Newton s Law of Universal Gravitation Every object in the universe is attracted to every other object. r 2

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1 6//01 Newton s Law of Univesal Gavitation Evey object in the univese is attacted to evey othe object. Cavendish poves the law in 1798 F= Gm 1 m G = 6.67 X N-m /kg m 1 = mass of one object m = mass of second object = distance fom cente of objects Eveything in the sola system pulls on eveything else. a. Calculate the foce of gavity between two 60.0 kg (1 lbs) people who standing.00 m apat. (6.00 X 10-8 N) b. Calculate the foce of gavity between a 60 kg peson and the eath? Assume the eath has a mass of 5.98 X 10 4 kg and a adius of 6,400,000 m (4,000 miles). (584 N) Sun pulls on Eath All the othe planets also pull on the Eath A planet has a mass fou times that of the eath, but the acceleation of gavity is the same as on the eath s suface. Relate the planet s adius to that of the eath (R e ). A 000-kg satellite obits the eath at an altitude of 680 km (the adius of the eath)above the eath s suface. What is the foce of gavity on the satellite? F= Gm 1 m = (6.67 X N m /kg )(000kg)(5.98 X 10 4 kg) (6,80,00 m + 6,80,00 m) F = 4900 N 1

2 6//01 A simple solution: At the eath s suface: F G = mg At twice that distance: F G = (1/) mg = ¼ mg F G = (¼ )(000 kg)(9.80 m/s ) F G = 4900 N What is the net foce on the moon when it is at a ight angle with the sun and the eath? Relevant Data: M M = 7.5 X 10 kg M E = 5.98 X 10 4 kg M S = 1.99 X 10 0 kg MS = 1.50 X m ME =.84 X 10 8 m Calculate each foce sepaately: F ME F ME = 1.99 X 10 0 N F MS = 4.4 X 10 0 N F R = F ME + F MS F R = 4.77 X 10 0 N tan q = opp = F ME adj F MS q = 4.6 o F MS q Sun F R Eath Conside an object (any object) Weight = mg (Foce due to gavity) mg = Gmm E E g = Gm E E m E =g E = (9.8 m/s )(6.8 X 10 6 m) = 5.98 X 10 4 kg G 6.67 X N-m /kg Calculating g g = Gm E E Calculate the value of g at the top of Mt. Eveest, 8848 m above the eath s suface. g = (6.67 X N-m /kg )(5.98 X 10 4 kg) (6.8 X 10 6 m) g = 9.80 m/s g = Gm E g = (6.67 X N-m /kg )(5.98 X 10 4 kg) (6.8 X 10 6 m m) g = 9.77 m/s

3 6//01 g vaies with: Altitude Location Eath is not a pefect sphee Diffeent mineal deposits can change density salt domes ae low density salt egions nea petoleum deposits Gavitational Potential Enegy mgh is not valid except nea the suface of the eath Zeo point is an infinite distance fom the eath Need calculus since the foce vaies with distance U = - F dx U = - Gm 1 m d U = -Gm 1 m A 1000 kg ocket is launched fom the suface of the eath. Calculate the escape velocity. Hint: set the final K and U equal to zeo. (11,00 m/s) A 1000 kg ocket is launched fom the suface of the moon. Calculate the escape velocity. The mass of the moon is 7.5 X 10 kg and the adius is 1.74 X 10 6 m. (8 m/s) A tile of a ocket falls off at an altitude of 1500 km above the suface of the eath. The ocket is tavelling upwads at 000 m/s. a. Calculate the speed of the tile as it hits the gound. (576 m/s) b. In eality. Could the tile actually each this speed? Pluto has a adius of 1150 km and a mass of 1.0 X 10 kg. a) Calculate the acceleation of gavity on Pluto (0.605 m/s ) b) Calculate the weight of a 70.0 kg peson on Pluto. (4.4 N) c) Calculate the acceleation of gavity on Pluto in tems of g s ( g s)

4 6//01 Suppose the eath stops obiting and falls into the sun (see page 95) a. Use consevation of enegy to calculate the speed of the eath as it just cashes into the sun. (6.1 X 10 5 m/s) b. Why doesn t ou eath get pulled into the sun? What speed should a satellite be launched if it needs to have a speed of 500 m/s at 400 km. Remembe to include the adius of the eath in you calculations. (755 m/s) Why don t satellites fall back onto the eath? Speed They ae falling due to the pull of gavity Can feel weightless if you ae on boad (just like in the elevato) F= GMm = mv GM = v v = \/GM/ Speed of a Satellite The Staship Entepise wishes to obit the eath at a 00 km height. a. Calculate the total distance of the Entepise fom the cente of the eath. (6.68 x 10 6 m) b. Calculate the pope obital speed fo the Entepise. (770 m/s) c. Would a heavie staship equie a geate obital speed? Keple s Laws ( ) 1. The obit of each planet is an ellipse, with the sun at one focus. Each planet sweeps out equal aeas in equal time. T 1 = 1 T 4

