a The obital peiod of Neptune is about 5.1 10 9 s o 165 yeas. Apply the law of univesal gavitation with Newton s second law, noting that the acceleation is centipetal acceleation. G M S F = G M Sm Neptune = m Neptune a = m v Neptune = v GM S 6.67 10 5.43 10 3 m s a The obital speed of Neptune is about 5.43 10 3 m s. 4.50 10 1 m 1.99 10 30 kg b The peiod of Neptune is 165 yeas. It was discoveed in 1846. Theefoe, it will complete its fist obit since its discovey in the yea: 1846 + 165 = 011. Accoding to efeence tables, the peiod of Neptune is 165 yeas, in ageement with the above calculation. Chapte 1 Univesal Gavitation MHR 348
Chapte 1 Review Answes to Poblems fo Undestanding Student Textbook page 597. Reducing the mass by half will educe the gavitational foce by half also. Doubling the distance will educe the foce to one quate. Togethe, the new foce will be educed to one eighth, o 10.0 N. F ga = G m 1Am A A F ga = 80 N G m 1Am A A = 80 N m 1B = 1 m 1A m B = m A B = A F gb = G m 1Bm B F gb = G F gb = 1 8 F ga B 1 m 1Am A A F gb = 1 80 N 8 F gb = 10 N 3. The coect answe is c F. Fom Newton s thid law: The gavitational foce is equal on both stas. 4. The coect answe is b a 3. Fom Newton s second law: a = F m. F = ma F 1 = F a = F m a 1 = F 1 m a = F 3m a = F 1 3m a = 1 F1 3 m a = 1 a 3 1 5. a Fom these equations, you obtain the expession Gm, whee m is the Sun s mass 1.99 10 30 kg and is Eath s obital adius 1.49 10 11 m. Using this equation gives a velocity of.98 10 4 m/s. F g = F c G m Em g = m Ev Gm 6.67 10 11 N m 1.99 10 30 kg v.98 10 4 m s 1.49 10 11 m b The equation fo centipetal acceleation gives 5.98 10 3 m/s. a c = v.9846 10 4 m a c = s 1.49 10 11 m a c 5.98 10 3 m s Chapte 1 Univesal Gavitation MHR 349
6. Fom the law of univesal gavitation, the foce of gavity on the Sun fom Eath is 3.56 10 N. Using this foce in Newton s second law, the Sun s acceleation is 1.80 10 8 m/s. F g = G m 1m F g = 6.67 10 F g 3.58 10 N F = ma a = F m a = 3.5753 10 N 1.99 10 30 kg a 1.80 10 8 m s 5.98 10 4 kg1.99 10 30 kg 1.49 10 11 m 7. By setting the foce of gavity equal to ma, you obtain an expession fo acceleation that yields a = 8.95 m/s. This value is only slightly less than the value of 9.81 m/s at Eath s suface. a = 6.67 10 a 8.95 m s F g = G m Em SS m SS a = G m Em SS 5.98 10 4 kg 6.38 10 6 m +.95 10 5 m 8. Setting the foce of gavity equal to the centipetal foce and solving fo m bh gives m bh = 4.1 10 36 kg appoximately two million times moe massive than the Sun. G m bhm gas = m gasv m bh = v G 3.4 10 4 m.365 m bh = s 10 17 m 6.67 10 m bh 4.1 10 36 kg 9. Using the law of univesal gavitation, the foce of gavity is.7 10 10 N. This value is in the ode of one million times smalle than the weight of a flea. F g = G m 1m F g = 6.67 10 F g.7 10 10 kg 1.0 kg1.0 kg 0.500 m Chapte 1 Univesal Gavitation MHR 350
30. a Setting the foce of gavity equal to the centipetal foce, solving fo, and subtacting Eath s adius of 6.38 10 6 m gives an altitude of 5.3 10 5 m. G m Em H = m Hv = G m E v = 6.67 10 b Solving fo T using d t d t π T T = π v = 6.9056 10 6 m 5.98 10 4 kg 7.6 10 3 m s = E + h h = E h = 6.9056 10 6 m 6.38 10 6 m h 5.3 10 5 m to obtain π T gives T = 5.7 103 s. T = π6.9056 106 m 7.6 10 3 m s T 5.7 10 3 s 31. Setting the foce of gavity equal to the centipetal foce and solving fo v gives 1.0 10 3 m/s. Solving fo T using d to obtain π t T gives T =.37 10 6 s. G m Mm E = m Mv d t G m π E T T = π 6.67 10 11 N m 5.98 10 4 kg v 3.84 10 8 m T = π3.84 108 m v 1.0 10 3 m s 1.019 10 3 m s T.37 10 6 s 3. a Calculating T 3 = k fo each satellite gives the same value fo k, o appoximately 1.04 10 15 s /m 3, veifying that they obey Keple s thid law. Detailed calculations fo Tethys appea below. Using this equation in all cases, you obtain the same value fo Dione, Titan, and Iapetus. Fo Rhea, the value is 1.05 10 15 s /m 3. T 3 = k 1.888 days 4 h 1 day k = 1.631 105 s.95 10 8 m 3 15 s k 1.04 10 m 3 3600 s = 1.631 10 1h 5 s Chapte 1 Univesal Gavitation MHR 351
b Combining Keple s thid law and Newton s law of univesal gavitation and solving fo m gives m = 5.69 10 6 kg. k = 4π Gm Sa m Sa = 4π Gk m Sa = 4π 15 s 6.67 10 1.04 10 m 3 m Sa 5.7 10 6 kg 33. a Using mass = density volume, and volume of a sphee = 4 3 π3, you obtain m = 5. 10 14 kg. m = ρv V = m = ρ 4 3 π 3 m = 4 3 π 3 ρ = 1.00 ρ = 1000 kg m 3 1000 kg 4 m 3 3 π5.0 103 m 3 m 5. 10 14 kg g 100 cm 3 1kg cm 3 1m 1000 g b Total mass = 5. 10 6 kg. m Oot = 1.0 10 1 cometsm comet m Oot = 1.0 10 1 comets 5. 10 14 m Oot = 5. 10 6 kg kg comet c The cloud s mass is 88 times lage than the mass of Eath and 3.6 times smalle than Jupite s mass. m Oot m E m Oot m E 88 = 5. 106 kg 5.98 10 4 kg m Oot m J = 5. 106 kg 1.90 10 7 kg m Oot m J 0.8 Chapte 1 Univesal Gavitation MHR 35