5 6//01 1. The obit of each planet is an ellipse, with the sun at one focus. Each planet sweeps out equal aeas in equal time Suppose the tavel time in both cases is thee days. Shaded aeas ae exactly the same aea Mas has a yea that is about 1.88 eath yeas. What is the distance fom Mas to the Sun, using the Eath as a efeence ( ES = X 10 8 m) M = T M E T E T 1 = 1 T M = (1.88y) (1.496 X 10 8 m) (1 y) T M = M T E E M = 1.18 X 10 5 m M =.8 X 10 8 m How long is a yea on Jupite if Jupite is 5. times fathe fom the Sun than the eath? T J = J T E E T J = J T E = (5.) (1 y) T J = 141 y E (1) T J = J T E E T J = 11.8 y 5

6 6//01 How high should a geosynchonous satellite be placed above the eath? Assume the satellite s peiod is 1 day, and compae it to the moon, whose peiod is 7 days. T s = s T M M s = M T s T M s = M T s = M (1 day) T M (7 day) S = M Take the d oot of 79 both sides s = M 9 The satellite must obit 1/9 the distance to the moon. Deiving the Thid Law To deive Keple s Law, we will need two fomulas. m J m 1 F= Gm 1 m J F=m 1 v GMm = mv GM = v GM = 4p T T = 4p GM Substitute v=p T T = 4p We can do this fo two GM diffeent moons T 1 = 4p T = 4p 1 GM GM T 1 = T 1 A Useful Fom This fom of the equation: T = 4p GM S could be the Sun, Eath, o othe body with satellites. Is useful fo detemining the mass of the cental planet, using only the peiod and distance of one of the satellites. 6

7 6//01 Calculate the mass of the sun, knowing that the eath is X 10 8 km fom the sun. (.0 X 10 0 kg) The mass of Mas is 6.40 X 10 kg. Calculate the peiod of its moon Phobos if Phobos has an obital adius of 977 km. (7.67 h) The Gemini 11 spacecaft sent two astonauts to a height of 174 km above the eath s suface. The adius of the Eath is 680 km and the mass of the Eath is 5.98 X 10 4 kg. a) Calculate the peiod of the satellite (1.89 hou) b) Calculate the speed of the obit. (717 m/s) Calculate the mass of Neptune if you know that the peiod of its moon Galatea is 0.49 days, and the adius is 61,95 km fom the cente of Neptune. (1.0 X 10 6 kg) A student is given the following data and asked to calculate the mass of Satun. The data descibes the obital peiods and adii of seveal of Satun s moons. Obital Peiod, T Obital Radius, R (seconds) 4p (m) 8.14 X X X X X X X X 10 8 Let s use this equation: T = 4p Gm S And eaange it: Gm S = 1 4p T Once moe: 1 = Gm S T 4p 7

8 6//01 Calculate the following values and gaph them. 1 G T 4p 1.60E E E E E E E-11.00E E E E E E-7.00E-7.50E-7.00E-7 Calculate the slope of the gaph y = m x + b 1 = m S G T 4p m s = 5.9 X 10 6 kg Obital Enegies GMm = mv GMm = mv 1 mv = GMm = K K = -½ U g E = K + U g E = -½ U g + U g E = ½ U g W = ½ DU g A 1500 kg satellite obits at an altitude of 4,000 km above the eath s suface. Calculate the wok equied to move that satellite to an obit of 1,000 km above the eath s suface. (1.5 X J) 8

9 6//01 Calculate the wok needed to boost a 1000 kg satellite fom an altitude of 00 km above the suface of the eath to a geosynchonous obit of 4. X 10 7 m fom the cente of the eath.. a) 6.7 X 10-9 N b) 6.81 X X a) 9.0 N b) feefall 8. a) 1.6 m/s b) 5.9 m/s R e 1. 9 m X 10 m/s 16. a) 1.80 X 10 7 m b) 9.41 km/s X 10 0 kg 0. a) 459 min b) 4440 N c) s, 91. min X 10 7 m 8. a) 6.4 o b) 85.1 o X 10-7 J. a) 1.8 X 10 7 m b) 4.45 km/s km a).05 X 10 8 yeas b) 4 obits c) 1.87 X kg d) 9.4 X X 10 4 m/s R X 10 8 J X J 50. a) 7.0 m/s b) 1. m/s 9

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